Class 11

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 10 The s-Block Elements Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 10 The s-Block Elements

PSEB 11th Class Chemistry Guide The s-Block Elements InText Questions and Answers

Question 1.
What are the common physical and chemical features of alkali metals?
Answer:
Physical properties of alkali metals are as follows :

1. They are quite soft and can be cut easily. Sodium metal can be easily cut using a knife.
2. They are light coloured and are mostly silvery white in appearance.
3. They have low density because of the large atomic sizes. The density increases down the group from Li to Cs. The only exception to this is K, which has lower density than Na.
4. The metallic bonding present in alkali metals is quite weak. Therefore they have low melting and boiling points.
5. Alkali metals and their salts impart a characteristic colour to flames. This is because the heat from the flame excites the electron present in the outermost orbital to a high energy level. When this excited electron reverts back to the ground state, it emits excess energy as radiation that falls in the visible region.
6. They also display photoelectric effect. When metals such as Cs and K are irradiated with light, they lose electrons.

Chemical properties of alkali metals are as follows :

Alkali metals are highly reactive due to their low ionization enthalpy. As we move down the group, the reactivity increases.
(1) They react with water to form respective oxides or hydroxides. As we move down the group, the reaction becomes more and more spontaneous.

(2) They react with water to form their respective hydroxides and dihydrogens. The general reaction for the same is given as :
2M + 2H²O → 2M⁺ + 2OH^– + H₂

(3) They react with dihydrogen to form metal hydrides. These hydrides are ionic solids and have high melting points.
2M + H₂ → 2M⁺ H^–

(4) Almost all alkali metals, except Li, react directly with halogens to form ionic halides.
2M + Cl₂ → 2MCl (M = Li, K, Rb, Cs)

Since Li⁺ ion is very small in size, it can easily distort the electron cloud around the negative halide ion. Therefore lithium halides are covalent in nature.

(5) They are strong reducing agents. The reducing power of alkali metals increases on moving down the group.
However, lithium is an exception. It is the strongest reducing agent among the alkali metals. It is because of its high hydration energy.

(6) They dissolve in liquid ammonia to form deep blue coloured solutions. Therefore, these solutions are conducting in nature.
M + (x + y) NH₃ → [M(NH₃)_x]⁺[M(NH₃)_y]^–

The ammoniated electrons cause the blue colour of the solution. These solutions are paramagnetic and if allowed to stand for some time, then they liberate hydrogen. This results in the formation of amides.
M⁺ + e^– + NH₃(l) → MNH + \(\frac{1}{2}\)H₂(g)

In a highly concentrated solution, the blue colour changes to bronze and the solution becomes diamagnetic.

Question 2.
Discuss the general characteristics and gradation in properties of alkaline earth metals.
Answer:
Physical and atomic properties of alkaline earth metals are as follows:

• The general electronic configuration of alkaline earth metals is (noble gas) ns².
• These metals lose two electrons to acquire the nearest noble gas configuration. Therefore, their oxidation state is +2.
• These metals have atomic and ionic radii smaller than that of alkali metals. Also when moved down the group, the effective nuclear charge decreases and this causes an increase in their atomic radii and ionic radii.
• Since the alkaline earth metals have large size, their ionization enthalpies are found to be fairly low. However, their first ionization enthalpies are higher than the corresponding group 1 metals.
• These metals are lustrous and silvery white in appearance. They are relatively less soft as compared to alkali metals.
• Atoms of alkaline earth metals are smaller than that of alkali metals. Also they have two valence electrons forming stronger metallic bonds. These two factors cause alkaline earth metals to have high melting and boiling points as compared to alkali metals.
• They are highly electropositive in nature. This is due to their low ionization enthalpies. Also the electropositive character increases on moving down the group from Be to Ba.
• Ca, Sr, and Ba impart characteristic colours to flames.
  Ca – Brick red Sr – Crimson red Ba – Apple green
  In Be and Mg, the electrons are too strongly bound to be excited. Hence, these do not impart any colour to the flame.

Chemical properties of alkaline earth metals are as follows :
The alkaline earth metals are less reactive than alkali metals and their reactivity increases on moving down the group.

(i) Reaction with air and water : Be and Mg are almost inert to air and water because of the formation of oxide layer on their surface.
(a) Powdered Be burns in air to form BeO and Be₃N₂.
(b) Mg, being more electropositive, burns in air with a dazzling sparkle to form MgO and Mg₃N₂.
(c) Ca, Sr, and Ba react readily with air to form respective oxides and nitrides.
(d) Ca, Ba, and Sr react vigorously even with cold water.

(ii) Alkaline earth metals react with halogens at high temperatures to form halides.
M + X₂ → MX₂ (X = F, Cl, Br, I)
(iii) All the alkaline earth metals, except Be, react with hydrogen to form hydrides.
(iv) They react readily with acids to form salts and liberate hydrogen gas.
M + 2HCl → MCl₂ + H₂(g) ↑
(v) They are strong reducing agents. However, their reducing power is less than that of alkali metals. As we move down the group, the reducing power increases.
(vi) Similar to alkali metals, the alkaline earth metals also dissolve in liquid ammonia to give deep blue coloured solutions.
M + (x – y) NH₃ → [M (NH₃)_x]²⁺ + 2 [e (NH₃)_y]^–

Question 3.
Why are alkali metals not found in nature?
Answer:
Alkali metals include lithium, sodium, potassium, rubidium, cesium, and francium. These metals have only one electron in their valence shell, which they lose easily, owing to their low ionization energies. Therefore, alkali metals are highly reactive and are not found in nature in their elemental state.

Question 4.
Find out the oxidation state of sodium in Na₂O₂.
Answer:
Let the oxidation state of Na be x. The oxidation state of oxygen, in case of peroxides, is -1.
Therefore, 2(x) + 2(-1) = 0 ⇒ 2x – 2 = 0 ⇒ 2x = 2 ⇒ x = ±1
Therefore, the oxidation state of sodium in Na₂O₂ is +1.

Question 5.
Explain why is sodium less reactive than potassium?
Answer:
In alkali metals, on moving down the group, the atomic size increases and the effective nuclear charge decreases.
Because of these factors, the outermost electron in potassium can be lost easily as compared to sodium. Hence, potassium is more reactive than sodium.

Question 6.
Compare the alkali metals and alkaline earth metals with respect to (i) ionization enthalpy (ii) basicity of oxides and (iii) solubility of hydroxides.
Answer:

• Ionization enthalpies : The first ionization enthalpies of the alkaline earth metals are higher than those of the corresponding alkali metals. This is due to their small size as compared to the corresponding alkali metals. But second ionization enthalpies of the alkaline earth metals are smaller than those of the corresponding alkali metals.
• Basicity of oxides : The oxides of the alkali and alkaline earth metals dissolves in water to form basic hydroxides. The alkaline earth metal hydroxides are however less basic and less stable than alkali metal hydroxides.
• Solubility of hydroxides : The solubility of hydroxides of alkaline earth metals is relatively less than their corresponding alkali metal hydroxides.

Question 7.
In what ways lithium shows similarities to magnesium in its chemical behaviour?
Answer:
Similarities between lithium and magnesium are as follows :
(i) Both Li and Mg react slowly with cold water.
(ii) The oxides of both Li and Mg are much less soluble in water and their hydroxides decompose at high temperature.
im 1

(iii) Both Li and Mg react withN2 to form nitrides.
im 2

(iv) Neither Li nor Mg form peroxides or superoxides.
(v) The carbonates of both are covalent in nature. Also, these decompose on heating.
im 3

(vi) Li and Mg do not form solid bicarbonates.
(vii) Both LiCl and MgCl₂ are soluble in ethanol owing to their covalent nature.
(viii) Both LiCl and MgCl₂ are deliquescent in nature. They crystallize from aqueous solutions as hydrates, for example, LiCl.₂H₂O and MgCl₂ . 8H₂O.

Question 8.
Explain why can alkali and alkaline earth metals not be obtained by chemical reduction methods?
Answer:
In the process of chemical reduction, oxides of metals are reduced using a stronger reducing agent. Alkali metals and alkaline earth metals are among the strongest reducing agents and the reducing agents that are stronger than them are not available. Therefore, they cannot be obtained by chemical reduction of their oxides.

Question 9.
Why are potassium and caesium, rather than lithium used in photoelectric cells?
Answer:
Potassium and caesium have much lower ionization enthalpy than that of lithium. Therefore, these metals on exposure to light emit electrons easily but lithium does not. That’s why K and Cs rather than Li are used in photoelectric cells.

Question 10.
When an alkali metal dissolves in liquid ammonia the solution can acquire different colours. Explain the reasons for this type of colour change.
Answer:
When an alkali metal is dissolved in liquid ammonia, it results in the formation of a deep blue coloured solution.
M + (x + y) NH₃ → [M (NH₃)_x]⁺ + [e^-1 (NH₃)_y] [Ammoniated electron]
The ammoniated electrons absorb energy corresponding to red region of visible light. Therefore, the transmitted light is blue in colour.
At a higher concentration (3 M), clusters of metal ions are formed. This causes the solution to attain a copper-bronze colour and a characteristic metallic lustre.

Question 11.
Beryllium and magnesium do not give colour to flame whereas other alkaline earth metals do so. Why?
Answer:
When an alkaline earth metal is heated, the valence electrons get excited to a higher energy level. When this excited electron comes back to its lower energy level, it radiates energy, which belongs to the visible region. Hence, the colour is observed. In Be and Mg, the electrons are strongly bound. The energy required to excite these electrons is very high. Therefore, when the electron reverts back to its original position the energy released does not fall in the visible region. Hence, no colour in the flame is seen.

Question 12.
Discuss the various reactions that occur in the Solvay process.
Answer:
In Solvay ammonia process, CO₂ is passed through brine, (a concentrated solution of NaCl) saturated with ammonia. The process involves the formation of a sparingly soluble sodium bicarbonate.
NaCl + NH₃ + CO₂ + H₂O → NaHCO₃↓ + NH₄Cl .
Sodium bicarbonate thus formed is filtered, dried and heated to obtain sodium carbonate.
im 4
CO₂ used in carbonating tower is prepared by heating calcium carbonate and the quicklime, CaO thus formed is dissolved in water to form slaked lime, Ca(OH)₂.
im 5
In ammonia recovery tower, NH3 is prepared by heating NH₄Cl with Ca(OH)₂.
im 6

Question 13.
Potassium carbonate cannot be prepared by Solvay process. Why?
Answer:
Solvay process cannot be used to prepare potassium carbonate. This is because unlike sodium bicarbonate, potassium bicarbonate is fairly soluble in water and does not precipitate out.

Question 14.
Why is Li₂CO₃ decomposed at a lower temperature whereas Na₂CO₃ at higher temperature?
Answer:
As we move down the alkali metal group, the electropositive character increases. This causes an increase in the stability of alkali carbonates.
However, lithium carbonate is not so stable to heat. This is because lithium carbonate is covalent. Lithium ion, being very small in size, polarizes a large carbonate ion, leading to the formation of more stable lithium oxide.
im 7
Therefore, lithium carbonate decomposes at a low temperature while a stable sodium carbonate decomposes at a high temperature.

Question 15.
Compare the solubility and thermal stability of the following compounds of the alkali metals with those of the alkaline earth metals (a) Nitrates (b) Carbonates (c) Sulphates.
Answer:
(a) Nitrates of alkali metals and alkaline earth metals :
(i) Nitrates of alkali metals are thermally not stable and decompose on heating to give MNO₂ and O₂ (except LiN03) whereas nitrates of alkaline earth metals decompose on heating give their oxides, nitrogen dioxide and oxygen gas.

im 8

(ii) Nitrates of alkali metals are highly soluble in water whereas alkaline earth metal nitrates are sparingly soluble and crystallise with six molecules of water.
(b) Carbonates of alkali metals and alkaline earth metals:
(i) Carbonates of alkali metals except Li are quite stable upto 1273 K and do not decompose, whereas carbonates of alkaline earth metals decompose at different temperatures, to give their oxides and carbon dioxide.

im 9

The thermal stability of carbonates of alkaline earth metals increase down the group.
BeCO₃ is least stable and BaCO₃ is most stable.

(ii) All the carbonates of alkali metals are generally soluble in water and their solubility increases rapidly on descending the group. This is due to the reason that their lattice energies decrease more rapidly than their hydration energies down the group. In the case of carbonates of alkaline earth metals they are sparingly soluble in water and their solubility decreases down the group from Be to Ba. For example, MgCO₃ is slightly soluble in water, but BaC03 is almost insoluble.

(c) Sulphates of alkali metals and alkaline earth metals :
(i) The sulphates of alkali metals are thermally quite stable whereas the sulphates of alkaline earth metals decompose on heating to give oxides and SO₃. The temperature of decomposition increases
down the group.
im 10

(ii) The sulphates of alkali metals Na and K are soluble in water. As far as the solubility of sulphates of alkaline earth metals in water is concerned, BeSO₄ and MgSO₄ are highly soluble, CaSO₄ is sparingly soluble, but the sulphates of Sr, Ba and Ra are virtually insoluble. Thus, the solubility of their sulphates in water decreases down the group.
BeSO₄ > MgSO₄ > CaSO₄ > SiSO₄ > BaSO₄

Question 16.
Starting with sodium chloride how would you proceed to prepare (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate?
Answer:
(i) Sodium metal from sodium chloride: Sodium is prepared from fused (molten) sodium chloride. Sodium chloride is mixed with CaCl₂ and KF [to lower the M.Pt. of NaCl to 850-875 K] and subjected to electrolysis (in DOWN’S CELL) when the following reactions occur :

im 11

Sodium, liberated at the cathode, is collected in kerosene oil and chlorine gas is liberated at the anode.
(ii) Sodium hydroxide from sodium chloride: Sodium hydroxide (caustic soda) is generally prepared by the electrolysis of brine solution (NaCl solution in water) in Castner Kellner cell. A mercury cathode and carbon anode are used. Sodium metal discharged at the cathode combines with mercury to form sodium amalgam. Cl₂ gas is evolved at the anode.

im 12

The amalgam is treated with water to give sodium hydroxide and hydrogen gas.
2Na – amalgam + 2H₂O → 2NaOH + 2Hg + H₂↑

(iii) Sodium peroxide from sodium chloride: Sodium metal obtained, by the electrolysis of molten sodium chloride is heated with O₂ at about 575 K when sodium forms mainly sodium peroxide.

im 13

(iv) Sodium carbonate from sodium chloride: Sodium carbonate is prepared from an aqueous solution of NaCl by SOLVAY PROCESS. In this process, CO₂ is passed through NaCl solution saturated with ammonia, when following reactions occur :

2NH₃ + H₂O + CO₂ → (NH₄)₂ CO₃
(NH₄)₂ CO₃ + H₂O + CO₂ → 2NH₄HCO₃
NH₄HCO₃ + NaCl → NH₄Cl + NaHCO₃
im 14

Question 17.
What happens when (i) magnesium is burnt in air (ii) quick lime is heated with silica (iii) chlorine reacts with slaked lime (iv) calcium nitrate is heated?
Answer:
(i) Magnesium bums in air with a dazzling light to form MgO and Mg3N₂.
im 15

(ii) Quick lime (CaO) combines with silica (SiO₂) to form slag.
im 16

(iii) When chlorine is added to slaked lime, it gives bleaching powder.
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(iv) Calcium nitrate, on heating decomposes to give calcium oxide.
im 18

Question 18.
Describe two important uses of each of the following (i) caustic soda (ii) sodium carbonate (iii) quicklime.
Answer:
(i) Uses of caustic soda
(a) It is used in soap industry.
(b) It is used as a reagent in laboratory.
(ii) Uses of sodium carbonate
(a) It is generally used in glass and soap industry.
(b) It is used as a water softener.
(iii) Uses of quick lime
(a) It is used as a starting material for obtaining slaked lime.
(b) It is used in the manufacture of glass and cement.

Question 19.
Draw the structure of (i) BeCl₂ (vapour) (ii) BeCl₂ (solid).
Answer:
(i) In the vapour state, BeCl₂ exists as a monomer with a linear structure.
im 19

(ii) In the solid state, BeCl₂ exists as a polymer in condensed phase.
im 20

Question 20.
The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding salts of magnesium and calcium are sparingly soluble in water. Explain.
Answer:
The atomic size of sodium and potassium is larger than that of magnesium and calcium. Thus, the lattice energies of carbonates and hydroxides formed by calcium and magnesium are much more than those of sodium and potassium.
Hence, carbonates and hydroxides of sodium and potassium dissolve readily in water whereas those of calcium and magnesium are only sparingly soluble.

Question 21.
Describe the importance of the following (i) limestone (ii) cement (iii) plaster of paris.
Answer:
(i) Importance of limestone
(a) It is used in the preparation of lime and cement.
(b) It is used as a flux during the smelting of iron ores.

(ii) Importance of cement
(a) It is used in plastering and in construction of bridges.
(b) It is used in concrete.

(iii) Importance of plaster of Paris
(a) It is used in surgical bandages.
(b) It is also used for making casts and moulds.

Question 22.
Why are lithium salts commonly hydrated and those of the other alkali ions usually anhydrous?
Answer:
Because of its smallest size among alkali metals, Li+ has the maximum degree of hydration. That’s why lithium salts are commonly hydrated and those of other alkali metal ions usually anhydrous.
im 21

Question 23.
Why is LiF almost insoluble in water whereas LiCl soluble not only in water but also in acetone?
Answer:
LiF is insoluble in water. On the contrary, LiCl is soluble not only in water, but also in acetone. This is mainly because of the greater ionic character of LiF as compared to LiCl. The solubility of a compound in water depends on the balance between lattice energy and hydration energy. Since fluoride ion is much smaller in size than chloride ion, the lattice energy of LiF is greater than that of LiCl. Also there is not much difference between the hydration energies of fluoride ion and chloride ion. Thus, the net energy change during the dissolution of LiCl in water is more exothermic than that during the dissolution of LiF in water. Hence, low lattice energy and greater covalent character are the factors making LiCl soluble not only in water, but also in acetone.

Question 24.
Explain the significance of sodium, potassium, magnesium and calcium in biological fluids.
Answer:
(i) Sodium (Na):
Sodium ions are found primarily in the blood plasma. They are also found in the interstitial fluids surrounding the cells.
(a) Sodium ions help in the transmission of nerve signals.
(b) They help in regulating the flow of water across the cell membranes.
(c) They also help in transporting sugars and amino acids into the cells.

(ii) Potassium (K):
Potassium ions are found in the highest quantity within the cell fluids.
(a) K⁺ ions help in activating many enzymes.
(b) They also participate in oxidising glucose to produce ATP.
(c) They also participate in transmitting nerve signals.

(iii) Magnesium (Mg) and calcium (Ca) :
Magnesium and calcium are referred to as macro-minerals. This term indicates their higher abundance in the human body system.
(a) Mg helps in relaxing nerves and muscles.
(b) Mg helps in building and strengthening bones.
(c) Mg maintains normal blood circulation in the human body system.
(d) Ca helps in the coagulation of blood.
(e) Ca also helps in maintaining homeostasis.

Question 25.
What happens when
(i) sodium metal is dropped in water?
(ii) sodium metal is heated in free supply of air?
(iii) sodium peroxide dissolves in water?
Answer:
(i) When sodium metal is dropped in water, it reacts violently to form sodium hydroxide and hydrogen gas. The chemical equation involved in the reaction is:
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
(ii) On being heated in air, sodium reacts vigorously with oxygen to form sodium peroxide. The chemical equation involved in the reaction is:
2Na(s) + O₂(g) → Na₂O₂(s)
(iii) When sodium peroxide is dissolved in water, it is readily hydrolysed to form sodium hydroxide and water. The chemical equation involved in the reaction is:
Na₂O₂(s) + 2H₂O(7) → 2NaOH(aq) + H₂O₂(aq)

Question 26.
Comment on each of the following observations:
(a) The mobilities of the alkali metal ions in aqueous solution are Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺
(b) Lithium is the only alkali metal to form a nitride directly.
(c) \(\mathbf{H}^{\ominus}\) for M²⁺ (aq) + 2e^– → M(s) (where M = Ca, Sr or Ba) is nearly constant.
Answer:
(a) On moving down the alkali group, the ionic and atomic sizes of the metals increase. The given alkali metal ions can be arranged in the increasing order of their ionic sizes as:
Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺

Smaller the size of an ion, the more highly is it hydrated. Since, Li⁺ is the smallest, it gets heavily hydrated in an aqueous solution. On the other hand, Cs+ is the largest and so it is the least hydrated. The given alkali metal ions can be arranged in the decreasing order of their hydrations as:
Li⁺ > Na⁺ > K⁺ > Rb⁺ > Cs⁺
Greater the mass of a hydrated ion, the lower is its ionic mobility. Therefore, hydrated Li⁺ is the least mobile and hydrated Cs⁺ is the most mobile. Thus, the given alkali metal ions can be arranged in the increasing order of their mobilities as:
Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺

(b) Unlike the other elements of group 1, Li reacts directly with nitrogen to form lithium nitride. This is because Li⁺ is very small in size and so its size is the most compatible with the N^3- ion. Hence, the lattice energy released is very high. This energy also overcomes the high amount of energy required for the formation of the N^3- ion.

(c) Electrode potential (\(\mathbf{E}^{\ominus}\)) of any M²⁺/M electrode depends upon three factors :
(i) Ionisation enthalpy
(ii) Enthalpy of hydration
(iii) Enthalpy of vaporisation
The combined effect of these factors is approximately the same for Ca, Sr, and Ba. Hence, their electrode potentials are nearly constant.

Question 27.
State as to why
(a) a solution of Na₂CO₃ is alkaline?
(b) alkali metals are prepared by electrolysis of their fused chlorides?
(c) sodium is found to be more useful than potassium?
Answer:
(a) When sodium carbonate is added to water, it hydrolyses to give sodium bicarbonate and sodium hydroxide (a strong base). As a result, the solution becomes alkaline.
Na₂CO₃ + H₂O → NaHCO₃ + NaOH

(b) It is not possible to prepare alkali metals by the chemical reduction of their oxides as they themselves are very strong reducing agents. They cannot be prepared by displacement reactions either (wherein one element is displaced by another). This is because these elements are highly electropositive. Electrolysis of aqueous solutions can neither be used to extract these elements. This is because the liberated metals react with water. Hence, to overcome these difficulties, alkali metals are usually prepared by the electrolysis of their fused chlorides.

(c) Blood plasma and the interstitial fluids surrounding the cells are the regions where sodium ions are primarily found. Potassium ions are located withimthe cell fluids. Sodium ions are involved in the transmission of nerve signals, in regulating the flow of water across the cell membranes, and in transporting sugars and amino acids into the cells. Hence, sodium is found to be more useful than potassium.

Question 28.
Write balanced equations for reactions between
(a) Na₂O₂ and water
(b) KO₂ and water
(c) Na₂O and CO₂
Answer:
(a) The balanced chemical equation for the reaction between Na202 and water is:
Na₂O₂(s) + 2H₂O(l) → 2NaOH(aq) + H₂O₂(aq)

(b) The balanced chemical equation for the reaction between K02 and water is:
2KO₂(s) + 2H₂O(l) → 2KOH(aq) + H₂O₂(aq) + O₂(g)
or 4KO₂(s) + 2H₂O(l) → 4KOH(aq) + 3O₂(g)

(c) The balanced chemical equation for the reaction between Na₂O and CO₂ is:
Na₂O(s) + CO₂(g) + Na₂CO₃

Question 29.
How would you explain the following observations?
(i) BeO is almost insoluble but BeSO₄ is soluble in water, ‘
(ii) BaO is soluble but BaSO₄ is insoluble in water,
(iii) Lil is more soluble than KI in ethanol.
Answer:
(i) BeO is almost insoluble in water and BeSO₄ is soluble in water. Be²⁺ is a small cation with a high polarising power and O^2- is a small anion. The size compatibility of Be²⁺ and O^2- is high. Therefore, the lattice energy released during their formation is also very high. When BeO is dissolved in water, the hydration energy of its ions is not sufficient to overcome the high lattice energy. Therefore, BeO is insoluble in water. On the other hand, \(\mathrm{SO}_{4}^{2-}\) ion is a large anion. Hence, Be²⁺ can easily polarise \(\mathrm{SO}_{4}^{2-}\) ions, making BeSO₄ unstable. Thus, the lattice energy of BeSO₄ is not very high and so it is soluble in water.

(ii) BaOis soluble in water, but BaSO₄ is not. Ba²⁺ is a large cation and O^2- is a small anion. The size compatibility of Ba²⁺ and O^2- is not high. As a result, BaO is unstable. The lattice energy released during its formation is also not very large. It can easily be overcome by the hydration energy of the ions. Therefore, BaO is, soluble in water. In BaSO₄, Ba+ and \(\mathrm{SO}_{4}^{2-}\) are both large-sized. The lattice energy released is high. Hence, it is not soluble in water.

(iii) Lil is more soluble than KI in ethanol. As a result, due to its small size, the lithium ion has a higher polarising power than the potassium ion. It polarises the electron cloud of the iodide ion to a much greater extent than the potassium ion. This causes a greater covalent character in Li than in KI. Hence, Lil is more soluble in ethanol.

Question 30.
Which of the alkali metal is having least melting point?
(a) Na (b) K (c) Rb (d) Cs
Answer:
(d) Atomic size increases as we move down the alkali group. As a result, the binding energies of their atoms in the crystal lattice decrease. Also, the strength of metallic bonds decreases on moving down a group in the periodic table.
This causes a decrease in the melting point. Among the given metals, Cs is the largest and has the least melting point.

Question 31.
Which one of the following alkali metals gives hydrated salts?
(a) Li (b) Na (c) K (d) Cs
Answer:
(a) Smaller the size of an ion, the more highly is it hydrated. Among the given alkali metals, Li is the smallest in size. Also, it has the highest charge density and highest polarising power. Hence, it attracts water molecules more strongly than the other alkali metals. As a result, it forms hydrated salts such as LiCl . 2 H2O. The other alkali metals are larger than Li and have weaker charge densities. Hence, they usually do not form hydrated salts.

Question 32.
Which one of the alkaline earth metal carbonates is thermally the most stable?
(a) MgCO₃
(b) CaCO₃
(c) SrCO₃
(d) BaCO₃
Answer:
(d) Thermal stability increases with the increase in the size of the cation present in the carbonate. The increasing order of the cationic size of the given alkaline earth metals is
Mg < Ca < Sr < Ba
Hence, the increasing order of the thermal stability of the given alkaline earth metal carbonates is
MgCO₃ < CaCO₃ < SrCO₃ < BaCO₃

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 9 Hydrogen Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 9 Hydrogen

PSEB 11th Class Chemistry Guide Hydrogen InText Questions and Answers

Question 1.
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer:
Hydrogen is the first element in the periodic table. It has the electronic configuration 1s¹. It is similar to alkali metal (ns¹) of group I. It shows resemblance with alkali metals of group I of the periodic table. So it can be placed above the alkali metals in group I of the periodic table.
On the other hand, the electronic configuration of hydrogen shows that it is short of one electron to the nearest noble gas configuration (He) having the electronic configuration 1s². Like halogens it forms covalent bonds (H₂, Cl₂, Br₂, etc.) as well as ionic bonds (e.g. Na⁺ H^–). It forms H⁺ ion by giving one electron and hydride ion (H^–) by gaining one electron. On the basis of its electronic configuration (1s²) hydrogen is placed with other ns¹ elements namely alkali metals in the group I as well as in group 17 of the periodic table. Thus, the position of hydrogen in the periodic table is anomalous.
Hydrogen with so many unique characteristics is, therefore best placed separately in the periodic table of elements.

Question 2.
Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
Answer:
Hydrogen has following three isotopes :
1. Protium,\({ }_{1}^{1} \mathrm{H}\),
2. Deuterium, \({ }_{1}^{2} \mathrm{H}\) or D, and
3. Tritium, \({ }_{1}^{3} \mathrm{H}\) or T
The mass ratio of protium, deuterium and tritium is 1.008 : 2.014 : 3.016 or 1:2:3.

Question 3.
Why does hydrogen occur in a diatomic form rather than in a monoatomic form under normal conditions?
Answer:
The ionization enthalpy of hydrogen atom is very high (1312 kJ mol^-1). Hence, it is very hard to remove its electron. As a result, its tendency to exist in the monoatomic form is rather low. Instead, hydrogen forms a covalent bond with another hydrogen atom and exists as a diatomic (H₂) molecule.

Question 4.
How can the production of dihydrogen, obtained from ‘coal gasification’, be increased?
Answer:
Dihydrogen, produced by coal gasification method as :

The yield of dihydrogen (obtained from coal gasification) can be increased by reacting carbon monoxide (formed during the reaction) with steam in the presence of iron chromate as a catalyst.

This reaction is called the water-gas shift reaction. Carbon dioxide is removed by scrubbing it with a solution of sodium arsenite.

Question 5.
Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in this process?
Answer:
Electrolysis of acidified water using platinum electrodes gives dihydrogen.

The role of an electrolyte is to make water conducting.

Question 6.
Complete the following reactions :

Answer:

Question 7.
Discuss the consequences of high enthalpy of H-H bond in terms of chemical reactivity of dihydrogen.
Answer:
The ionization enthalpy of H-H bond is very high (1312 kJ mol^-1). This indicates that dihydrogen has a low tendency to form H+ ions. Its ionization enthalpy value is comparable to that of halogens. Hence, it forms diatomic molecules (H₂), hydrides with elements, and a large number of covalent bonds.
Since ionization enthalpy is very high, dihydrogen does not possess metallic characteristics (lustre, ductility, etc.) like metals.

Question 8.
What do you understand by (i) electron-deficient,
(ii) electron-precise, and (iii) electron-rich compounds of hydrogen? Provide justification with suitable examples.
Answer:
(i) An electron-deficient hydride : It has very few electrons, less than that required for representing its conventional Lewis structure e.g., diborane (B₂H₆). In B₂H₆, there are six bonds in all, out of which only four bonds are regular i.e., two electrons are shared by two atoms.
The remaining two bonds are three centered-two electron bonds i.e., two electrons are shared by three atoms. Hence, its conventional Lewis structure cannot be drawn.

(ii) An electron-precise hydride : It has sufficient number of electrons to be represented by its conventional Lewis structure e.g., CH₄. The Lewis structure can be written as :

Four regular bonds are formed where two electrons are shared by two atoms.

(iii) An electron-rich hybride : It contains excess electrons as lone pair e.g., NH₃

There are three regular bonds in all with a lone pair of electrons on the nitrogen atom.

Question 9.
What characteristics do you expect from an electron-deficient hydride with respect to its structure and chemical reactions?
Answer:
These hydrides do not have sufficient number of electrons to form normal covalent bonds, e.g., B in BF₃ has 6 electrons in its valence shell. These hydrides are trigonal planar in shape.

These hydrides act as Lewis acids, i.e., electron pair acceptor e.g.,

To make up the deficiency of electrons, these hydrides exist in polymeric forms e.g., B₂H₆, B₄H₁₀, etc. Electron deficient hydrides are very reactive. These reacts readily with metals and non-metals and their compounds, e.g.,

B₂H₆ + 3O₂(g) → B₂O₃(s) + 3H₂O(g)

Question 10.
Do you expect the carbon hydrides of the type (CraH2jt + 2) to act as ‘Lewis’ acid or base? Justify your answer.
Answer:
Carbon hydrides of the type C_nH_2n+2 are CH₄, C₂H₆ etc. in which number of electrons present are just sufficient to write down their conventional Lewis structures.

C has neither extra electrons nor less electrons. Such compounds of the formula C_nH_2n+2 are called ELECTRON-PRECISE compounds. They will act neither as Lewis acids nor as Lewis bases.

Question 11.
What do you understand by the term “non-stoichiometric hydrides”? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer.
Answer:
Non-stoichiometric hydrides are hydrogen-deficient compounds formed by the reaction of dihydrogen with d-block and f-block elements. These hydrides do not follow the law of constant composition.

For example :
LaH₂₈₇, YbH_2.55, TiH_{1.5 – 1.8} etc.
Alkali metals form stoichiometric hydrides. These hydrides are ionic in nature. Hydride ions have comparable sizes (208 pm) with alkali metal ions. Hence, strong binding forces exist between the constituting metal and hydride ion. As a result, stoichiometric hydrides are formed. Alkali metals will not form non-stoichiometric hydrides.

Question 12.
How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.
Answer:
Metallic hydrides are hydrogen deficient, i.e., they do not hold the law of constant composition. It has been established that in the hydrides of Ni, Pd, Ce and Ac hydrogen occupies the interstitial position in lattices allowing further absorption of hydrogen on these metals. Metals like Pd, Pt, etc. have the capacity to accommodate a large volume of hydrogen. Therefore, they are used for the storage of hydrogen and serve as a source of energy.

Question 13.
How does the atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes? Explain.
Answer:
Atomic hydrogen atoms are produced by the dissociation of dihydrogen with the help of an electric arc. This releases a huge amount of energy (435.88 kJ mol^-1). This energy can be used to generate a temperature of 4000 K, which is ideal for welding and cutting metals. Hence, atomic k hydrogen or oxy-hydrogen torches are used for these purposes. For this reason, atomic hydrogen is allowed to recombine on the surface to be welded to generate the desired temperature.

Question 14.
Among NH₃, H₂O and HF, which would you expect to have highest magnitude of hydrogen bonding and why?
Answer:
The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen atoms available for bonding. Among nitrogen, fluorine, and oxygen, the increasing order of their electronegativities are N < O < F. Hence, the order of the extent of hydrogen bonding is HF > H₂O > NH₃.

Question 15.
Saline hydrides are known to react with water violently producing fire. Can CO₂, a well known fire extinguisher, be used in this case? Explain.
Answer:
Saline hydrides (i.e., NaH, LiH, etc.) react with water to form a base and hydrogen gas. The chemical equation used to represent the reaction can be written as:
NaH(s) + H₂O(7) → NaOH(aq) + H₂(g)
This reaction is violent and produces fire.
This type of fire cannot be extinguished by CO₂ because it gets reduced by the hot metal hydride to form sodium format.
NaH + CO₂ → HCOONa

Question 16.
Arrange the following
(i) CaH₂, BeH₂ and TiH₂ in order of increasing electrical conductance.
(ii) LiH, NaH and CsH in order of increasing ionic character.
(iii) H-H, D-D and F-F in order of increasing bond dissociation enthalpy.
(iv) NaH, MgH₂ and H₂O in order of increasing reducing property.
Answer:
(i) BeH₂ < CaH₂ < TiH₂
(ii) LiH < NaH < CsH
(iii) F—F < H—H < D—D
(iv) H₂O < MgH₂ < NaH

Question 17.
Compare the structures of H₂O and H₂O₂.
Answer:
In gaseous phase, water molecule has a bent form with a bond angle of 104.5°. The O—H bond length is 95.7 pm. The structure can be shown as:

Hydrogen peroxide has a non-planar structure both in gas and solid phase. The dihedral angle in gas and solid phase is 111.5° and 90.2° respectively.

Question 18.
What do you understand by the term ‘auto-protolysis’ of water? What is its significance?
Answer:
Auto-protolysis (self-ionization) of water is a chemical reaction in which two water molecules react to produce a hydroxide ion (OH^–) and a hydronium ion (H₃O⁺).

The reaction involved can be represented as :

Auto-protolysis of water indicates its amphoteric nature i.e., its ability to act as an acid as well as a base. The acid-base reaction can be written as :

Question 19.
Consider the reaction of water with F₂ and suggest, in terms of oxidation and reduction, which species are oxidized/reduced.
Answer:

In these reactions, water acts as a reducing agent and hence itself gets oxidised to either oxygen or ozone. Fluorine acts as an oxidising agent and hence itself reduced to F^– ion.

Question 20.
Complete the following chemical reactions.
(i) PbS(s) + H₂O₂(ciq) →
(ii) MnO^–₄(aq) + H₂O₂(aq) →
(iii) CaO(s) + H₂O(g) →
(iv) AlCl₃(g) + H₂O(Z) →
(v) Ca₃N₂(s) + H₂O(l) →
Classify the above into (a) hydrolysis, (b) redox and (c) hydration reactions.
Answer:
(i) PbS(s) + 4H₂O₂(aq) → PbSO₄(s) + 4H₂O(l)
H₂O₂ is acting as an oxidizing agent in the reaction. Hence, it is a redox reaction.

(ii) 2MnO₄^– (aq) + 5H₂O₂(Z) + 6H⁺(aq) → 2Mn²⁺(aq) + 8H₂O(Z) + 5O₂(g)
H₂O₂ is acting as a reducing agent in the acidic medium, thereby oxidizing MnO^–₄(aq). Hence, the given reaction is a redox reaction.

(iii) CaO(s) + H₂O(g) → Ca(OH)₂(aq)
The reactions in which a compound reacts with water to produce other compounds are called hydrolysis reactions. The given reaction is hydrolysis reaction.

(iv) AlCl₃(g) + 6H₂O(Z) → [Al(OH₂)₆]³⁺ (aq) + 3Cl^–(aq)
It is a hydration reaction, because A1C13 is hydrated to [Al(OH₂)₆]³⁺.

(v) Ca₃N₂(s) + 6H₂O(Z) → 3Ca(OH)₂(aq) + 2NH₃(aq)
The reactions in which a compounds reacts with water to produce other compounds are called hydrolysis reactions. The given reaction represents hydrolysis of Ca₃N₂.

Question 21.
Describe the structure of the common form of ice.
Answer:
Ice is the crystalline form of water. It takes a hexagonal form if crystallized at atmospheric pressure, but condenses to cubic form if the temperature is very low.
The three-dimensional structure of ice is represented as :

The structure is highly ordered and has hydrogen bonding. Each oxygen atom is surrounded tetrahedrally by four other oxygen atoms at a distance of 276 pm. The structure also contains wide holes that can hold molecules of appropriate sizes interstitially.

Question 22.
What causes the temporary and permanent hardness of water?
Answer:
Temporary hardness of water is due to the presence of soluble salts of magnesium and calcium in the form of hydrogen carbonates (MHCO₃, where M = Mg, Ca) in water.
Permanent hardness of water is due to the presence of soluble salts of calcium and magnesium in the form of chlorides in water.

Question 23.
Discuss the principle and method of softening of hard water by synthetic ion-exchange resins.
Answer:
The process of treating permanent hardness of water using synthetic resins is based on the exchange of cations (e.g., Na⁺, Ca²⁺, Mg²⁺ etc.) and anions (e.g.,Cl^– , \(\mathrm{SO}_{4}^{2-}\), \(\mathrm{HCO}_{3}^{-}\) etc.) present in water by H⁺ and OH^– ions respectively.

Synthetic resins are of two types :
1. Cation exchange resins
2. Anion exchange resins

Cation exchange resins are large organic molecules that contain the -SO₃H group. The resin is firstly changed to RNa (from ROS₃H) by treating it with NaCl. This resin then exchanges Na⁺ ions with Ca²⁺ and Mg²⁺ ions, thereby making the water soft.

2RNa + M²⁺ (aq) → R₂M(s) + 2Na⁺ (aq)

There are cation exchange resins in H+ form. The resins exchange H⁺ ions for Na⁺, Ca²⁺ and Mg²⁺ ions.

2RH + M²⁺ (aq) ⇌ MR₂(s) + 2H⁺(aq)

Anion exchange resins exchange OH- ions for anions like Cl^–, \(\) and \(\) present in water.

During the complete process, water first passes through the cation exchange process. The water obtained after this process is free from mineral cations and is acidic in nature.
This acidic water is then passed through the anion exchange process where OH^– ions neutralize the H⁺ ions and de-ionize the water obtained.

Question 24.
Write chemical reactions to show the amphoteric nature of water.
Answer:
Water is amphoteric in character. It behaves both as an acid as well as a base. With acids stronger than itself, it behaves as a base and with bases stronger than itself, it acts as an acid.

Question 25.
Write chemical reactions to justify that hydrogen peroxide can function as an oxidizing as well as reducing agent.
Answer:
Hydrogen peroxide, H₂O₂ acts as an oxidizing as well as a reducing agent in both acidic and alkaline media. Reactions involving oxidizing actions are :

Question 26.
What is meant by ‘demineralised’ water and how can it be obtained?
Answer:
Water which does not contain cations and anions is called ‘demineralised’ water. It is soft water. Demineralised water is obtained the same way as soft water is obtained from hard water. Demineralised or deionised water is obtained by passing hard water first through a cation exchange resin (RCOOH or RSO₃H) which removes Ca²⁺ and Mg²⁺ ions from hard water by exchanging them with H⁺ ions and then passing through an anion exchange resin which removes Cl^– and \(\mathrm{SO}_{4}^{2-}\) ions present in hard water by exchanging them with OH^– ions.

Question 27.
Is demineralised or distilled water useful for drinking purposes? If not, how can it be made useful?
Answer:
Water is an important part of life. It contains several dissolved nutrients that are required by human beings, plants and animals for survival.
Demineralised water is free from all soluble minerals. Hence, it is not fit for drinking.
It can be made useful only after the addition of desired minerals in specific amounts, which are important for growth.

Question 28.
Describe the usefulness of water in biosphere and biological systems.
Answer:
Water is essential for all forms of life. It constitutes around 65% of the human body and 95% of plants. Water plays an important role in the biosphere owing to its high specific heat, thermal conductivity, surface tension, dipole moment and dielectric constant.
The high heat of vaporization and heat of capacity of water helps in moderating the climate and body temperature of all living beings.
It acts as a carrier of various nutrients required by plants and animals for various metabolic reactions.
Water is also required for photosynthesis in plants which releases O₂ into the atmosphere.

Question 29.
What properties of water make it useful as a solvent? What types of compound can it (i) dissolve, and (ii) hydrolyse?
Answer:
Water because of its high dielectric constant (78.39) has the ability to dissolve most of the inorganic (ionic) compounds and is, therefore, regarded as a universal solvent. Whereas the solubility of ionic compounds takes place due to ion-dipole interactions (i.e., solvation of ions) the solubility of covalent compounds such as alcohols, amines, urea, glucose, sugar etc. takes place due to tendency of these molecules to form hydrogen bonds with water.

(i) It can dissolve both ionic compounds as well as covalent compounds which can form hydrogen bonds with water.
Ionic compounds whose lattice energy is lower than hydration energy get dissolved in water.
(ii) Water can hydrolyse many oxides (metallic and non-metallic), hydrides, carbides, nitrides, phosphides and many other salts e.g.,

CaO(s) + H₂O(l) → Ca(OH)₂
SO₂(g) + H₂O(l) → H₂SO₃(aq)
CaH₂(s) + 2H₂O(l) → Ca(OH)₂(uq) + 2H₂(g)
SiCl₄(l) + 4H₂O(l) → SiO₂ -2H₂O(s) + 4HCl(aq)
Al₄C₃ + 12H₂O(l) → 4Al(OH)₃ + 3CH₄
Ca₃P₂(s) + 6H₂O → 3Ca(OH)₂ + 2PH₃

Question 30.
Knowing the properties of H20 and DaO, do you think that D20 can be used for drinking purposes?
Answer:
Heavy water (D₂O) acts as a moderator, i.e., it slows the rate of a reaction. Due to this property of D₂O, it cannot be used for drinking purposes because it will slow down anabolic and catabolic reactions takes place in the body and lead to a casualty.

Question 31.
What is the difference between the terms ‘hydrolysis’ and ‘hydration’?
Answer:
Hydrolysis is defined as a chemical reaction in which hydrogen and hydroxide ions (H+ and OH- ions) of water molecule react with a compound to form products. For example :
NaH + H₂O → NaOH + H₂
Hydration is defined as the addition of one or more water molecules to ions or molecules to form hydrated compounds. For example :
CUSO₄ + 5H₂O → CUSO₄ . 5H₂O

Question 32.
How can saline hydrides remove traces of water from organic compounds? ’
Answer:
Saline hydrides are ionic in nature. They react with water to form a metal hydroxide along with the liberation of hydrogen gas. The reaction of saline hydrides with water can be represented as :
AH(s) + H20(Z) → AOH(aq) + H2(g)
(where, A= Na, Ca, )

When added to an organic solvent, they react with water present in it. Hydrogen escapes into the atmosphere leaving behind the metallic hydroxide.Then, the dry organic solvent distills over.

Question 33.
What do you expect the nature of hydrides is, if formed by elements of atomic number 15, 19, 23 and 44 with dihydrogen? Compare their behaviour towards water,
Answer:
The elements of atomic number 15, 19, 23 and 44 are nitrogen, potassium, vanadium and ruthenium respectively.
1. Hydride of nitrogen
Hydride of nitrogen (NH₃) is a covalent molecule. It is an electron-rich hydride owing to the presence of excess electrons as a lone pair on nitrogen.

2. Hydride of potassium
Dihydrogen forms an ionic hydride with potassium owing to the high electropositive nature of potassium. It is crystalline and non-volatile in nature.

3. Hydrides of Vanadium and Ruthenium
Both vanadium and ruthenium belong to the d-block of the periodic table. The metals of d-block form metallic or non-stoichiometric hydrides. Hydrides of vanadium and ruthenium are therefore, metallic in nature having a deficiency of hydrogen.

4. Behaviour of hydrides towards water
Potassium hydride reacts violently with water as :
KH(s) + H₂O(aq) → KOH(aq) + H₂(g)
Ammonia (NH₃) behaves as a Lewis base and reacts with water as :
H₂O(Z) + NH₃(aq) ⇌ OH^–(aq) + \(\mathrm{NH}_{4}^{+}\)(aq)
Hydrides of vanadium and ruthenium do not react with water. Hence, the increasing order of reactivity of the hydrides is as: H < NH₃ < KH.

Question 34.
Do you expect different products in solution when aluminium (III) chloride and potassium chloride treated separately with
(i) normal water (ii) acidified water, and (iii) alkaline water? Write equations wherever necessary.
Answer:
Potassium chloride (KCl) is the salt of a strong acid (HCl) and strong base (KOH). Hence, it is neutral in nature and does not undergo hydrolysis in normal water. It dissociates into ions as follows :

In acidified and alkaline water, the ions do not react and remain as such. Aluminium (III) chloride is the salt of a strong acid (HCl) and weak base [Al(OH)₃]. Hence, it undergoes hydrolysis in normal water.

In acidified water, H⁺ ions react with Al(OH)₃ forming water and giving Al³⁺ ions. Hence, in acidified water, AlCl₃ will exist as Al³⁺(aq) and Cl^–(aq) ions.

In alkaline water the following reaction takes place :

Question 35.
How does H₂O₂ behave as a bleaching agent?
Answer:
H₂O₂ acts as a bleaching agent due to the nascent oxygen.
H₂O₂ → H₂O + O
Coloured matter + [O] → Colourless matter
It bleaches materials like silk, hair, ivory, cotton, wool, etc.

Question 36.
What do you understand by the terms :
(i) hydrogen economy (ii) hydrogenation (iii) ‘syngas’ (iv) water-gas shift reaction (v) fuel-cell?
Answer:
(i) Hydrogen economy : Hydrogen economy is a technique of using dihydrogen in an efficient way. It involves transportation and storage of dihydrogen in the form of liquid or gas.
Dihydrogen releases more energy than petrol and is more eco-friendly. Hence, it can be used in fuel cells to generate electric power. Hydrogen economy is about the transmission of this energy in the form of dihydrogen.

(ii) Hydrogenation : It refers to the addition of dihydrogen to another reactant. This process is used to reduce a compound in the presence of a suitable catalyst. For example, hydrogenation of vegetable oil using nickel as a catalyst gives edible fats such as vanaspati ghee etc.

(iii) Syngas : Syngas is a mixture of carbon monoxide and dihydrogen. Since the mixture of the two gases is used for the synthesis of methanol, it is called syngas, synthesis gas, or water gas.

Syngas is produced on the action of steam with hydrocarbons or coke at a high temperature in the presence of a catalyst.

(iv) Water gas shift reaction : The production of hydrogen by reacting carbon monoxide (CO) of syngas mixtures with steam in the presence of iron chromate as catalyst is called water-gas shift reaction.

CO₂ is removed by scrubbing with sodium arsenite solution.
(v) Fuel cell: It is a device which converts the energy produced during the combustion of a fuel directly into electrical energy. One such fuel cell is hydrogen-oxygen fuel cell. It does not cause any pollution. Fuel cells generated electricity with conversion efficiency of 70-85%.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 8 Redox Reactions Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

PSEB 11th Class Chemistry Guide Redox Reactions InText Questions and Answers

Question 1.
Assign oxidation numbers to the underlined elements in each of the following species:

Answer:
(a) Let the oxidation number of P be x.
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = – 2
In neutral compounds, the sum of the oxidation numbers of all the atoms is zero.
1 (+1) + 2 (+1) +1 00 + 4 (-2) = 0
1 + 2 + x – 8 = 0
3 + x + (-8) – 0
x = 8 – 3
⇒ x = + 5
Hence, the oxidation number of P is +5.

(b) Let the oxidation number of S be x.

1 (+1) +1 (+1) +1 (x) + 4 (-2) = 0
⇒ 1 + 1 + x – 8 = 0
⇒ x = + 6
Hence, the oxidation number of S is +6.

(c) Let the oxidation number of P be x.

4 (+1) + 2 (x) + 7 (-2) = 0
⇒ 4 + 2x – 14 = 0
⇒ 2x = +10
⇒ x = + 5
Hence, the oxidation number of P is + 5.

(d) Let the oxidation number of Mn is x.

2 (+1) + x + 4 (-2) = 0
⇒ 2 + x — 8 = 0
⇒ x = + 6
Hence, the oxidation number of Mn is + 6.

(e) Let the oxidation number of O be x.

1 (+ 2) + 2 (x) = 0
⇒ 2 + 2x = 0
⇒ x = – 1
Hence, the oxidation number of O is -1.

(f) Let the oxidation number of B be x.

1 (+1)+1 (x) + 4 (-1) = 0
⇒ 1 + x – 4 = 0
⇒ x = + 3
Hence, the oxidation number of B is + 3.

(g) Let the oxidation number of S is x.

2 (+1) + 2 (x) + 7 (-2) = 0
⇒ 2 + 2x -14 = 0
⇒ 2x = +12
x = +6
Hence, the oxidation number of S is + 6.

(h) Let the oxidation number of S be x.

1 (+1) +1 (+ 3) + 2 (x) + 8 (-2) +12 (2 x 1 + (-2)) = 0
⇒ 1 + 3 + 2x -16 + 24 – 24 = 0
⇒ 2x = 12
x = + 6
Hence, the oxidation number of S is + 6.

Question 2.
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?


Answer:
(a) In

the oxidation number (O. N.) of K is +1. Hence, the average oxidation number of I is \(\frac{-1}{3}\). However, O.N. cannot be fractional.
Therefore, we will have to consider the structure of KI₃ to find the oxidation states.
In a KI₃ molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI₃ molecule, the O.N. of the two I atoms forming the I₂ molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is -1.

(b) Let the oxidation number of S be x.

2 (+1) + 4 (x) + 6 (-2) = 0
=» 2 + 4x – 12 = 0
⇒ 4x = +10
⇒ x = + 2 \(\frac{1}{2}\)
However, O.N. cannot be fractional. Hence S must be present in different oxidation states in the molecule.

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c) Let the oxidation number of Fe be x.

3(x) + 4(-2) = 0
3x – 8 = 0
x = \(+\frac{8}{3}\)
However O.N. cannot’be fractional. Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of+3

(d) Let oxidation number of C be x.

2 (x) + 6 (+1) + 1 (-2) = 0
2x + 4 = 0
x = -2
Hence, the O.N. of C is – 2

(e) Let the oxidation number of C be x.

2 (x) + 4 (+1) + 2 (-2) = 0
2x = 0
x = 0
Therefore, the average oxidation number of C is zero.
Let us consider the structure of CH₃COOH

Oxidation number of atom = 1(+1) + x+1(-2) +1(-1) = 0
x = +2
Similarly, oxidation number of C₂ atom
3(+1) + x+ 1(-1) = 0
x = -2.

Question 3.
Justify that the following reactions are redox reactions:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆(g) + 3LiCl (s) + 3 AlCl₃(S)
(d) 2K (s) + F₂ (g) → 2K⁺F^–(s)
(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O (g)
Answer:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
Let us write the oxidation number of each element involved in the given reaction as :

Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H₂ to +1 in H₂O i.e., H₂ is oxidized to H₂O. Hence, this reaction is a redox reaction.

(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
Let us write the oxidation number of each element in the given reaction as:

Here, the oxidation number of Fe decreases from +3 in Fe₂O₃ to 0 in Fe i.e., Fe₂O₃ is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO₂ i.e., CO is oxidized to CO₂.
Hence, the given reaction is a redox reaction.

(c) 4BCl₃ (g) + 3LiAlH₄ (s) → 2B₂H₆(g) + 3LiCl (s) + 3 AlCl₃(S)
The oxidation number of each elements in the given reaction can be represented as:

In this reaction, the oxidation number of B decreases from +3 in BCl₃ to -3 in B₂H_6- i.e., BCl₃ is reduced to B₂H₆. Also, the oxidation number of H increases from -1 in LiAlH₄ to +1 in B₂H₆ i.e., LiAlH₄ is oxidized to B₂H₆. Hence, the given reaction is a redox reaction.

(d) 2K (s) + F₂ (g) → 2K⁺F^–(s)
The oxidation number of each element in the given reaction can be represented as:

In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F₂ to -1 in KF i.e., F₂ is reduced to KF.
Hence, the above reaction is a redox reaction.

(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O (g)
The oxidation number of each elements in the given reaction can be represented as:

Here, the oxidation number of N increases from -3 in NH₃ to +2 in NO. On the other hand, the oxidation number of O₂ decreases from 0 in O₂ to -2 in NO and H₂O i.e.,O₂ is reduced. Hence, the given reaction is a redox reaction.

Question 4.
Fluorine reacts with ice and results in the change:
H₂O(S) + F₂(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Ans. Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

Here, we have observed that the oxidation number of F increases from 0 in F₂ to +1 in HOF. Also, the oxidation number decreases from 0 in F₂ to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

Question 5.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) and \(\mathrm{NO}_{3}^{-}\). Suggest structure of these compounds. Count for the fallacy.
Answer:

2 (+1) +1 (x) + 5 (-2) = 0
⇒ 2 + x -10 = 0
⇒ x = + 8
However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6. The structure of H2S05 is shown as follows :

Now, 2 (+1) +1 (x) + 3 (-2) + 2 (-1) – 0
⇒ 2 + x – 6 – 2 = 0
⇒ x = + 6
Therefore, the O.N. of S is +6.

2 (x) + 7 (-2) = – 2 ⇒ 2x -14 = – 2
⇒ x = + 6
The structure of Cr2Oy is shown as follows:

Let the oxidation number of each Cr atom be
4(-2) + (-2) +1(-2) + 2x = 0
– 8 – 2 – 2 + 2x = 0
2x = +12
x = +6

Oxidation number of Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is same.
Hence, there is no fallacy.

1 (x) + 3(-2) = -1
= x – 6 = -1
x = +5
The structure of \(\mathrm{NO}_{3}^{-}\) is shown as follows:

Let the oxidation number of N be x.
1(-1) + x + 1(-2) + 1(-2) = 0
x = + 5
Oxidation number of N in \(\mathrm{NO}_{3}^{-}\) ion is same.
Hence, there is no fallacy.

Question 6.
Write formulas for the following compounds:
(a) Mercury (II) chloride
(b) Nickel (II) sulphate
(c) Tin (IV) oxide
(d) Thallium (I) sulphate
(e) Iron (III) sulphate
(f) Chromium (III) oxide
Answer:
(a) Mercury (II) chloride: Hg (II) Cl₂
(b) Nickel (II) sulphate: Ni (II) SO₄
(c) Tin (IV) oxide : Sn (IV) O₂
(d) Thallium (I) sulphate: Tl₂ (I) SO₄
(e) Iron (III) sulphate: Fe₂(III) (SO₄)₃
(f) Chromium (III) oxide: Cr₂(III)O₃

Question 7.
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
Answer:
The substances where carbon can exhibit oxidation states from -4 to +4 are listed in the following table :

Substance	Formula	Oxidation State of Carbon
Methane	CH4	-4
Ethane	C2H6	-3
Ethene	C2H4	-2
Ethyne	C2H2	-1
Dichloromethane	CH2Cl2	0
Hexachlorobenzone	C6Cl6	+1
Carbon monoxide	CO	+2
Oxalic acid	(COOH)2	+3
Carbon dioxide	CO2	+4

The substances where nitrogen can exhibit oxidation sates from -3 to +5 are listed in in the following table.

Substance	Formula	Oxidation State of Nitrogen
Ammonia	NH3	-3
Hydrazine	N2H4	-2
Hydride	N2H2	-1
Dinitrogen gas	N2	0
Nitrous oxide	N2O	+1
Nitric oxide	NO	+2
Dinitrogen trioxide	N2O3	+3
Nitrogen dioxide	NO2	+4
Nitrogen pentoxide	N2O5	+5

Question 8.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Answer:
(i) In sulphur dioxide (SO₂), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to -2.
Therefore, SO₂ can act as an oxidising as well as reducing agent.
(ii) In hydrogen peroxide (H₂O₂), the O.N. of O is -1 and the range of the O.N. that O can have is from 0 to -2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H₂O₂ can act as an oxidising as well as reducing agent.
(iii) In ozone (O₃) the O.N. of O is zero and the range of the O.N. that O can have is from 0 to -2. Therefore, the O.N. of O can only decrease in this case. Hence,O₃ acts only as an oxidant.
(iv) In nitric acid (HNO₃) the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to -3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO₃ acts only as an oxidizing agent.

Question 9.
Consider the reactions:
(a) 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆ (aq) + 6O₂(g)
(b) O₃(g) + H₂O₂(Z) → H₂O(1) + 2O₂(g)
Why it is more appropriate to write these reactions as:
(a) 6CO₂ (g) + 12H₂O(1) → C₆H₁₂O₆ (aq) + 6H₂O(l) + 6O₂ (g)
(b) O₃(g) + H₂O₂(Z) → + H₂O(Z) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer:
(a) The process of photosynthesis involves two steps :
Step 1: H₂O decomposes to give H₂ and O₂.
2H₂O(l) → 2H₂(g) + O₂(g)
Step 2: The H₂ produced in step 1 reduces CO₂ thereby producing glucose (C₆H₁₂O₆) and H₂O.
6CO₂(g) + 12H₂(g) → C₆H₁₂O₆(S) + 6H₂O(l)
Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.
The path of this reaction can be investigated by using radioactive H₂O¹⁸ in place of H₂O

(b) O₂ is produced from each of the two reactants O₃ and H₂O₂. For this reason O₂ is written twice.
The given reaction involves two steps. First O₃ decomposes to form O₂ and O. In the second step H₂O₂ reacts with the O produced in the first step thereby producing H₂O and O₂.

The path of this reaction can be investigating by using \(\mathrm{H}_{2} \mathrm{O}_{2}^{18}\) or \(\mathrm{O}_{3}^{18}\).

Question 10.
The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Answer:
AgF₂ → Ag + F₂
The oxidation state of Ag in Ag F₂ is +2. But +2 is an unstable oxidation state of Ag. Therefore, whenever Ag F₂ is formed, silver readily accepts an electron to form Ag⁺. This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, Ag F₂ acts as a very strong oxidizing agent.

Question 11.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving two illustrations.
Answer:
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. This can be illustrated as follows:
(i) P₄ and F₂ are reducing and oxidising agents respectively.
If an excess of P₄ is treated with F₂ then PF₃ will be produced, where in the oxidation number (O.N.) of P is +3.

However, if P₄ is treated with an excess of F₂, then PF₅ will be produced, wherein the O.N. of P is +5.

(ii) K acts as a reducing agent, whereas O₂ is an oxidising agent.
If an excess of K reacts with O₂, then K₂O will be formed, wherein the O.N. of O is -2.

However, if K reacts with an excess of O₂, then K₂O₂ will be formed, wherein the O.N. of O is -1.

Question 12.
How do you account for the following observations?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer:
(a) Oxidation of toluene to benzoic acid in acidic medium.

Oxidation of toluene to benzoic acid in basic and neutral medium.

On industrial scale, alcoholic potassium permanganate is preferred to acidic or alkaline potassium permanganate because in the presence of alcohol, both the reactants KMnO₄ and C₆H₅CH₃ are mixed very well and form homogeneous solution and in homogeneous medium reaction takes place faster than in heterogeneous medium. Further more in neutral medium, OH^– ions are produced in the reaction itself.

(b)

HCl is a weak reducing agent. k cannot reduce H₂SO₄ to SO₂ that’s why pungent smelling gas HCl is obtained.

HBr is a strong reducing agent, it reduces H₂S0₄ to SO₂ and is itselfoxidised to Br₂. That’s why we get red vapours of bromine when conc.
H₂SO₄ reacts with inorganic mixture containing bromide salt.

Question 13.
Identify the substance oxidised, reduced, oxidising agent andreducing agent for each of the following reactions:
(a) 2AgBr(s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)
(b) HCHO(l) +2 [Ag (NH₃)₂]⁺ (aq) + 3OH^–(aq) → 2Ag(s) + HCOO^–(aq) + 4NH₃(aq) + 2H₂O (l)
(c) HCHO(l) + 2Cu²⁺ (aq) + 5 OH^– → Cu₂O (s) +HCOO^–(aq) + 3H₂O(l)
(d) N₂H₄ (Z) + 2H₂O₂ (1) → N₂ (g) + 4H₂O (l)
(e) Pb (s) +PbO₂ (s) + 2H₂SO₄ (ag) → 2PbSO₄(s) + 2H₂O(l)
Answer:
(a) Oxidised substance -C₆H₆O₂
Reduced substance – AgBr
Oxidising agent – AgBr
Reducing agent -C₆H₆O₂

(b) Oxidised substance – HCHO
Reduced substance – [Ag (NH₃)₂]⁺
Oxidising agent – [Ag (NH₃)₂]⁺
Reducing agent – HCHO

(c) Oxidised substance – HCHO
Reduced substance -Cu²⁺
Oxidising agent – Cu²⁺
Reducing agent – HCHO

(d) Oxidised substance – N₂H₄
Reduced substance -H₂O₂
Oxidising agent-H₂O₂
Reducing agent -N₂H₄

(e) Oxidised substance – Pb
Reduced substance – PbO₂
Oxidising agent – Pb O₂
Reducing agent – Pb

Question 14.
Consider the reactions:

Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
The average oxidation number (O.N.) of S in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is +2. Being a stronger oxidising agent than I₂, Br₂ oxidises \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) in which the O.N. of S is +6. However I₂ is a weak oxidising agent. Therefore, it oxidises \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) in which the average O.N. of S is only +2.5. As a result, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) reacts differently with iodine and bromine. ,

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
(i) F₂ can oxidize Cl^– to Cl₂, Br^– to Br₂ and I^– to I₂ as:
F₂(aq) + 2Cl^–(s) → 2F^–(aq) + Cl₂(g)
F₂(aq) + 2Br^–(aq) → 2F^–(aq) + Br₂(7)
F₂(aq) + 2I^–(aq) → 2F^–(aq) + I₂(s)

On the other hand, Cl₂, Br₂ and I₂ cannot oxidize F^– to F₂. The oxidizing power of halogens increases in the order of I₂ < Br₂ < Cl₂ < F₂. Hence fluorine is the best oxidant among halogens.

(ii) HI and HBr can reduce H₂SO₄ to SO₂, but HCl and HF cannot. Therefore HI and HBr are stronger reductants than HCl and HF.

2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O
2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O

Again, I- can reduce Cu²⁺ to Cu⁺ but Br^– cannot.

4I^–(aq) + 2Cu²⁺(aq) → Cu₂I₂(s) + I₂(aq)

Hence, hydroiodic acid is the best reductant among hydrohalic compounds.
Thus, the reducing power of hydrohalic acids increases in the order of
HF < HCl < HBr < HI.

Question 16.
Why does the following reaction occur?
\(\mathrm{XeO}_{6}^{4-}\)(aq) + 2F^– (aq) + 6H⁺(aq) → XeO₃(g) +F₂(g) + 3H₂O(i)
What conclusion about the compound Na₄XeO₆ (of which \(\mathrm{XeO}_{6}^{4-}\) is a part) can be drawn from the reaction.
Answer:
The given reaction occurs because \(\mathrm{XeO}_{6}^{4-}\) oxidizes being an oxidizing agent and F^– reduces \(\mathrm{XeO}_{6}^{4-}\)

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in \(\mathrm{XeO}_{6}^{4-}\) to +6 in XeO₃ and the O.N. of F increases from -1 in F^– to 0 in F₂.
Hence, we can conclude that Na₄XeO₆ is a strong oxidizing agent thanF^–.

Question17.
Consider the reactions:
(a) H₃PO₂ (aq) + 4 AgNO₃ (aq) + 2H₂O(l) → H₃PO₄ (aq) + 4Ag(s) + 4HNO₃ (aq)
(b) H₃PO₂ (oqr) + 2CuSO₄ (aq) + 2 H₂O(0 → H₃PO₄ (aq) + 2Cu(s) +H₂SO₄ (aq)
(c) C₆H₅CHO(l) + 2[Ag (NH₃)₂]+ (aq) + 3OH^– (aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃ (aq) + 2H₂O(l)
(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^–(aqf) → No change observed.
What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?
Answer:
Ag⁺ and Cu²⁺ act as an oxidizing agents in reactions (a) and (b) respectively.
In reaction (c), Ag⁺ oxidizes C₆H₅CHO to C₆H₅COO^–, but in reaction (d) Cu²⁺ cannot oxidize C₆H₅CHO. Hence, we can say that Ag⁺ is a stronger oxidising agent than Cu²⁺.

Question 18.
Balance the following redox reactions by ion-electron method:


Answer:
(a) Step 1 : The two half reactions involved in the given reaction are:
Oxidation half reaction :

Reduction half reaction:

Step 2 : Balancing I in the oxidation half reaction, we have
2I^– (aq) → I₂(s)
Now, to balance the charge, we add 2 e“ to the RHS of the reaction.
2I^–(aq) → I₂(s) + 2e^–

Step 3 : In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.
\(\mathrm{MnO}_{4}^{-}\) + 3e^– → MnO₂(aq)
Now, to balance the charge, we add 4 OH“ ions to the RHS of the reaction as the reaction is taking place in a basic medium.
\(\mathrm{MnO}_{4}^{-}(a q)^{-}\) + 3e^– → MnO₂(aq) + 4OH^–

Step 4 : In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.
\(\mathrm{MnO}_{4}^{-}\) (aq) + 2H₂O + 3e^– → MnO₂(aq) + 4OH^–

Step 5 : Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
6I^–(aq) → 3I₂(s) + 6e^–
2 \(\mathrm{MnO}_{4}^{-}\)(aq) + 4H₂O + 6e^– → 2MnO₂(s) + 8OH^–(aq)

Step 6 : Adding the two half reactions, we have the net balanced redox reaction as:
6I^–(aq) + 2\(\mathrm{MnO}_{4}^{-}\)(aq) + 4H₂O(l) → 3I₂(s) + 2MnO₂(s) + 8OH^–(aq)

(b) Following the steps as in part (a) we have the oxidation half reaction as:
SO₂(g) + 2H₂O(Z) → \(\mathrm{HSO}_{4}^{-}\)(aq) + 3H⁺ (aq) + 2e^– (aq)
And the reduction half reaction as:
\(\mathrm{MnO}_{4}^{-}\)(aq) + 8H⁺(aq) + 5e^– → Mn²⁺(aq) + 4H₂O(l)

Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as :

2Mno₄^–(aq) + 5SO₂(g) + 2H₂O(Z) + H⁺(aq) → 2Mn(aq) + 5HSO₄^–((aq)

(c) Following the steps as in part (a), we have the oxidation half reaction as:
Fe²⁺(aq) → Fe³⁺(aq) + e^–

And the reduction half reaction as:
H₂O₂(aq) + 2H⁺ (aq) + 2e^– → 2H₂O(l)

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
H₂O₂(aq) + 2Fe²⁺(aq) + 2H⁺(aq) → 2Fe³⁺ (aq) + 2H₂O(Z)

(d) Following the steps as in part (a), we have the oxidation half reaction as:

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P₄(s) + OH^–(aq) → PH₃(g) + H₂PO₂^–(aq)
(b) N₂H₄(l) + CIO₃^–(aqr) → NO(g) + Cl^–(g)
(c) Cl₂O₇(g) + H₂O₂(aqr) → ClO^–₂(aq) + O₂(g) + H⁺
Answer:

O.N. increases by 1 per P atom.
P₄ acts both as an oxidising as well as a reducing agent.

Oxidation number method:
Total decrease in O.N. of P₄ in PH₃ = 3 x 4 = 12
Total increase in O.N. of P₄ in H₂PO₂ = 1 x 4 = 4
Therefore, to balance increase/decrease in O.N. multiply PH₃ by 1 and H₂PO₂^– by 3, we have,

P₄(s) + OH^–(aq) → PH₃(g) + 3H₂PO₂^–(aq)

To balance O atoms, multiply OH- by 6, we have,

P₄(s) + 6OH^–(aq) → PH₃(g) + 3H₂PO₂^–(aq)

To balance H atoms, add 3H₂O to L.H.S. and 3OH^– to the R.H.S. we have,

P₄ (s) + 6OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq) + 3OH^–(aq)
or P₄(s) + 3OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq) …(1)

Thus, Eq. (1) represents the correct balanced equation.

Ion-electron method : The two half reactions are:

Oxidation number method :
Total increase in O.N. of N = 2 x 4 = 8
Total decrease in O.N. of Cl = 1 x 6 = 6
Therefore, to balance increase/decrease in O.N. multiply N₂H₂ by 3 and \(\mathrm{ClO}_{3}^{-}\) by 4, we have,

3N₂H₄(l) + 4ClO₃^– (aq) → NO(g) + Cl^–(aq)

To balance N and Cl atoms, multiply NO by 6 and Cl^– by 4, we have,

3N₂H₄(l) + 4ClO₃^–(aq) → 6NO(g) + 4Cl^–(aq)

Balance O atoms by adding 6H₂O, in R.H.S.

3N₂H₄(l) + 4ClO₃^–(aq) → 6NO(g) + 4Cl^–(aq) + 6H₂O (l) …(1)

H atoms get automatically balanced and thus Eq. (1) represents the correct balanced equation.

Ion electron method :

Oxidation number method :
Total decrease in O.N. of Cl₂O₇ = 4 x 2 = 8
Total increase in O.N. of H₂O₂ = 2 x 1 = 2
∴ To balance increase/decrease in O.N. multiply H₂O₂ and O₂ by 4, we have,

Question 20.
What type of information can you draw from the following reaction?
(CN)₂(g) + 2OH^–(aq) → CN^–(aq) + CNO^–(aq) + H₂O(l)
Answer:
The oxidation number of carbon in (CN)₂ CN^– and CNO^– is +3, +2 and +4 respectively. These are obtained as shown below:
Let the oxidation number of C be x.
(i) (CN)₂
2 (x – 3) = 0
∴ x = +3

(ii) CN^–
x -3 = -1
∴ x = +2

(iii) CNO^–
x – 3 – 2 = -1
∴ x = + 4

The oxidation number of carbon in the various species is:

The following information we can drawn from the above reaction :
(i) Decomposition of cyanogen in the cyanide ion (CN^–) and cyanate ion (CNO^–) occurs in basic medium.
(ii) Cyanogen (CN)₂ acts as both reducing agent as well as oxidising agent.
(iii) The reaction is an example of disproportionation reaction.
(iv) Cyanogen (CN)₂ is called pseudohalogen while CN^–, CNO^– ions are called pseudohalide ions.

Question 21.
The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write a balanced ionic equation for the reaction.
Answer:
The given reaction can be represented as:

Mn³⁺ (aq) → Mn²⁺(aq) + MnO₂(s) + H⁺(aq)

The oxidation half equation is:

The oxidation number is balanced by adding one electron as:
Mn³⁺(aq) → MnO₂(s) + e^–

The charge is balanced by adding 4H+ ion as :
Mn³⁺(aq) → MnO₂(s) + 4H⁺ (aq) + e^–

The O atoms and H+ ions are balanced by adding 2H20 molecules as:
Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + 4H⁺ (aq) + e^– …(1)

The reduction half equation is:
Mn³⁺(aq) → Mn²⁺(aq)

The oxidation number is balanced by adding one electron as :
Mn³⁺ (aq) + e^– → Mn²⁺ (aq) … (2)

The balanced chemical equation can be obtained by adding equation (1) and (2) as :
2Mn³⁺(aq) + 2H₂O(l) → MnO₂(s) + Mn²⁺(aq) + 4H⁺(aq)

Question 22.
Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor the positive oxidation state.
Answer:
(a) F exhibits only negative oxidation state of-1.
(b) Cs exhibits only positive oxidation state of +1
(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of-1, 0, +1, + 3, + 5, and + 7.
(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.

Question 23.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Answer:
The given redox reaction can be represented as:

Question 24.
Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non-metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.
Answer:
In disproportionation reaction, one of the reacting substances always contains an element that can exist in at least three oxidation states.

Question 25.
In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with lO.OOg of ammonia and 20.00 g of oxygen?
Answer:
The balanced chemical equation for the given reaction is given as:

68 g of NH₃ reacts with 160 g of O₂
Therefore 10 g of NH₃ reacts with \(\frac{160 \times 10}{68}\). g of O₂ or 23.53 g of O₂
But the available amount of O₂ is 20 g which is less than the amount required to react with 10 g NH₃. So, O₂ is the limiting reagent and it limits the amount of NO produced. From the above balanced equation.
160 g of O₂; produces 120 g NO.
Therefore, 20 g of O₂; produces = \(\frac{120 \times 20}{160}\) = 15 g NO

Question 26.
Using the standard electrode potentials given in the table 8.1,
predict if the reaction between the following is feasible: *
(a) Fe³⁺ (aq) and I^– (aq)
(b) Ag⁺ (aq) and Cu (s)
(c) Fe³⁺ (aq) and Cu(s)
(d) Ag (s) and Fe³⁺ (aq)
(e) Br₂ (aq) and Fe²⁺(aqr)
Answer:

Question 27.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO₃ with silver electrodes.
(ii) An aqueous solution of AgNO₃ with platinum electrodes
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes,
Answer:
(i) AgNO₃ ionizes in aqueous solutions to form Ag⁺ and \(\mathrm{NO}_{3}^{-}\) ions.
On electrolysis, either Ag⁺ ions or H₂O molecules can be reduced at the cathode. But the reduction potential of Ag⁺ ions is higher than that of H₂O.

Hence, Ag+ ions are reduced at the cathode. Similarly, Ag metal or H20 molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H20 molecules.

Ag(s) → Ag⁺(aq) + e^– ;\(E^{\ominus}\) = -0.80V

2H₂O₂(g) → O₂(g) + 4H⁺(aq) + 4e^– ;\(E^{\ominus}\) = -1.23V
Therefore, Ag metal gets oxidized at the anode.

(ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O₂. At the cathode, Ag⁺ ions are reduced and get deposited.

(iii) H₂SO₄ ionizes in aqueous solutions to give H⁺ and \(\mathrm{SO}_{4}^{2-}\) ions.

H₂SO₄ (aq) → 2H⁺(aq) + \(\mathrm{SO}_{4}^{2-}\) (aq)

On electrolysis, either of H⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of H⁺ ions is higher than that of H₂O molecules.

2H⁺(aq) + 2e^– → H₂(g); \(E^{\ominus}\) = 0.0 V
2H₂O(aq) + 2e^– → H₂(g) + 2OH^–(aq): \(E^{\ominus}\) = – 0.83V

Hence, at the cathode, H⁺ ions are reduced to liberate H₂ gas.
On the other hand, at the anode, either of \(\mathrm{SO}_{4}^{2-}\) ions or H₂O molecules can get oxidized. But the oxidation of \(\mathrm{SO}_{4}^{2-}\) involves breaking of more bonds than that of H₂O molecules. Hence, \(\mathrm{SO}_{4}^{2-}\) ions have a lower oxidation potential than H₂O. Thus, H₂O is oxidized at the anode to liberate O₂ molecules.

(iv) In aqueous solutions, CuCl₂ ionizes to give Cu²⁺ and Cl^– ions as
CuCl₂ (aq) → Cu²⁺ (aq) + 2Cl^– (aq)

On electrolysis either of Cu²⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of Cu²⁺ is more than that of H₂O molecules.

Cu²⁺(aq) + 2e^– → Cu(aq) ;\(E^{\ominus}\) = + 0.34V ;
H₂O(l) + 2e^– → H₂(g) + 2OH^– ;\(E^{\ominus}\) = – 0.83V

Hence, Cu²⁺ ions are reduced at the cathode and get deposited.
Similarly, at the anode, either of Cl^– or H₂O is oxidized. The oxidation , potential of H₂O is higher than that of Cl^–

2Cl^–(aq) → Cl₂(g) + 2e^– ;\(E^{\ominus}\) = -1.36V :K
2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e^– ;\(E^{\ominus}\) = -1.23V

But oxidation of H₂O molecules occurs at a lower electrode potential . than that of Cl^– ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl^– ions are oxidized at the anode to liberate Cl₂ gas. :

Question 28.
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn. ‘
Answer:
A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.
The order of the increasing reducing power of the given metals is
Cu < Fe < Zn < Al < Mg. Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is given below:
Mg > Al > Zn > Fe > Cu

Question 29.
Given the standard electrode potentials:
K⁺ / K = – 2.93V, Ag⁺ / Ag = 0.80 V, Hg²⁺ /Hg = 0.79 V
Mg²⁺/ Mg = -2.37 V, Cr³⁺/Cr = -0.74 V Arrange these metals in their increasing order of reducing power.
Answer:
The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metal is
Ag < Hg < Cr < Mg < K.

Question 30.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aqr) → Zn²⁺(aqr) + 2Ag(s) takes place, further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode.
Answer:
The galvanic cell corresponding to the given redox reaction can be represented as:
Zn / Zn²⁺ (aq) | | Ag⁺(aq) / Ag
(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn²⁺ and the leaving electrons accumulate on this electrode.
(ii) Ions are the carriers of current in the cell.
(iii) The reaction taking place at Zn electrode can be represented as:
Zn(s) → Zn²⁺ (aq) + 2e^–
and the reaction taking place at Ag electrode can be represented as:
Ag⁺(aq) + e^– → Ag(s)

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 7 Equilibrium Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 7 Equilibrium

PSEB 11th Class Chemistry Guide Equilibrium InText Questions and Answers

Question 1.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure?
(b) How do rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer:
(a) If the volume of the container is suddenly increa50sed, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

Question 2.
What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂] = 0.60M,[O₂] = 0.82M and [SO₃] = 1.90 M?
2SO₂(g) + O₂(g) ↔ 2SO₃(g)
Answer:
The given reaction is
2SO₂(g) + O₂(g) ↔ 2SO₃(g)
Equilibrium constant
K_c = \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}=\frac{(1.90 \mathrm{M})^{2}}{(0.60 \mathrm{M})^{2}(0.82 \mathrm{M})}\)
= 12.238 M^-1

Question 3.
At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms I₂(g) ⇌ 2I(g)
Calculate K_p for the equilibrium.
Answer:
Given, I₂(g) ⇌ 2I(g)
I atoms in iodine vapours = 40% by volume
So, iodine vapours of I₂ molecules = 60% by volume

Question 4.
Write the expression for the equilibrium constant, Kc for each of the following reactions:
(i) 2NOCl(g) ⇌ 2NO (g) + Cl₂(g)
(ii) 2CU(NO₃)₂(S) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g)
(iii) CH₃COOC₂H₅(oq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(ag)
(iv) Fe³⁺(aq) + 3OH^–(aq) ⇌ Fe(OH)₃(s)
(v) I₂(s) + 5F₂ ⇌ 2IF₅
Answer:

Question 5.
Find out the value of K_c for each of the following equilibria from the value of Kp.
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 x 10^-2 at 500 K
(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Answer:
The relation between K_p and K_c is given as
K_p = K_c(RT)^Δn
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 x 10^-2 at 500 K.
Δn = 3 – 2 = 1
R = 0.0831 bar L mol^-1K^-1
T = 500 K
K_p =1.8 x 10^-2
K_p = K_c( RT)^Δn
1.8 x 10^-2 = K_c(0.0831 x 500)¹
K_c = \(\frac{1.8 \times 10^{-2}}{0.0831 \times 500}\) = 4.33 x 10^-4

(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Δn = 2 -1 = 1
R = 0.0831 bar L mol^-1K^-1
T = 1073 K
K_p =167
Now, K_p = K_c(RT)^Δn
⇒ 167 = K_c(0.0831 x 1073)¹
⇒ K_c = \(\frac{167}{0.0831 \times 1073}\) = 1.87

Question 6.
For the following equilibriuih, K_c = 6.3 x 10¹⁴ at 1000 K
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?
Answer:
It is given that K_c for the forward reaction is 6.3 x 10¹⁴ at 1000 K.
Then, K_c for the reverse reaction will be,
NO₂(g) + O₃(g) ⇌ NO(g) + O₃(g)
K_c = \(\frac{1}{K_{c}}=\frac{1}{6.3 \times 10^{14}}\) = 1.59 x 10^-15

Question 7.
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Answer:
For a pure substance (both solids and liquids),

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

Question 8.
Reaction between N₂ and O₂ takes place as follows:
2N₂(g) + O₂(g) ⇌ 2N₂O(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 x 10^-37 , determine the composition of equilibrium mixture.
Answer:
Let the concentration of N₂O at equilibrium be x.
The given reaction is :

The value of equilibrium constant i.e., K_c = 2.0 x 10^-37 is very small which means negligible amounts of N₂ and O₂ react.

Question 9.
Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:
2NO(g) + Br₂(g) ⇌ 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO andBr₂.
Answer:
The balanced chemical equation is

Given, 2x = 0.0518
x = 0.0259 mol
Moles of NO at equilibrium = 0.087 – 2x
= 0.087-0.0518
= 0.0352 mol
Moles of Br2 at equilibrium = 0.0437 – x
= 0.0437 – 0.0259
= 0.0178 mol

Question 10.
At 450 K, K_p = 2.0 x 10¹⁰/bar for the given reaction at equilibrium.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
What is K_c at this temperature?
Answer:
The given reaction is
2SO₂(g) + O₂(g) ⇌ 2SO₃Cg)
Δn = 2 – 3 = -1
T = 450 K
R = 0.0831 bar L K^-1 mol^-1
K_p = 2.0 x 10¹⁰ bar^-1
We know that,
K_p = K_c(RT)^Δn
=> 2.0 x 10¹⁰ bar^-1 = k_c(0.0831 L bar K^-1 mol^-1 x 450 K)^-1
\(K_{c}=\frac{2.0 \times 10^{10} \mathrm{bar}^{-1}}{\left(0.0831 \mathrm{~L} \mathrm{barK} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 450 \mathrm{~K}\right)^{-1}}\)
K_c = (2.0 x 10¹⁰ bar^-1) (0.0831 L bar K^-1mol^-1450 K)
= 74.79 x 10¹⁰ L mol^-1 = 7.48 x 10¹¹ L mol^-1

Question 11.
A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium?
2HI(g) ⇌ H₂(g) + I₂(g)
Answer:
The given reaction is

∵ Decrease is pressure of HI = 0.2 – 0.04 = 0.16 atm;
So equilibrium pressure of H₂ is \(\frac{0.16}{2}\) = 0.08 atm and for I₂ is \(\frac{0.16}{2}\) = 0.08 atm
as two moles of HI on dissociation gives 1 mol of H₂ and 1 mol of I₂.
Therefore,
K_p = \(\frac{p_{\mathrm{H}_{2}} \times p_{\mathrm{I}_{2}}}{\left(p_{\mathrm{HI}}\right)^{2}}=\frac{0.08 \times 0.08}{(0.04)^{2}}=\frac{0.0064}{0.0016}\) = 4.0
Hence, the value of K_p is 4.0.

Question 12.
A mixture of 1.57 mol of N₂,1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1. 7 x 10².
Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Answer:
The given reaction is :
N₂(g) + 3H₂(g) 2NH₃(g)
Given, [N₂] = \(\frac{1.57}{20}\) = 0.0785 M
[H₂] = \(\frac{1.92}{20}\) = 0.096 M
[NH₃] = \(\frac{8.13}{20}\) = 0.4065 M
Now, reaction quotient Q_c. is :
Q_c = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}=\frac{(0.4065)^{2}}{(0.0785)(0.096)^{3}}\) = 2.4 x 10³M^-2
Since Q_c ≠ K_c the reaction mixture is not in equilibrium.
Again Q_c > K_c. Hence, the reaction will proceed in the reverse direction.

Question 13.
The equilibrium constant expression for a gas reaction is, K_c = \(\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{5}}{\left[\mathrm{NO}^{4}\left[\mathrm{H}_{2} \mathrm{O}\right]^{6}\right.}\)
Write the balanced chemical equation corresponding to this expression.
Answer:
The balanced chemical equation corresponding to the given expression can be written as :
4NO(g) + 6H₂O(l) ⇌ 4NH₃(g) + 5O₂(g)

Question 14.
One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,
H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Calculate the equilibrium constant for the reaction.
Answer:
The given reaction is

H₂O reacted = 40% of 1 mol of H₂O = 0.4 mol
x = 0.4 mol 1 – x = 1 – 0.4 = 0.6 mol
Therefore, the equilibrium constant for the reaction,
K_c = \(\) = 0.444

Question 15.
At 700 K, equilibrium constant for the reaction
H₂(g) + I₂(g) ⇌ 2HI(g)
is 54.8. If 0.5 mol L^-1 of Hl(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Answer:
The given reaction is
H₂(g) + I₂(g) ⇌ 2HI(g); K_c = 54.8
Or the reaction
2HI(g) ⇌ H₂(g) + I₂(g); K_c‘ = \(\)
Given, [HI] = 0.5 mol L^-1
According to equation
[H₂] = [I₂] = x mol L^-1
Therefore,
\(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}=K_{c}^{\prime}\)
⇒ \(\frac{x \times x}{(0.5)^{2}}=\frac{1}{54.8}\)
⇒ x² = \(\frac{0.25}{54.8}\)
⇒ x = 0.06754
x = 0.068 mol L^-1
Hence, at equilibrium, [H₂] = [I₂] = x = 0.068 mol L^-1

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of IC1 was 0.78 M?
2ICl(g) ⇌ I₂(g) + Cl₂(g); K_c = 0.14
Answer:

Question 17.
K_p = 0.04 atm at9 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
Answer:

Question 18.
Ethyl acetate is formed by the reaction between ethanol acid and acetic acid and the equilibrium is represented as :
CH₃COOH(l) + C₂H₅OH (l) ⇌ CH₃COOC₂H₅(Z) + H₂O(l)
(i) Write the concentration ratio (reaction quotient), Q_c, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Answer:

Question 19.
A sample of pure PCl₅ was introduced into an evacuated vessel at 473K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 x 10^-1 mol L^-1. If value of K is 8.3 x 10^-3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g)
Answer:
The given reaction is

It is given that the value of equilibrium constant, K = 8.3 x 10^-3.
K_c = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
[Given, [PCl₅]_equili = 0.5 x 10^-1 mol L^-1]
\(\frac{x \times x}{0.5 \times 10^{-1}}\) = 8.3 x 10^-3
⇒ x² = 4.15 x10^-4
⇒ x = 2.04 x 10^-2 = 0.0204 mol L^-1 = 0.02 mol L^-1
Therefore, at equilibrium,
[Pcl₃] = [Cl₂] = 0.02 mol L^-1

Question 20.
One of the reaction that takes place in producing steel from iron ore is the reduction of iron (H) oxide by carbon monoxide to give iron metal and CO₂.
FeO(s) + CO(g) ⇌ Fe(s) + CO₂ (g); K_p = 0.265 atm at 1050 K. What are the equilibrium partial pressures of CO and ₂ at 1050 K if the initial partial pressures are P_Co = 1.4 atm and pCO₂ = 0.80 atm?
Answer:
(i) The given reaction is

Since Q_p > K_p, the reaction will proceed in the backward direction.
Therefore, we can say that the pressure of CO will increase while the pressure of CO₂ will decrease.
Now,let the increase in pressure of CO = decrease in pressure of CO₂ be p.
Hence pCO₂ = 0.80 – p and P_CO = 1.4 + p
and K_p = \(\frac{p_{\mathrm{CO}_{2}}}{p_{\mathrm{CO}}}\)
0.265 = \(\frac{0.80-p}{1.4+p}\)
0.371 + 0.265p = 0.80 — p= 1.265p= 0.429
p = 0.339atm
Hence, at equilibrium
P_CO2 = 0.80 – 0.339 = 0.461 atm
And, equilibrium partial pressure of
P_CO = 1.4 + 0.339 = 1.739 atm.

Question 21.
Equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K is 0.06 1.
At a particular time, the analysis shows that composition of the
reaction mixture is 3.0 mol L^-1 N₂,2.0 mol L^-1 H₂ and 0.5 mol L^-1 NH₃ Is the reaction at equilibrium? if not in which direction
does the reaction tend to proceed to reach equilibrium?
Answer:
The given reaction is
N₂(g) + 3H₂(g) ⇌ 2NH₃(g);K_c = 0.061 at 500K
Given, [N₂] = 3.0mol L^-1, [H₂] = 2.0 mol L^-1, [NH₃] = 0.5 mol L^-1
So, Q = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}=\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}\) = 0.0104
It is given that K_c = 0.06 1
Since Q_c ≠ K_c, the reaction is not at equilibrium.
Since Q_c < K_c, the reaction will proceed in the forward direction to reach equilibrium.

Question 22.
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl(g) ⇌ Br₂(g) + Cl₂(g)
for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 x 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium?
Answer:

x = 11.312(3.30 x 10^-3 – x)
x = 0.03732 – 11.312x
x + 11.312x = 0.03732
x = \(\frac{0.03732}{12.312}\)= 3.0321 x 10^-3 mol L^-1
[BrCl]_equili = (3.30 x 10^-3 – 3.032 x 10^-3) mol L^-1
= 2.68 x 10^-4 mol L^-1

Question 23.
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass
C(s) + CO₂(g) ⇌ 2CO(g)
Calculate K_c for this reaction at the above temperature.
Answer:
Let the total mass of the gaseous mixture be 100g.
Mass of CO = 90.55 g
and, mass of CO₂ = (100 – 90.55) = 9.45 g
Now, number of moles of CO,
n_CO = \(\frac{90.55}{28}\) = 3.234 mol
(Molar mass of CO = 28 g mol^-1 )
Now, number of moles of CO₂,
n_CO = \(\)
(Molar mass of CO₂ = 44 g mol^-1 )

For the given reaction, Δn = 2 -1 = 1
We know that,
K_p = K_c(RT)^Δn
⇒ 14.19 = K_c(0.0831 x 1127)¹
⇒ K_c = 0.154 (approximately)

Question 24.
Calculate (a) \(\Delta \boldsymbol{G}^{\ominus}\) and (b) the equilibrium constant for the formation of N02 from NO and Oa at 298K
NO(g) + \(\frac{1}{2}\)O₂(g) ⇌ NO₂(g)
where \(\Delta_{f} \boldsymbol{G}^{\ominus}\) (N02)= 52.0 kJ/mol; \(\Delta_{f} \boldsymbol{G}^{\ominus}\) (NO) = 87.0kJ/mol;
\(\Delta_{f} \boldsymbol{G}^{\ominus}\) (O2) = 0 kJ/mol
Answer:
(a) The given reaction is
N0(g) + \(\frac{1}{2}\)O₂(g) ⇌ NO₂(g)

= (52.0 – 87.0 + \(\frac{1}{2}\) x 0 )kJ mol^-1 = -35.0 kJ mol^-1

(b) \(\Delta_{r} G^{\ominus}\) = – 2.303 RT logK_c
-35.0 = – 2.303 x 0.0831 x 298 log K_c
∴ log K_c= \(\frac{35}{5.7058}\)= 6.134
∴ K_c = antilog 6.134 = 1.361 x 10⁶.

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) PCl₅(g) ⇌ PCl₃(g) +Cl₂(g)
(b) CaO (s) + CO₂ (g) ⇌ CaCO₃ (s)
(c) 3Fe(s) + 4H₂O (g) ⇌ Fe₃O₄ (s) + 4H₅(g)
Answer:
(a)The number of moles of reaction products will increase. According to Le-Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.
(b) The number of moles of reaction products will decrease.
(c) The number of moles of reaction products remains the same

Question 26.
Which of the following reactions will get affected hy increasing the pressure?
Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) COCl₂(g) ⇌ CO(g) +Cl₂(g)
(ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
(iii) CO₂(g) + C(S) ⇌ 2CO(g)
(iv) 2H₂(g) +CO(g) ⇌ CH₃OH(g)
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Answer:
In all the above reactions, the reaction no. (ii) proceeds with the same no. of moles on both sides
i.e., n_p = n_r = 3 .
∴ This reaction will not be affected by the increase in pressure i. e., the direction of equilibrium will not be affected by the increase in pressure. All other reactions will be affected by the increase in pressure.
(i) COCl₂(g) ⇌ CO(g) +Cl₂(g)
n_p > n_r , n_p = 2; nr = 1
∴ Equilibrium will shift to the left increasing pressure.
(iii) CO₂(g) + C(S) ⇌ 2CO(g)
Here, n_r – 1; n_p = 2, therefore n_p > n_r
∴ Equilibrium will go to left on increase of pressure.
(iv) 2H₂(g) +CO(g) ⇌ CH₃OH(g)
Here, n_r = 3; n_p = 1 therefore n_p < n_r
∴ Equilibrium will shift to the right on increasing pressure.
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Here n_r = 0; n_p = 1, therefore n_p > n_r
∴ Equilibrium will shift backwards (left) on increasing the pressure.
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Here n_r = 9; n_p = 10, therefore n_p > n_r
∴ Equilibrium will shift backwards on increasing the pressure.

Question 27.
The equilibrium constant for the following reaction is 1.6 x 10⁵ at 1024 K.
H₂(g) + Br₂(g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
Answer:
Given reaction is H₂(g) + Br₂(g)⇌ 2HBr(g); K_p = 1.6 x 10⁵ at 1024 K
Therefore, for the reaction 2HBr(g) ⇌ H₂(g)+Br₂(g), the equilibrium constant will be,

p = 2.5 x 1^-2(5.0 x 10^-3)p
p+(5.0 x 10^-3)p = 2.5 x 10^-2
(1005 x 10^-3)p = 2.5 x 10^-2
p = 2.49 x 10^-2 bar = 2.5 x 10^-2 bar
rherefore, at equilibrium,
[H₂] = [Br₂] = 2.49 x 10^-2 bar
[HBr] =10 — 2 x (2.49 x 10^-2) bar
= 9.95 bar = 10 bar

Question 28.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
(a) Write an expression for K_p for the above reaction.
(b) How will the values of k_p and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) Using a catalyst?
Answer:
(a) The given reaction is
CH₄(g) + H₄O(g) ⇌ CO(g) + 3H₂(g)
\(K_{p}=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_{2}}^{3}}{p_{\mathrm{CH}_{4}} \times p_{\mathrm{H}_{2} \mathrm{O}}}\)
(b) (1) According to LeChatelier’s principle, the equilibrium will shift in the backward direction.
(ii) According to Le-Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.
(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

Question 29.
Describe the effect of:
(a) addition of H₂
(b) addition of CH₃OH
(e) removal of CO (d) removal of CH₃OH
on the equilibrium of the reaction:
2H₂(g) + CO (g) ⇌ CH₃OH(g)
Answer:
2H₂(g) + CO(g) ⇌ CH₃OH(g)
According to Le Chatelier’s principle,
(a) Addition of H₂ (increase in concentration of reactants) shifts the equilibrium in forward direction (more product is formed).
(b) Addition of CH₃OH (increase in concentration of product) shifts the equilibrium in backward direction.
(c) Removal of CO also shifts the equilibrium in backward direction.
(d) Removal of CH₃OH shifts the equilibrium in forward direction.

Question 30.
At 473K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 x 10^-3. If decomposition is depicted as,
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = 1240 kJ mol^-1
(a) Write an expression for K_c for the reaction.
(b) What is the value of K_c for the reverse reaction at the same temperature?
(c) What would be the effect on K_c if
(i) more PCl₅ is added
(ii) pressure is increased?
(iii) the temperature is increased?
Answer:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); Kc = 8.3 x 10^-3
(a) K_c = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
(b) Value of K_c for the reverse reaction at the same temperature is
K’_c = \(\frac{1}{K_{c}}=\frac{1}{8.3 \times 10^{-3}}\) = 1.2048 x 10² = 120.48
(c) (i) Addition pf PCl₅ have no effect on K_c because K_c is constant at constant temperature.
(ii) K_c does not change with pressure.
(iii) The given reaction is endothermic, hence on increasing the temperature, K_c will increase.

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO(g) +H₂O (g) ⇌ CO₂(g) + H₂(g)
If a reaction vessel at 400° C is charged with an equimolar mixture of CO and steam such that p_CO = P_H2O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_P = 10.1 at 400°C
Answer:
The given reaction is
CO(g) + H20(g) ⇌ C02(g) + H2(g)

p = 12.71 – 3.17p
4.17 p = 12.71
p = \(\frac{12.71}{4.17}\) = 3.04 bar
Hence PH₂ = 3.04 bar

Question 32.
Predict which of the following reaction will have appreciable concentration of reactants and products:
(a) Cl₂(g) ⇌ 2Cl(g);K_c = 5 x 10^-39
(b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g); Kc = 3.7 x 10⁸
(c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g); K_c = 1.8
Answer:
Following conclusions can be drawn from the values of K_c:
(a) Since the value of K_c is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of K_c is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of K_c is 1.8, this means that both the products and reactants have appreciable concentration.

Question 33.
The value of K_c for the reaction
3O₂(g) ⇌ 2O₃(g)
is 2.0 x 10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 x 10^-2, what is the concentration of O₃?
Answer:
The given reaction is
3O₂(g) ⇌ 2O₃(g)
Then K
It is given that Kc = 2.0 x 10^-50 and [02(g)] = 1.6 x 10^-2
Then, we have,
\(2.0 \times 10^{-50}=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[1.6 \times 10^{-2}\right]^{3}}\)
⇒ [O₃]² = 2.0 x 10^-50 x (1.6 x 10^-2)³
⇒ [O₃]² = 8.192 x 10^-56
⇒ [O₃] = 2.86 x 10^-28 M
Hence, the concentration of O₃ is 2.86 x 10^-28 M.

Question 34.
The reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H20 and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Answer:
The given equation is
CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)
Therefore,
\(\frac{\left[\mathrm{CH}_{4}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2}\right]^{3}}=K_{c}\)
Given, K_c = 3.90, [CO] = 0.30 mol, [H₂] = 0.10 mol and [H₂O] \(\frac{\left[\mathrm{CH}_{4}\right] \times 0.02}{0.3 \times(0.1)^{3}}\) = 3.90
[CH₄] = \(\frac{3.90 \times 0.3 \times(0.1)^{3}}{0.02}=\frac{0.00117}{0.02}\)
= 0.0585 M= 5.85 x 10^-2M
Hence, the concentration of CH4 at equilibrium is 5.85 x 10^-2 M.

Question 35.
What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species :
\(\mathrm{HNO}_{2}, \mathrm{CN}^{-}, \mathrm{HClO}_{4}, \mathrm{~F}^{-}, \mathrm{OH}^{-}, \mathrm{CO}_{3}^{2-} \text { and } \mathrm{S}^{2-}\)
Answe:
A conjugate acid-base pair is a pair that differs only by one proton.
The conjugate acid-base for the given species is mentioned in the table below:

Question 36.
Which of the followings are Lewis acids
\(\mathbf{H}_{2} \mathbf{O}, \mathbf{B F}_{3}, \mathrm{H}^{+} \text {and } \mathrm{NH}_{4}^{+}\)
Answer:
Lewis acids are those acids which can accept a pair of electrons. For example, BF₃, H⁺ and \(\mathrm{NH}_{4}^{+}\) are Lewis acids.

Question 37.
What will be the conjugate bases for the Bronsted acids : HF, H₂SO₄ and \(\mathrm{HCO}_{3}^{-}\)?
Answer:
The table below lists the conjugate bases for the given Bronsted acids :

Question 38.
Write the conjugate acids for the following Bronsted bases: \(\mathbf{N H}_{2}^{-}\), NH₃ andHCOC^–.
Answer:
The table below lists the conjugate acids for the given Bronsted bases : Bronsted base Conjugate acid

Question 39.
The species : H₂O, HCO^–₃, HSO^–₄ and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.
Answer:
The table below lists the conjugate acids and conjugate bases for the given species :

Question 40.
Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH^– (b)F^– (c)H⁺ (d) BCl₃
OH^– and F^– are electron rich species and can donate electron pair. Hence, these act as Lewis base.

H⁺ and BCl₃ are electron deficient species and can accept electron pair. Hence, these act as Lewis acid.

Question 41.
The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10^-3 M. What is its pH?
Answer:
Given,
[H⁺] = 3.8 x 10^-3 M
∴ pH value of soft drink = – log[H⁺] = – log(3.8 x 10^-3)
= – log3.8 – log10^-3 = – log3.8 + 3 log10
= – log3.8 + 3
= -0.58 + 3
= 2.42

Question 42.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
Given, pH = 3.76
We know that,
pH = – log[H⁺]
⇒ log[H⁺] = -pH
⇒ [H⁺] = antilog (-pH)
= antilog (-3.76) -1 +1 = antilog \(\overline{4} .24\) = 1.74 x 10-4 M Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 x 10^-4 M.

Question 43.
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 x 10^-4, 1.8 x 10^-4 and 4 8 x 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Answer:
If K_a is the ionization constant of a weak acid and Kb is the ionization constant of its conjugate base then K_a.K_b = K_w
or K_b = \(\frac{K_{w}}{K_{a}}\)
Given, K_a of HF = 6.8 x 10^-4
Hence, K_b of its conjugate base F^–
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{6.8 \times 10^{-4}}\)= 1.5 x 10^-11
(K_w = ionic product of water =1 x 10^-14 at 298 K)
Given, K_a of HCOOH = 1.8 x 10^-4
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-4}}\) = 5.6 x 10^-11
Hence, K_b of its coagulate base CN^–
Given, K_a of HCN = 4.8 x 10^-9
Hence, K_b of its coagulate base HCOO^–
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{4.8 \times 10^{-9}}\) = 2.08 x 10^-6

Question 44.
The ionization constant of phenol is 1.0 x 10^-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?
Answer:
Ionization of phenol :
C₆H₅OH + H₂O ⇌ C₆H₅O^– + H₃O⁺

Question 45.
The first ionization constant of H2S is 9.1 x 10^-8. (i) Calculate the concentration of HS^– ion in its 0.1 M solution. (ii) How will this concentration be affected if the solution is 0.1 M in HCI also? (ifi) If the second dissociation constant of H₂S is 1.2 x 10^-13, calculate the concentration of S^2- under both conditions.
Answer:

Hence, concentration of [HS^–] is decreased in the presence of 0.1 M
HCI due to common-ion effect.
(iii) For second dissociation constant,
HS + H₂O ⇌ H₃O⁺ + S^2- (In absence of HCl)
[HS^–] = 9.54 x 10^-5 M
\(K_{a_{2}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{HS}^{-}\right]}\)

Question 46.
The ionization constant of acetic acid is 1.74 x 10^-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Answer:
CH₃COOH CH₃COO^– + H⁺
K_a for CH₃COOH = 1.74 x 10^-5
[CH₃COOH] = c = 0.05 M
CH₃COOH CH₃COO^– + H⁺ [where α = degree of dissociation and c = molar concentration]

[CH₃COO^–] = 0.933 x 10^-3 = 9.33 x 10^-4 M
pH = – log[H⁺] = – log (9.33 x 10^-4)
= – (-4) – log9.3 = 4 – 0.9 = 3.03

Question 47.
It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.
Answer:
Let the organic acid be HA.
⇒ HA ⇌ H⁺ + A^–
Concentration of HA = 0.01 M
pH = 4.15
-log[H⁺] = pH= 4.15
log[H⁺] = – 4.15
log[H⁺] = 5.85
[H⁺] = antilog \(\overline{5} .85\)
= 7.080 x 10^-5
K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
Now, [H⁺] = [A^–] = 7.08 x 10^-5 M
Then K_a = \(
K_a = 5.01 x 10^-7
PK_a = – logK_a = – log(5.01 x 10^-7)
pK_a = 7 – 0.699 = 6.301

Question 48.
Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH
Answer:
(a) HCl (aq) ⇌ H⁺ (aq) + Cl^–(aq)
[HCl]= 0.003 M
As HC1 is completely dissociated into H⁺ ions
∴ [H⁺] = [HCl] = 0.003 M
pH = – log[H⁺] = – log [3 x 10^-3]
= 3 + (-0.4771) = 2.523
(b) NaOH(aq) ⇌ Na⁺(aq) + OH^– (aq)
[NaOH] = 0.005 = 5 x 10^-3 M
[OH^–] = [NaOH] = 5 x 10^-3 M
∴ [latex]\left[\mathrm{H}^{+}\right]=\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{5.0 \times 10^{-3}}\)
[H⁺]= 2.0 x 10^-12
∴ pH = – log(2 x 10^-12) = – (-12) – log2
= 12 – 0.30 = 11.70
[log2 = 0.30]

(c)

[HBr] = 0.002 M
[H⁺]= [HBr] = 0.002 M= 2.0 x 10^-3 M
pH = – log[H⁺] = – log[2 x 10^-3]
=- (-3) – log2
= 3 – log2
= 3 – 0.3 = 2.70

(d)

[OH^–] = 0.002 M
[H⁺] = \(\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.002}\) = 2 x 10^-12
pH = – log[H⁺] = -(-12) – log 5 = 12 – 0.70 = 11.30

Question 49.
Calculate the pH of the following solutions:
(a) 2 g of TIOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 niL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 mL of 13.6 M HC1 is diluted with water to give 1 litre of solution.
Answer:

(d) 1 mL of 13.6 M HC1 is diluted with water to give 1 litre of solution HC1 is completely dissociated to give H+ ions
[HCl] = ?
M₁V₁ = M₂V₂
1 mL of 13.6 M HCl = 1000 mL of M₂
M₂ = \(\frac{1 \times 13.6}{1000}\) = 0.0136 M
[HC1] = [H+] = 0.0136 M pH = – log[H+] = – log(1.36 x 10-2)
= – (-2) – log 1.36 = 2 – 0.13 = 1.87

Question 50.
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.
Answer:
α (Degree of ionization) = 0.132
c (molar cone.) = 0.1 M

∴ H⁺ = c x α = 0.1 x 0.132 = 0.0132
pH = – logH⁺ = – log(1.32 x 10^-2)
= – (-2) – log 1.32 = 2 – 0.12 = 1.88
pK_a = -logK_a
Now, K_a = cα²
K_a = 0.1 x (0.132)² = 1.74 x 10^-3
∴ pK_a = – log (1.74 x 10^-3) = – (-3) – log1.74 = 3 – 0.24 = 2.76

Question 51.
The pH of 0.005 M codeine (C₁₈H₂₁NO₃) solution is 9.95.
Calculate its ionization constant and pK_b.
Answer:
Molar cone, of codeine, c = 0.005 = 5 x 10^-3
pH = 9.95
pOH = 14 – 9.95 = 4.05 (∵ pH + pOH = 14)
pOH = – log [OH^–]
log[OH^–] = -4.05= \(\overline{5} .95\)
[OH^–] = antilog \(\overline{5} .95\)
= 8.91 x 10-5
k_b = \(\left(\frac{8.91 \times 10^{-5}}{5 \times 10^{-3}}\right)^{2}\) = 1.588 x 10^-6
pK_b = – logK_b = – log(1.588 x 10^-6)
= 6 + (-0.2009) = 5.7991 = 5.80

Question 52.
What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from table 7.7 (427 x 10^-10). Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer:
Given, K_b = 4.27 x 10^-10, c = 0.001 M
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-}\)

[0H] = 6.534 x 10^-7
pOH = — log(6.534 x 10^-7)
= 7+ (-0.8152)= 6.18
pH + pOH =14
pH 14—6.18=7.82

Thus, the ionization constant of the conjugate acid of aniline is 2.34 x 10^-5.

Question 53.
Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74.
How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?
Answer:
PK_a = – log K_a,
4.74 =-logK_a
log K_a = -4.74 = \(\overline{5} .26\)
Ka = antilog \(\overline{5} .26\) = 1.82 x 10^-5

[CH₃COOH is a weak acid and HC1 is a strong acid, so we can assume
that (cα + 0.01) 0.01]

In the presence of strong acid, dissociation of weak acid i.e., CH₃COOH decreases due to common ion effect.

Q.54. The Ionization constant of dimethylanilne is 54 x 10.
Calculate Its degree of ionization in its 0.02 M solution. What
percentage of dimethylamine is ionized if the solution is also
0.1MInNaOH?
Ans. Given, Kb = 5.4 x 10
c=0.02M

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

Question 55.
Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below :
(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2
(c) Human blood, 7.38 (d) Human saliva, 6.4.
Answer:
(a) pH of Human muscle fluid = 6.83
pH = – log[H+] log[H⁺] = -6.83 = \(\overline{7} .17\)
[H⁺] = antilog \(\overline{7} .17\)
[H⁺] = 1.48 x 10^-7 M

(b) pH of Human stomach fluid =1.2
log[H⁺] = -1.2 = \(\overline{2} .80\)
[H⁺] = antilog \(\overline{2} .80\)
.-. [H⁺] = 6.3 x 10^-2 M

(c) pH of Human blood = 7.38
log[H⁺] = – 7.38 = \(\overline{8} .62\)
.-. [H⁺] = antilog \(\overline{8} .62\) = 4.17 x 10^-8 M

(d) pH of Human saliva = 6.4
log[H⁺] =-6.4 = \(\overline{7} .60\)
[H⁺] = antilog \(\overline{7} .60\) = 3.98 x 10^-7 M

Question 56.
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
Answer:
The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH = – log[H⁺]
(i) pH of milk = 6.8
Since, pH = -log[H⁺]
6.8 = -log[H+] log[H⁺] = -6.8 = \(\overline{7} .20\)
[H+] = antilog (\(\overline{7} .20\)) = 1.5 x 10~7 M

(ii) pH of black coffee = 5.0
Since, pH = – log[H⁺]
5.0 = – log[H+] log[H⁺] = – 5.0
[H⁺] = antilog (-5.00) = 10-5 M

(iii) pH of tomato juice = 4.2
Since, pH = – log[H⁺]
4.2 = – log[H⁺]
log[H⁺] = – 4.2 = \(\overline{5} .80\)
[H⁺] = antilog (\(\overline{5} .80\)) = 6.31 x 10^-5M

(iv) pH of lemon juice = 2.2
Since, pH = – log[H⁺]
2.2 = – log[H+]
log[H⁺] = -2.2 = \(\overline{3} .8\)
[H⁺] – antilog (\(\overline{3} .8\)) = 6.31 x 10^-3 M

(v) pH of egg white = 7.8
Since, pH = -log[H⁺]
7.8 = – log[H⁺]
log[H⁺]= -7.8 = \(\overline{8} .20\)
[H⁺] = antilog (\(\overline{8} .20\)) = 1.58 x 10^-8 M

Question 57.
0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Answer:
Molar cone, of KOH = \(\frac{0.561 \times 1000}{56.1 \times 200}\) = 0.05M
56.1 x 200
KOH being a strong electrolyte, is completely ionized in aqueous solution.
KOH(aq) ⇌ K⁺(aq) + OH^–(aq)
[OH^–] = 0.05 M = [K⁺]
[H⁺][OH^–] = k_w
[H⁺] = \(\) = 2 x 10^-13
pH = – log[H⁺] = – log[2 x 10^-13]
= – (-13) – log2 = 13 – 0.03
∴ pH = 12.70

Question 58.
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Answer:
Solubility of Sr(OH)₂ = 19.23 g/L
Then, concentration of Sr(OH)₂ = \(\frac{19.23}{121.63 \times 1}\) M = 0.1581 M
Sr(OH)₂(aq) Sr²⁺(aq) + 2(OH^–)(aq)
∴ [Sr²⁺] = 0.1581 M
[OH^–] – 2 x 0.1581 M = 0.3162 M
Now
K_w = [OH^–] [H⁺]
\(\frac{1 \times 10^{-14}}{0.3162}\) = [H⁺]
[H⁺] = 3.16 x 10^-14
pH = – log[H⁺]
pH = 14 – 0.4997 = 13.5003 ≈ 13.5

Question 60.
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Answer:
Given, pH = 2.34
Molar cone, (c) = 0.1 M
HCNO H⁺ + CNO^–
pH = – log[H⁺]
2.34 = -log[H⁺]
log[H⁺] =-2.34 = \(\overline{3} .66\)
.-. [H⁺] = antilog \(\overline{3} .66\) = 4.57 x 10^-3 M
[H⁺] = \(\sqrt{K_{a}^{c}}\)
4.57 x 10^-3 = \(\sqrt{K_{a}^{c}}\)
Ionization constant,
K_a = 2.088 x 10^-4
Degree of ionization α = \(\sqrt{\frac{K_{a}}{c}}=\sqrt{\frac{2.088 \times 10^{-14}}{0.1}}\)
α = 0.0457

Question 61.
The ionization constant of nitrous acid is 45 x 104. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Answer:
Hydrolysis constant K_h = \(\frac{K_{w}}{K_{a}}\)
where K_w = Ionic product of water, K_a = Ionisation constant of the acid

pOH = -log(9.42 x 10^-7)= 7-0.97= 6.03
∴ pH = 14 – pOH = 14 – 6.03 = 7.97

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Given, pH = 3.44
We know that,
PH = – log[H⁺]
.-. [H⁺]= 3.63 x 10^-4
Then K_h = \(\frac{\left(3.63 \times 10^{-4}\right)^{2}}{0.02}\) (Concentration = 0.02M)
=> K_h = 6.6 x 10^-6
Now, K_h = \(\frac{K_{w}}{K_{a}}\)
K_a = \(\frac{K_{w}}{K_{h}}=\frac{1 \times 10^{-14}}{6.6 \times 10^{6}}\)
= 1.51 x 10^-9

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF
Answer:

Question 64.
The ionization constant of chloroacetic acid is 1.35 x 10^-3 . What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?
Answer:
Given that Ka = 1.35 x 10^-3.
=> Ka = cα²
α = \(\sqrt{\frac{K_{a}}{c}}=\sqrt{\frac{1.35 \times 10^{-3}}{0.1}}\)
(∵ Concentration of acid = 0.1 M)
= \(\sqrt{1.35 \times 10^{-2}}\) =0.116
.-. [H⁺]= cα = 0.1 x 0.116 = 0.0116
=> pH = – log[H⁺] = – log[0.0116] = 1.94
To find pH of 0.1 M sodium salt, we use the formula
pH = – \(\frac{1}{2}\)[log C_w + log K_a – log c]
= –\(\frac{1}{2}\)[log1 x 10^-14 + log(1.35 x 10^-3) – log(0.1)]
= –\(\frac{1}{2}\)[-14 + (-3 + 0.1303) – (-1)]
= – \(\frac{1}{2}\) [-15.8697] = 7.93485 ≈ 7.94

Question 65.
Ionic product of water at 310 K is 2.7 x 10^-14. What is the pH of neutral water at this temperature?
Answer:
Ionic product,
K_w = [H₃O⁺] [OH^–]
= 2.7 x 10^-14 at 310 K
H₂O + H₂O *=* [H₃0⁺][OH^–]
[H₃0⁺]= [OH^–]
Therefore, [H₃0⁺] = \(\sqrt{2.7 \times 10^{-14}}\)
⇒ = 1.64 x 10^-7 M
⇒ [H₃0⁺] = 1.64 x 10^-7
⇒ pH = – log[H₃0⁺] = – log[1.64 x 10^-7 = 6.78
Hence, the pH of neutral water is 6.78.

Question 66.
Calculate the pH of the resultant mixtures :
(a) 10 mL of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HC1
(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
(c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH
Answer:

Question 67.
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants.
[K_sp(Ag₂CrO₄) = 1.1 x 10^-12, K_sp(BaCrO₄) = 1.2 x 10^-10,
K_sp[Fe(OH)₃] = 1.0 x 10^-38, K_sp(PbCl₂) = 1.6 x 10^-5
K_sp(Hg₂I₂) = 4.5 x 10^-29]
Determine also the molarities of individual ions.
Answer:
(i) Silver chromate : Ag₂CrO₄⇌ 2Ag⁺ + CrO₄^2-, K_sp = 1.1 x 10^-12
Then, K_sp = [Ag⁺]²[CrO₄^2-]
Let the solubility of Ag₂CrO₄ be s.
⇒ [Ag⁺] = 2s and [CrO₄^2-] = s
Then, K_sp = (2s)² s – 4s³
⇒ 1.1 x 10^-12 = 4s³
0.275 x 10^-12 = s³
s = 0.65 x 10^-4 M
Molarity of Ag⁺ = 2s = 2 x 0.65 x 10^-4
= 1.30 x 10^-4 M
Molarity of CrO₄^2- = s = 0.65 x 10^-4 M

(ii) Barium chromate : BaCrO₄ ⇌ Ba²⁺ + \(\mathrm{CrO}_{4}^{2-}\); K_sp = 1.2 x 10^-10
Then, K_sp = [Ba²⁺] [latex]\mathrm{CrO}_{4}^{2-}[/latex]
Let s be the solubility of BaCrO₄.
⇒ [Ba²⁺] = s and [latex]\mathrm{CrO}_{4}^{2-}[/latex] = s
⇒ K_sp = s²
⇒ 1.2 x 10^-10 = s²
⇒ s = 1.09 x 10^-5 M
Molarity of [Ba²⁺] = Molarity of [latex]\mathrm{CrO}_{4}^{2-}[/latex] = s = 1.09 x 10^-5 M

(iii) Ferric hydroxide: Fe(OH)₃ ⇌ Fe²⁺ + 3OH^–; K_sp = 1.0.x 10^-38

K_sp = [Fe²⁺][OH^–]³
Let s be the solubility of Fe(OH)₃
⇒ [Fe³⁺] = s and [OH^–] = 3s
⇒ K_sp = s. (3s)³ = s x 27x³
K_sp = 27x⁴
1.0 x 10^-38 = 27x⁴
0.037 x 10^-38 = s⁴
0.00037 x 10^-36 = s⁴
s = 1.39 x 10^-10 M
Molarity of [Fe³⁺] = s = 1.39 x 10^-10 M
Molarity of [OH^–] = 3s = 4.17 x 10^-10 M

(iv) Lead chloride : PbCl₂ ⇌ Pb²⁺ + 2Cl^–; K_sp = 1.6 x 10^-5
K_sp = [Pb²⁺][Cl^–]²
Let s be the solubility of PbCl₂.
⇒ [Pb²⁺] = s and [Cl^–] = 2s
Thus, K_sp = s. (2s)² = 4s³
⇒ 1.6 x 10^-5 = 4s³
⇒ 0.4 x 10^-5 = s³
4 x 10^-6 = s³
⇒ s = 1.59 x 10^-2 M
Molarity of [Pb²⁺] = s = 1.59 x 10^-2 M
Molarity of [Cl^–] = 2s = 3.18 x 10^-2 M

(v) Mercurous iodide : Hg₂I₂ ⇌ \(\mathrm{Hg}_{2}^{2+}\) + = 4.5 x 10^-29
K_sp = [\(\mathrm{Hg}_{2}^{2+}\) ] []I^–]
Let s be the solubility of [Hg²I²]
⇒ [Hg₂] = s and [I^–] = 2s
K_sp = (s).(2s)² = 4s³
⇒ 4s³ = 4.5 x 10^-29
⇒ s³ = 1.125 x 10^-29
s = 2.24 x 10^-10 M
Molarity of [ \(\mathrm{Hg}_{2}^{2+}\) ] = s = 2.24 x 10^-10 M
Molarity of [I^–] = 2s = 4.48 x 10^-10 M.

Question 68.
The solubility product constant of Ag₂CrO₄ and AgBr are
1.1 x 10^-12 and 5.0 x 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:
Let s be the solubility of Ag₂CrO₄
Thus, Ag₂CrO₄ ⇌ 2Ag²⁺ + \(\mathrm{CrO}_{4}^{-}\); K_sp = 1.1 x 10^-12
K_sp = [Ag²⁺]². [latex]\mathrm{CrO}_{4}^{-}[/latex]
=> [Ag²⁺] = (2s)² and [latex]\mathrm{CrO}_{4}^{-}[/latex] = s
K_sp = (2s)². s= 4s³
1.1 x 10^-12 = 4s³
s = 6.5 x 10^-5 M

Let s be the solubility of AgBr.
AgBr(s) ⇌ Ag⁺ + Br^–; K_sp = 5.0 x 10^-13
K_sp = s² = 5.0 x 10^-13
s = \(\sqrt{5.0 \times 10^{-13}}\)
∴ s = 7.07 x 10^-7 M

Therefore, the ratio of the molarities of their saturated solution is
\(\frac{s\left(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\right)}{s(\mathrm{AgBr})}=\frac{6.5 \times 10^{-5} \mathrm{M}}{7.07 \times 10^{-7} \mathrm{M}}\) = 9.19

Question 69.
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, K_sp = 7.4 x 10^-8).
Ans. 2NaIO₃ + Cu(ClO₃)3 → 2NaClO₃ + Cu(IO₃)₂
Molar cone, of both solutions before mixing = 0.002 M
Molar cone, of both solution after mixing
\(\left[\mathrm{IO}_{3}^{-}\right]=\left[\mathrm{Cu}^{2+}\right]=\frac{0.002}{2}\)= 0.001 M
Cu(IO₃)₂ ⇌ Cu²⁺ + \(2 \mathrm{IO}_{3}^{-}\)
[Cu²⁺] = 0.001 M
\(\) = 0.001 M
Ionic product = [Cu²⁺]\(\left[\mathrm{IO}_{3}^{-}\right]^{2}\)
– 1 x 10^-3 x [1 x 10^-3]² = 1 x 10^-9
K_sp = 7.4 x 10^-8
Cu(IO₃)₂ is precipitated if [Cu²⁺] .\(\left[\mathrm{IO}_{3}^{-}\right]^{2}\) > K_sp
Since, the ionic product is less than the solubility product. Hence there will be no precipitation.

Question 70.
The ionization constant of benzoic acid is 6.46 x 10^-5 and Ksp for silver benzoate is 2.5 x 10 . How many times is silver benzoate
more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
Answer:

where s is the solubility of C₆H₅COOAg
pH = 3.19
pH = -log[H⁺]
log[H⁺] = – pH = -3.19 = \(\overline{4} .81\)
[H⁺] = antilog \(\overline{4} .81\) = 6.46 x 10^-4

C₆H₅COOAg is 3.2 times more soluble in buffer than in pure water.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 6 Thermodynamics Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

PSEB 11th Class Chemistry Guide Thermodynamics InText Questions and Answers

Question 1.
Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
Answer:
(ii) A thermodynamic state function is a quantity whose value is independent of path. Functions like p, V, T etc., depend only on the state of a system and not on the path.

Question 2.
For the process to occur under adiabatic conditions, the correct condition is:
(i) ΔT = 0 (ii) Δp = 0
(iii) q = 0 (iv) w = 0
Answer:
(iii) A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0.

Question 3.
The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Answer:
(ii) The enthalpies of all elements in their standard states are zero.

Question 4.
\(\Delta \boldsymbol{U}^{\ominus}\) of combustion of methane is – X kJ mol^-1. The value of \(\Delta \boldsymbol{H}^{\ominus}\) is
(i) = ΔU
(ii) > ΔU
(iii) \(\Delta \boldsymbol{U}^{\ominus}\)
(iv) = 0
Answer:
(iii) CH₄(g) + 2O₂(g) > CO₂(g) + 2H₂O(l)
Δ_ng = n_p-n_r = 1 – 3 =-2
Hence,\(\Delta \boldsymbol{H}^{\ominus}\) = \(\Delta \boldsymbol{U}^{\ominus}\) + Δ n_gRT
\(\Delta \boldsymbol{H}^{\ominus}\) = – X – 2RT
Hence, \(\Delta \boldsymbol{H}^{\ominus}\) < \(\Delta \boldsymbol{U}^{\ominus}\)

Question 5.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol^-1 – 393.5 kJ mol^-1 and -285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄(g) will be
(i) -748kJmol^-1
(ii) -52.27kJ mol^-1
(iii) +748kJ mol^-1
(iv) +52.26kJ mol^-1
Answer:
According to the equation
(i) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH = -890.3 kJ mol^-1

(ii) C(s) + O₂(g) → CO₂(g); ΔH = – 393.5 kJ mol^-1

(iii) H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l); ΔH = -285.8 kJ mol^-1
Multiplying equation (iii) by 2, we get equation (iv).

(iv) 2H₂(g) + O₂(g) → 2H₂O(l); ΔH = – 571.6 kJ mol^-1
Adding eqs. (ii) and (iv), we get

(v) (C(s) +2H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH = – 965.1 kJ mol^-1
Reversing eqs. (i)

(vi) CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g); ΔH = +890.3 kJ mol^-1
Adding eqs. (v) and (vi), we get
C(s) + 2H₂(g) → CH₄(g); ΔH = – 74.8 kJ mol^-1
Hence, option (i) is correct.

Question 6.
A reaction, A + B → C + D +q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Answer:
(iv) For a reaction to be spontaneous, AG should be negative.
ΔG = ΔH – TΔS
According to the question, for the given reaction,
ΔS = positive
ΔH = negative (since heat is evolved)
=> ΔG = negative
Therefore, the reaction is spontaneous at any temperature.
Hence, option (iv) is correct.

Question 7.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
Answer:
According to the first law of thermodynamics,
ΔU = q + W …(i)
Given,
q = +701 J (heat is absorbed here, q is positive)
W=- 394 J
(work is done by the system hence W is negative)
Substituting the values in expression (i), we get
ΔU= +701 J + (-394 J)
ΔH = 307 J
Hence, the change in internal energy for the given process is 307 J.

Question 8.
The reaction of cyanamide, NH₂CN(s) with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be – 742.7 kJ mol^-1 at 298K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN(s) + \(\frac{3}{2}\)O₂(g) → N₂(g) + CO₂(g) + H₂O(l)
Answer:
The given reaction is
NH₂CN(s) + \(\frac{3}{2}\)O₂(g) → N₂(g) + CO₂(g) + H₂O(l)
Difference of moles of gaseous products and reactants,
Δ_ng = n_p – n_r = 2 – \(\frac{3}{2}=\frac{1}{2}\) = 0.5 mol
Given, ΔU=- 742.7kJ mol^-1
Enthalpy change ΔH = ΔU + Δ_ngRT
= – 742.7 + (0.5 x 8.314 x 10^-3 x 298)
= – 742.7 +1238.786 x 10^-3
= – 741.46 kJ mol^-1

Question 9.
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1K^-1.
Answer:
Given, mass of Al = 60.0 g
ΔT= 55 – 35 = 20°C
No.of moles of Al = \(\frac{60.0}{27}\) mol
Molar heat capacity of Al
= 24 J mol^-1K^-1
Heat, q = n.C . ΔT
= [ \(\frac{60}{27}\) mol )(24 J mol^-1K^-1)(20 K)
q= 1066.7 J
q = 1.07 kJ

Question 10.
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. Δ_fusH = 6.03 kJ mol^-1 at 0°C.
C_p[H₂O(l)] – 75.3 J mol^-1K^-1
C_p[H₂O(s)] = 36.8J mol^-1K^-1
Answer:
(i) Heat change required to lower the temperature of water from 10.0°Cto0°C
ΔH₁= n x C_p x ΔT= 1.0 x 75.3 x (-10) = – 753 J mol^-1
(ii) Heat change required to convert 1 mol of H₂O(l) at 0°C to H₂O(s) at 0°C
ΔH₂ = ΔH_fusion = – 6.03 kJ mol^-1 as heat is given out
(iii) Heat change required to change 1 mole of ice from 0°C to -10.0° C
ΔH₃ = – 36.8 x 10 x 1 = – 368 J mol^-1
Total heat change
= ΔH₁ + ΔH₂ + ΔH₃ = (- 0.753 – 6.03 – 0.368) kJ mol^-1
∴ Total enthalpy change = – 7.151 kJ mol^-1
As in each step, heat is evolved, each step will have a negative sign with Δ_H

Question 11.
Enthalpy of combustion of carbon to CO₂ is -393,5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Answer:
The reaction for the combustion of carbon into CO₂ is
C(s) + O₂(g) → CO₂(g); ΔH = – 393.5 kJ mol^-1 (1 mole CO₂ – 44g)
Heat released in the formation of 44 g CO₂ = 393.5 kJ
∴ Heat released in the formation of 35.2 g CO₂
\(\frac{393.5 \mathrm{~kJ}}{44 \mathrm{~g}}\) x 5.2g = 314.8 kJ mol^-1

Question 12.
Enthalpies of formation of CO(g), CO₍₂₎(g), N₂O(g) and N₂O₄(g) are -110, -393, 81and 9.7 kJ mol^-1 respectively. Find the value of Δ_rH for the reaction:
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
Answer:
Δ_fH (CO) = -110 kJ mol^-1
ΔH_f(CO₂) = -393 kJ mol^-1
Δ_fH (N₂O)= 81 kJ mol^-1
Δ_fH (N₂O₄) = 9.7 kJ mol^-1
The given reaction is
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g); Δ_rH = ?

= [81 + 3 (-393)] – [9.7 + 3 (-110)] kJ = [81 -1179]-[9.7-330] kJ
Δ_rH = – 777.8 kJ.

Question 13.
Given : N₂(g) + 3H₂(g) → 2NH₃(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = -92.4kJ mol^-1.
What is the standard enthalpy of formation of NH₃ gas?
Answer:
Given, N₂(g) + 3H₂(g) → 2NH₃(g); Δ_rH = – 92.4 kJ mol^-1
Chemical reaction for the enthalpy of formation of NH3(g) is as follows :
\(\frac{1}{2}\)N₂(g) + \(\frac{3}{2}\)H₂(g) → NH₃(g)
∴ Standard enthalpy of formation of NH₃(g)
= \(\frac{1}{2}\)\(\Delta H^{\theta}\) = \(\frac{1}{2}\) (-92.4 kJ mol^-1) = -46.2 kJ mol^-1

Question 14.
Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
CH₃OH(l) + \(\frac{3}{2}\)O₂(g) → CO₂(g) + 2H₂O(l); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = -726 kJ mol^-1
C_(graphite) + O₂(g) → CO₂(g); \(\Delta_{\boldsymbol{c}} \boldsymbol{H}^{\ominus}\) = – 393 kJ mol^-1
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l); \(\Delta_{\boldsymbol{f}} \boldsymbol{H}^{\ominus}\) = – 286 kJ mol^-1
Answer:

Question 15.
Calculate the enthalpy change for the process
CCl₄(g) → C(g) + 4Cl(g)
and calculate bond enthalpy of C—Cl in CCl₄(g).
\(\Delta_{\mathbf{v a p}} \boldsymbol{H}^{\ominus}\)(CCl₄) = 30.5kJ mol^-1. \(\Delta_{\boldsymbol{f}} \boldsymbol{H}^{\ominus}\)(CCl₄) = – 135.5kJ mol^-1.
\(\Delta_{a} \boldsymbol{H}^{\ominus}(\mathbf{C})\) = 715.0kJ mol^-1, \(\Delta_{\boldsymbol{a}} \boldsymbol{H}^{\ominus}\left(\mathbf{C l}_{\mathbf{2}}\right)\)(Cl2) = 242 kJ mol^-1; where \(\Delta_{\boldsymbol{a}} \boldsymbol{H}^{\ominus}\) is enthalpy of atomisation.
Answer:
Given,
(i) CCl₄(Z) → CCl₄(g); \(\Delta_{\mathrm{vap}} H^{\ominus}\) = +30.5 kJ mol^-1
(ii) C(s) → C(g); \(\Delta_{a} H^{\ominus}\) = 715.0 kJ mol^-1
(iii) Cl₂(g) → 2Cl(g); \(\Delta_{a} H^{\ominus}\) = 242 kJ mol^-1
(iv) C(s) + 2Cl₂(g) → CCl₄ a); \(\Delta_{f} H^{\ominus}\) = – 135.5KJ mol^-1

Enthalpy change for the given process
CCl₄(g) → C(g) + 4Cl(g); \(\Delta_{r} H^{\ominus}\) = ?

Add (i) and (iv) and subtract (ii) and (iii) x 2
CCl₄(l) + C(s) + 2Cl₂(g) – C(s) – 2Cl₂ (g) → CCl₄(g) + CCl₄(g) – C(g) – 4Cl(g)
or \(\Delta_{r} H^{\ominus}\) = 30.5 -135.5 – 715 – 484 = -1304 kJ
0 = CCl₄(g) – C(g) + 4Cl(g); \(\Delta_{r} H^{\ominus}\) = -1304kJ
CCl₄(g) → C (g) + 4Cl(g); \(\Delta_{r} H^{\ominus}\) = 1304 kJ
There are four bonds of C—Cl in CCl₄.
Bond enthalpy of C—Cl bond
= \(\frac{1304}{4}\) mol^-1
= 326 kJmol^-1

Question 16.
For an isolated system, ΔU = 0, what will be ΔS ?
Ans. For an isolated system, ΔU = 0 and for a spontaneous process, total entropy change must be positive. For example, consider the diffusion of two gases A and B into each other in a closed container which is isolated from the surroundings.

The two gases A and B are separated by a movable partition. When partition is removed, the gases begin to diffuse into each other and the system becomes more disordered. It shows that ΔS > 0 and ΔU = 0 for this process.
Moreover, ΔS =\(=\frac{q_{\mathrm{rev}}}{T}=\frac{\Delta H}{T}=\frac{\Delta U+p \Delta V}{T}=\frac{p \Delta V}{T}\) (∴ ΔU = 0)
i.e., TΔS or ΔS > 0

Question 17.
For the reaction at 298 K, 2A + B > C
ΔH = 400kJ mol^-1 and ΔS = 0.2kJK^-1mol^-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
Answer:
Given, ΔH = 400 kJ mol^-1, ΔS = 0.2 kJ K^-1mol^-1
Gibbs free energy, ΔG = ΔH – TΔS
0 = 400 kJ mol^-1 – T x 0.2 kJ K^-1mol^-1
(ΔG = 0 at equilibrium)
T = \(\frac{400 \mathrm{~kJ} \mathrm{~mol}^{-1}}{0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}\)
T = 2000 K
For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

Question 18.
For the reaction,
2Cl(g) → Cl₂(g), what are the signs of ΔH and ΔS?
Answer:
ΔH and ΔS are negative.
The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, ΔH is negative.
Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔS is negative for the given reaction.

Question 19.
For the reaction : 2A(g) + B(g) → 2D(g)
\(\Delta \boldsymbol{U}^{\ominus}\) = – 10.5 kJ and ΔS = – 441 JK^-1.
Calculate \(\Delta \boldsymbol{G}^{\ominus}\) for the reaction, and predict whether the reaction may occur spontaneously.
Answer:
For the given reaction,
2A(g) + B(g) → 2D(g)
Δn_g = 2 – 3 = -1 mol
Substituting the value of \(\Delta \boldsymbol{U}^{\ominus}\) in the expression ofΔH
\(\Delta H^{\ominus}=\Delta U^{\ominus}+\Delta n_{g} R T\)
= (-10.5 kJ) + (-1) x (8.314 x 10^-3kJK^-1 mol^-1) x (298 K)
= -10.5 kJ-2.48 kJ
\(\Delta H^{\ominus}\) = -12.98 kJ
We know that,
\(\Delta G^{\ominus}=\Delta H^{\ominus}-T \Delta S^{\ominus}\)
= -12.98 kJ – (298K) x ( – 44.1 JK^-1)
= -12.98 kJ + 13.14kJ
\(\Delta G^{\ominus}\) = +0.16 kJ
Since, \(\Delta G^{\ominus}\) is positive, the reaction will not occur spontaneously.

Question 20.
The equilibrium constant for a reaction is 10. What will be the value of \(\Delta G^{\ominus}\)? R = 8.314 JK^-1mol^-1, T = 300 K.
Answer:
\(\Delta G^{\ominus}\) = – 2303RT log K_c
Given, K_c = 10, T = 300 K, R = 8.314 J K^-1 mol^-1
\(\Delta G^{\ominus}\) = (- 2.303) (8.314 JK^-1mol^-1) (300K) (loglO) (∵ log 10 = 1)
= – 5744.14 Jmol^-1 = – 5.744 kJmol^-1

Question 21.
Comment on the thermodynamic stability of NO(g), given
\(\frac{1}{2}\)N₂(g) + \(\frac{1}{2}\)O₂(g) → NO(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = 90 kJ mol^-1
NO(g) + \(\frac{1}{2}\)O₂(g) → NO₂(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = – 74 kJ mol^-1
Answer:
NO(g) is unstable because formation of NO is endothermic (energy is absorbed) but NO₂(g) is formed because its formation is exothermic (energy is released).
Hence, unstable NO(g) changes to stable NO₂(g).

Question 22.
Calculate the entropy change in surroundings when 1.00 mol of H₂O(Z) is formed under standard conditions. \(\Delta_{\boldsymbol{f}} \boldsymbol{H}^{\ominus}\) = – 286
kJ mol^-1.
Answer:
Enthalpy change for the formation of 1 mol of H₂O(Z)
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l); \(\Delta_{f} H^{\mathrm{s}}\) = -286 kJ mol^-1
Energy released in the above reaction is absorbed by the surroundings. * It means,
q_surr = + 286 kJ mol^-1
Entropy change ΔS_surr = \(=\frac{q_{\mathrm{surr}}}{T}=\frac{286 \mathrm{~kJ} \mathrm{~mol}^{-1}}{298 \mathrm{~K}}\)
ΔS_surr = 0.95973 kJ K^-1 mol^-1 = 959.73 J mol^-1K^-1

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 5 States of Matter Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 5 States of Matter

PSEB 11th Class Chemistry Guide States of Matter InText Questions and Answers

Question 1.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm³ at 30° C?
Answer:
Given,
Initial pressure, P₁ =1 bar
Initial volume, V₁ = 500 dm³
Final volume, V₂ = 200 dm³
Final pressure, P₂ = ?
Sipce the temperature remains constant at30°C.
According to Boyle’s law,
P₁V₁ = P₂V₂
⇒ P₂ = \(\frac{p_{1} V_{1}}{V_{2}}=\frac{1 \times 500}{200}\) bar = 2.5 bar
Therefore, the minimum pressure required is 2.5 bar.

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure?
Answer:
Given,
Initial pressure, P₁ =1.2 bar
Initial volume, V₁ = 120 mL
Final volume, V₂ = 180 mL
Final pressure, P₂ = ?
Since the temperature remains constant at 35°C.
According to Boyle’s law,
⇒ P₂ = \(\frac{p_{1} V_{1}}{V_{2}}=\frac{1.2 \times 120}{180}\) bar = 0.8 bar
Therefore, the pressure would be 0.8 bar.

Question 3.
Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure p.
Answer:
The equation of state is given by
pV = nRT

or d = \(\frac{p M}{R T}\)
If T constant, then d ∝ p.

Question 4.
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer:
Density (d) of the substance at temperature (T)
d = \(\frac{p M}{R T}\)
When T and d are same and R is constant then
p₁M₁ = p₂M₂
Given, P₁ = 2 bar
p₂ = 5 bar
Molecular mass of nitrogen, M₂ = 28g/mol
Now, M₁ = \(\frac{M_{2} p_{2}^{2}}{p_{1}}=\frac{28 \times 5}{2}\) = 70 g/mol
Hence, the molecular mass of the oxide is 70 g/ mol.

Question 5.
Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
Answer:
For ideal gas A, the ideal gas equation is given by
p_AV – n_ART …(i)
Where, p_A and n_A represent the pressure and number of moles of gas A. For ideal gas B, the ideal gas equation is given by
p_BV = n_BRT …(ii)
Where, p_B and n_B represent the pressure and number of moles of gas B.
Number of moles of A gas, n_A = \(\frac{1}{M_{A}}\) [M_A = molar mass of gas A]
Number of moles of B gas, n_B = \(\frac{2}{M_{B}}\) [M_B = molar mass of gas B]
Pressure of gas A, p_A = 2 bar
Total pressure P_total = p_A + p_B = 3 bar
Pressure of gas B, p_B = p_total – p_A = 3 – 2 = 1 bar
V, R and T are same for both the gases.
Hence from eqs. (i) and (ii)

Question 6.
The drain cleaner, Drainex contains small hits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15g of aluminum reacts?
Answer:

Question 7.
What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4g of carbon dioxide contained in a 9 dm³ flask at 27°C?
Answer:
Mass of CH₄ = \(\frac{\text { Mass of } \mathrm{CH}_{4}}{\text { Molar mass of } \mathrm{CH}_{4}}\)
[Molar mass of CH₄ = 12 + 4 x 1 = 16]
= \(\frac{3.2}{16}\) = 0.2 mol
Mass of CO₂ = \(\frac{4.4}{44}\) = 0.1 mol
[Molar mass of CO₂ = 12 + 2 x 16 = 44]
Total moles = 0.2 + 0.1 = 0.3 mol
Pressure P = \(\frac{n R T}{V}=\frac{0.3 \times 8.314 \times 300}{9 \times 10^{-3}}\) = 8.314 x 10⁴ Pa

Question 8.
What will be the pressure of the gaseous mixture when o.5L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?
Answer:
To calculate the partial pressure of H₂, Le., pH₂
V₁ = 0.5L, V₂ = 1L, p₁ = 0.8 bar, p₂ = ?
Temperature remaining constant, applying Boyl&s Law
p₁V₁=p₂V₂
0.8 x 0.5= p₂ x 1
or P₂ = 0.4 bar
To calculate the partial pressure of O₂ i.e., PO₂
V₁ = 2.0L, V₂ = 1L, p₁ = 0.7 bar, P₂ = ?
Applying Boyle’s Law
p₁V₁ = p₂V₂
0.7 x 2.0 = 1 x p₂
or p₂ = 1.4 bar.
If p is final pressure of the gas mixture, then according to Dalton’s Law of partial pressures
p = pH₂ + pO₂ = (0.4 +1.4)
= 1.8 bar

Question 9.
Density of a gas is found to be 5.46 g/dm³ at 27°C at 2 bar pressure. What will be its density at STP?
Answer:
Given,
d₁ = 5.46 g/dm³
p₁ = 2 bar
T₁ = 27°C = (27 + 273)K = 300 K
P₂ = 1 bar
T₂ = °C = 273 K
d₂ = ?
Density a = \(\frac{M p}{R T}\)
For same gas at different temperatures and pressures

Question 10.
34.05 ml. of phosphorus vapour weighs 0.0625g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?
Answer:
Given,
p= 0.1 bar
V= 34.05 mL = 34.05 x 10^-3L
R = 0.083 bar dm³K^-1 mol^-1
T = 546°C = (546 + 273) K = 819 K
Mass, M = 0.0625 g
Now, pV = nRT
n = \(\frac{p V}{R T}=\frac{0.1 \times 34.05 \times 10^{-3}}{0.083 \times 819}\)= 5.0 x 10^-5 mol^-1
∴ Molar mass of phosphorus = \(\frac{0.0625}{5.0 \times 10^{-5}}\) = 1250 g mol^-1
Hence, the molar mass of phosphorus is = 1250 g mol^-1

Question 11.
A student forgot to add the reaction nixture to the round bottomed flask at 27°C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Answer:
Let the volume of the round bottomed flask = V cm3 at 27°C = 300 K
V₁ = V, T₁ = (27 + 273)K = 300K, V₂ =?,
T₂ = 477° C = (477 + 273)K
According to Charles’s law,
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
V₂ = \(\frac{V_{1} T_{2}}{T_{1}}=\frac{750 \mathrm{~V}}{300}\) = 2.5V
Therefore, volume of air expelled out = 2.5 V – V = 1.5 V
Hence, fraction of air expelled out = \(\frac{1.5 \mathrm{~V}}{2.5 \mathrm{~V}}\) = 0.6

Question 12.
Calculate the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar. (R = 0.083 bar dm³K^-1 mol^-1).
Answer:
Given,
n = 4.0 mol, V = 5 dm³, p = 3.32 bar, R = 0.083 bar dm³K^-1mol^-1
Applying ideal gas equation ‘
pV = nRT
T = \(\frac{p V}{n R}=\frac{3.32 \times 5}{4 \times 0.083}\) = 50K
Hence, the required temperature is 50 K.

Question 13.
Calculate the total number of electrons present in 1.4g of dinitrogen gas.
Answer:
Molar mass of dinitrogen (N₂) = 28 g mol^-1 Mass
Moles = \(\frac{Mass}{Molar mass}\)
ⁿN₂ = \(\frac{1.4}{28}\) = 0.05 mol
1 mol = 6.022 x 10²³ molecules
0.05 mol = 0.05 x 6.022 x 10²³ molecules
= 0.3011 x 10²³ molecules
Now, 1 molecule of N₂ contains = 14 electrons.
Therefore, 0.3011 x 10²³ molecules will contain = 14 x 0.3011 x 10²³
= 4.214 x 10²³ electrons

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains, if 10¹⁰ grains are distributed each second?
Answer:
Avogadro number, N_A = 6.022 x 10²³

Hence, the time taken would be 1.909 x 10⁶ years.

Question 15.
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm³ at 27°C.
R 0.083 bar dm³ K^-1 mol^-1
Answer:
Moles of O₂ = \(\frac{\text { Mass }}{\text { Molar weight }}=\frac{8}{32}\) = 0.25 mol
Moles of H₂ = \(\frac{4}{2}\) = 2 mol
Therefore, total number of moles in the mixture = 0.25 + 2- 2.25 mol
Given, V = 1 dm³
n = 2.25 mol
R = 0.083 bar dm³K^-1mol^-1
T = 27°C = 300K
pV= nRT
⇒ Pressure, p = \(\frac{n R T}{V}=\frac{2.25 \times 0.083 \times 300}{1}\) = 56.025 bar.
Hence, the total pressure of the mixture is 56.025 bar.

Question 16.
Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10m, mass 100kg is filled with helium at 1.66 bar at 27°C.(Density of air = 1.2 kg m^-3 and R = 0.083 bar dm³K^-1 mol^-1)
Answer:
Given,
Radius of the balloon, r = 10m
∴ Volume of the balloon = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7}\) x 10³ = 4190.5 m3(approx)
∴ Mass of displaced air = V_{displaced air} x density of air
= 4190.5 x 1.2 kg = 5028.6 kg
Now, mass of helium (m) filled in balloon
\(m_{\mathrm{He}}=\frac{M p V}{R T}\)
Here, M = 4 x 10^-3 kg mol^-1
p= 1.66 bar
V = Volume of the balloon = 4190.5 x 10³ dm³
R = 0.083 bar dm³K^-1mol^-1
T = 27°C = 300 K
Then m_He = \(\frac{4 \times 10^{-3} \times 1.66 \times 4190.5 \times 10^{3}}{0.083 \times 300}\) = 1117.5kg(approx)
Now, total mass of the balloon filled with helium
= (100 +1117.5)kg = 1217.5kg
Hence, pay load = mass of displaced air – mass of balloon
= (5028.6 -1217.5) kg = 3811.1 kg
Hence, the pay load of the balloon is 3811.1 kg

Question 17.
Calculate the volume occupied by 8.8g of CO2 at 31.1°C and 1 bar pressure. (R = 0.083 bar LK^-1 mol^-1).
Answer:
We know that,
pV = nRT m
pV = \(\frac{m}{M} R T\)
V = \(\frac{m R T}{M p}\)
Here, m = 8.8 g
R = 0.083 bar LK^-1mol^-1
T = 31.1°C = 304.1 K
M = 44 g mol^-1
p = 1 bar
Then, volume V = \(\frac{8.8 \times 0.083 \times 304.1}{44 \times 1}\) = 5.048 L
Hence, the volume occupied is 5.048 L.

Question 18.
2.9 g of a gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Answer:
Applying the ideal gas equation pV = nRT

Hence, the molar mass of the gas is 40 g mol^-1 .

Question 19.
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer:
Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80g.
Moles of dihydrogen,
ⁿH₂ = 20/2 = 10 mol
Moles of dioxygen,
ⁿO₂ = 80/32 = 2.5 mol
Given,
Total pressure of the mixture, p_total = 1 bar
Then, partial pressure of dihydrogen,
^pH₂ = \(\frac{n_{\mathrm{H}_{2}}}{n_{\mathrm{H}_{2}}+n_{\mathrm{O}_{2}}} \times p_{\text {total }}=\frac{10}{10+2.5} \times 1\) = 0.8 bar
Hence, the partial pressure of dihydrogen is 0.8 bar.

Question 20.
What would be the SI unit for the quantity pV² T² / n?
Answer:
The SI unit of \(\frac{p V^{2} T^{2}}{n}\) is given by
= \(\frac{\left(\mathrm{Nm}^{-2}\right)\left(\mathrm{m}^{3}\right)^{2}(\mathrm{~K})^{2}}{\mathrm{~mol}}\) = Nm⁴K²mol^-1

Question 21.
In terms of Charles’ law explain why -273°C is the lowest possible temperature.
Answer:
According to Charles’ law,
V_t = V₀ [1 + \(\frac{t}{273}\) ]
At t = -273° C
V_t = V₀ = [1 – \(\frac{273}{273}\) ] = 0
Thus, at -273° C, volume of a gas becomes zero and below this temperature the volume becomes negative, which is meaningless.

Question 22.
Critical temperature for carbon dioxide and methane are 31.1° C and -81.9°C respectively. Which of these has stronger intermolecular forces and why?
Answer:
Higher is the critical temperature of a gas, easier is its liquefaction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of C02.

Question 23.
Explain the physical significance of van der Waals parameters.
Answer:
Significance of constant ‘a’ : The value of constant ‘a’ is a measure of the magnitude of intermolecular forces between the molecules of the gas. Its units are atm L mol^-2 . Larger the value of ‘a’ larger will be the intermolecular forces among the gas molecules.
Significance of constant ‘b’ : The constant ‘b’ is called co-volume or excluded volume per mol of a gas. Its units are litre mol^-1.The volume of V is four times the actual volume of the molecules. It is measure of effective size of the gas molecules.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 4 Chemical Bonding and Molecular Structure Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

PSEB 11th Class Chemistry Guide Chemical Bonding and Molecular Structure InText Questions and Answers

Question 1.
Explain the formation of a chemical bond.
Answer:
A chemical bond is defined as an attractive force that holds the constituents (atoms, ions etc.) together in a chemical species.
Various theories have been suggested for the formation of chemical bonds such as the electronic theory, valence shell electron pair repulsion theory, valence bond theory, and molecular orbital theory.

A chemical bond formation is attributed to the tendency of a system to attain stability. It was observed that the inertness of noble gases was because of their fully filled outermost orbitals. Hence, it was postulated that the elements having incomplete outermost shells are unstable (reactive). Atoms, therefore, combine with each other and complete their respective octets or duplets to attain the stable configuration of the nearest noble gases. This combination can occur either by sharing of electrons or by transferring one or more electrons from one atom to another. The chemical bond formed as a result of sharing of electrons between atoms is called a covalent bond. An ionic bond is formed as a result of the transference of electrons from one atom to another.

Question 2.
Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Answer:
Mg : There are two valence electrons in Mg atom (2, 8, 2). Hence, the Lewis dot symbol of Mg is :

Na : There is only one valence electron in an atom of sodium (2, 8,1). Hence, the Lewis dot Symbol is :

B : There are three valence electrons in Boron atom (2, 3). Hence, the Lewis dot symbol is :

0 : There are six valence electrons in an atom of oxygen (2, 6). Hence, the Lewis dot symbol is :

N: There are five valence electrons in an atom of nitrogen (2,5). Hence, the Lewis dot symbol is :

Br : There are seven valence electrons in bromine (2, 8, 18, 7). Hence, the
Lewis dot symbol is :

Question 3.
Write Lewis symbols for the following atoms and ions: S and S^2-; Al and Al³⁺; H and H^–
Answer:

Question 4.
Draw the Lewis structure for the following molecules and ions :
H₂S, SiCl₄, BeF₂, \(\mathrm{CO}_{3}^{2-}\), HCOOH
Answer:

Question 5.
Define octet rule. Write its significance and limitations.
Answer:
Octet rule : Atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shell.
Significance of octet rule : It help to explain why different atom combine with each other to form ionic or covalent compounds. Limitations of the Octet Rule
Although octet rule can explain the formation of a large number of compounds but it has many exceptions also, which are discussed below :

(i) Electron deficient molecules : There are some molecules in which the central atom is surrounded by less than eight electrons, i.e., their octet is incomplete. Elements having less than four valence electrons generally form molecules of this category.
e.g., BeCl₂, BF₃, AlCl₃, LiCl, BeH₂ etc.

(ii) Odd electron molecules : Molecules like NO, NO₂, O₂ etc., are examples of such molecules in which bonded atoms have odd number of electron (usually 3) in between them. That’s why these are called odd electron molecules.
In case of these molecules, the octet rule is not satisfied for all the atoms, e.g.,

Species with one unpaired electron are called free radicals. These are paramagnetic and most of them are generally unstable.

(iii) Electron rich molecules : Elements of the third and higher periods of the periodic table, because of the availability of d orbitals can expand their covalency and -can accommodate more than eight valence electrons around the central atom. This is referred as expanded octet. Here, also the octet rule is not applicable, e.g., PF₅ (10 electrons around P atom), SF₆ (12 electrons around S atom), H₂SO₄ (12 electron around S atom).
Compounds having expanded octet are also termed as hypervalent compounds.

(iv) Other drawbacks : Other drawbacks of this theory are as follows:
1. Octet rule is based on the inertness of noble gases but some noble gases like xenon and krypton form several compounds with oxygen and fluorine like. XeF₂, XeF₄, XeF₆, XeOF₄, XeO₂F₂, KrF₂ etc.
2. It does not tell anything about the shapes of molecules and their relative stabilities.
3. It fails to explain the paramagnetic behaviour of oxygen. (Which should be diamagnetic according to this rule but it is infact paramagnetic in nature).

Question 6.
Write the favourable factors for the formation of ionic bond.
Answer:
The favourable factors for ionic bond formation are as follows :

(i) Low ionization enthalpy of element forming cation.
(ii) More negative electron gain enthalpy of element forming anion.
(iii) High lattice energy of the compound formed.

Question 7.
Discuss the shape of the following molecules using the VSEPR model:
BeCl₂, BCl₃, SiCl₄, ASF₅, H₂S, PH₃
Answer:
According to VSEPR theory, the shape of a molecule depends upon the number of valence shell electron pairs (bonded or non-bonded) around the central atom. Pairs of electrons in the valence shell repel each other. The order of their repulsion is as follows :
Ip -Ip > Ip -bp > bp – bp
(i) BeCl₂ or

The central atom Be has only 2 valence electrons which are bonded to Cl, so there are only 2 bond pairs and no lone pairs. It is of the type AB₂ and hence, the shape is linear.

(ii) BCl₃:

The central atom B has only 3 valence electrons which are bonded with three Cl atoms, so it contains only 3 bond pairs and no lone pair. It is of the type AB₃ and hence, the shape is trigonal planar.

(iii) SiCl₄ :

Similarly, the central atom Si has only 4 bond pairs and no lone pair. It is of the type AB₄ and hence, the shape is tetrahedral.

(iv) AsF₅:

The central atom As has only 5 bond pairs and no lone pair. It is of the type AB₅ and hence, the shape is trigonal bipyramidal.

(v) H₂S:

The central atom S has 6 valence electrons. Out of these only two are used in bond formation with two H-atoms while four (two pairs) remains as non-bonding electrons (i.e., lone pairs). So, it contains 2 bond pairs and 2 lone pairs. It is of the type AB₂E₂ and hence, the shape
is bent or V-shaped.

(vi) PH₃

The central atom P has 5 valence electrons. Out of which three are utilised in bonding with H atoms and one pair remains as lone pair. So, it contains 3 bond pairs and one lone pair. It is of the type AB₃E and hence the shape is pyramidal.

Question 8.
Although geometries of NH₃ and H₂O molecules are distorted
tetrahedral, bond angle in water is less than that of ammonia.
Discuss.
Answer:

In H₂O molecule there, is lone pair-lone pair repulsion due to the presence of two lone pairs of electrons while in NH₃ molecule there are only lone pair-bond pair repulsion. According to VSEPR theory the former one is more stronger and hence the bond angle in water is less than that of ammonia.

Question 9.
How do you express the bond strength in terms of bond order?
Answer:
Bond strength represents the extent of bonding between two atoms forming a molecule. The larger the bond energy, the stronger is the bond and the greater is the bond order.

Question 10.
Define the bond length.
Answer:
Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
Bond lengths are expressed in terms of Angstrom (10^-10 m) or picometer (10^-12 m) and are measured by spectroscopic X-ray diffractions and |
electron-diffraction techniques.

Question 11.
Explain the important aspects of resonance with reference to \(\mathrm{CO}_{3}^{2-}\) the ion.
Answer:
According to experimental findings, all carbon to oxygen bonds in \(\mathrm{CO}_{3}^{2-}\) are equivalent. Hence, it is inadequate to represent \(\mathrm{CO}_{3}^{2-}\) ion by a single Lewis structure having two single bonds and one double bond.
The \(\mathrm{CO}_{3}^{2-}\) ion is best described as a resonance hybrid of the canonical forms I, II and III.

All canonical forms have similar energy, same positions of atoms and same number of bonded and non-bonded pairs of electrons.

Question 12.
H₃PO₃ can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons for the same.

Answer:
The given structures cannot be taken as the canonical forms of the resonance hybrid of H₃PO₃, because the positions of the atoms have been changed.

Question 13.
Write the resonance structures for SO₃, NO₂ and \(\mathrm{NO}_{3}^{-}\)
Answer:
The resonance structures are :
(a) SO₃:

(b) NO₂:

(c) \(\mathrm{NO}_{3}^{-}\)

Question 14.
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b) Ca and O (c) Al and N.
Answer:
(a) K and S :
The electronic configurations of K and S are as follows :
K : 2, 8, 8, 1
S : 2, 8, 6

Sulphur (S) requires 2 more electrons to complete its octet. Potassium (K) requires one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as:

(b) Ca and O :
The electronic configurations of Ca and O are as follows :
Ca : 2, 8, 8, 2
O : 2, 6
Oxygen requires two electrons more to complete its octet, whereas calcium has two electrons more than the nearest noble gas i.e., Argon. Hence, the electron transfer takes place as :

(c) Al and N :
The electronic configurations of Al and N are as follows :
A1: 2, 8, 3
N : 2, 5
Nitrogen is three electrons short of the nearest noble gas (Neon), whereas aluminium has three electrons more than neon. Hence, the electron transference can be shown as :

Question 15.
Although both CO₂ and H₂O are triatomic molecules, the shape of H₂O molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.
Answer:
According to experimental results, the dipole moment of carbon dioxide is zero. This is possible only if the molecule is linear so that the dipole moments of C—O bonds are equal and opposite to nullify each other.

Resultant µ = 0 D
On the other hand, H₂O molecule is found to have a net dipole moment value of 1.84 D (thoughit is a triatomic molecule as CO₂). The value of the dipole moment suggests that the structure of H20 molecule is bent where the dipole moment of O—H bonds are unequal.

Question 16.
Write the significance/applications of dipole moment.
Answer:
The applications of dipole moment are as follows :
1. In determining the polarity of bonds : As µ = e x d, obviously greater is the magnitude of dipole moment, higher will be the polarity of the bond. This is applicable to molecules containing only one polar bond like HC1, HBr etc. In non-polar molecules like, H₂, O₂, N₂ the dipole moment is zero. It is because there is no charge separations in these molecules [e = 0]. Thus, dipole moment can also be used to distinguish between polar and non-polar molecules.

2. In the calculation of percentage ionic character :
Take the example of HCl. Its µ = 1.03 D
If HCl is 100% ionic, each end would carry charge of one unit
i.e., 4.8 x 10^-10 e.s.u.
d (bond length) in H—Cl = 1.275Å
∴ for 100% ionic character, dipole moment will be
µ_ionic = e x d
= 4.8 x 10^-10 e.s.u x 1.275 x 10^-8 cm
= 6.12 x 10^-18 e.s.u cm = 6.12 D
∴ Percentage of ionic character =

= \(\frac{1.03}{6.12}\) x 100 = 16.83.

3. In determining the symmetry (or shape) of the molecules : Dipole moment is an important property in determining the shape of molecules containing 3 or more atoms. For instant if any molecule possesses two or more polar bonds, it will not be symmetric if it possesses some net molecular dipole moment as in case of water (\(\mu_{\mathrm{H}_{2} \mathrm{O}}\) = 1-84 D) and ammonia (\(\mu_{\mathrm{NH}_{3}}\) = 1.49 D). But if a molecule contains a number of similar atoms linked to a central atom the overall dipole moment of the molecule is found out to be zero, this will imply that the molecule is symmetrical as in the case of CO₂, BF₃, CH₄,CCl₄,etc.

Question 17.
Define electronegativity. How does it differ from electron gain enthalpy?
Answer:
Electronegativity is the ability of an atom in a chemical compound to attract a bond pair of electrons towards itself. Electronegativity of any given element is not constant. It varies according to the element to which it is bound. It is not a measurable quantity. It is only a relative number.

On the other hand, electron gain enthalpy is the enthalpy change that takes place when an electron is added to a neutral gaseous atom to form an anion. It can be negative or positive depending upon whether the electron is added or removed. An element has a constant value of the electron gain enthalpy that can be measured experimentally.

Question 18.
Explain polar covalent bond with the help of suitable example.
Answer:
When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of electrons is not shared equally. The bond pair shifts towards the nucleus of the atom having greater electronegativity. As a result, electron distribution gets distorted and the electron cloud is displaced towards the electronegative atom.

As a result, the electronegative atom becomes slightly negatively charged while the other atom becomes slightly positively charged. Thus, opposite poles are developed in the molecule and this type of a bond is called a polar covalent bond.

For example-In HF, the electron pair is attracted more towards F atom due to its higher electronegativity. HF may be written as

Question 19.
Arrange the bonds in order of increasing ionic character in the molecules : LiF, K₂O, N₂, SO₂ and ClF₃.
Answer:
More the difference of electronegativity, more the ionic character of the molecules
N₂ < SO₂ < ClF₂ < K₂O < LiF.

Question 20.
The skeletal structure of CH₃COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.

Answer:
The correct Lewis structure for acetic acid is given below :

Question 21.
Apart from tetrahedral geometry, another; possible geometry for CH₄ is square planar with the four H atoms at the comers of the square and the C atom at its centre. Explain why CH₄ is not square planar?
Answer:
Electronic configuration of carbon atom :
₆C : 1s² 2s² 2p²
In the excited state, the orbital picture of carbon can be represented as:

Hence, carbon atom undergoes sp³ hybridization in CH₄ molecule and takes a tetrahedral shape.

For a square planar shape, the hybridization of the central atom has to be dsp. However, an atom of carbon does not have d-orbitals to undergo dsp² hybridization. Hence, the structure of CH₄ cannot be square planar. Moreover, with a bond angle of 90° in square planar, the stability of CH₄ will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for CH₄.

Question 22.
Explain why BeH₂ molecule has a zero dipole moment although the Be—H bonds are polar.
Answer:
BeH₂ molecule is linear. The two equal bond dipoles point in opposite
directions and cancel the effect of each other.
That is why its dipole moment is zero.

Question 23.
Which out of NH₃ and NF₃ has higher dipole moment and why?
Answer:
In both molecules i.e., NH₃ and NF₃, the central atom (N) has a lone pair electron and there are three bond pairs. Hence, both molecules have a pyramidal shape. Since fluorine is more electronegative than hydrogen, it is expected that the net dipole moment of NF₃ is greater than NH₃. However, the net dipole moment of NH₃ (1.46 D) is greater than that of NF₃ (0.24 D).
This can be explained on the basis of the directions of the dipole moments of each individual bond in NF₃ and NH₃. These directions can be shown as :

Thus, the resultant moment of the N—H bonds add up to the bond moment of the lone pair (the two being in the same direction), whereas that of the three N—F bonds partly cancels the moment of the lone pair.
Hence, the net dipole moment of NF₃ is less than that of NH₃.

Question 24.
What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp² , sp³ hybrid orbitals.
Answer:
Hybridisation : It is defined as the mixing of the atomic orbitals belonging to the same atom but having slightly different energies so that a redistribution of energy takes place between them resulting in the formation of new orbitals of equal energies and identical shapes. The new orbitals thus formed are known as Hybrid Orbitals. sp Hybridisation : Here one s and one p orbitals of same atom mix up to
give two sp hybrid orbitals with \(\frac{1}{2}\)s and \(\frac{1}{2}\)p character and linear shape with
bond angle of 180° between them. For example, in BeH₂, BeF₂ and C₂H₂, Be and C are sp-hybridised.

sp hybridization is also called diagonal hybridization.

sp² Hybridisation : Here one s and two p-orbitals of same atom mix up to form three sp²hybrid orbitals with \(\frac{1}{3}\)s and \(\frac{2}{3}\)p character. They form Trigonal Planar shapes with an angle of 120° with themselves. For example, in BH₃ and BF₃, boron is sp² hybridised and in C₂H₄, carbon is sp² hybridised.

sp³ Hybridisation : Here one s and three p orbitals of same atom mix up to give four sp³ hybrid orbitals with \(\frac{1}{4}\) s character and \(\frac{3}{4}\)p character. They form tetrahedral shapes with angles of 109°, 28′ with themselves. For example, in methane (CH₄), ethane (C₂H₆) and all compounds of carbon containing C—C single bonds, carbon is sp³ hybridised.

Question 25.
Describe the change in hybridisation (if any) of the A1 atom in the following reaction.
AlCl₃ + Cl^– → \(\mathbf{A l C l}_{\mathbf{4}}^{-}\)
Answer:
Electronic configuration of A1 in ground state is \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{\prime}{ }_{x}\) and it is \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{\prime}, 3 p_{x}^{\prime}, 3 p^{\prime} y\) in excited state.

In the formation of AlCl₃ Al undergoes sp²-hybridisation and it is trigonal planar in shape. While in the formation of \(\mathrm{AlCl}_{4}^{-} \), Al undergoes sp³-hybridisation. It means empty 3p_z-orbital also involved in hybridisation. Thus, the shape of \(\mathrm{AlCl}_{4}^{-} \) ion is tetrahedral.

Question 26.
Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF₃ + NH₃ → F₃B . NH₃
Answer:
In BF₃, B is sp² hybridised and in NH₃, N is sp³ hybridised. After the reaction hybridisation of B changes to sp³ but that of N remains unchanged.

Question 27.
Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.
Answer:
Formation of C₂H₄ (ethylene)

Formation of C₂H₂ (acetylene)

Question 28.
What is the total number of sigma and pi bonds in the following molecules?
(a) C₂H₂ (b) C₂H₄
Answer:
(a) The structure of C₂H₂ can be represented as:

Hence, there are three sigma and two pi-bonds in C₂H₂.

(b) The structure of C₂H₄ can be represented as:

Hence, there are five sigma bonds and one pi-bond in C2H4.

Question 29.
Considering x-axis as the intemuclear axis which out of the following will not form a sigma bond and why?
(a) 1s and 1s (b) 1s and 2p_x, (c) 2p_y and 2p_y (d) Is and 2s.
Answer:
(c) 2p_y and 2p_y orbitals will not a form a sigma bond. Taking x-axis as
the intemuclear axis, 2p_y and 2p_y orbitals will undergo lateral overlapping, thereby forming a pair bond.

Question 30.
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH₃— CH₃; (b) CH₃—CH = CH₂; (c) CH₃—CH₂—OH;
(d) CH₃—CHO; (e) CH₃COOH
Answer:

Question 31.
What do you understand by bond pan’s and lone pah’s of electrons? Illustrate by giving one example of each type.
Ans. When two atoms combine by sharing their one or more valence electrons, a covalent bond is formed between them. The shared pairs of electrons present between the bonded atoms are called bond pairs. All valence electrons may not participate in bonding. The electron pairs that do not participate in bonding are called lone pairs of electrons.
For example, in C₂H₆ (ethane), there are seven bond pairs but no lone pair present.

In H₂O, there are two bond pairs and two lone pairs on the central atom (oxygen).

Question 32.
Distinguish between a sigma and a pi bond.
Answer:
The following are the differences between sigma and pi-bonds :

	Sigma (σ) Bond	Pi (π) Bond
(a)	It is formed by the end to end over lapping (axial over lapping) of atomic orbitals.	It is formed by the lateral overlapping (sideway overlapping) of atomic orbitals.
(b)	The orbitals involved in the overlapping are s—s, s—p or p—p.	These bonds are formed by the overlapping of p—p orbitals only.
(c)	It is a strong bond.	It is a weak bond.
(d)	The electron cloud is symmetrical about the line joining the two nuclei.	The electron cloud is not symmetrical.
(e)	It consists of one electron cloud, which is symmetrical about the internuclear axis.	There are two electron clouds lying above and below the plane of the atomic nuclei.
(f)	Free rotation about σ bonds is possible.	Rotation is restricted in case of pi-bonds.

Question 33.
Explain the formation of H₂ molecule on the basis of valence bond theory.
Answer:
Consider two hydrogen atoms A and B are approaching each other. Their nuclei are N_A and N_B and electrons present in them are represented by e_A and e_B. When the two atoms are far apart, there is no interaction between them but as these approach each other, some new ‘ attractive and repulsive force begin to operate.
Attractive forces generated between

(i) nucleus of one atom and its own electron i.e., N_A – e_A and N_B – e_B.
(ii) nucleus of one atom and electron of other atom i.e., N_A – e_B and N_B-e_A.
Similarly, repulsive forces originated in between
1. electrons of two atoms i.e., e_A – e_B
2. nuclei of two atoms N_A – N_B.
Attractive forces tend to bring the combining atoms close to each other “ whereas repulsive forces tend to push them apart as shown in the figure

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is reached when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.

Question 34.
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
Conditions required for the Combination of Atomic Orbitals
The linear combination of atomic orbitals to form molecular orbitals is possible only when they satisfied the following conditions :
(i) Similar energy of combining atomic orbitals : The combining atomic orbitals must possess the same or nearly the same energy. It means that Is orbital can combine with another Is orbital but not with 2s orbital because the energy of 2s orbitals is appreciable higher than that of Is orbital. However, it is not true in case of very different atoms.
(ii) Similar symmetry of combining atomic orbitals : The combining atomic orbitals must possess the same symmetry about the molecular axis along with the same energy. If the orbitals have same energy but their symmetry is not same, they will not combine e.g., 2p_z orbital of one atom can combine with 2p_z orbital or 2s orbital of the other atom but not with the 2p_x or 2p_y orbitals as their symmetries are different.
(iii) Maximum overlap : The combining atomic orbitals must overlap to the maximum extent. Higher the extent of overlapping, more will be the electron-density between the nuclei of a molecular orbital.

Question 35.
Use molecular orbital theory to explain why the Be₂ molecule does not exist.
Answer:
The electronic configuration of Beryllium is 1s² 2s²
The electronic configuration of Be₂ molecule (4 + 4 = 8),
σ1s², σ* 1s², σ2s²s², σ* 2s²
Hence, the bond order of Be₂ is -(N_b – N_a).
where,
N_b = Number of electrons in bonding orbitals.
N_a – Number of electrons in anti-bonding orbitals.
∴ Bond order of Be₂ = \(\frac{1}{2}\) (4 – 4) = 0
A negative or zero bond order means that the molecule is unstable. Hence, Be2 molecule does not exist.

Question 36.
Compare the relative stability of the following species and indicate their magnetic properties ; \(\mathbf{O}_{2}, \mathbf{O}_{2}^{+}, \mathbf{O}_{2}^{-}\)(superoxide), \(\mathrm{O}_{2}^{2-}\) (peroxide)
Answer:

Question 37.
Write the significance of a plus and a minus sign shown in representing the orbitals.
Answer:
Molecular orbitals are represented by wave function. A plus sign in an orbital indicates a positive wave function while a minus sign in an orbital represents a negative wave function. Combination of two wave functions having similar sign gave bonding molecular orbital while that having opposite sign gave antibonding molecular orbital.

Question 38.
Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?
Answer:
The ground state and excited state outer electronic configurations of phosphorus (Z = 15) are as follows :

Phosphorus atom is sp³ d hybridised in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as :

The five sp³d hybrid orbitals are directed towards the five corners of the trigonal bipyramidals. Hence, the geometry of PCl₅ can be represented as :

There are five P—Cl sigma bonds in PCl₅. Three P—Cl bonds lie in one plane and make an angle of 120° with each other. These bonds are called equatorial bonds.
The remaining two P—Cl bonds lie above and below the equatorial plane and make an angle of 90° with the plane. These bonds are called axial bonds.

As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.

Question 39.
Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Answer:
A hydrogen bond is defined as an attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule (may be of the same kind). Due to a difference between electronegativities, the bond pair between hydrogen and the electronegative atom gets drifted far away from the hydrogen atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a positive charge.

The magnitude of H-bonding is maximum in the solid state and minimum in the gaseous state.
There are two types of H-bonds:
(i) Intermolecular H-bonds e.g., HF, H₂O etc
(ii) Intramolecular H-bonds e.g., o-nitrophenol
Hydrogen bonds are stronger than Van der Waals forces since hydrogen bonds are regarded as an extreme form of dipole-dipole interaction.

Question 40.
What is meant by the term bond order? Calculate the bond order of: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}\) and \(\mathbf{O}_{2}^{-}\).
Answer:
Bond order is defined as half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.
Bond order = \(\frac{1}{2}\) (N_b – N_a)
If N_b > N_a, then the molecule is said be stable. However, if N_b ≤ N_a, then the molecule is considered to be unstable.
Bond order values 1, 2 or 3 correspond to single, double or triple bonds respectively.

Calculation of the bond order of \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}\) and \(\mathbf{O}_{2}^{-}\).
Electronic configuration of N₂

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 3 Classification of Elements and Periodicity in Properties Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

PSEB 11th Class Chemistry Guide Classification of Elements and Periodicity in Properties InText Questions and Answers

Question 1.
What is the basic theme of organisation in the periodic table?
Answer:
The basic theme of organisation of elements in the periodic table is to classify the elements in periods and groups according to their properties. This arrangement makes the study of elements and their compounds simple and systematic. In the periodic table, elements with similar properties are placed in the same group.

Question 2.
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that? ‘
Answer:
Mendeleev arranged the elements in his periodic table ordered by atomic weight or mass. He arranged the elements in periods and groups in order of their increasing atomic weight. He placed the elements with similar properties in the same group.
However, he did not stick to this arrangement for long. He found out that if the elements were arranged strictly in order of their increasing atomic weights, then some elements did not fit within this scheme of classification. Therefore, he ignored the order of atomic weights in some cases. For example, the atomic weight of iodine is lower than that of tellurium. Still Mendeleev placed tellurium (in Group VI) before iodine (in Group VII) simply because iodine’s properties are so similar to fluorine, chlorine and bromine.

Question 3.
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modem Periodic Law?
Answer:
Mendeleev’s periodic law : It states that the properties of the elements are a periodic function of their atomic weights.
Modern periodic law : It states that the properties of the elements are a periodic function of their atomic numbers.
Thus, change in the base of classification of elements from atomic weight to atomic number is the basic difference between Mendeleev’s periodic law and the modern periodic law.

Question 4.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer:
In the periodic table of the elements, a period indicates the value of the principal quantum number (n) for the outermost shells. Each period begins with the filling of principal quantum number (n). The value of n for the sixth period is 6. For n = 6, azimuthal quantum number (0 can have values of 0, 1, 2, 3, 4, 5.
According to Aufbau’s principle, electrons are added to different orbitals in order of their increasing energies. The energy of the 6d subshell is even higher than that of the 7s subshell.

In the 6th period, electrons can be filled in only 6s, 4f, 5d and 6p subshells. Now 6s has one orbital, 4f has seven orbitals, 5d has five orbitals, and 6p has three orbitals. Therefore, there are a total of sixteen (1+ 7 + 5 + 3 = 16) orbitals available. According to Pauli’s exclusion principle, each orbital can accommodate a maximum of 2 electrons. Thus, 16 orbitals can accommodate a maximum of 32 electrons.

Hence, the sixth period of the periodic table should have 32 elements.

Question 5.
In terms of period and group where would you locate the element with Z = 114?
Answer:
₁₁₄Z = ₈₆[Rn] 7s², 5f¹⁴, 6d¹⁰, 7p²
In the periodic table the element with Z = 114 is located in
Block : p-block (as last electron enters in p-subshell).
Period : 7th (as n = 7 for valence shell).
Group : 14th (for p-block elements, group number = 10 + number of electrons in the valence shell).

Question 6.
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer:
There are two elements in the 1st period and eight elements in the 2nd period. The third period starts with the element with Z = 11. Now, there are eight elements in the third period. Thus, the 3rd period ends with the element with Z = 18 i.e., the element in the 18th group of the third period has Z = 18. Hence, the element in the 17th group of the third period has atomic number Z = 17.

Question 7.
Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group.
Answer:
(i) Lawrencium (Lr) with Z = 103 and Berkelium (Bk) with Z = 97
(ii) Seaborgium (Sg) withZ = 106 ,

Question 8.
Why do elements in the same group have similar physical and chemical properties?
Answer:
Same group elements have similar electronic configuration therefore, have similar physical and chemical properties.

Question 9.
What does atomic radius and ionic radius really mean to you?
Answer:
Atomic radius is the radius of an atom. It measures the size of an atom. If the element is a metal, then the atomic radius refers to the metallic radius, and if the element is a non-metal, then it refers to the covalent radius.

Metallic radius is calculated as half the internuclear distance separating the metal cores in the metallic crystal. For example, the internuclear distance between two adjacent copper atoms in solid copper is 256 pm. Thus, the metallic radius of copper is taken as \(\frac{256}{2}\) pm = 128 pm.

Covalent radius is measured as the distance between two atoms when they are found together by a single bond in a covalent molecule. For example, the distance between two chlorine atoms in chlorine molecule is 198 pm. Thus, the covalent radius of chlorine is taken as \(\frac{198}{2}\)pm = 99 pm.

Ionic radius means the radius of an ion (cation or anion). The ionic radii can be calculated by measuring the distances between the cations and anions in ionic crystals.
Since a cation is formed by removing an electron from an atom, the cation has fewer electrons than the parent atom resulting in an increase in the effective nuclear charge. Thus, a cation is smaller than the parent atom. For example, the ionic radius of Na+ ion is 95 pm, whereas the atomic radius of Na atom is 186 pm. On the other hand, an anion is larger in size than its parent atom. This is because an anion has the same nuclear charge, but more electrons than the parent atom resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. For example, the ionic radius of F^– ion is 136 pm, whereas the atomic radius of F atom is 64 pm.

Question10.
How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer:
Atomic radius generally decreases from left to right across a period. This is because within a period, the outer electrons are present in the same valence shell and the atomic number increases from left to right across a period, resulting in an increased effective nuclear charge. As a result, the attraction of electrons to the nucleus increases.
On the other hand, the atomic radius generally increases down a group. This is because down a group, the principal quantum number (n) increases which results in an increase of the distance between the nucleus and valence electrons.

Question 11.
What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F^– (ii) Ar (iii) Mg²⁺ (iv) Rb⁺
Answer:
Atoms and ions having the same number of electrons but different nuclear charges are called isoelectronic species. In case of isoelectronic species, as the nuclear charge increases their size decreases.
(i) F^– ion has 9+1 = 10 electrons.
(ii) Ar has 18 electrons.
(iii) Mg²⁺ ion has 12 – 2 = 10 electrons.
(iv) Rb⁺ ion has 37 -1 = 36 electrons.

Question12.
Consider the following species :
N^3-, 0^2-, F^–, Na⁺, Mg²⁺ and Al³⁺
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Ans. (i) All the given species have same number of electrons (10e^–). Therefore, all are isoelectronic.
(ii) The ionic radii of isoelectronic species decreases with increase in atomic number (as magnitude of the nuclear charge increases with increase in atomic number). Therefore, their ionic radii increase in the order.
Isoelectronic ions = Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2- < N^3-
Atomic number =13 12 11 9 8 7

Question 13.
Explain why cations are smaller and anions larger in radii than their parent atoms?
Answer:
A cation has a fewer number of electrons than its parent atom, while its nuclear charge remains the same. As a result, the attraction of electrons to the nucleus is more in a cation than in its parent atom. Therefore, a cation is smaller in size than its parent atom.

On the other hand, an anion has one or more electrons than its parent atom, resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. As a result, the distance between the valence electrons and the nucleus is more in anions than in its parent atom. Hence, an anion is larger in radius than its parent atom.

Question 14.
What is the significance of the terms-‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
Answer:
Ionization enthalpy : It is the minimum amount of energy required to remove an electron from an isolated gaseous atom (A) in its ground state.
X(g) → X⁺ (g) + e^–
The force by which an electron is attracted by nucleus is also affected by the presence of other atoms within its molecule or in the neighbourhood. Therefore, ionization enthalpy is determined in gaseous state because in gaseous state interatomic distances are larger and interatomic forces of attractions are minimum. Further more, ionization enthalpy is determined at a low pressure because it is not possible to isolate a single atom but interatomic attractions can be further reduced by reducing pressure. Due to these reasons, the term isolated gaseous atom in ground state has been included in definition of ionization enthalpy.

Electron gain enthalpy : It is the energy released when an isolated gaseous atom (X) in ground state gains an electron to form gaseous anion.
X(g) + e^– → X^– (g)
The most stable state of an atom is ground state. If isolated gaseous atom is in excited state, comparatively lesser energy will be released on addition of an electron. So, electron gain enthalpies of gaseous atoms must be determined in their ground states. Therefore, the terms ground state and isolated gaseous atom (explained above) has been also included in the definition of electron gain enthalpy.

Question 15.
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^-1.
Answer:
It is given that the energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18 J.
Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18 x 10^-18 J.
∴ Ionization enthalpy of atomic hydrogen = 2.18 x 10^-18 J
Hence, ionization enthalpy per mol of hydrogen atoms
= 2.18 x 10^-18 x 6.022 x 10²³ J mol^-1
= 1.31 x 10⁶ J mol^-1

Question16.
Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < Cl < N < F < Ne. Explain why
(i) Be has higher A^H than B
(ii) O has lower AjH than N and F?
Answer:
(i) Δ_iH of Be is higher than that of B .
Electronic configuration of Be is 1s², 2s²
whereas that of B is 1s², 2s², 2p_x¹
In the case of Be, electron has to be removed from an s-orbital whereas in the case of B, it has to be removed from a p-orbital. It is difficult to remove an s-electron because it is closer to the nucleus than a p-electron hence more energy is required to remove an electron from 2s and Be, than 2p-electron in the case of B. Hence, ionization enthalpy of Be is higher than that of B.

(ii) Electronic configuration of O is
1s², 2s², \(2 p_{x}^{2} 2 p_{y}^{1} 2 p_{z}^{1}\) (neither exactly half-filled nor completely filled)
whereas N is 1s², 2s², \(2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}\) (exactly half-filled)
It is difficult to remove an electron from the valence shell of N because its p-subshell is exactly half-filled and so has more stability whereas O has electronic configuration which is neither completely filled nor exactly half-filled. Therefore, it is easier to remove one electron from O atom. F, due to increased nuclear charge, has more ionization enthalpy than either O or N.

Question17.
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer:
The first electron in both the cases has to be removed from 3s orbital, but nuclear charge of Na is less than that of Mg. Hence ionization enthalpy of Na is lower than that of Mg.

After the loss of first electron the electronic configuration of Na⁺ is 1s², 2s², 2p⁶, i.e., that of noble gas which is very stable and hence the removal of second electron from Na+ is very difficult. In the case of Mg after the loss of first electron, electronic configuration of Mg⁺ ion is 1s² , 2s² 2p⁶ , 3s¹ . The second electron to be removed is from 3s orbital which is easier.
Hence, second ionization enthalpy of sodium is much larger than second . ionization enthalpy of Mg.

Question 18.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer:
The factors responsible for the ionization enthalpy of the main group elements to decrease down a group are listed below:
(i) Atomic size : On moving down the group atomic size increases due to addition of new higher energy shell. As a result, force of attraction of nucleus for valence electrons decreases and ionization enthalpy also decreases.
(ii) Screening effect : On moving down the group, screening effect or shielding effect increases so ionization enthalpy decreases.

Question19.
The first ionization enthalpy values (in kJ mol^-1) of group 13 elements are :

How would you explain this deviation from the general trend?
Answer:
On moving down a group, ionization enthalpy generally decreases due to an increase in the atomic size and shielding. Thus, on moving down group 13, ionization enthalpy decreases from B to Al. But, Ga has higher ionization enthalpy than Al. Al follows immediately after s-block elements, whereas Ga follows after d-block elements. The shielding provided by d-electrons is not very effective. These electrons do not shield the valence electrons very effectively. As a result, the valence electrons of Ga experience a greater effective nuclear charge than those of Al. Further, moving from Ga to In, the ionization enthalpy decreases due to an increase in the atomic size and shielding. But, on moving from In to Tl, the ionization enthalpy again increases. In the periodic table, Tl follows after 4f and Ad electrons. The shielding provided by the electrons in both these orbitals is not very effective. Therefore, the valence electron is held quite strongly by the nucleus. Hence, the ionization energy of Tl is on the higher side.

Question20.
Which of the following pairs of elements would have more negative electron gain enthalpy?
(i) OorF (ii) F or Cl
Answer:
(i) O and F are present in the same period of the periodic table. An F atom has one proton and one electron more than O and as an electron is being added to the same shell, the atomic size of F is smaller than that of O. As F contains one proton more than O, its nucleus can attract the incoming electron more strongly in comparison to the nucleus of O atom. Also, F needs only one more electron to attain the stable noble gas configuration. Hence, the electron gain enthalpy of F is more negative (- 328 kJ mol^-1) than that of O (-141 kJ mol^-1).

(ii) F and Cl belong to the same group of the periodic table. The electron gain enthalpy usually becomes less negative on moving down a group.
However, in this case, the value of the electron gain enthalpy of Cl is more negative than that of F. This is because the atomic size of F is smaller than that of Cl. In F, the electron will be added to quantum level n = 2, but in Cl, the electron is added to quantum level n = 3. Therefore, there are less electron-electron repulsion in Cl and an additional electron can be accommodated easily. Hence, the electron gain enthalpy of Cl is more negative (- 349kJ mol^-1) than that of F (-328kJ mol^-1).

Question 21.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Answer:
When an electron is added to O atom to form O^– ion, energy is released. Thus, the first electron gain enthalpy of O is negative.
0(g) + e^– → O^– (g); Δ_egH = -141 kJ mol^-1

On the other hand, when an electron is added to O ion to form O ion, energy has to be given out in order to overcome the strong electronic repulsion. Thus, the second electron gain enthalpy of O is positive.
O^–(g) + e^– → O^2-(g); Δ_egH = +780 kJ mol^-1

Question 22.
What is the basic difference between the terms electron gain enthalpy and electronegativity? •
Answer:
Electron gain enthalpy is the measure of the tendency of an isolated gaseous atom to accept an electron, whereas electronegativity is the measure of the tendency of an atom in a chemical compound to attract a shared pair of electrons.

Question 23.
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Answer:
On Pauling scale, the electronegativity of nitrogen, (3.0) indicates that it is sufficiently electronegative. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It depends upon its state of hybridisation in a particular compound, greater the percentage of s-character, more will be the electronegativity of the element. Thus, the electronegativity of nitrogen increases in moving from sp³ hybridised orbitals to sp hybridised orbitals i.e., as sp³ < sp² < sp.

Question 24.
Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
Answer:
(a) When an atom gains an electron, its size increases. When an electron is added, the number of electrons goes up by one. This results in an increase in repulsion among the electrons. However, the number of protons remains the same. As a result, the effective nuclear charge of the atom decreases and the radius of the atom increases. For example, ionic radius of Cl^– ion is greater than the radius of its parent atom Cl.

(b) When an atom loses an electron, the number of electrons decreases by one while the nuclear charge remains the same. Therefore, the interelectronic repulsions in the atom decrease. As a result, the effective nuclear charge increases. Hence, the radius of the atom decreases.
For example, ionic radius of Na⁺ is smaller than the radius of its parent atom Na.

Question 25.
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Answer:
The ionization enthalpy of an atom depends on the number of electrons and protons (nuclear charge) of that atom. Now, the isotopes of an element have the same number of protons and electrons. Therefore, the first ionization enthalpy for two isotopes of the same element should be the same.

Question 26.
What are the major differences between metals and non-metals?
Answer:

Metals	Non-metals
1. Metals can lose electrons easily.	Non-metals cannot lose electrons easily.
2. Metals cannot gain electrons easily.	Non-metals can gain electrons easily.
3. Metals generally form ionic compounds.	Non-metals generally form covalent compounds.
4. Metal oxides are basic in nature.	Non-metal oxides are acidic in nature.
5. Metals have low ionization enthalpies.	Non-metals have high ionization enthalpies.
6. Metals have less negative electron gain enthalpies.	Non-metals have high negative electron gain enthalpies.
7. Metals are less electronegative. They are rather electropositive elements.	Non-metals are electronegative.
8. Metals have a high reducing power.	Non-metals have a low reducing power.

Question27.
Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal liquid as well as gas at the room temperature.
Answer:
(a) General electronic configuration of elements having five electrons in the outer sub shell is ns2 np . This configuration belongs to halogen family, i.e., F, Cl, Br, I, At.
(b) Elements of second group are known as alkaline earth metals (Mg, Ca, Sr, Ba, etc). Their general electronic configuration for valence shell is ns2. These elements form dipositive cations by the lose of two electrons easily.
(c) 16th group elements such as O, S, Se, etc., have a tendency to accept two electrons because by the gain of two electrons they attain noble gas configuration. Their general electronic configuration for valence shell is ns² np⁴.
(d) Group 1 or 17 of the periodic table contains metal, non-metal, liquid as well as gas at the room temperature, e.g., H2 is a non-metal and in gaseous state at room temperature. All other elements of this group are metals. Cs is a liquid metal. Similarly, Br₂ is a liquid non-metal while other elements of this group are gaseous non-metals. Iodine can form I⁺ so it consists some metallic properties.

Question 28.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Answer:
The elements present in group 1 have only 1 valency electron, which they tend to lose. Group 17 elements, on the other hand, need only one electron to attain the noble gas configuration. On moving down group 1, the ionization enthalpies decrease. This means that the energy required to lose the valence electron decreases. Thus reactivity increases on moving down a group. Thus, the increasing order of ( reactivity among group 1 elements is as follows :
Li < Na < K < Rb < Cs In group 17, as we move down the group from Cl to I, the electron gain enthalpy becomes less negative i. e., its tendency to gain electrons decreases on moving down a group. Thus, reactivity decreases down a group. The electron gain enthalpy of F is less negative than Cl. Still, it is the most reactive halogen. This is because of its low bond dissociation energy. Thus, the decreasing order of reactivity among group 17 elements is as follows : F > Cl > Br > I.

Question 29.
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Answer:

Element	General outer electronic configuration
s-block	ns1-2, where n = 2 – 7
p-block	ns2 np1-6      , where n = 2 – 6
d-block	(n – 1)d1–10ns(0 – 2), where n = 4 – 7
f-block	(n – 2)f1–14 (n – 1)d0- 1ns2, where n = 6 – 7

Question30.
Assign the position of the element having outer electronic configuration
(i) ns² np⁴ for n = 3 (ii) (n – 1 )d²ns² for n = 4, and
(iii) (n – 2)f⁷(n – 1)d¹ns² for n = 6, in the periodic table.
Answer:
(i) ns²np⁴ for n = 3; it is 3s²3p⁴
The complete electronic configuration is 1s², 2s², 2p⁶, 3s², 3p⁴
Atomic number = 2 + 2+ 6+ 2 + 4 = 16
The element is sulphur in the 3rd period and in Group 16 (p-block)

(ii) (n – 1)d²ns² for n = 4; it is 3d²4s²
The complete electronic configuration is
1s², 2s²,2p⁶, 3s², 3p⁶, 3d², 4s² Atomic number is 22, the element is Titanium.
It is a transition element present in the 4th period and in Group 4.

(iii) (n – 2)f⁷(n – l)d¹ns² for n = 6; it is 4f⁷5d¹6s²
Its complete electronic configuration is
1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶, 4d¹⁰, 4f⁷, 5s², 5p⁶, 5d¹, 6s²
It is Gadolinium (Gd)
It is an inner transition element, belongs to Lanthanoid series or 4f series. It is an f-block element.

Question 31.
The first (Δ_i,H₁) and the second (Δ_iH₂) ionization enthalpies (in kJ mol^-1) and the (Δ_egH) electron gain enthalpy (in kJ mol^-1) of a few elements are given below:

Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX₂ (X = halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?
Answer:
(a) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy (Δ_i,H₁) and a positive electron gain enthalpy (Δ_eg,H₁).
(b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (Δ_i,H₁) and the highest negative electron gain enthalpy (AegH).
(c) Element III is likely to be the most reactive non-metal as it has a high first ionization enthalpy Δ_i,H₁) and the highest negative electron gain enthalpy (Δ_eg,H).
(d) Element IV is likely to be the least reactive non-metal since it has a very high first ionization enthalpy (Δ_i,H₂) and a positive electron gain enthalpy (Δ_eg,H).
(e) Element VI has a low negative electron gain enthalpy (Δ_eg,H). Thus it is a metal. Further, it has the lowest second ionization enthalpy (Δ_i,H₂). Hence, it can form a stable binary halide of the formula MX₂ ( X = halogen).
(f) Element I has low Δ_i,H₁ but a very high Δ_i,H₂. It has less negative electron gain enthalpy. So, element (I) is alkali metal. The given values for element I match with Li. Lithium firms predominantly stable covalent halide of the formula Mx.

Question32.
Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Answer:
(a) Li₂0
(b) Mg₃N₂
(c) AlI₃
(d) SiO₂
(e) PF₃ or PF₅
(f) The element with the atomic number 71 is Lutetium (Lu). It has valency 3. Hence, the formula of the compound is LuF₃.

Question 33.
In the modem periodic table, the period indicates the value of:
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.
Answer:
(c) The value of the principal quantum number (n) for the outermost shell or the valence shell indicates a period in the Modern periodic table.

Question 34.
Which of the following statements related to the modem periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The df-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
Answer:
The statement (b) is incorrect. The correct statement (b) is that the d-block has 10 columns because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.
All other given statements are correct.

Question 35.
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Answer:
(c) Nuclear mass (protons + neutrons) does not affect the valence shell, only protons i. e., nuclear charge affects the valence shell.

Question 36.
The size of isoelectronic species F^–, Ne and Na⁺ is affected by
(a) Nuclear charge (Z)
(b) Valence principal quantum number (n)
(c) Electron-electron interaction in the outer orbitals.
(d) None of the factors because their size is same.
Answer:
(a) The size of an isoelectronic species increases with a decrease in the nuclear charge (Z).

Question 37.
Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Answer:
(d) Electrons in orbitals bearing a lower n value are more attracted to the nucleus than electrons in orbitals bearing a higher n value. Hence, the removal of electrons from orbitals bearing a higher n value is easier than the removal of electrons from orbitals having a lower n value.

Question 38.
Considering the elements B, Al, Mg and K the correct order of their metallic character is:
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer:
(d) In a group, metallic character increases from top to bottom as ; ionisation energy decreases and in a period metallic character decreases from left to right as tendency to lose electron decreases. Therefore, the correct order is K > Mg > Al > B.

Question 39.
Considering the elements B, C, N, F and Si the correct order of their non-metallic character is
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Answer:
(c) The non-metallic character of elements increases from left to right across a period. Thus, the decreasing order of non-metallic character is F > N > c > B.
Again, the non metallic character of elements decreases down a group. ;
Thus, the decreasing order of non-metallic characters of C and Si are C > Si. However, Si is less non-metallic than B i.e., B > Si.
Hence, the correct order of non-metallic characters is F > N > C > B > Si.

Question 40.
Considering the elements F, Cl, O and N the correct order of their chemical reactivity in terms of oxidizing property is :
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Answer:
(b) In a group, oxidising power decreases from top to bottom as the size increases but when we move left to right in a period it increases because size decreases.
Therefore, among F, Cl, O and N, the oxidising power decreases in the order F > O > Cl > N.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 2 Structure of Atom Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

PSEB 11th Class Chemistry Guide Structure of Atom InText Questions and Answers

Question 1.
(i) Calculate the number of electrons that will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Answer:
(i) Mass of one electron = 9.1 × 10^-31 g
Number of electrons in 1.0 g = \(\frac{1}{9.1 \times 10^{-31}}\) = 1.098 x 10³⁰ electrons

(ii) One mole of electron = 6.022 × 10²³ electrons
Mass of 1 electron = 9.1 × 10^-31 kg
Mass of 6.022 × 1023 electrons – (9.1 x 10^-31 kg) × (6.022 x 10²³)
= 5.48 × 10^-7 kg
Charge on one electron = 1.602 × 10^-19 coulomb
Charge on one mole of electrons = 1.602 × 10^-19 × 6.022 × 10²³
= 9.65 × 10⁴ coulomb.

Question 2.
(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C.
(Assume that mass of a neutron = 1.675 × 10^-27 kg)
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP.
Will the answer change if the temperature and pressure are changed?
Answer:
(i) Number of electrons present in 1 molecule of methane (CH₄)
= {1(6) + 4(1)} = 10 electrons
Number of electrons present in 1 mole i.e., 6.022 × 10²³ molecules of
methane = 6.022 × 10²³ × 10 = 6.022 × 10²⁴ electrons
(ii) (a) Mass of a neutron = 1.675 × 10^-27 kg
14.0 g of ¹⁴C contains 1 mole of atoms of C-14
∴ 14.0 g of 14C contains 8 × 6.022 × 10²³ neutrons
[∵ each atom of C-14 contains 8 neutrons]
7 mg (= 7 × 10^-3 g) of ¹⁴C contains
= \(\frac{6.022 \times 10^{23} \times 8 \times 7 \times 10^{-3}}{14}\) = 2.4092 × 10²¹ neutrons

(b) Mass of one neutron = 1.675 × 10²⁷ kg
Mass of 2.4092 × 10²¹ neutrons
= (2.4092 × 10²¹)(1.675 × 10^-27 kg) = 4.035 × 10^-6 kg

(iii) 1 mole of NH₃ contains protons = 7 + 3 = 10 moles of protons
= 6.022 × 10²³ × 10 protons = 6.022 × 10²⁴ protons

(a) 17 gNH₃ contains 6.022 × 10²⁴ protons.
Number of protons in 34 mg NH₃ will contain
= \(6.022 \times 10^{24} \times 34 \times 10^{-3}\) = 1.2046 x 10²² Protons

(b) Mass of one proton = 1.675 x 10^-27 kg
Total mass of protons in 34 mg of NH₃
– (1.675 × 10^-27 kg) (1.2046 × 10²²)
= 2.0177 × 10^-5 kg
The number of protons, electrons and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Question 3.
How many neutrons and protons are there in the following nuclei?
\({ }_{6}^{13} \mathrm{C},{ }_{8}^{16} \mathrm{O},{ }_{12}^{24} \mathrm{Mg},{ }_{26}^{56} \mathrm{Fe},{ }_{38}^{88} \mathrm{Sr}\)
Answer:
\({ }_{6}^{13} \mathrm{C}\)C : Atomic mass =13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) – (Atomic number)
= 13 – 6 = 7
\({ }_{8}^{16} \mathrm{O}\) :
Atomic mass =16
Atomic number = 8
Number of protons = 8
Number of neutrons = (Atomic mass) – (Atomic number)
= 16 – 8 = 8
\({ }_{12}^{24} \mathrm{Mg}\) : Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) – (Atomic number)
= 24-12 = 12
\({ }_{26}^{56} \mathrm{Fe}\) : Atomic mass = 56
Atomic number = Number of protons = 26
Number of neutrons = (Atomic mass) – (Atomic number)
= 56 – 26 = 30
\({ }_{38}^{88} \mathrm{Sr}\) : Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) – (Atomic number)
= 88 – 38 = 50

Question 4.
Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9
Answer:
(i) Z = 17, A = 35
Since the no. of protons = 17 – no. of electrons
∴ The atom is chlorine, Cl: \({ }_{17}^{35} \mathrm{Cl}\)

(ii) Z = 92, A = 233
No. of protons = 92
∴ The atom is uranium, U : \({ }_{92}^{233} U\)

(iii) Z = 4, A = 9
No. of protons = 4
∴ The atom is Beryllium, Be : \({ }_{4}^{9} \mathrm{Be}\)

Question 5.
Yellow light emitted from a sodium lamp has a wavelength (A.) of 580 nm. Calculate the frequency (υ) and wave number (\(\overline{\mathbf{v}}\)) of the ‘ yellow light.
Answer:
We know that,
Frequency, v = \(\frac{c}{\lambda}\)
1 nm = 10^-9 m
580 nm = 580 × 10^-9 m = 580 x 10^-7 cm
v = \(\) = 5.17 × 10¹⁴s^-1
(velocity of light = 3 × 10⁸ m/s)
Wave number, \(\bar{v}=\frac{1}{\lambda}\)
= \(\frac{1}{580 \times 10^{-7}}\) = 1.72 × 10⁴ cm^-1

Question 6.
Find energy of each of the photons which
(i) correspond to light of frequency 3 × 10¹⁵ Hz.
(ii) have wavelength of 0.50 Å
Answer:
(i) Energy E = hv
where, h = Planck’s constant = 6.626 × 10^-34 Js
and v = frequency of light = 3 × 10¹⁵ Hz
E= (6.626 × 10^-34)(3 × 10¹⁵)
E = 1.988 × 10^-18 J

(ii) Energy E = \(\frac{h c}{\lambda}\)
where, h = Planck’s constant = 6.626 × 10^-34 Js
c = velocity of light in vacuum = 3 × 10⁸ m/s
= wavelength = 0.50Å = 0.50 × 10^-10
E = \(\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{0.50 \times 10^{-10}}\) = 3.976 × 10^-15 J

Question 7.
Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10^-10 s.
Answer:
Frequency of light (v) = \(\frac{1}{\text { Time period }}=\frac{1}{2.0 \times 10^{-10} \mathrm{~s}}\) = 5.0 × 10⁹s^-1
Wavelength of light (λ) = \(\frac{c}{v}\)
where, c = velocity of light in vacuum = 3 × 10⁸ m/s
λ = \(\frac{3 \times 10^{8}}{5.0 \times 10^{9}}\) = 6.0 × 10^-2 m
Wave number(\(\bar{v}\)) of light
= \(\frac{1}{\lambda}=\frac{1}{6.0 \times 10^{-2}}\) = 1.66 × 10¹m^-1 = 16.66m^-1

Question 8.
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Answer:
Energy (E) of a photon = hv = \(\frac{h c}{\lambda}\)
where, = wavelength of light = 4000 pm = 4000 × 10^-12 m
(∵ 1 pm = 10^-12 m)
c = velocity of light in vacuum = 3 × 10⁸ m/s
h = Planck’s constant = 6.626 × 10^-34 Js
E = \(\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-12}}\) = 4.9695 × 10^-17J
Number of protons, N = \(\frac{1}{4.9695 \times 10^{-17}}\) = 2.0122 × 10¹⁶ protons.

Question 9.
A photon of wavelength 4 × 10^-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 × 10^-19 J).
Answer:
(i) Energy (E) of a photon = hv = \(\frac{h c}{\lambda}\)
where,
h = Planck’s constant = 6.626 × 10^-34 Js
c = velocity of light in vacuum = 3 × 10⁸ m/s
λ = wavelength of photon =4 × 10^-7 m
E = \(\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{4 \times 10^{-7}}\)
= 4.9695 × 10^-19j = \(\frac{4.9695 \times 10^{-19}}{1.602 \times 10^{-19}}\) [∵ 1 eV = 1.602 x 10^-19J]
= 3.10 eV
Hence, the energy of the photon is 3.10 eV.

(ii) The kinetic energy of emission E_k is given by
= hv – hv₀ = (E – W) eV
= 3.10 – 2.13 eV = 0.97 eV
Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron is given by
KE = \(\frac{1}{2}\) mv² – hv – hv₀ = 0.97 eV
\(\frac{1}{2}\)mv² = 0.97 × 1.602 × 10^-19 J
(∵ 1 eV = 1.602 x 10^-19 J)
\(\frac{1}{2}\) × 9.11 × 10^-31 kg × v² = 0.97 × 1.602 × 10^-19 J
(∵ mass of electron = 9.11 × 10^-31 kg)
v² = \(\frac{0.97 \times 1.602 \times 10^{-19} \times 2}{9.11 \times 10^{-31}}\) = 0.341 × 10¹²
v = 0.584 × 10⁶ = 5.84 × 10⁵ m/s
Hence, the velocity of the photoelectron is 5.84 × 10⁵ ms^-1.

Question 10.
Electromagnetic radiation of wavelength 242 run is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^-1.
Answer:
Energy (E) = \(\frac{h c}{\lambda}\)
= \(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{242 \times 10^{-9} \mathrm{~m}}\)
= 0.0821 × 10^-17 J
This energy is sufficient for ionisation of one Na atom.
E = 6.02 × 10²³ × 0.0821 × 10^-17 J/mol
E = 4.945 × 10⁵ J/mol = 4.945 × 10² kJ/mol

Question 11.
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 (am. Calculate the rate of emission of quanta per second.
Answer:
Power of bulb, P = 25 watt = 25 Js^-1
Wavelength,
λ = 0.57 µm = 0.57×10^-6 m
Energy of one photon,

Question 12.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800Å. Calculate threshold frequency (v₀) and work function (W₀) of the metal.
Answer:
Threshold wavelength of radiation
λ₀ = 6800Å = 6800 × 10^-10 m
Threshold frequency (v₀) of the metal
= \(\frac{c}{\lambda_{0}}=\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{6800 \times 10^{-10} \mathrm{~m}}\) = 4.41 × 10¹⁴s^-1
Thus, the threshold frequency (v₀) of the metal is 4.41 × 10¹⁴s^-1.
Hence, work function (W₀) of the metal = hv₀
= (6.626 × 10^-34Js)(4.41 × 10¹⁴s^-1)= 2.922 × 10^-19 J

Question 13.
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2 ?
Answer:
The n_i = 4 to n_f= 2 transition will give rise to a spectral line of the
Balmer series. The energy involved in the transition is given by the relation,
E = 2.18 × 10^-18 \(\left[\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right]\)

Question 14.
How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n – 1 orbit).
Answre:
Energy change, ΔE = E_f – E_i

Hence, far higher energy (25 times) is required to remove an electron from first orbit.

Question 15.
What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Answer:
The maximum number of emission lines
= \(\frac{n(n-1)}{2}=\frac{6(6-1)}{2}\) = 3 × 5 = 15

Question 16.
(i) The energy associated with the first orbit in the hydrogen atom is -2.18 × 10^-18 J atom^-1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Answer:
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as :
E₅ = \(\frac{-\left(2.18 \times 10^{-18}\right)}{(5)^{2}}=\frac{-2.18 \times 10^{-18}}{25}\)
E₅ = -8.72 × 10^-20 J
(ii) Radius of Bohr’s nth orbit for hydrogen atom is given by,
r_n = (0.0529 nm) n²
For, n= 5
r₅ = (0.0529 nm) (5)²
r₅ = 1.3225 nm

Question 17.
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer:
For the Balmer series, a transition from n₁= 2 to n₂ = 3 is allowed.

Question 18.
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18 × 10^-11 ergs.
Answer:
ΔE = E₅ – E₁ = 2.18 × 10^-11 (\(\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right)\))erg
(n_i = 1st orbit and n_f = 5th orbit)

Question 19.
The electron energy in hydrogen atom is given by E_n = (-218 × 10^-18)/n²J. Calculate the energy required to remove an electron completely from n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:

Question 20.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ ms^-1.
Answer:
According to de Broglie’s equation,
λ = \(\frac{h}{m v}\)
where, λ = wavelength of moving electron
m = mass of electron = 9.1 × 10^-31 kg
h= Planck’s constant – 6.626 × 10^-34 Js
v = velocity of electron = 2.05 × 10⁷ ms^-1
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(2.05 \times 10^{7} \mathrm{~ms}^{-1}\right)}\)
λ = 3.548 × 10^-11 m
Hence, the wavelength of the electron moving with a velocity of
2.5× 10⁷ ms^-1 is 3.548 x 10^-11m.

Question 21.
The mass of an electron is 9.1 x 10^-31 kg. If its K.E. is 3.0 x 10^-25 J, calculate its wavelength.
Answer:
Kinetic energy, K.E. = \(\frac{1}{2}\) mv²
∴ Velocity(v) = \(\sqrt{\frac{2 \mathrm{KE}}{m}}=\sqrt{\frac{2\left(3.0 \times 10^{-25}\right)}{9.1 \times 10^{-31} \mathrm{~kg}}}=\sqrt{0.6593 \times 10^{6}}\)
_ 6.626 x 10~34 Js
~ (9.1 x 10-31 kg)(811.579 ms-1)
λ = 8.968 x 10^-7 m
Hence, the wavelength of the eliectron is 8.968 x 10^-7 m or 8968 Å.

Question 22.
Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2-, Ar
Ans. No. of electrons in Na⁺ = 10 [11-1]
No. of electrons in K⁺ = 18 [19-1]
No. of electrons in Mg²⁺ = 10 [12-2]
No. of electrons in Ca²⁺ =18 [20 – 2]
No. of electrons in S^2- = 18 [16 + 2]
No. of electrons in Ar = 18
∴ Na⁺ , Mg²⁺ are isoelectronic (10 e^– each)
∴ Ca²⁺, K⁺, S^2-, Ar are isoelectronic (18 e^– each).

Question 23.
(i) Write the electronic configurations of the following ions:
(a) H^– (b) Na⁺ (c)0^2- (d)F^–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by
(a) 3s¹ (b) 2p³ (c) 3p⁵?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s¹ (b) [Ne] 3s² 3p³ (c) [Ar] 4s²3d¹.
Answer:
(i) (a) H^– ion
The electronic configuration of H atom is 1s¹.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of H^– = 1s²

(b) Na⁺ ion
The electronic configuration of Na atom is 1s² 2s² 2p⁶ 3s¹.
A positive charge on the species indicates the loss of an electron by it.
∴ Electronic configuration of Na⁺ = 1s² 2s² 2p⁶ 3s° or 1s² 2s² 2p⁶

(c) O^2- ion
The electronic configuration of O atom is 1s²2s² 2p⁴
A dinegative charge on the species indicates that two electrons are gained by it.
∴ Electronic configuration of 0^2- ion is 1s² 2s² 2p⁶

(d) F^– ion
The electronic configuration of F atom is 1s²2s²2p⁵.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of F“ ion is 1s² 2s² 2p⁶.

(ii) (a) 3s¹
Completing the electronic configuration of the element as 1s² 2s² 2p⁶ 3s1. .-. Number of electrons present in the atom of the element = 2 +2 + 6 +1=11
∴ Atomic number of the element = 11

(b) 2p³
Completing the electronic configuration of the element as 1s² 2s² 2p³
∴ Number of electrons present in the atom of the element = 2+ 2+ 3 = 7
∴ Atomic number of the element = 7

(c) 3p⁵
Completing the electronic configuration of the element as 1s² 2s² 2p⁶ 3s² 3 P⁵
∴ Number of electrons present in the atom of the element
2 + 2 + 6 + 2 + 5 = 17
∴ Atomic number of the element = 17

(iii) (a) [He] 2s¹
The electronic configuration of the element is [He]2s¹ = 1s²2s¹.
∴ Atomic number of the element = 3
Hence, the element with the electronic configuration [He] 2s¹ is lithium (Li).

(b) [Ne] 3s² 3p³
The electronic configuration of the element is
[Ne] 3s² 3p³ = 1s² 2s² 2p⁶ 3s² 3p³
∴ Atomic number of the element = 15
Hence, the element with the electronic configuration [Ne] 3s² 3p³ is phosphorus (P).

(c) [Ar] 4s² 3d¹
The electronic configuration of the element is [Ar] 4s² 3d¹ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹.
∴ Atomic number of the element = 21
Hence, the element with the electronic configuration [Ar] 4s² 3d² is scandium (Sc).

Question 24.
What is the lowest value of n that allows g orbitals to exist?
Answer:
For g-orbitals, l = 4.
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n —1).
∴ For l = 4, minimum value of n = 5

Question 25.
An electron is in one of the 3d orbitais. Give the possible values of n, l and m_l for this electron.
Answer:
For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (L) = 2
Magnetic quantum number (m_l) = – l to + l including 0 = -2 -1, 0, 1, 2

Question 26.
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Answer:
No. of protons in a neutral atom = No. of electrons = 29
Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.

Question 27.
Give the number of electrons in the species : H₂⁺, H₂ and 0₂⁺.
Answer:
H₂⁺ = one ; H₂ = two ; 0₂⁺ = 15

Question 28.
(i) An atomic orbital has n = 3. What are the possible values of l and m_l ?
(ii) List the quantum numbers m_l and l of electron in 3rd orbital.
(iii) Which of the following orbitals are possible ?
1p, 2s, 2p and 3f.
Answer:
(i) For n = 3; l = 0, 1 and 2.
For l = 0 ; m_l = 0
For l = 1; m_l = +1, 0, -1
For l = 2 ; m_l = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; m_l can have any of the values -2, -1, 0,
+ 1, +2.
(iii) 1p and 3f orbitals are not possible.

Question 29.
Using s, p and d notations, describe the orbitals with follow ing quantum numbers :
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2
Answer:
(a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital

Question 30.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.
(i) n = 0, l = 0, m_l = 0, m_s = +1/2
(ii) n = 1, l = 0, m_l = 0, m_s – – 1/2
(iii) n = 1, l = 1, m_l = 0, m_s= +1/2
(iv) n = 1, l = 0, m_l = +1, m_s= +1/2
(v) n = 3, l = 3, m_l = -3, m_s = +1/2
(vi) n = 3, l = 1, m_l = 0, m_s= +1/2
Answer:
(i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(ii) The set of quantum numbers is possible.
(iii) The set of quantum numbers is not possible because, for n = 1, l can not be equal to 1. It can have 0 value.
(iv) The set of quantum numbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantum numbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantum numbers is possible.

Question 31.
How many electrons in an atom may have the following quantum numbers ?
(a) n = 4 ; m_s = -1/2
(b) n = 3, l = 0.
Answer:
(a) For n = 4
Total number of electrons = 2n² = 2 × 16 = 32
Half out of these will have ms = —1/2
∴ Total electrons with ms (-1/2) = 16
(b) For n = 3
l= 0 ; m_l = 0, m_s +1/2, -1/2 (two e^–)

Question 32.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:

Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.

Question 33.
Calculate the number of atoms present in :
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He.
Answer:

Question 34.
Calculate the energy required for the process :
He⁺fe) → He²⁺(g) + e^–
The ionisation energy’ for the H atom in the ground state is 2.18 × 10^-18  J atom^-1
Answer:

For H atom (Z = 1), E_n =2.18 × 10^-18 × (l)² J atom^-1 (given)
For He⁺ ion (Z = 2), E_n =2.18 × 10^-18 × (2)² = 8.72 × 10^-18 J atom^-1 (one electron species)

Question .35.
If the diameter of carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm long.
Answer:

Question 36.
2 × 10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
Answer:
The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 ×10⁸

Radius of each carbon atom = 12(1.2 × 10^-8) = 6.0 × 10^-9cm = 0.06 nm

Question 37.
The diameter of zinc atom is 2.6 Å. Calculate :
(a) the radius of zinc atom in pm
(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side length wise.
Answer:
(a) Radius of zinc atom =Undefined control sequence = 1.3 Å = 1.3 × 10^-10m = 130 × 10^-12m = 130 pm
(b) Length of the scale = 1.6 cm = 1.6 × 10¹⁰ pm
Diameter of zinc atom = 260 pm

Question 38.
A certain particle carries 2.5 x 10^-16 C of static electric charge. Calculate the number of electrons present in it.
Answer:

Question 39.
In Millikan’s experiment, the charge on the oil droplets was found to be – 1.282 x 10^-18C. Calculate the number of electrons present in it.
Answer:

Question 40.
In Rutherford experiment, generally the thin foil of heavy atoms like gold, platinum etc. have been used to be bombarded by the a-particles. If a thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
Answer:
We have studied that in the Rutherford’s experiment by using heavy metals like gold and platinum, a large number of a-particles sufferred deflection while a very few had to retrace their path.

If a thin foil of lighter atoms like aluminium etc. be used in the Rutherford experiment, this means that the obstruction offered to the path of the fast moving a-particles will be comparatively quite less.

As a result, the number of a-particles deflected will be quite less and the particles which are deflected back will be negligible.

Question 41.
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br can be written whereas symbols \(_{ 79 }^{ 35 }{ Br }\) and ₃₅Br are not accepted. Answer in brief.
Answer:
In the symbol \(_{ A }^{ B }{ X }\) of an element :
A denotes the atomic number of the element
B denotes the mass number of the element.
The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus,
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br are accepted because atomic number of Br will remain 35 even if not mentioned. Symbol \(_{ 79 }^{ 35 }{ Br }\) is not accepted because atomic number of Br cannot be 79 (more than the mass number = 35). Similarly, symbol 35Br cannot be accepted because mass number has to be mentioned. This is needed to differentiate the isotopes of an element.

Question 42.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the symbol to the element.
Answer:
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + [\(\frac { x\times 31.7 }{ 100 } \) = (x × 0.317x)
Now, Mass no. of element = no. of protons =no. neutrons
81 = x + x + 0-317 x = 2.317 x or x = \(\frac { 81 }{ 2.317 } \) = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine \(_{ 35 }^{ 81 }{ Br }\).

Question 43.
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11 -1% more neutrons than the electrons, find the symbol of the ion.
Answer:
Let the no. of electron in the ion = x
∴ the no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x + \(\frac { x\times 11.1 }{ 100 } \) = 1.111 x
Mass no. or mass of the ion = No. of protons + No. of neutrons
(x – 1 + 1.111 x)
Given mass of the ion = 37
∴ x- 1 + 1.111 x = 37 or 2.111 x = 37 + 1 = 38
x = \(\frac { 38 }{ 2.111 } \) = 18
No. of electrons = 18 ; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17 ; Atom corresponding to ion = Cl
Symbol of the ion = \(_{ 17 }^{ 37 }{ Cl }\)^–

Question 44.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
Answer:
Let the no. of electrons in the ion = x
∴ the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac { x\times 31.7 }{ 100 } \) = xc + 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons
= (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x) or 2.304x = 56 – 3 = 53
x = \(\frac { 53 }{ 2.304 } \) = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Question 45.
Arrange the following type of radiations in increasing order of frequency:
(a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.
Answer:
The increasing order of frequency is as follows :
Radiation from FM radio < Radiation from microwave oven < Amber light from traffic signal < X-rays < Cosmic rays from outer space.

Question 46.
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 10²⁴ , calculate the power of this laser.
Answer:
Power of laser (E) = \(\frac{n h c}{\lambda}\)
= \(\frac{\left(5.6 \times 10^{24}\right)\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(337.1 \times 10^{-9} \mathrm{~m}\right)}\)
= 0.3302 x 10⁷ J = 3.33 x 10⁶ J
Hence, the power of the laser is 3.33 x 10⁶ J.

Question 47.
Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance travelled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.
Answer:
(a) Frequency (v) of emission can be calculated as: c = v x λ
where, λ – wavelength = 616 nm = 616 x 10^-9 ms^-1
c = velocity of light = 3.0 x 10⁸ ms^-1
v = \(\frac{c}{\lambda}=\frac{3.0 \times 10^{8}}{616 \times 10^{-9}}\) = 4.87 x 10¹⁴s^-1

(b) Distance travelled in 1 second = 3.0 x 10⁸ m
Distance travelled in 30 seconds = 30 x 3.0 x 10⁸ m = 9 x 10⁹ m
(c) Energy of the photon (quantum)
E= hv = 6.626 x 10^-34 x 4.87 x 10¹⁴J
= 32.26 x 10^-20 J

(d) Number of quanta = \(\frac{Total energy produced}
{Energy of one quantum}\)
= \(\frac{2}{32.26 \times 10^{-20}}\) = 6.2 x 10¹⁸

Question 48.
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10^-18 J from the radiations of 600 rnn, calculate the number of photons received by the detector.
Answer:
Energy of one photon, E = \(\frac{h c}{\lambda}\)
= \(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(600 \times 10^{-9} \mathrm{~m}\right)}\) = 3.313 x 10^-19 J
Number of photons = \(\frac{Total energy received}{Energy of one photon}\)
= \(\frac{3.15 \times 10^{-18} \mathrm{~J}}{3.313 \times 10^{-19} \mathrm{~J}}\)

Question 49.
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 10¹⁵, calculate the energy of the source.
Answer:
Duration of radiation source = 2 ns = 2 x 10^-9 s
No. of photons (n) = 2.5 x 10¹⁵
Frequency of radiation
v = \(\frac{1}{2.0 \times 10^{-9} \mathrm{~s}}\) = 5.0 x 10⁸s^-1
Energy (E) of source = n x hv
E= (2.5 x 10¹⁵)(6.626 x 10^-34Js) (5.0 x 10⁸)
E = 8.282 x 10^-10J
Hence, the energy of the source (E) is 8.282 x 10^-10 J.

Question 50.
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Answer:
For λ₁ = 589 nm = 589 x 10^-9
Frequency of transition (v₁) = \(\) = 5.093 x 10¹⁴s^-1
Similarly, for λ₂ = 589.6 nm = 589.6 x 10^-9m
Frequency of transition (v₂) = \(\frac{c}{\lambda_{2}}=\frac{3.0 \times 10^{8} \mathrm{~ms}^{-1}}{589.6 \times 10^{-9} \mathrm{~m}}\) = 5.088 x 10¹⁴s^-1
Energy difference (ΔE) between excited states = E₁ – E₂
ΔE = h(v₁ – v₂)
= (6.626 × 10^-34Js) (5.093 × 10¹⁴ – 5.088 × 10¹⁴)s^-1
= (6.626 × 10^-34Js) (0.005 × 10^-14) s^-1
ΔE = 3.31 × 10^-22

Question 51.
The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Answer:
It is given that the work function (W₀) for caesium atom is 1.9 eV.
(a) Work function,

λ₀ = 6.53 × 10^-7 m
= 653 × 10^-9m = 653nm.
Hence the threshold wave length is 653nm

(b) threshold frequency
v₀ = \(\frac{W_{0}}{h}\)
v₀ = \(\frac{1.9 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}\) = 4.59 x 10¹⁴s^-1
Hence the threshold frequency of radiation is 4.59 x 10¹⁴s^-1
According to the question :
wavelength used in irradiation = 500nm
Kinetic Energy = h(v – v₀)
= hc( \(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\) )

Question 52.
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

Answer:

Mass of electron, m = 9.1 × 10^-31 kg
Here, λ = 500 × 10^-9 m
v_max = 2.55 ×10⁵ms^-1
λ₀ = ?

Question 53.
The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Answer:
h = 6.626 × 10^-34 Js
V₀ = 0.35 volt
e = 1.6 × 10^-19 C
λ = 256.7 nm = 256.7 × 10^-9 m
W₀ = ? (eV)

Question 54.
If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 x 10⁷ms^-1, calculate the energy with which it is bound to the nucleus.
Answer:
λ = wavelength = 150 pm = 150 × 10^-12 m
v = velocity = 1.5 × 10⁷ ms^-1
Kinetic energy of the ejected electron
= \(\frac{1}{2}\) mv⁷ = \(\frac{1}{2}\) x 9.1 x 10^-31 x (1.5 x 10⁷ )² J = 0.102 x 10^-15 J
Energy of the incident radiation
hv= hv₀ +\(\frac{1}{2}\)mv²
E = hv = \(\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{150 \times 10^{-12}}\)
= 1.325 × 10^-15 J
Minimum energy required to eject the electron
E₀ = E – K.E.
= (1.325 × 10^-15 – 0.102 × ^-15 ) J
= 1.223 × 10^-15J
= \(\frac{1.223 \times 10^{-15}}{1.602 \times 10^{-19}} \mathrm{eV}\)
[ 1J = 1.602 × 10^-19 eV]
= 7.6 × 10³ eV

Question 55.
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 x 10¹⁵ (Hz) \(\left[\frac{1}{3^{2}}-\frac{1}{n^{2}}\right]\)
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Answer:

Question 56.
Calculate the wavelength for the emis8ion transition if it starts from the orbit having radius 1.3225 mn and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:
The radius of the nth orbit of hydrogen like particles is given by

Thus, the transition ocurs from the 5th orbit to the 2nd orbit. It belongs to the Balmer series.
Wave number ( \(\bar{V}\) ) for the transition is given by,
\(\bar{v}=\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right)=1.097 \times 10^{7}\left(\frac{21}{100}\right)\)
= 2.303 × 10⁶ m^-1
∴ Wavelength ) associated with the emission transition is given by,
λ = \(\frac{1}{\bar{v}}=\frac{1}{2.303 \times 10^{6} \mathrm{~m}^{-1}}\)
= 0.434 × 10^-6 = 434 nm
∴ This transition belongs to Balmer series and comes in the visible region of the spectrum.

Question 57.
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 10⁶ ms^-1, calculate de Broglie wavelength associated with this electron.
Answer:
From Broglie’s equation,
λ = \(\frac{h}{m v}\)
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(1.6 \times 10^{6} \mathrm{~ms}^{-1}\right)}\) (1J = 1 kgm²s^-2)
= 4.55 × 10^-10 m = 455 pm
∴ de Broglie’s wavelength associated with the electron is 455 pm.

Question 58.
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Answer:
From de Broglie’s equation,
λ = \(\frac{h}{m v}\)
v = \(\frac{h}{m \lambda}\)
mass of neutron, m = 1.675 × 10^-27 kg
λ = 800pm = 800 × 10^-12 m
v = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(1.675 \times 10^{-27} \mathrm{~kg}\right)\left(800 \times 10^{-12} \mathrm{~m}\right)}\) = 4.94 × 10²ms^-1
= 494 ms^-1
∴ Velocity associated with the neutron is 494 ms^-1.

Question 59.
If the velocity of the electron in Bohr’s first orbit is 2.19 × 10⁶ms^-1, calculate the de Broglie wavelength associated with it.
Answer:
According to de Broglie’s equation,
λ = \(\frac{h}{m v}\)
Mass of electron = 9.1 × 10^-31 kg, h = 6.626 × 10^-34 Js,
velocity = 2.19 × 10⁶ ms^-1
= \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(2.19 \times 10^{6} \mathrm{~ms}^{-1}\right)}\) = 3.32 × 10^-10m
= 3.32 × 10^-10m × \(\frac{100}{100}\) = 332 × 10^-12 m
λ = 332pm
∴ Wavelength associated with the electron is 332 pm.

Question 60.
The velocity associated with a proton moving in a potential difference of 1000 V is 437 × 10⁵ ms^-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Answer:
Wavelength associated with the velocity of hockey ball,
λ = \(\frac{h}{m v}\)
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{(0.1 \mathrm{~kg})\left(4.37 \times 10^{5} \mathrm{~ms}^{-1}\right)}\)
λ = 1.516 × 10^-38 m

Question 61.
If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is \(\frac{h}{4 \pi_{m}}\) × 0.05 nm, is there any problem in defining this value.
Answer:
From Heisenberg’s uncertainty principle,

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

Question 62.
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these comhination(s) has/have the same energy lists.
1. n = 4, l = 2, m_l = -2, m_s = – \(\frac{1}{2}\)
2. n = 3, l = 2, m_l= 1, m_s = + \(\frac{1}{2}\)
3. n = 4, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
4. n = 3, l – 2, m_l = -2, m_s = –\(\frac{1}{2}\)
5. n = 3, l = 1, m_l = -1, m_s = + \(\frac{1}{2}\)
6. n = 4, l = 1, m_l = 0, m_s = + \(\frac{1}{2}\)
Answer:
For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5 and 6 are present in the 4d, 3d, 4p, 3d, 3p and 4p orbitals respectively.
Therefore, the increasing order of energies are
5(3p) < 2(3d) = 4(3d) < 6(4p) = 3(4p) < 1(4d).

Question 63.
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital,6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge.
Answer:
Effective nuclear charge decreases as the distance of the orbitals increases from the nucleus. Hence, 4p electrons experience the lowest effective nuclear charge.

Question 64.
Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.
Answer:
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.
(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.
(ii) 4d will experience greater nuclear charge than 4/ since 4d is closer to the nucleus.
(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3/.

Question 65.
The unpaired electrons in A1 and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?
Answer:
₁₃Al = 1s²2s²2p⁶3s²3p¹
₁₄Si = 1s²2s²2p⁶3s²3p²
Si (+4) has greater nuclear charge than aluminium (+3). Hence, 3p unpaired electrons of Si experience greater effective nuclear charge than Al.

Question 66.
Indicate the number of unpaired electrons in
(a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Answer:
(a) Phosphorus (P) : Atomic number = 15
The electronic configuration of P is : 1s² 2s² 2p⁶ 3s² 3p³

The orbital picture of P can be represented as :

From the orbital picture, phosphorus has three unpaired electrons
(b) Silicon (Si) : Atomic number = 14
The electronic configuration of Si is : 1s² 2s² 2p⁶ 3s² 3p²
The orbitai picture of Si can be represented as :

From the orbital picture, silicon has two unpaired electrons.
(c) Chromium (Cr) : Atomic number = 24
The electronic configuration of Cr is : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
The orbital picture of chromium is :

From the orbital picture, chromium has six unpaired electrons.
(d) Iron (Fe) : Atomic number = 26

The electronic configuration of Fe is : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
The orbital picture of Fe is:

From the orbital picture, iron has four unpaired electrons.
(e) Krypton (Kr) : Atomic number = 36
The electronic configuration of Kr is : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
The orbital picture of krypton is :

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

Question 67.
(a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having m_s value of \(\) 1/2 for n – 4?
Answer:
(a) n = 4 (Given)
For a given value of ‘rt, V can have values from zero to n -1.
l = 0, 1, 2, 3
Thus, four sub-shells are associated with n = 4, which are s, p, d and f
(b) Number of orbitals in the nth shell = n²
For n = 4
Number of orbitals = 16
If each orbital is taken fully, then it will have 1 electron with m_s value of \(-\frac{1}{2}\)
Number of electrons with ms value of (\(-\frac{1}{2}\)) is 16.