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

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.
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(iii) Both Li and Mg react withN2 to form nitrides.
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(iv) Neither Li nor Mg form peroxides or superoxides.
(v) The carbonates of both are covalent in nature. Also, these decompose on heating.
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(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.
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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.
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In ammonia recovery tower, NH3 is prepared by heating NH₄Cl with Ca(OH)₂.
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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.
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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.

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(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.

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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.
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(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 :

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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.

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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.

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(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
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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.
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(ii) Quick lime (CaO) combines with silica (SiO₂) to form slag.
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(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.
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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.
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(ii) In the solid state, BeCl₂ exists as a polymer in condensed phase.
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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.
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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