Important Questions – Page 16

PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

June 21, 2022 by Bhagya

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry: Some Basic Principles and Techniques Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

Very Short Answer Type Questions

Question 1.
How many σ bonds and π bonds are present in the second member of the alkene series?
Answer:
The second member of the alkene series is propene. The structual formula of the propene is
PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry Some Basic Principles and Techniques 1

Question 2.
Show the polarisation of carbon-magnesium bond in the following structure.
CH₃ —CH₂ —CH₂ —CH₂ —Mg —X
Answer:
Carbon is more electronegative than magnesium. Therefore, Mg acquires a partial positive charge and carbon acquires a partial negative charge.

Question 3.
What are primary and secondary suffixes as applied to IUPAC nomenclature?
Answer:
The primary suffix indicates whether the carbon chain is saturated or unsaturated while the secondary suffix indicates the functional group present in the molecule.

Question 4.
Draw all position isomers of an alcohol with molecular formula, C3HgO.
Answer:

Question 5.
CH₂ = CH^– is more basic than HC = C^– . Explain why?
Answer:
PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry Some Basic Principles and Techniques 4
Since, sp-carbon is more electronegative than sp²-carbon, therefore, CH ≡ C^– is less willing to donate a pair of electrons than H₂C = CH^–. In other words, H₂C = CH^– is more basic than HC ≡ C^–.

Question 6.
Why does SO₃ act as an electrophile? [NCERT Exemplar]
Answer:
In SO₃, three highly electronegative oxygen atoms are attached to sulphur atom. It makes sulphur atom electron deficient. Further, due to resonance, sulphur acquires a positive charge. Both these factors, make SO₃ an electrophile.

Question 7.
How will you separate a mixture of o-nitro phenol and p-nitrophenol?
Answer:
A mixture of o-nitrophenol and p-nitrophenol can be separated by steam distillation, o-nitrophenol being less volatile distils over along with water while p-nitrophenol being non-volatile remains in the flask.

Question 8.
In DNA and RNA, nitrogen atom is present in the ring system. Can Kjeldahl’s method be used for the estimation of nitrogen present in these? Give reason.
Answer:
DNA and RNA have nitrogen in the heterocyclic rings. Nitrogen present in rings, azo and nitro groups cannot be converted into (NH₄)₂SO₄. That’s why Kjeldahl’s method cannot be used for the estimation of nitrogen present in these.

Question 9.
Lassaigne’s test is not shown by diazonium salts, though they contain nitrogen. Why?
Answer:
Diazonium salts (C₆H₅N₂⁺X^–) readily lose N₂ on heating before reacting with fused sodium metal. Therefore, these salts do not give positive Lassaigne’s test for nitrogen.

Question 10.
Write three-dimensional wedge-dashed or wedge-line representations for the following:
(a) CH₃CH₂OH
(b) CH₂FCl
Answer:
PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry Some Basic Principles and Techniques 5

Short Answer Type Questions

Question 1.
Draw all polygon formula for the molecular formula C₅H₁₀.
Answer:
The different polygon formula of the compound having molecular formula C₅H₁₀ are :
PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry Some Basic Principles and Techniques 6

Question 2.
An alkane has a molecular mass of 72. Draw all its possible chain isomers and write their IUPAC names.
Answer:
First of all, we will derive the molecular formula.
Molecular formula of alkane is C_nH_2n+2
∴ Molecular mass = 72
∴ 12n + 2n + 2 = 72
n = 5
The alkane is C₅H₁₂. The possible chain isomers are
PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry Some Basic Principles and Techniques 7

Question 3.
Arrange the following
PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry Some Basic Principles and Techniques 8 1
Answer:

Question 4.
Suggest a method to purify
(i) camphor containing traces of common salt.
(ii) kerosene oil containing water.
(iii) a liquid which decomposes at its boiling point.
Answer:
(i) Sublimation-camphor sublimes while common salt remains as residue in the China dish.
(ii) Since the two liquids are immiscible, the technique of solvent extraction with a separating funnel is used. The mixture is thoroughly shaken and the separating funnel is allowed to stand. Kerosene being lighter than water forms the upper layer while water forms the lower layer.
The lower water layer is run off using the stop cork of the funnel and kerosene oil is obtained. It is dried over anhydrous CaCl₂ or MgSO₄ and then distilled to give pure kerosene oil.
(iii) Distillation under reduced pressure.

Question 5.
The structure of triphenylmethyl cation is given below. This is very stable and some of its salts can he stored for months. Explain the cause of high stability of this cation.
PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry Some Basic Principles and Techniques 9
Answer:
In triphenylmethyl cation, due to resonance, the positive charge can move at both the o-and p-position of each benzene ring. This is illustrated below.

Since, there are three benzene rings, hence, there are, in all, nine resonance structures.
Thus, triphenylmethyl cation is highly stable due to these nine resonance structures.

Long Answer Type Questions

Question 1.
Consider structures I to VII and answer the following question (i) to (iv).
PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry Some Basic Principles and Techniques 11

(i) Which of the above compounds form pairs of metamers?
(ii) Identify the pairs of compounds which are functional group isomers.
(iii) Identify the pairs of compounds that represent position isomerism.
(iv) Identify the pairs of compounds that represent chain isomerism.
Answer:
(i) V and VI or VI and VII form a pair of metamers since they differ in the number of carbon atoms on the either side of the functional group, i.e., O-atom.
(ii) I and V, I and VI, I and VII; II and V, II and VI, II and VII; III and V, III and VI; III and VII; IV and V; IV and VI and IV and VII are all functional group isomers.
(iii) I and II, III and IV and, VI and VII represent position isomerism.
(iv) I and III, I and IV, II and III and II and IV represent chain isomerism.

Question 2.
Write structural formulae for all the isomeric amines with molecular formula C^^N.
Answer:
PSEB 11th Class Chemistry Important Questions Chapter 12 Organic Chemistry Some Basic Principles and Techniques 13

PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements

June 21, 2022 by Bhagya

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements Important Questions

Very Short Answer Type Questions

Question 1.
What is the formula of kemite, an ore of boron?
Answer:
Formula of kernite, Na₂[B₄O₅(OH)₄] or Na₂B₄O₇.2H₂O

Question 2.
Complete the following chemical equations :
Z + 3LiAlH₄ → X + 3LiF + 3AlF₃
X + 6H₂O → Y + 6H₂
X + 3O₂ → B₂O₃ + 3H₂O
Answer:
PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements 1

Question 3.
Tl(NO₃)₃ acts as an oxidising agent. Explain.
Answer:
Due to inert pair effect, Tl in +1 oxidation state is more stable than that of +3 oxidation state. Therefore, Tl(NO₃)₃ acts as an oxidising agent and readily reduced to TlNO₃.

Question 4.
Why is B—X bond distance in BX₃ shorter than the theoretically expected value?
Answer:
This is due to pπ—pπ back bonding of the completely filled p-orbital of halogen X into the empty p-orbital of boron.

Question 5.
Are all the B—H bonds in diborane equivalent?
Answer:
No, there are two types of bonds in diborane two electron normal bonds and three centred two electron bonds.

Question 6.
Name the member of group 14 that forms the most acidic oxide.
Answer:
Among monoxides, CO is neutral and GeO is acidic while among dioxides, CO₂, SiO₂ are acidic, GeO₂ is also acidic but less acidic than Si02.

Question 7.
Silicones are used for making waterproof fabrics. Give reason.
Answer:
Silicones are synthetic polymers containing repeated units of R₂SiO where R is alkyl group.
Therefore, these are water repellants i.e., do not absorb water and are used for making waterproof fabrics.

Question 8.
Atomic radius of gallium (135 pm) is less than that of aluminium (143 pm).
Answer: It is due to poor shielding effect of 3d-electrons due to which effective nuclear charge increases in Ga, therefore, it is smaller than Al.

Question 9.
AlF₃ is high melting solid but AlCl₃ is low melting. Explain.
Answer: AlF3 is high melting solid because it is ionic in nature. On the other hand, A1C13 is covalent in nature and hence is a low melting solid.

Question 10.
How will you prepare an ahuninosilicate?
Answer:
Aluminosilicate is prepared by substituting some of the Si atoms in the three dimensional network of SiO₂ by Al atoms.

Short Answer Type Questions

Question 1.
Like CO, why its analog of SiO is not stable?
Answer:
CO is a resonance hybrid of the following two structures :

Thus, CO contains pπ-pπ multiple bonds. This is due to the reason that carbon has a strong tendency to form pπ—pπ multiple bonds due to its small size and high electronegativity. Silicon, on the other hand, because of its _ bigger size and lower electronegativity has no tendency to form pπ—pπ multiple bonds and hence, Si does not form SiO molecule analogous to CO molecule.

Question 2.
Account for the following:
(i) Graphite is used as lubricant.
(ii) Diamond is used as an abrasive.
Answer:
(i) Graphite has layered structure. Layers are held together by weak van der Waals’ forces and hence can be made to slip over one another. Therefore, graphite acts as a dry lubricant.
(ii) In diamond, each sp3 hybridised carbon atom is linked to four other carbon atoms. It has three-dimensional network of carbon atoms. It is very difficult to break extended covalent bonding and therefore diamond is a hardest substance on the earth. That’s why it is used as an abrasive.

Question 3.
Which one is more soluble in diethyl ether, anhydrous AlCl₃ or hydrated AlCl₃? Explain in terms of bonding.
Answer:
Anhydrous AlCl₃ is an electron-deficient compound while hydrated AlCl₃ is not. Therefore, anhydrous A1C13 is more soluble in diethyl ether because the oxygen atom of ether donates a pair of electrons to the vacant p-orbital on the Al atom in AlCl₃ forming a coordinate bond.
PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements 3

In case of hydrated AlCl₃, Al is not electron deficient since H₂O has already donated a pair of electrons to it.

Question 4.
BCl₃ is trigonal planar while AlCl₃ is tetrahedral in dimeric state. Explain.
Or
BCl₃ exists as monomer whereas AlCl₃ is dimerised through halogen bridging. Give reason. Explain the structure of the dimer of AlCl₃ also.
Answer:
Both BCl₃ and AlCl₃ are electron deficient molecules having six electrons in the valence shell of their respective central atoms. To complete their octets, the central atom in each case can accept a pair of electrons from the chlorine atom of another molecule forming dimeric structures. However, because of small size of B, it cannot accommodate four big sized Cl atoms around it. Therefore, BCl₃ prefers to exist as a monomeric planar molecule in which B atom is sp² -hybridised.
PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements 4

On the other hand, Al because of its bigger size can easily accommodate four Cl atoms around it. As a result, AlCl₃ exists as a dimer. In this dimer, since the covalency of Al has increased to 4.

Therefore, Al is sp³-hybridised and the four Cl atoms are held tetrahedrally around it.
PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements 5

Question 5.
Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 element in more stable oxidation state. Give reason for your choice. State the nature of bonding also.
(i) TICl₃, TlCl
(ii) AlCl₃, AlCl
(iii) InCl₃, InCl
Answer:
(i) Due to strong inert pair effect, +1 oxidation state of T1 is more stable than +3. Since, compounds in lower oxidation state are ionic but covalent in higher oxidation state, therefore, TlCl₃ is less stable and covalent in nature but TlCl is more stable and is ionic in nature.

(ii) Due to absence of d-orbitals, Al does not show inert pair effect. Therefore, it is most stable than A1C1. Further, in the solid or the vapour state, AlCl₃ is covalent in nature but in aqueous solution, it ionises to form Al³⁺ (aq) and Cl^–(aq) ions.
(iii) Due to inert pair effect, indium exists in both +1 and +3 oxidation states, out of which +3 oxidation state is more stable than +1 oxidation
state. In other words, InCl₃ is more stable than InCl. Being unstable, InCl undergoes disproportionation reactions.
3InCl(aq) → 2In(s) + In³⁺(aq) + 3Cl^– (aq)

Long Answer Type Questions

Question 1.
(i) What are silicones? State the uses of silicones.
(ii) What are boranes? Give chemical equation for the preparation of diborane.
Answer:
(i) Silicones are a group of organosilicon polymers, which have (R2SiO) as a repeating unit. These may be linear silicones, cyclic silicones and cross-linked silicones. These are prepared by the hydrolysis of alkyl or aryl derivatives of SiCl₄ like RSiCl₃, R₂SiCl₂ and R₃SiCl and polymerisation of alkyl or aryl hydroxy derivatives obtained by hydrolysis.

PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements 6

Uses : These are used as sealant, greases, electrical insulators and for water proofing of fabrics. These are also used in surgical and cosmetic plants.
(ii) Boron forms a number of covalent hydrides with general formulae B_nH_n+4 and B_nH_n+6. These are called boranes. B₂H₆ and B₄H₁₀ are the representative compounds of the two series respectively.

Preparation of diborane : It is prepared by treating boron trifluoride with LiAlH4 in diethyl ether.
4BF₃ + 3LiAlH₄ → 2B₂H₆ + 3LiF + 3AlF₃
On industrial scale it is prepared by the reaction of BF₃ with sodium hydride.

Question 2.
Account for the following observations :
(i) AlCl₃ is a Lewis acid.
(ii) Though fluorine is more electronegative than chlorine yet BF₃ is weaker Lewis acid than BCl₃.
(iii) PbO₂ is stronger oxidising agent than SnO₂.
(iv) The +1 oxidation state of thallium is more stable than its +3 state.
(i) In AlCl₃, Al has only six electrons in its valence shell. It is an electron deficient species. Therefore, it acts as a Lewis acid (electron acceptor).
(ii) In BF₃, boron has a vacant 2p-orbital and fluorine has one 2p completely filled unutilized orbital. Both of these orbitals belong to same energy level therefore, they can overlap effectively and form pπ—pπ bond. This type of bond formation is known as back bonding. While back bonding is not possible in BCl₃ because there is no effective overlapping between the 2p-orbital of boron and 3p-orbital of chlorine. Therefore, electron deficiency of B is higher in BCl₃ than that of BF₃. That’s why BF₃ is a weaker Lewis acid than BCl₃.

PSEB 11th Class Chemistry Important Questions Chapter 11 The p-Block Elements 8

(iii) Pb⁴⁺ is less stable thanPb²⁺, due to inert pair effect therefore, Pb⁴⁺ salts act as strong oxidising agents. Sn²⁺ is also less stable than Sn⁴⁺, thus Sn⁴⁺ can also act as an oxidising agent. But Pb⁴⁺ is a stronger oxidising agent than Sn⁴⁺ because inert pair effect increases down die group.
(iv) Tl⁺ is more stable than Tl³⁺ because of inert pair effect.

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

June 21, 2022 by Bhagya

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements Important Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
Complete the following reactions,
(i) \(\mathbf{O}_{2}^{2-}\) +H₂O →
(ii) O₂ +H₂O →
Answer:
(i) Peroxide ions react with water to form H202.
\(\mathbf{O}_{2}^{2-}\) + 2H₂O→ 20H’ + H₂O₂
(ii) Superoxides react with water to form H202 and 02.
\(2 \mathrm{O}_{2}^{-}\) + 2H₂O → 20H^– + H₂O₂ + O₂

Question 2.
(i) Predict giving reason, the outcome of the reaction
PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements 1
Answer:
(i) Large cation (K⁺) can stabilise large anion (I^–).
(ii) This is because the larger cation (K⁺) can stabilise larger anion (Cl^–).

Question 3.
Which colours are imparted to flame when the following elements are introduced in the flame one by one?
(i) Strontium (ii) Barium (iii) Calcium
Answer:
(i) Strontium — Brick red
(ii) Barium — Grassy green
(iii) Calcium — Crimson red

Question 4.
Sodium fire in the laboratory should not be extinguished by pouring water. Why?
Answer:
This is because sodium produces hydrogen gas with water which catches fire because of the exothermic nature of the reaction.

Question 5.
What is light soda ash? Wliy is it called so?
Answer:
Light soda ash is anhydrous Na₂CO₃. It is called so because it is fluffy solid with a low packing density of about 0.5 g cm^-3.

Question 6.
What is baryta water? Give its one use.
Answer:
Baryta is an aqueous solution of barium hydroxide. It can also be used for detection of CO₂.

Question 7.
Give the chemical formula of quick lime, slaked lime and lime water.
Answer:
Quick lime is CaO, slaked lime is Ca(OH)₂ and lime water is an aqueous solution of Ca(OH)₂.

Question 8.
Which magnesium compounds are the constituents of toothpaste?
Answer:
Mg(OH)₂ and MgCO₃ are the constituents of toothpaste.

Question 9.
What is the mixture of CaCN₂ and carbon known as?
Answer:
A mixture of calcium cyanamide (CaCN₂) and carbon is known as nitrolim. It is used as a fertiliser.

Question 10.
It is necessary to add gypsum in the final stages of preparation of cement. Explain why?
Answer:
Gypsum (CaSO₄ . 2H₂O) is added in the final stages of preparation of cement because it slows down the process of setting of cement so that it gets sufficiently hardened thereby imparting greater strength to it.

Short Answer Type Questions

Question 1.
How would you distinguish between
(i) Be(OH)₂ and Ba(OH)₂
(ii) BeSO₄ and BaSO₄
Answer:
(i) Be(OH)₂, beryllium hydroxide is soluble in aqueous sodium hydroxide solution whereas Ba(OH)₂, (barium hydroxide) does not, because Be(OH)₂ is amphoteric in nature and Ba(OH)₂ is basic in nature. Be(OH)₂ + 2NaOH → Na2[Be(OH)₄] (Sodium beryllate)
(ii) BeSO₄ is soluble in water whereas BaSO₄ is insoluble in water.

Question 2.
Element A bums in nitrogen to give an ionic compound B. Compound B reacts with water to give C and D. A solution of C becomes milky on bubbling carbon dioxide. Identify A, B, C and D.
Answer:
Element A is calcium
PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements 2
Compounds = Ca₃N₂ ; CompoundC = Ca(OH)₂ and Compound D = NH₃

Question 3.
What happens when
(i) chlorine gas is passed through a cold and dilute solution of NaOH?
(ii) yellow phosphorus is heated with NaOH solution?
(iii) carbon dioxide is passed through ammonical brine solution?
(iv) sodium hydrogen carbonate is heated?
Answer:
(i) Sodium hypochlorite and sodium chloride are obtained.
Cl2 + 2NaOH → NaCl + NaClO + H₂O
(ii) Phosphine gas is obtained.
P4 + 3NaOH + 3H₂O → 3NaH₂PO₂ + PH₃
(iii) Sodium hydrogen carbonate is precipitated.
NH₃ + H₂O + CO₂ + NaCl → NH₄Cl + NaHCO₃↓
(iv) Sodium ash is obtained.

Question 4.
Mention the various sources of sodium chloride and explain the preparation of sodium chloride from sea-water and salt mines.
Answer:
NaCl occurs abundantly in nature. Its major sources are (a) Sea water which contains 2.7 to 2.9 % NaCl.
(b) Water of inland lakes such as Sambhar Lake in Rajasthan.
(c) Salt-mines which contain rock salt are located in England, Australia, and Himachal Pradesh.
Preparation
(i) From sea water : Sea water is filled in big tanks where it slowly evaporates, leaving behind solid salt. In cold countries, where temperatures are very low, pure water get freeze. Ice formed is removed and concentration of NaCl in solution increases. The concentrated sodium can be separated and evaporated to get NaCl.
(ii) From salt-mines : Salt mines are located deep under the surface of the earth. Holes are made into these mines with the help of drillers and broken pieces of salt rocks are taken out by suitable means.

Question 5.
When water is added to compound (A) of calcium, solution of compound (B) is formed. When carhon dioxide is passed into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution, milkiness disappears due to the formation of compound (D). Identify the compound A, B, C and D. Explain why the milkiness disappears in the last step? [NCERT Exemplar]
Answer:
Appearance of milkiness on passing CO₂ in the solution of compound B indicates that compound B is lime water and compound C is CaCO₃. Since, compound B is obtained by adding H₂O to compound A, therefore compound A is quicklime, CaO. The reactions are as follows :
PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements 4

Long Answer Type Questions

Question 1.
Ions of an element of group 1 participate in the transmission of nerve signals and transport of sugars and amino acids into cells. This element imparts yellow colour to the flame in flame test and forms an oxide and a peroxide with oxygen. Identify the element and write chemical reaction to show the formation of
its peroxide. Why does the element impart colour to the flame?
Answer:
Yellow colour flame in flame test indicates that the alkali metal must be sodium. It reacts with O₂ to form a mixture of sodium peroxide, Na₂O₂ and sodium oxide, Na₂O.

PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements 6

Ionization enthalpy of sodium is low. When sodium metal or its salt is heated in Bunsen flame, the flame energy causes an excitation of the outermost electron which on reverting back to its initial position gives out the absorbed energy as visible light. That’s why sodium imparts yellow colour to the flame.

Question 2.
The stability of peroxide and superoxide of alkali metals increase as we go down the group. Explain giving reason.
Answer:
The stabilit-y of peroxides or superoxides increases as the size of metal ion increases, i.e., KO₂ < RbO₂ < CsO₂.
The reactivity of alkali metals towards oxygen to form different oxides is due to strong positive field around each alkali metal cation. Li⁺ is smallest, it does not allow 0^2- ion to react with O₂ further. Na⁺ is larger than Li, its positive field is weaker than Li⁺. It cannot prevent the conversion of O^2- into \(\mathrm{O}_{2}^{2-}\). The larger K⁺, Rb⁺ and Cs⁺ ions permit \(\mathrm{O}_{2}^{2-}\)ion to react with O₂ further forming superoxide ion (\(\mathrm{O}_{2}^{-}\)).

PSEB 11th Class Chemistry Important Questions Chapter 10 The s-Block Elements 7

Further more, increased stability of the peroxide or superoxide with increase in the size of metal ion is due to the stabilisation of large anions by larger cations through lattice energy effect.

PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen Important Questions

June 21, 2022 by Bhagya

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen Important Questions

Very Short Answer Type Questions

Question 1.
Name the isotope of hydrogen which contains equal number of protons and neutrons.
Answer:
Deuterium (\({ }_{1}^{2} \mathrm{H}\))
Number of protons (p) = number of electrons
= atomic number = 1
Number of neutrons (n) = mass number – atomic number
= 2 – 1 = 1 .

Question 2.
Why is the ionisation enthalpy of hydrogen higher than that of sodium?
Answer:
Both H and Na contain one electron in the valence shell. But the size of H is much smaller as compare to that of Na and hence, the ionisation enthalpy of hydrogen is much higher (1312 kJ mol^-1) than that of Na (496 kJ mol^-1).

Question 3.
What do you mean by 15 volume H2O2 solution?
Answer:
‘15 volume H₂O₂’ means 1 mL of a 15 volume H₂O₂ solution gives 15 mL of O₂ at NTP.

Question 4.
Which isotope of hydrogen is radioactive?
Answer:
Tritium

Question 5.
Arrange H₂, D₂ and T₂ in the decreasing order of their
(i) boiling points
(ii) heat of fusion
Answer:
(i) T₂ > D₂ > H₂
(ii) T₂ > D₂ > H₂

Question 6.
Write the Lewis structure of hydrogen peroxide.
Answer:
The Lewis structure of hydrogen peroxide is :
PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen 1

Question 7.
Write one chemical reaction for the preparation of D₂O₂.
Answer:
D₂O₂ is prepared by distillation of potassium persulphate (K₂S₂O₈) with D₂O.

Question 8.
Suggest a method to show the electronegative nature of hydrogen.
Answer:
When sodium hydride is electrolysed, hydrogen is evolved at anode, which shows its electronegative nature.

Question 9.
What type of bonds are broken when water evaporates.
Answer:
Intermolecular hydrogen bonds are broken when water evaporates.

Short Answer Type Questions

Question 1.
Describe the industrial applications of hydrogen dependent on
(i) the heat liberated when its atoms are made to combine on the surface of a metal.
(ii) its effect on the unsaturated organic systems in the presence of a catalyst.
(iii) its ability to combine with nitrogen under specific conditions.
Answer:
(i) Due to this property, hydrogen is used in atomic hydrogen welding/cutting torch.
(ii) Due to this property hydrogen is used for the manufacture of vanaspati ghee from edible oils such as cotton-seed oil, soyabean oil, corn oil etc.

(iii) Due to this property dihydrogen is used for the manufacture of ammonia (Haber’s process).
PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen 3 - 1

Question 2.
Why does water show high boiling point as compared to hydrogen sulphide? Give reasons for your answer.
Answer:
Water show high boiling point as compared to hydrogen sulphide due to high electronegativity of oxygen (EN = 3.5), water undergoes extensive H-bonding as a result of which water exists as associated molecule.

PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen 4

For breaking these hydrogen bond, a large amount of energy is needed and hence the boiling point of H₂O is high. In other words, due to lower electronegativity of S (EN =2.5), hydrogen sulphide do not undergo H-bonding. Consequently, H₂S exists as discrete molecule and hence its boiling point is much lower than that of H₂O. That is why H₂S is a gas at room temperature.

Question 3.
If a given sample of water has degree of hardness equal to 46 ppm. If entire hardness is due to MgSO₄, how much MgSO₄ is present per kg of water?
Answer:
Given, degree of hardness = 46 ppm
Which means that 106 g of sample require 46 g of CaCO₃
∴ CaCO₃ present in 1000 g of water = \(\frac{46 \times 1000}{10^{6}}\) = 46 x 10^-3 g
1 mol (or 100 g) of CaC0₃ = 1 mol (or 120 g) of MgSO₄
∴ 46 x 10^-3 g of CaCO₃ = \(\frac{120 \times 46 \times 10^{-3}}{100}\)g = 0.055 g or 55 mg

Question 4.
What are the advantages in using hydrogen as a fuel?
Answer:
Hydrogen as a fuel has the following advantages :

It has high calorific value.
During combustion, it does not produce smoke or any unpleasant fumes.
It leaves no ash after burning. The only product of combustion is water.
It does not pollute the air because no pollutant is produced during its combustion.
It can be used in a fuel cell to generate electricity.
It can be used in the internal combustion engines with slight modifications.

Question 5.
Calculate the volume strength of a 3% solution of H₂O₂
Answer:
100 mL of H₂O₂ solution contains H₂O₂ = 3 g
∴ 1000 mL of H₂O₂ solution will contains
H₂O₂ = \(\frac{3}{100}\) x 1000 = 30 g
Consider the chemical equation,

Now 68 g of H₂O₂ gives O₂ at NTP = 22.7 L
∴ 30 g of H₂O₂ will give 02 at NTP = \(\frac{22.7}{68}\) x 30 = 10.014
But 30 g of H₂O₂ are present in 1000 mL of H₂O₂.
Hence, 1000 mL of H₂O₂ solution gives 02 at NTP = 1.0014 mL
∴ 1 mL of H₂O₂ solution will give O₂ at NTP = \(\frac{10014}{1000}\)= 10.01 mL
Hence, the volume strength of 3% H202 solution = 10.01

Long Answer Type Questions

Question 1.
(i) (a) How would you prepare dihydrogen from water by using a reducing agent?
(b) How would you prepare dihydrogen from a substance other than water?
(c) How would you prepare very pure dihydrogen in the laboratory?
(ii) Write a short note on hydrogenation of vegetable oils.
Answer:
(i) (a) Sodium metal is a good reducing agent. It reduces water to hydrogen (or dihydrogen).
2H₂O + 2Na → 2NaOH + H₂(g)
(b) Dihydrogen can be obtained by treating zinc with dilute HCl
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
(c) Highly pure dihydrogen (hydrogen gas) can be prepared by the following methods :
I. Fairly pure hydrogen can be obtained by treating pure magnesium or pure aluminium with chemically pure H₂SO₄ or HCl diluted with distilled water. The gas is passed over P₂O₅ and is collected by the displacement of mercury.
Mg(s) + H₂SO₄(aq) > MgSO₄(aq) + H₂(g)

II. Highly pure hydrogen gas can be obtained by electrolysing a warm solution of Ba(OH)2 in a U-tube using nickel electrodes. The gas is purified by passing it over heated platinum gauze when traces of oxygen combine with hydrogen forming water. The gas is then dried by passing it over caustic potash sticks and phosphorus pentoxide. Hydrogen is finally adsorbed in palladium and the impurities remain unadsorbed. On heating palladium under reduced pressure pure hydrogen is liberated.

(ii) When oils like groundnut oil or cotton seed oil (which are unsaturated compound i.e., have double bond) are treated with hydrogen in the presence of nickel as catalyst, they get converted into edible fats like margarine and vanaspati ghee (which are saturated compounds). This reaction is called hydrogenation of vegetable oils or hardening of oils.

Question 2.
Give ion electron equations for the following reactions :
(i) Oxidation of ferrous ions to ferric ions by hydrogen peroxide both in acidic and basic media.
(ii) Oxidation of iodide ion to iodine by hydrogen peroxide in acidic medium.
(iii) Reduction of acidified potassium dichromate solution.
(iv) Oxidation of sulphurous acid to sulphuric acid.
(v) Oxidation of ferrocyanide ions to ferricyanide ions in acidic medium.
Answer:
PSEB 11th Class Chemistry Important Questions Chapter 9 Hydrogen 6

PSEB 11th Class Chemistry Important Questions Chapter 8 Redox Reactions

June 21, 2022 by Bhagya

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 8 Redox Reactions Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 8 Redox Reactions

Very Short Answer Type Questions

Question 1.
What are spectator ions? Give one example.
Answer:
Spectator ions are ions that stay unaffected during a chemical reaction. They appear both as reactant and as product in an ionic equation. For example, in the following ionic equation, the sodium and nitrate ions are spectator ions.
Ag⁺ (aq) + NO₃^–(aq) + Na⁺ (aq) + Cl^– (aq) → AgCl(s) + Na⁺ (aq) + NO^–₃ (aq)

Question 2.
Why is anode called oxidation electrode, whereas cathode is called reduction electrode?
Answer:
At anode, loss of electrons takes place, i.e., oxidation takes place, whereas at cathode, gain of electrons takes place, i.e., reduction takes place.
Therefore, cathode is called reduction electrode and anode is called oxidation electrode.

Question 3.
Can we use KCl as electrolyte in the salt bridge of the cell?
Answer:
KCl cannot be used as electrolyte in the salt bridge because Cl^– ions will combine with Ag⁺ ions to form white precipitates of AgCl.

Question 4.
What would happen if no salt bridge were used in the electrochemical cell (e.g., Zn – Cu cell)?
Answer:
If no salt bridge is used, the positive ions (i.e., Zn²⁺ ) formed by loss of electrons will accumulate around the zinc electrode and negative ions (i.e., \(\mathrm{SO}_{4}^{2-}\)) left after reduction of Cu²⁺ ions will accumulate around the copper electrode. Thus, the solution will develop charges and the current stops flowing. Further, since the inner circuit is not complete, the current stops flowing.

Question 5.
Zn rod is immersed in CUSO₄ solution. What will you observe after an hour? Explain your observation in terms of redox reaction.
Answer:
The blue colour of CuSO₄ solution will get discharged and reddish brown copper metal will be deposited on Zn rod. This is because blue colour Cu²⁺ (in CuSO₄) gets reduced to Cu by accepting two electrons from Zn, which gets oxidised to colourless ZnSO₄.

PSEB 11th Class Chemistry Important Questions Chapter 8 Redox Reactions 1 - 1

Question 6.
What is the most essential conditions that must be satisfied in a redox reaction?
Answer:
In a redox reaction, the total number of electrons lost by the reducing agent must be equal to the number of electrons gained by the oxidising agent.

Question 7.
Find the value of n in \(\mathrm{MnO}_{4}^{-}\) + 8H⁺ + ne^– → Mn²⁺ + 4H₂O
Answer:
\(\mathrm{MnO}_{4}^{-}\) + 8H⁺ + ne^– → Mn²⁺ + 4H₂O
-1 + 8 + n = + 2
-1 – 2 + 8 + n = 0
n = – 5 or 5e^–

Question 8.
Can Fe³⁺ oxidise Br^– to Br₂ at 1 M concentrations?
\(\boldsymbol{E}^{\ominus}\)(Fe³⁺ /Fe²⁺) – 0.77 V and \(\boldsymbol{E}^{\ominus}\)(Br/Br^– ) = 1.09 V
Answer:
E^s( Fe³⁺ / Fe²⁺) is lower than that of E^s(Br / Br^–).
Therefore, Fe²⁺ can reduce Br₂ but Br^– cannot reduce Fe³⁺. Thus, Fe³⁺ cannot oxidise Br^– to Br₂.

Question 9.
Identify the substance that get reduced in the following reaction:
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
Answer:
In the reaction, Fe₂O₃ loses oxygen and is reduced to Fe.

Question 10.
Can the following reaction, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons 2 \mathrm{CrO}_{4}^{2-}+2 \mathrm{H}^{+}\) be regarded as a redox reaction?
Answer:
In this reaction, oxidation number of Cr in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is +6 and oxidation number of Cr in \(\mathrm{CrO}_{4}^{2-}\) is +6. Since, during the reaction, the oxidation number of Cr has neither decreased nor increased, therefore, the above reaction is not a redox reaction.

Short Answer Type Questions

Question 1.
2Cu₂S + 3O₂ ⇌ 2Cu₂O + 2SO₂
In this reaction which substance is getting oxidised and which substance is getting reduced? Name the reducing agent and oxidising agent.
Answer:
Since, oxygen is being added to Cu, therefore, Cu₂S is oxidised to Cu₂O and the other reactant i.e., O₂ is getting reduced. Hence, Cu₂S is a reducing agent and O₂ is an oxidising agent.

Question 2.
One mole of N₂H₄ loses 10 moles electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation number of N in Y? There is no change in oxidation state of H.
Answer:
Suppose the oxidation number of N in Y is x
(N^2-)₂ → (2N)^x + 10e^–
(as N₂H₄ → Y +10e^–)
Therefore, 2x -10 = – 4, which gives x = + 3. Hence, oxidation number of N in Y = 3.

Question 3.
What are the net charges on the left and right side of the following equations? Add electrons as necessary to make each of them balanced half reactions.
(i) \(\mathrm{NO}_{3}^{-}+\mathbf{1 0 H}^{+} \longrightarrow \mathbf{N H}_{4}^{+}+3 \mathrm{H}_{2} \mathrm{O}\)
(ii) \(\mathrm{Cl}_{2}+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathbf{2 C l O}_{2}^{-}+8 \mathrm{H}^{+}\)
Answer:
(i) +9 charge on the left, +1 charge on the right; add 8 electrons to the left side.
(ii) 0 charge on the left, +6 charge on the right; add 6 electrons on the right side.

Question 4.
An iron rod is immersed in solution containing 1.0 M NiSO₄ and 1.0 M ZnSO₄. Predict giving reasons which of the following reactions is likely to proceed?
(i) Fe reduces Zn²⁺ ions,
(ii) Iron reduces Ni²⁺ ions. Given
\(E_{\mathbf{Z n}^{2+} / \mathbf{Z n}}^{\ominus}=-0.76 \mathrm{~V}, E_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{=}}=-0.44 \mathrm{~V}\)
\(E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\ominus}=-0.25 \mathrm{~V}\)
Answer:
(i) Since \(E^{\ominus}\) of Zn is more negative than that of Fe, therefore, Zn will be oxidised to Zn²⁺ ions while Fe²⁺ ions will be reduced to Fe. In other words, Fe will not reduced Zn²⁺ ions.
(ii) Since, \(E^{\ominus}\) of Fe is more negative than that of Ni, therefore, Fe will be oxidised to Fe²⁺ ions while Ni²⁺ ions will be reduced to Ni. Thus, Fe reduces Ni²⁺ ions.

Question 5.
Copper dissolves in dilute nitric acid but not in dilute HC1. Explain.
Answer:
Since, \(E^{\ominus}\) of Cu²⁺/Cu electrode (+ 0.34 V) is higher than that of H⁺/H₂
electrode (0.0 V), therefore, H⁺ ions cannot oxidise Cu to Cu²⁺ ions and hence, Cu does not dissolve in dil. HCl.

In contrast, the electrode potential of \(\mathrm{NO}_{3}^{-}\) ion, i.e.\(\mathrm{NO}_{3}^{-}\) /NO electrode (+0.97 V) is higher than that of copper electrode and hence, it can oxidise Cu to Cu²⁺ ions and hence Cu dissolves in dil.HNO₃ due to oxidation of Cu by \(\mathrm{NO}_{3}^{-}\) ions and not by H⁺ ions.
Using standard electrode potential, the oxidative and reductive strength of a variety of substances can be composed.

Long Answer Type Questions

Question 1.
Why does fluorine doesn’t show disproportionation reaction?
Answer:
In a disproportionation reaction, the same species is simultaneously oxidised as
well as reduced. Therefore, for such a redox reaction to occur, the reacting species must contain an element which has atleast three oxidation states. The element, in reacting species, is present in an intermediate state while lower and higher oxidation states are available for reduction and oxidation to occur (respectively).
Fluorine is the strongest oxidising agent. It does not show positive oxidation state. That’s why fluorine does not show disproportionation reaction.

Question 2.
Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example.
Answer:
Measure the electrode potential of the given species by connecting the redox couple of the given species with standard hydrogen electrode. If it is positive, the electrode of the given species acts as reductant and if it is negative, it acts as an oxidant. Find the electrode potentials of the other given species in the same way, compare the values and determine their comparative strength as an reductant or oxidant.
Examples : Measurement of standard electrode potential of Zn⁺/Zn electrode using SHE as a reference electrode.

PSEB 11th Class Chemistry Important Questions Chapter 8 Redox Reactions 1

The EMF of the cell comes out to be 0.76 V. (reading of voltmeter is 0.76 V). Zn²⁺/Zn couple acts as anode and SHE acts as cathode.

PSEB 11th Class Chemistry Important Questions Chapter 8 Redox Reactions 2

PSEB 11th Class Chemistry Important Questions Chapter 7 Equilibrium

June 21, 2022 by Bhagya

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 7 Equilibrium Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 7 Equilibrium

Very Short Answer Type Questions

Question 1.
A tank is full of water. Water is coming in as well as going out at same rate. What will happen to level of water in a tank? What is name given to such state?
Answer:
It will remain the same because rate of inflow is equal to rate of outflow. This state is called state of ‘equilibrium’.

Question 2.
The ionization of hydrogen chloride in water is given t
HCl(aq) + H₂O(l) ⇌ H₃O⁺+(aq) + Cl^–(aq)
Label two conjugate acid-base pairs in this ionization.
Answer:
PSEB 11th Class Chemistry Important Questions Chapter 7 Equilibrium 1

Question 3.
Why solution of sugar in water does not conduct electricity whereas that of common salt in water does?
Answer:
Common salt (NaCl) is an electrolyte which gives Na⁺ and Cl^– ions in the aqueous solution. Hence, it conducts electricity. Sugar is sucrose (C₁₂H₂₂O₁₁) which is a non-electrolyte and does not give ions in the solution. Hence, it does not conduct electricity.

Question 4.
Why is ammonia termed as a base though it does not contain OH^– ions?
Answer:
Ammonia is termed as a base due to its tendency to donate electron pair. Therefore it is a Lewis base.

Question 5.
K_b for NH₄O, H is 1.8 x 10^-5 and for CH₃NH₂ is 44 x 10^-4. Which of them is strongest base and why?
Answer:
CH₃NH₂ is strongest base because it has high value of base dissociation constant.

Question 6.
pK_a value of acids A, B, C, D are 1.5, 3.5, 2.0 and 5.0. Which of them is strongest acid?
Answer:
Acid A with pK_a = 1.5 is strongest acid, lower the value of pK_a stronger will be the acid.

Question 7.
What will be the pH of 1M Na₂SO₄ solution?
Answer:
Na₂SO₄ is salt of strong acid and strong base, thus its aqueous solution will be neutral. Therefore, its pH will be 7.

Question 8.
Is it possible to get precipitate of Fe(OH)₃ at pH = 2? Give reason.
Answer:
No, because Fe(OH)₃ will dissolve in strongly acidic medium.

Question 9.
What happens to ionic product of water if some acid is added to it?
Answer:
Ionic product will remain unchanged.

Question 10.
How does common ion affect the solubility of electrolyte?
Answer:
Solubility of electrolyte decreases due to common ion effect.

Short Answer Type Questions

Question 1.
A certain buffer is made by mixing sodium form ate and formic acid in water. With the help of equations explain how this buffer neutralises addition of a small amount of an acid or a base?
Answer:
HCOONa → HCOO^– + Na⁺
HCOOH ⇌ HCOO^– + H⁺

HCOO^– is common ion in the above acidic buffer. When small amount of H⁺ ions is added, these H⁺ ions combine with HCOO^– which are in excess to form HCOOH back and [H⁺] remains practically same, so pH remains constant. When small amount of OH^– ions are added, OH^– ions will take up H⁺ and association of HCOOH will increase so as to maintain concentration of H⁺ ions. So, pH would not be affected.

Question 2.
How much volume of 0.1 M CH₃COOH should he added to 50 ml of 0.2 M CH₃COONa solution to prepare a buffer solution of pH 4.91. (pA_a of AcH is 4.76).
According to Henderson’s equation
PSEB 11th Class Chemistry Important Questions Chapter 7 Equilibrium 2
Required volume of 0.1 M acetic acid = 70.92 mL

Question 3.
Some processes are given below. What happens to the process if it is subjected to a change given in the brackets?

(ii) Dissolution of NaOH in water (Temperature is increased)
(iii) N₂(g) + O₂(g) ⇌ 2NO(g) -180.7 kJ (Pressure is increased and temperature is decreased.)
Answer:
(i) Equilibrium will shift in the forward direction, i.e., more ice will melt.
(ii) Solubility will decrease because it is an exothermic process.
(iii) Pressure has no effect. Decrease of temperature will shift the equilibrium in the backward direction.

Question 4.
50.0 g of CaCO₃ are heated to 1073 K in a 5 L vessel. What percent of the CaCO₃ would decompose at equilibrium? K_p for the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g) is 1.15 atm at 1073 K.
Answer:
The reaction is : CaCO₃(s) ⇌ CaO(s) + CO₂(g)
K_p = P_Co2 = 1.15 atm, pV = nRT
\(\mathrm{n}_{\mathrm{CO}_{2}}=\frac{p_{\mathrm{CO}_{2}} \mathrm{~V}}{R T}=\frac{1.15 \times 5}{0.082 \times 1073}\) = 0.065 mol

1 mole of CO₂ is obtained by decomposition of 1 mole CaCO₃. Therefore, moles of CaCO₃ decomposed is equal to the moles of CO₂ = 0.065 mol.
Mole of CaCO₃ initially present = \(\frac{50}{100}\) = 0.5 mol
[Molecular mass of CaCO₃ = 100]
Per cent of CaCO₃ decomposed = \(\frac{0.065}{0.5}\) x 100 = 13%

Question 5.
Arrange the following in increasing order of pH.
KNO₃(aqr), CH₃COONa(aq), NH₄Cl(aq), C₆H₅COONH₄(aq)
Answer:
(i) KNO₃ is a salt of strong acid-strong base, hence its aqueous solution is neutral; pH = 7
(ii) CH₃COONa is a salt of weak acid and strong base, hence, its aqueous solution is basic; pH < 7.
(iii) NH₄Cl is a salt of strong acid and weak base, hence its aqueous solution is acidic; pH < 7.
(iv) C₆H₅COONH₄ is a salt of weak acid, C₆H₅COOH and weak base, NH4OH. ButNH4OH is slightly stronger than C6H5COOH. Hence, pH is slightly greater than 7.
Therefore, increasing order of pH of the given salts is,
NH₄Cl < KNO₃ < C₆H₅COONH₄ < CH₃COONa

Long Answer Type Questions

Question 1.
Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionisation constants of acetic acid is 1.75 x 10^-5. Also calculate the change in pH of the buffer if the following adds in 1 L of the buffer (i) 1 cc of 1 M NaOH. (ii) 1 cc of 1 M HC1. Assume that the charge in volume is negligible, (iii) What will be the buffer index of the above buffer?
Answer:
pH = pK_a + log\(\frac{Salt}{Acid}\) = – log(1.75 x 10^-5) + log
\(\frac{0.15}{0.10}\)
= (5 – 0.2430) + 0.1761 = 4.757 + 0.1761 = 4.933.

(i) 1 cc of 1M NaOH contains NaOH = 10^-3 mol. This will convert 10^-3 mol of acetic acid into the salt so that salt formed = 10^-3 mol.
[Acid] = 0.10 – 0.001 = 0.099 M
[Salt] = 0.15 + 0.001 = 0.151 M
pH =. 4.757 + log \(\frac{0.151}{0.099}\)
= 4.757 + 0.183 = 4.940
∴ Increase in pH = 4.940 – 4.933 = 0.007 which is negligible.

(ii) 1 cc of 1 M HC1 contains HCl = 1CF₃ mol. This will convert 10^-3 mol CH₃COONa into CH₃COOH.
Now, [Acid] = 0.10 + 0.001 = 0.101 M
[Salt] = 0.15 – 0.001 = 0.149 M 0.149
∴ pH = 4.757 + log\(\frac{0.149}{0.101}\) = 4.757 + 0.169 = 4.925
∴ Decrease in pH = 4.933 = 0.007 which is again negligible.

(iii) Calculation of buffer index No. of moles of HC1 or NaOH added = 0.001 mol
Change in pH = 0.007
Hence, buffer index = \(\frac{\Delta n}{\Delta \mathrm{pH}}=\frac{0.001}{0.007}=\frac{1}{7}\)= 0.143

Question 2.
On the basis of Le-Chatelier’s principle, explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction?
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
What will be the effect of addition of argon to the above reaction mixture at constant volume?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92.38 kJ mol^-1

It is an exothermic process. According to Le-Chatelier’s principle, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction. So, optimum temperature, 700 K is favourable in attainment of equilibrium.

Similarly, high pressure about 200 atm is favourable for high yield of ammonia. On increasing pressure, reaction goes in the forward direction because the number of moles decreases in the forward direction.

At constant volume, addition of argon does not affect the equilibrium because it does not change the partial pressures of the reactants or products involved in the reaction and the equilibrium remains undisturbed.

PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics

June 21, 2022 by Bhagya

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics

Very Short Answer Type Questions

Question 1.
Identify the state functions and path functions out of the following.
Enthalpy, entropy, heat, temperature, work, free energy.
Answer:
State function Enthalpy, entropy, temperature, free energy.
Path function Heat, work

Question 2.
At 1 atm will the Δ_fH⁰ be zero for Cl₂(g) and Br₂(g)? Explain.
Answer:
Δ_fH⁰ for Cl₂(g) will be zero but Δ_fH⁰ for Br₂(g) will not be zero because liquid bromine state is elementary state not gaseous.

Question 3.
Why for predicting the spontaneity of a reaction, free energy criteria is better than the entropy criteria?
Answer:
Criteria of free energy change is better because it requires free energy change of the system only whereas the entropy change requires the total entropy change of the system and the surroundings.

Question 4.
Water can be lifted into the water tank at the top of the house with the help of a pump. Then why is it not considered to be spontaneous?
Answer:
A spontaneous process should occur continuously by itself after initiation. But this is not so in the given case because water will go up so long as the pump is working.

Question 5.
Given that ΔH = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
Answer:
It is a spontaneous process because although ΔH = 0, i.e., energy factor has no role to play but randomness increases, i.e., randomness factor favours the process.

Question 6.
Under what condition, the heat evolved or absorbed in a reaction is equal to its free energy change?
Answer:
As ΔG = ΔH – TΔS. Thus, ΔG = ΔH only when either the reaction is carried out at 0 K or the reaction is not accompanied by any entropy change, i.e., ΔS = 0.

Question 7.
In the equation, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) what would be the sign of work done?
Answer:
The sign of work done will be positive, i.e., work will be done on the system due to decrease in volume.

Question 8.
The molar enthalpy of vaporisation of acetone is less than that of water. Why?
Answer:
Enthalpy of vaporisation of water is more than that of acetone because there is strong hydrogen bonding in H₂O molecules.

Question 9.
One mole of acetone requires less heat to vaporise than 1 mole of water. Which of the two liquids has higher enthalpy of vaporisation?
Answer:
Less the heat required to vaporise 1 mole of a liquid, less is its enthalpy of vaporisation. Hence, water has higher enthalpy of vaporisation.

Question 10.
Which quantity out of Δ_rG and Δ_rG° will be zero at equilibrium?
Answer:
Δ_rG = Δ_rG° + RT In K
At equilibrium, 0 (zero) = Δ_rG° + RT In K
(v Δ_rG = 0)
or Δ_rG° = -RT In It;
Δ_rG° = 0 when it = 1
For all other values of K, Δ_rG° will be non-zero.

Short Answer Type Questions

Question 1.
Define the following :
(i) First law of thermodynamics.
(ii) Standard enthalpy of formation.
Answer:
(i) First law of thermodynamics : It states that energy can neither be created nor be destroyed. The energy of an isolated system is constant. ΔU = q + w
(ii) Standard Enthalpy of Formation : It is defined as the amount of heat evolved or absorbed when one mole of the compound is formed from its constituent elements in their standard states.

Question 2.
Give reason for the following:
(i) Neither q nor w is a state function but q + w is a state function.
(ii) A real crystal has more entropy than an ideal crystal.
Answer:
(a) q + w = ΔU
As ΔU is a state function hence, q + w is a state function.
(b) A real crystal has some disorder due to the presence of defects in its structural arrangement whereas ideal crystal does not have any disorder. Hence, a real crystal has more entropy than an ideal crystal.

Question 3.
Represent the potential energy/enthalpy change in the following processes graphically
(i) Throwing a stone from the ground to roof.
(ii) \(\frac{1}{2}\)H₂(g) + \(\frac{1}{2}\)Cl₂(g) ⇌ HCl(g); Δ_rH^s = – 92.32kJ mol^-1
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Answer:
PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics 1

Energy increases in (a) and it decreases in (b). Hence, in process (b), enthalpy change is the contributing factor to the spontaneity.

Question 4.
A man takes a diet equivalent to 10000 kJ per day and does work, by expending his energy in all forms equivalent to 12500 kJ per day. What is change in internal energy per day? If the energy lost was stored as sucrose (1632 kJ per 100 g), how many days should it take to lose 2 kg of his weight? (Ignore water loss)
Answer:
Energy taken by a man = 10000 kJ
Change in internal energy per day = 12500 -10000 = 2500 kJ
The energy is lost by the man as he expends more energy than he takes.
Now 100 g of sugar corresponds to energy = 1632 kJ loss in energy.
2000 g of sugar corresponds to energy = \(\frac{1632 \times 2000}{100}\) = 32640 kJ
∴ Number of days required to lose 2000 g of weight or 32640 kJ of energy = \(\frac{32640}{2500}\) = 13 days

Question 5.
Give the appropriate reason :
(i) It is preferable to determine the change in enthalpy rather than the change in internal energy.
(ii) It is necessary to define the ‘standard state’.
(iii) It is necessary to specify the phases of the reactants and products in a thermochemical equation.
Answer:
(i) Because it is easier to make measurement under constant pressure than under constant volume conditions.
(ii) Enthalpy change depends upon the conditions in which a reaction is carried out. For making the comparison of results obtained by different people meaningful, the reaction conditions must be well-defined.
(iii) Because enthalpy depends upon the phase of reactants and products.

Long Answer Type Questions

Question 1.
(i) A cylinder of gas supplied by a company is assumed to contain 14 kg of butane. If a normal family requires 20000 kJ of energy per day for cooking, how long will the cylinder last?
(ii) If the air supplied to the burner is insufficient, a portion of gas escapes without combustion. Assuming that 25% of the gas is wasted due to this inefficiency, how long will the cylinder last (Heat of combustion of butane = 2658kJ!mol.)?
Answer:
(i) Molecular formula of butane = C₄H₁₀
Molecular mass of butane = 4 x 12 +10 x 1 = 58
Heat of combustion of butane 2658 kJ mol^-1
1 mole.or 58 g of butane on complete combustion gives heat = 2658 kJ
∴ 14 x 10³ g of butane on complete combustion will give heat
= \(\frac{2658 \times 14 \times 10^{3}}{58}\) = 641586 kJ
The family needs 20000 kJ of heat per day.
∴ 20000 kJ of heat is used for cooking by a family in 1 day.
∴ 641586 kJ of heat will be used for cooking by a family in
= \(\frac{641586}{20000}\) = 32days
The cylinder will last for 32 days

(ii) 25 per cent of the gas is wasted due to inefficiency. This means that only 75% of butane gets combusted. Therefore, the energy produced by
75% combustion of butane = \(\frac{641586 \times 75}{100}\) = 481189.5 kJ
∴ The number of days the cylinder will last = \(\frac{481189.5}{20000}\) = 24 days.

Question 2.
10 moles of an ideal gas expand isothermally and reversibly from a pressure of 5 atm to 1 atm at 300 K. What is the largest mass that can be lifted through a height of 1 m by this expansion?
Answer:
W_exp = -2.303 nRT log \(\frac{p_{1}}{p_{2}}\)
= -2.303(10) x (8.314)(300) log \(\frac{5}{1}\) = – 40.15 x 10³ J
If M is the mass that can be lifted by this work through a height of 1 m, then work done = Mgh
40.15 x 10³ J = M x 9.81 ms^-1 x 1 m
or M = \(\frac{40.15 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}}{9.81 \mathrm{~m} \mathrm{~s}^{-2} \times 1 \mathrm{~m}}\) [∵ J = kg m²s^-2]
= 4092.76 kg

PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter

June 21, 2022 by Bhagya

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter

Very Short Answer Type Questions

Question 1.
Name two intermolecular forces that exist between HF molecules in a liquid state.
Answer:
HF are polar covalent molecules. In the liquid state, there are dipole-dipole interactions and H-bonding.

Question 2.
Explain why Boyle’s law cannot be used to calculate the volume of a real gas when it is converted from its initial state to the final state by an adiabatic expansion.
Answer:
During adiabatic expansion, the temperature is lowered, and therefore, Boyle’s law cannot be applied.

Question 3.
Boyle’s law states that at constant temperature, if pressure is increased on a gas, volume decreases and vice-versa. But when we fill air in a balloon, volume as well as pressure increase. Why?
Answer:
The law is applicable only for a definite mass of the gas. As we fill air into the balloon, we are introducing more and more air into the balloon.
Thus, we are increasing the mass of air inside. Hence, the law is not applicable.

Question 4.
What will be the molar volume of nitrogen and argon at 273.15 K and 1 atm?
Answer:
Every gas has 22.4 L molar volume at 273.15 K and 1 atm pressure (STP).

Question 5.
A gas that follows Boyle’s law, Charles’ law and Avogadro’s law is called an ideal gas. Under what conditions a real gas would behave ideally?
Answer:
At low pressure and high temperature, a real gas behaves as an ideal gas.

Question 6.
Explain why temperature of a boiling liquid remains constant?
Answer:
This is because at the boiling point, the heat supplied is used up in breaking off the intermolecular forces of attraction of the liquid to change it into vapour and not for raising the temperature of the liquid.

Question 7.
Assuming C02 to be van der Waals’ gas, calculate its Boyle temperature.
Given a = 3.59 L² atm mol^-2 and b = 0.0427 L mol^-1.
\(T_{b}=\frac{a}{R b}=\frac{3.59 \mathrm{~L}^{2} \mathrm{~atm} \mathrm{} \mathrm{mol}^{-2}}{\left(0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\left(0.0427 \mathrm{~L} \mathrm{~mol}^{-1}\right)}\) = 1025.3 K

Question 8.
Name two phenomena that can be explained on the basis of surface tension.
Answer:
Surface tension can explain
(i) capillary action, i.e., rise or fall of a liquid in capillary,
(ii) spherical shape of small liquid drops.

Question 9.
Why are the gases helium and hydrogen not liquefied at room temperature by applying very high pressure?
Answer:
Because their, critical temperature is lower than room temperature. (Gases cannot be liquefied above the critical temperature by applying even very high pressure).

Question 10.
What would have happened to the gas if the molecular collisions were not elastic?
Answer:
On every collision, there would have been loss of energy. As a result, the molecules would have slowed down and ultimately settle down in the vessel. Moreover, the pressure would have gradually reduced to zero.

Short Answer Type Questions

Question 1.
(i) What do you mean by ‘Surface Tension’ of a liquid?
(ii) Explain the factors which can affect the surface tension of a liquid.
Answer:
(i) Surface tension : It is defined as the force acting per unit length perpendicular to the line drawn on the surface. It’s unit is Nm-1.
(ii) Surface tension of a liquid depends upon the following factors :
(a) Temperature : Surface tension decreases with rise in temperature. As the temperature of the liquid increases, the average kinetic energy of the molecules increases. Thus, there is a decrease in intermolecular force of attraction which decrease the surface tension.
(b) Nature of the liquid : Greater the magnitude of
intermolecular forces of attraction in the liquid, greater will be the value of surface tension.

Question 2.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of gases in the cylinder is 25 bar.
What is the partial pressure of dioxygen and neon in the mixture?
Answer:
PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter 1
Alternatively, mole fraction of neon = 1 – 0.21 = 0.79
Partial pressure of a gas = mole fraction x total pressure
⇒ Partial pressure of oxygen = 0.21 x (25bar) = 5.25bar
Partial pressure of neon = 0.79 x (25bar) = 19.75bar

Question 3.
Give reasons for the following:
(i) The size of weather balloon becomes larger and larger as it ascends into higher altitudes.
(ii) Tyres of automobiles are inflated to lesser pressure in summer than in winter.
Answer:
(i) As we go to higher altitudes, the atmospheric pressure decreases.
Thus, the pressure outside the balloon decreases. To regain equilibrium with the external pressure, the gas inside expands to decrease its pressure, Hence, the size of the balloon increases.
(ii) In summer, due to higher temperature, the average kinetic energy of the air molecules inside the tyre increases, i.e., molecules start moving faster. Hence, the pressure on the walls of the tube increases. If pressure inside is not kept low at the time of inflation, at higher temperature, the pressure may become so high that the tyre may burst.

Question 4.
On the basis of intermolecular forces and thermal energy, explain why .
(i) a solid has rigidity but liquids do not have rigidity?
(ii) gases have high compressibility but liquids and solids have poor compressibility?
Answer:
(i) It is because in solids, the intermolecular forces are very strong and predominate over thermal energy but in liquid, these forces are no longer strong enough.
(ii) Because of very weak intermolecular forces and high thermal energy, molecules of gases are far apart. That is why gases are highly compressible.

Question 5.
A gas is enclosed in room. The temperature, pressure, density and number of moles respectively are t°C,p atm, g cm^-3 and n moles.
(i) What will be the pressure, temperature, density and number of moles in each compartment, if room is partitioned into four equal compartments?
(ii) What will be the value of pressure, temperature, density and number of moles in each compartment if the walls between the two compartments (say 1 and 2) are removed?
(iii) What will be the values of pressure, temperature, density and number of moles, if an equal volume of gas at pressure
(p) and temperature (t) is let inside the same room? .
Answer:
(i) (a) Pressure in each compartment is same, (p atm)
(b) Temperature will remain same (t°C).
(c) Density will remain same (d g cm^-3).
(d) Because of partition, volume of each compartment becomes 1/4 and the number of molecules also become 1/4. The number of moles in each compartment will be n/4.

(ii) (a) Pressure will remain same (p atm).
(b) Temperature will remain same (t°C).
(c) Density will remain same (d g cm^-3).
(d) The number of moles in each compartment will be n/2.

(iii) (a) Pressure will be doubled (2p atm).
(b) Temperature will remain same.
(c) Density will remain same (d g cm^-3) ,
(d) Number of moles will be doubled i.e., 2n.

Long Answer Type Questions

Question 1.
Explain the following:
(i) The boiling point of a liquid rises on increasing pressure. ;
(ii) Drops of liquid assume spherical space.
(iii) The boiling point of water (373 K) is abnormally high when compared to that of H₂S (211.2 K).
(iv) The level of mercury in capillary tube is lower than the s level outside when a capillary tube is inserted in the mercury.
(v) Tea or coffee is sipped from a saucer when it is quite hot.
Answer:
(i) A liquid boils when its vapour pressure becomes equal to the atmospheric pressure. An increase in pressure on liquid, therefore, causes a rise in the boiling temperature of the liquids.
(ii) Liquids have a property, called surface tension, due to which liquids tend to contract (to decrease the surface area). For a given volume of a liquid, since a sphere has the least surface area, hence the liquids tend to form spherical droplet.
(iii) The extensive hydrogen bonding in water gives a polymeric structure. This makes the escape of molecules from the liquid more difficult. Therefore, water requires higher temperature to bring its vapour pressure equal to the atmospheric pressure.
On the other hand, sulphur being less electronegative, does not form hydrogen bonds with H of H2S. As a result, H2S has low boiling point.
(iv) The cohesive forces in mercury are much stronger than the force of adhesion between glass and mercury. Therefore, mercury-glass contract angle is greater than 90°C.
As a result, the vertical component of the surface tension forces acts vertically downward, thereby lowering the level of mercury column in the capillary tube.
(v) Evaporation causes cooling and the rate of evaporation increases with an increase in the surface area. Since, saucer has a large surface area, hence tea/coffee taken in a saucer cools quickly.

Question 2.
Nitrogen molecule (N₂) has radius of about 0.2 nm. Assuming that nitrogen molecule is spherical in shape, calculate
(i) volume of a single molecule of N₂.
(ii) the percentage of empty space in one mole of N₂ gas at STP.
Answer:
(i) The volume of a sphere = \(\frac{4}{3}\)πr³ nr where Volume of a molecule of N₂
= \(\frac{4}{3} \times \frac{22}{7}\) x (2 x 10^-8)³ cm³ – 3.35 x 10^-23 cm³

(ii) To calculate the empty space, let us first find the total volume of 1 mole (6.022 x 10²³ molecules) of N₂.
Volume of 6.022 x 10²³ molecules of N₂
= 3.35 x 10^-23 x 6.022 x 10²³ = 20.17 cm³
Now, volume occupied by 1 mole of gas at STP
= 22.4 litre = 22400 cm³
Empty volume = Total volume of gas – Volume occupied by molecules
= (22400 – 20.17) cm³ – 22379.83 cm³
∴ Percentage of empty space = \(\frac{Empty space}{Total volume}\) x 100
= \(\frac{22379.83}{22400}\) x 100 = 99.9%
Thus, 99.9% of space of 1 mole of N₂ at STP is empty.

PSEB 11th Class Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure

June 21, 2022 by Bhagya

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure

Very Short Answer Type Questions

Question 1.
In \(\mathrm{PO}_{4}^{3-}\) ion formal charge on the oxygen atom of P—O bond is
Answer:
In \(\mathrm{PO}_{4}^{3-}\) ion, formal charge on each O-atom of P—O bond
= \(\frac{\text { total charge }}{\text { Number of O-atoms }}=-\frac{3}{4}\) = -0.75

Question 2.
Which of the following molecules show super octet?
CO₂, CIF₃, SO₂, IF₅
Answer:
ClF₃ and IF₅ are super octet molecules.

Question 3.
Which of the following has highest lattice energy and why?
CsF, CsCl, CsBr, Csl
Answer:
CsF has highest lattice energy because ‘F’ is smallest in size and is more electronegative, therefore, it has maximum ionic character and maximum force of attraction, hence, highest lattice energy.

Question 4.
Account for the following:
The experimentally determined N—F bond length in NF₃ is greater than the sum of the single covalent radii of N and F.
Answer:
This is because both N and F are small and hence, have high electron density. So, they repel the bond pairs thereby making the N—F bond length larger.

Question 5.
What is valence bond approach for the formation of covalent * bond?
Answer:
A covalent bond is formed by the overlapping of half-filled atomic orbitals.

Question 6.
Why axial bonds of PCI5 are longer than equatorial bonds?
Answer:
This is due to greater repulsion on the axial bond pairs by the equatorial bond pairs of electrons.

Question 7.
Which type of atomic orbitals can overlap to form molecular orbitals?
Answer:
Atomic orbitals with comparable energies and proper orientation can overlap to form molecular orbitals.

Question 8.
Why KHF₂ exists but KHCl₂ does not?
Answer:
Due to H-bonding in HF, we have

This can dissociate to give \(\mathrm{HF}_{2}^{-}\) ion and hence, KHF₂ exists but there is no H-bonding in H-Cl. So, \(\mathrm{HCl}_{2}^{-}\) ion does not exist and hence, KHCl₂ also does not exist.

Question 9.
How many nodal planes are present in n(2px) and n(2px) molecular orbitals?
Answer:
One and two respectively.

Question 10.
What is the magnetic character of the anion of K02?
Answer:
Anion of KO₂ is \(\mathrm{O}_{2}^{-}\) (superoxide ion) which has one unpaired electron and hence is paramagnetic.

Short Answer Type Questions

Question 1.
Describe the change in hybridisation (if any) of the Al-atom in
the following reaction:
AlCl₃ + Cl^– → \(\mathrm{AlCl}_{\mathbf{4}}^{-}\)
Answer:
Electronic configuration of Al in ground state,
₁₃Al = 1s², 2s²,2p⁶,3s²,3p¹_x
In excited state = 1s², 2s², 2p⁶, 3s¹, 3p¹_x, 3p¹_y
In the formation of AlCl₃, Al undergoes sp² hybridisation and it is trigonal
planar in shape. While in the formation of AlCl₄^–, Al undergoes sp3
hybridisation.
It means empty 3P_z orbital also involved in hybridisation.
Thus, the shape of AlCl₄^– ion is tetrahedral.

Question 2.
Arrange the following in order of decreasing bond angle, with appropriate reason
\(\mathrm{NO}_{2}, \mathrm{NO}_{2}^{+}, \mathrm{NO}_{2}^{-}\)
Answer:
\(\mathrm{NO}_{2}, \mathrm{NO}_{2}^{+}, \mathrm{NO}_{2}^{-}\). This is because \(\mathrm{NO}_{2}^{+}\) has no lone pair of electrons
(i.e., has only bond pairs on two sides) and hence it is linear.

NO₂ has one unshared electron while \(\mathrm{NO}_{2}^{-}\) has one unshared electron pair. There are greater repulsion on N—O bonds in case of \(\mathrm{NO}_{2}^{-}\) than in case of NO₂

PSEB 11th Class Chemistry Solutions Chapter 4 Chemical Bonding and Molecular Structure 2

Question 3.
Among the molecules, \(\mathbf{O}_{2}^{-}, \mathbf{N}_{2}^{+}, \mathrm{CN}^{-}\) and \(\mathbf{O}_{2}^{+}\) identify the species which is isoelectronic with CO.
Answer:
Isoelectronics species are those species which have the same number of electrons. CO in total has 14 electrons (6 from carbon and 8 from oxygen). Out of the given ions CN is the ion which has 14 electrons (6 from carbon 7 from Nitrogen and 1 from the negative charge). Thus CN^– ion is isoelectronic with CO.

Question 4.
Which is more polar : COa or N20? Give reason.
Answer:
N₂Ois more polar than CO₂. This is because CO₂ is linear and symmetrical.
Its net dipole moment is zero im
on the other hand, is linear but unsymmetrical. It is considered as a resonance hybrid of the following two structures

It.has a net dipole moment of 0.116 D.

Question 5.
Aluminium forms the ion Al3+, but not Al4+ why?
Answer:
Aluminium [Ne]3s² 3p¹ can achieve the electronic configuration of the nearest noble gas (Ne) by losing only three electrons. : Al³⁺ = 1s² 2s² 2p⁶.
Aluminium will not form Al⁴⁺ ion because an extremely high amount of energy would be required to remove an electron from the stable noble gas configuration.

Long Answer Type Questions

Question 1.
On the basis of VSEPR theory, predict the shapes of the following
(i) \(\mathbf{N H}_{2}^{-}\) (ii) O₃
Answer:
(i) Shape of \(\mathbf{N H}_{2}^{-}\)
Number of valence electrons on central N atom = 5 + 1 (due to one unit negative charge) = 6
Number of atoms linked to it = 2
∴ Total number of electron pairs around N
= \(\frac{6+2}{2}\) = 4 and number of bond pairs = 2
∴ Number of lone pairs = 4 —2 = 2. Thus, the ion is of the type AB₂E₂
Hence, it has bent shape (V-shape).

(ii) Shape of O₃
While predicting geometry of molecules containing the double (or multiple) bond is considered as one electron pair, e.g., in case of ozone, its two resonating structures are

Thus, the central O-atom is considered to have two bond pairs and one lone pair, i.e., it is of the type AB₂E. Hence, it is a bent molecule. Thus, the two resonating structures will be

PSEB 11th Class Chemistry Solutions Chapter 4 Chemical Bonding and Molecular Structure 5

Question 2.
In each of the following pairs of compounds, which one is more covalent and why?
(i) AgCl, Agl
(ii) BeCl₂,MgCl₂
(iii) SnCl₂, SnCl₄
(iv) CuO, CuS
Answer:
Applying Fajans’ rules, the result can be obtained in each case as follows :
(i) Agl is more covalent than AgCl. This is because I^– ion is larger in size than Cl^– ion and hence is more polarised than Cl^– ion.
(ii) BeCl₂ is more covalent thanMgCl₂. This is because Be²⁺ ion is smaller in size than Mg₂ ion and hence has the greater polarising power.
(iii) SnCl₄ is more covalent than SnCl₂. This is because Sn⁴⁺ ion has greater charge and smaller size than Sn²⁺ ion and hence has greater polarising power.
(iv) CuS is more covalent than CuO. This is because S^2- ion has larger size
than O^2- ion and hence is more polarised than O^2-ion.

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties

June 21, 2022 by Bhagya

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

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties

Very Short Answer Type Questions

Question 1.
Which properties of the elements depend on’ the electronic configuration of the atoms and which do not?
Answer:
Chemical and many physical properties of the elements depends on the electronic configuration of the atoms, whereas the nuclear properties do not.

Question 2.
Write the number designation of a group that has 2 electrons beyond a noble gas configuration.
Answer:
The number designation of a group that has 2 electrons beyond a noble gas configuration will be 2 which means it will belong to group 2 of periodic table.

Question 3.
Why is it more logical to call the atomic radius as the effective atomic radius?
Answer:
This is because the size of atom is very small and it has no sharp boundaries.

Question 4.
A boy has reported the radii of Cu, Cu⁺ and Cu²⁺ as 0.096 nm, 0.122 nm and Question072 nm respectively. However, it has been noticed that he interchanged the values by mistake. Assign the correct values to different species.
Answer:
Cu [0.122 nm], Cu⁺ [0.096 nm], Cu²⁺ [0.072 nm].
∵ Size ∝ \(\frac{1}{\text { positive charge }}\)

Question 5.
Atomic radii of fluorine is 72 pm where as atomic radii of neon is 160 pm. Why? [NCERT Exemplar]
Answer:
Atomic radius of F is expressed in terms of covalent radius while, atomic radius of neon is usually expressed in terms of van der Waals’ radius, van der Waals’ radius of an element is always larger than its covalent radius.
Therefore, atomic radius of F is smaller than atomic radius of Ne (F = 72 pm, Ne = 160 pm)

Question 6.
Arrange the following elements in order of decreasing electron gain enthalpy : B, C, N, O.
Answer:
N has positive electron gain enthalpy while all others have negative electron gain enthalpies. Since size decreases on moving from B → C → O, therefore, electron gain enthalpies become more and more negative from B → C → O. Thus, the overall decreasing order of electron gain enthalpies is N, B, C, O.

Question 7.
Which of the following atoms would most likely form an anion (i) Be, (ii) Al, (iii) Ga, (iv) I ?
Answer:
I, because of high electron gain enthalpy, it can accept an electron readily to form an anion F < Cl < Br > I.

Question 8.
Explain why chlorine can be converted into chloride ion more easily as compared to fluoride ion from fluorine.
Answer:
Electron gain enthalpy of Cl is more negative than that of F.

Question 9.
Among alkali metals which element do you expect to be least electronegative and why? [NCERT Exemplar]
Answer:
On moving down the group, electronegativity decreases because atomic size increases. Fr has the largest size, therefore it is least electronegative.

Question 10.
Arrange the following elements in the increasing order of non-metallic character. B, C, Si, N, F
Answer:
The given non-metals are arranged in the increasing order of non-metallic character as follows:

Short Answer Type Questions

Question 1.
What would be IUPAC names and symbols for elements with atomic numbers 122, 127, 135, 149 and 150? .
Answer:
The roots 2, 7, 5, 9 and 0 are referred as bi, sept, pent, enn and nil respectively. Therefore, their names and symbol are

Z (Atomic number)	Name	Symbol
122	Unbibium	Ubb
127	Unbiseptium	Ubs
135	Untripentium	Dtp
149	Unquadennium	Uqe
150	Unpentnilium	Upn

Question2.
All transition elements are d-block elements, but all d-block elements are not transition elements. Explain.
Answer:
Elements in which the last electron enters in the d-orbitals, are called d-block elements or transition elements. These elements have the general outer electronic configuration (n – 1)d^1-10ns^0-2 Zn, Cd and Hg having the electronic configuration, (n – l1)d¹⁰ns² do not show most of the properties of transition elements. The d-orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. Thus, on the basis of properties, all transition elements are d-block elements but on the basis of electronic configuration, all d-block elements are not transition elements.

Question 3.
Arrange the elements N, P, O and S in the order of
(i) increasing first ionisation enthalpy.
(ii) increasing non-metallic character.
Give reason for the arrangement assigned.
Answer:

	Group 15	Group 16
2nd period	N	0
3rd period	P	S

(i) Ionisation enthalpy of nitrogen (₇N = 1s², 2s², 2p³) is greater than oxygen (₈O = 1s² , 2s² , 2p⁴ ) due to extra stable half-filled 2p-orbitals. Similarly, ionisation enthalpy of phosphorus (₁₅P = 1s², 2s², 2p⁶, 3s², 3p³) is greater than sulphur (₁₆S = 1s², 2x², 2p⁶, 3s², 3p⁴).
On moving down the group, ionisation enthalpy decreases with increasing atomic size. So, the increasing order of first ionisation enthalpy is S < P < O < N.

(ii) Non-metallic character across a period (left to right) increases but on moving down the group it decreases. So, the increasing order of non-metallic character is P < S < N < 0.

Question 4.
What do you understand by exothermic reaction and endothermic reaction? Give one example of each type.
Answer:
Exothermic reactions : Reactions which are accompanied by evolution of heat are called exothermic reactions. The quantity of heat produced is shown either along with the products with a ‘+’ sign or in terms if ΔH with a sign, e.g.,

C(s) + O₂(g) → CO₂(g) + 393.5 kJ
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l) ΔH = -285.8 kJ mol^-1

Endothermic reactions : Reactions which proceed with absorption of heat are called endothermic reactions. The quantity of heat absorbed is shown either alongwith the products with a sign or in terms of ΔH with a ‘-‘ sign, e.g.,

C(s) + H₂O(g) → CO(g) + H₂(g) -131.4 kJ
N₂(g) + 3H₂(g) → 2NH₃(g); ΔH = +92.4 kJ mol^-1

Question5.
How does the metallic and non-metallic character vary on moving from left to right in a period?
Answer:
As we move from left to right in a period, the number of valence electrons increases by one at each succeeding element but the number of shells remains same. Due to this, effective nuclear charge increases.

More is the effective nuclear charge, more is the attraction between nuclei and electron.
Hence, the tendency of the element to lose electrons decreases, this results in decrease in metallic character.
Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non-metallic character increases on moving from left to right in a period.

Long Answer Type Questions

Question 1.
Write the drawbacks in Mendeleev’s Periodic Table that led to its modification.
Answer:
The main drawbacks of Mendeleev’s Periodic Table are:
(i) Some elements having similar properties were placed in different groups whereas some elements having dissimilar properties were placed in the same group. For example alkali metals such as Li, Na, K, etc., (I A group) are grouped together with coinage metals such as Cu, Ag, Au (I B group) though their properties are quite different. Chemically similar elements such as Cu (I B group) and Hg (II B group) have been placed in different groups.

(ii) Some elements with higher atomic weights are placed before the elements with lower atomic weights in order to maintain the similar chemical nature of elements. For example,

(iii) Isotopes did not find any place in the Periodic Table. However, according to Mendeleev’s classification, these should be placed at different places in the Periodic Table.
(All the above three defects were however removed when modern periodic law based on atomic number was given.)

(iv) Position of hydrogen in the Periodic Table is not fixed but is
controversial. ,
(v) Position of elements of group VIII could not be made clear which have been arranged in three triads without any justification.
(vi) It could not explain the even and odd series in IV, V and VI long periods.
(vii) Lanthanides and actinides which were discovered later on, have not been given proper positions in the main frame of Periodic Table.

Question 2.
p-block elements form acidic, basic and amphoteric oxides. Explain each property by giving two examples and also write the reactions of these oxides with water.
Answer:
In p-block, when we move from left to right in a period, the acidic character of the oxides increases due to increase in electronegativity, e.g.,

(i) 2nd period
B₂O₃ < CO₂ < N₂O₃ acidic character increases.

(ii) 3rd period
Al₂O₃ < SiO₂ < P₄O₁₀ < SO₃ < Cl₂O₇ acidic character increases.
On moving down the group, acidic character decreases and basic character increases, e.g.,

(a) Nature of oxides of 13 group elements

(b) Nature of oxides of 15 group elements

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties 4

Among the oxides of same element, higher the oxidation state of the element, stronger is the acid. e.g., SO₃ is a stronger acid than SO₂. B₂O₃ is weakly acidic and on dissolution in water, it forms orthoboric acid. Orthoboric acid does not act as a protonic acid (it does not ionise) but acts as a weak Lewis acid.

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties 5

Al₂O₃ is amphoteric in nature. It is insoluble in water but dissolves in alkalies and react with acids.

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties 6

Tl₂O is as basic as NaOH due to its lower oxidation state (+1).
Tl₂O + 2HCl → 2TlCl + H₂O

P₄O₁₀ on reaction with water gives orthophosporic acid.

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties 7

Cl₂O₇ is strongly acidic in nature and on dissolution in water, it gives perchloric acid.

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties 8