NCERT Exemplar Problems Class 12 Chemistry The p-Block Elements

NCERT Exemplar Problems Class 12 Chemistry Chapter 7 The p-Block Elements

Multiple Choice Questions

Single Correct Answer Type
Question 1. On addition of cone. H₂SO₄ to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because

Solution: (c) HI formed during reaction is oxidized to I₂ which is violet in colour.

Question 2. In qualitative analysis when H₂S is passed through an aqueous solution of
salt acidified with dil. HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO3, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives

Solution: (b) Black precipitate of copper sulphide is formed which gives blue colour of copper nitrate on boiling with dilute HNO₃. When aqueous solution of ammonia is added to it, deep blue colour of

Question 3. In a cyclotrimeta phosphoric acid molecule, how many single and double bonds are present? .
(a) 3 double bonds; 9 single bonds
(b) 6 double bonds; 6 single bonds
(c) 3 double bonds; 12 single bonds
(d) Zero double bonds; 12 single bonds
Solution: (c) Cyclotrimetaphosphoric acid contains three double bonds and 9 single bonds as shown below

a, b, and c are three pi-bonds and numerics 1 to 12 are sigma bonds.

Question 4. Which of the following element can be involved in pπ-dπ bonding ?
(a) Carbon (b) Nitrogen (c) Phosphorus (d) Boron
Solution: (c) Phosphorus can be involved in pπ-dπ bonding due to presence of vacant d orbitals C, N, and B do not have o’ orbitals.

Question 5. Which of the following pairs of ions are isoelectronic and isostructural?

Solution: (a) CO^2- and NO₃ are isoelectronic with 32 electrons and sp2 hybridisation
hence trigonal planar structure.

Question 6. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy?
(a) HF (b) HCl (c) HBr (d) HI
Solution: (a)
HF On moving top to bottom HCl Size of halogen atom increases HBr H-X bond length increases HI Bond dissociation enthalpy decreases

Question 7. Bond dissociation enthalpy of E – H (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent?

Solution: (d) SbH₃ will act as strongest reducing agent due to minimum bond enthalpy.

Question 8. On heating with concentrated NaOH solution in an inert atmosphere of CO2, white phosphorus gives a gas. Which of the following statement is incorrect about the gas?
(a) It is highly poisonous and has smell like rotten fish.
(b) Its solution in water decomposes in the presence of light.
(c) It is more basic than NH₃  
(d) It is less basic than NH₃
Solution:

Question 9. Which of the following acid forms three series of salts?
(a) H₂PO₂ (b) H₃BO₃ (C)H3PO₄(d)H₃PO₃
Solution:

H3PO₄ has 3 -OH groups i.e., has three ionisable H-atoms and hence forms three series of salts. These three possible series of salts of H3PO₄ are as follows: .
NaH2PO₄, NaHPO₄ and Na₃PO₄

Question 10. Strong reducing behaviour of H₃PO₂ is due to
(a) low oxidation state of phosphorus
(b) presence of two -OH groups and one P – H bond
(c) presence of one -OH group and two P – H bonds
(d) high electron gain enthalpy of phosphorus
Solution: (c) H₃PO₂  is a monobasic acid. It acts as a reducing agent as it has two P – H bonds.

Question 11. On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed
are
(a)NO₂ , PbO (b) NO₂, PbO (c) NO, PbO (d) NO, PbO₂
Solution: (b) 2Pb(NO₃)₂ —> 2PbO + 4NO₂ + O₂

Question 12. Which of the following element does not show allotropy?
(a) Nitrogen (b) Bismuth (c) Antimony (d) Arsenic
Solution: (a) Nitrogen does not show allotropy due to its weak N – N single bond.
Therefore, ability of nitrogen to form polymeric structure or more than one structure become less. Hence, nitrogen does not show allotropy.

Question 13. Maximum covalency of nitrogen is
(a) 3 (b) 5 (c) 4 (d) 6
Solution: (c) Maximum covalency of nitrogen is four as it cannot extend its valency beyond four [ NH₄, R₄N] due to absence of vacant d orbitals.

Question 14. Which of the following statement is wrong?
(a) Single N – N bond is stronger than the single P – P bond.
(b) PH₃ can act as a ligand in the formation of coordination compound with transition elements.
(c) NO₂ is paramagnetic in nature.
(d) Covalency of nitrogen in N₂O₂ is four.
Solution: (a) Nitrogen forms pπ-pπ multiple bond and the bond strength is very high. Single N – N bond is weaker than single P – P bond.

Question 15. A brown ring is formed in the ring test for NO₃ ion. It is due to the formation of

Solution: (a) When freshly prepared solution of FeSO₄ is added in a solution containing NO₃ ion, it leads to formation of a brown coloured complex. This is known as brown ring test of nitrate.

Hence, 2 moles of ammonia will produce 2 moles of NO.

Question 16. Elements of group 15 form compounds in +5 oxidation state. However, bismuth forms only one well characterized compound in +5 oxidation state. The compound is
(a) Bi₂O₅ (b) BiF₅ (c) BiCl₅ (d)Bi₂S₅
Solution: (b) Bismuth commonly shows +3 oxidation state instead of +5 oxidation state due to inert pair effect. But Bi forms BiF₅ compound due to high electronegativity and small size of fluorine atom.

Question 17.On heating ammonium dichromate and barium azide separately we get
(a) N₂ in both cases
(b) N₂ with ammonium dichromate and NO with barium azide
(c) N₂O with ammonium dichromate and N₂ with barium azide
(d) N₂O with ammonium dichromate and N₂O with barium azide
Solution: (a)

Question 18. In the preparation of HNO₃, we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of NH₃ will be (a) 2 (b) 3 (c) 4 (d) 6
Solution: (a) Two moles of NH₃ will produce 2 moles of NO on catalytic oxidation of ammonia in preparation of nitric acid

Question 19. The oxidation state of central atom in the anion of compound NaH₂PO₂ will be
(a) +3 (b) +5 (c) +1 (d) -3
Solution:

Question 20. Which of the following is not tetrahedral in shape?
(a) NH₄ (b) SiCl₄ (c) SF₄ (d) S0₂
Solution: (c) SF₄ trigonal bipyramidal structure.

Question 21. Which of the following pairs are’peroxoacids of sulphur?
(a) H₂S0₄ and H₂S₂O₈ (b) H₂SO₅ and H₂S₂O₇
(c) H₂S₂O₇ and H₂S₂O₈ (d) H₂S₂O₆ and H₂S₂O₇
Solution: (a) Peroxoacids of sulphur must contain one -O – O – bond as shown below

Question 22. Hot cone. H₂SO₄ acts as moderately strong oxidizing agent. It oxidises both metals and non-metals. Which of the following element is oxidised by cone. H₂SO₄ into two gaseous products?
(a) Cu (b) S (c) C (d) Zn
Solution:

Question 23. A black compound of manganese reacts with a halogen acid to give greenish
yellow gas. When excess of this gas reacts with NH₃ an unstable trihalide is formed. In this process, the oxidation state of nitrogen changes from
(a) -3 to +3 (b) -3 to 0 (c) -3 to +5 (d) 0 to -3
Solution: (a) Mn0₂ + 4HCl -» MnCl₂ + 2H₂O + Cl
(greenish yellow gas)
When excess of Cl2 reacts with NH₃ the products are NCl₃ and HCl.
NH₃ + 3Cl₂ —> NCl₃ +3HCl
O.S. (-3) O.S.(+3)

Question 24. In the preparation of compounds of Xe, Bartlett had taken O₂⁺Pt F₆^– as a base compound. This is because
(a) both O₂ and Xe have same size
(b) both O₂ and Xe have same electron gain enthalpy
(c) both O₂ and Xe have almost same ionization enthalpy
(d) both Xe and O₂ are gases.
Solution: (c) Bartlett had taken 02 Pt as a base compound because 02 and Xe both have almost same ionization enthalpy. The ionization enthalpies of noble gases are the highest in their respective periods due to their stable electronic configurations.

Question 25. In solid state PCl₅ is a
(a) covalent solid
(b) octahedral structure
(c) ionic solid with [PCl₆]⁺ octahedral and [PCl₄]^– tetrahedral
(d) ionic solid with [PCl₄]⁺ tetrahedral and [PCl₆]^– octahedral
Solution: (d) Structure of PCl₅ in solid state

Question 26. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidizing power.

Solution: (c) The higher the reduction potential, the higher is its tendency to get reduced. Hence, the order of oxidizing power is
BrO₄^– > IO₄^– > CIO₄^–

Question 27. Which of the following is isoelectronic pair?
(a) ICl₂, ClO₂ (b) BrO₂^–,BrF₂⁺ (c) ClO₂, BrF (d) CN^–, O₃
Solution: (b) Isoelectronic pair have same number of electrons

More than One Correct Answer Type

Question 28. If chlorine gas is passed through hot NaOH solution, two changes are observed in the oxidation number of chlorine during the reaction. These are
—— and ——-
Solution: (a) 0 to +5

Question 29. Which of the following options are not in accordance with the properly mentioned against them?

Solution: (b, c) Bond dissociation enthalpy order is Cl₂ > Br₂ > F₂ > I₂
Ionic character of metal halides increases as electronegativity of halogen increases
MI < MBr < MCI < MF

Question 30. Which of the following is correct for P₄ molecule of white phosphorus?
(a) It has 6 lone pairs of electrons (b) It has six P – P single bonds
(c) It has three P – P single bonds (d) It has four lone pairs of electrons,
Solution: (b, d) Structure of P₄molecule can be represented as

It has total four lone pairs of electrons situated at each P-atom.
It has six P – P single bond.

Question 31. Which of the following statements are correct?
(a) Among halogens, radius ratio between iodine and fluorine is maximum.
(b) Leaving F – F bond, all halogens have weaker X – X bond than X – X’ bond in interhalogens.
(c) Among interhalogen compounds maximum number of atoms ate present in iodine fluoride.
(d) Interhalogen compounds are more reactive than halogen compounds.
Solution: (a, c, d) Among halogens radius ratio between iodine and fluorine is maximum
because iodine has maximum radius while fluorine has minimum radius. Also, due to highest radius ratio maximum number of atoms are present in iodine fluoride.
Interhalogen compounds are more reactive than halogen compounds because A – B bond of dissimilar halogen is weaker than A – A or B – B bond of halogens.

Question 32. Which of the following statements are correct for SO₂ gas?
(a) It acts as a bleaching agent in moist conditions.
(b) Its molecule has a linear geometry.
(c) Its dilute solution is used as disinfectant.
(d) It can be prepared by the reaction of dilute H₂SO₄ with metal sulphide.
Solution: (a, c) SO₂ acts as a bleaching agent in presence of moisture. This is due to reducing nature of SO₂
SO₂ + 2H₂O ->• H₂SO₄ + 2H
Coloured matter + H –> colourless matter. Its dilute solution is used as disinfectant.

Question 33. Which of the following statements are correct?
(a) All three N – O bond lengths in HNO₃ are equal.
(b) All P – Cl bond lengths in PCl₅ molecule in gaseous state are equal.
(c) P₄ molecule in white phosphorus have angular strain therefore white phosphorus is very reactive.
(d) PCl₅ is ionic in solid state in which cation is tetrahedral and anion is octahedral.
Solution: (c, d)
(a) All the three N – O bond lengths in HNO₃ are not equal.
(b) All P – Cl bond lengths in PCl₅ molecule in gaseous state are not equal. Axial bond is longer than equatorial bond.
(c) P₄ molecule in white phosphorus have angular strain therefore white phosphorus is very reactive.
(d) PCl₅ is ionic in solid state in which cation is tetrahedral and anion is octahedral.

Question 34. Which of the following orders are correct as per the properties mentioned against each?

Solution: (a, d) Acidic strength of oxides in a group decreases down the group and increases along a period from left to right.
Thermal stability of hydrides of group 16 decreases down the group.

Question 35. Which of the following statements are correct?
(a) S – S bond is present in H₂S₂O₆.
(b) In peroxosulphuric acid (H₂SO₅) sulphur is in +6 oxidation state.
(c) Iron powder along with Al₂O₃ and K₂O is used as a catalyst in the preparation of NH₃ by Haber’s process
(d) Change in enthalpy is positive for the preparation of SO₃ by catalytic
oxidation of SO₂.
Solution: (a, b)

Question 36. In which of the following reactions cone. H₂S0₄ is used as an oxidizing reagent?

Solution: (b, c) In the above given four reactions, (b) and (c) represent oxidizing behaviour of H₂S0₄. As we know that oxidizing agent reduces itself as oxidation state of central atom decreases.
Here,

Question 37. Which of the following statements are true?
(a) Only type of interactions between particles of noble gases are due to weak dispersion forces.
(b) Ionisation enthalpy of.molecular oxygen is very close to that of xenon.
(c) Hydrolysis of XeF₆ is a redox reaction.
(d) Xenon fluorides are not reactive.
Solution: (a, b) Weak dispersion forces are present between particles of noble gases. Ionization enthalpy of molecular oxygen is very close to that of xenon.

Short Answer Type Questions

Question 38. In the preparation of H₂S0₆ by contact process, why is S0₃ not absorbed directly in water to form H₂S0₄?
Solution: The S0₃ is absorbed directly in H₂S0₄, acid fog is formed which is difficult to condense.

Question 39. Write a balanced chemical equation, for the reaction showing catalytic oxidation of NH₃ by atmospheric oxygen.
Solution:

Question 40. Write the structure of pyrophosphoric acid.
Solution: H₄P₂O₇ is pyrophosphoric acid. Its structure is

Question 41. PH₃ forms bubbles when passed slowly in water but NH₃ dissolves. Explain.
Solution: NH₃ forms hydrogen bonds with water and therefore soluble in water but
PH₃ cannot form hydrogen bonds with water and, therefore, is not soluble in water. It escapes as gas.

Question 42. In PCl₅, phosphorus is in sp₃d hybridised state but all its five bonds are not equivalent. Justify your answer with reason.
Solution: In PCl₅, phosphorus undergoes sp₃d hybridisation and a trigonal bipyramidal configuration comes into existence.

The equatorial P–Cl bonds are equivalent while two axial bonds are different and larger than equatorial bonds.

Question 43. Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling is diamagnetic?
Solution: Nitric oxide (NO) in the gaseous state is monomer having one unpaired electron as it is paramagnetic in nature.

Question 44. Give reason to explain why ClF₃ exists but FCl₃ does not exist.
Solution: ClF₃ is known because chlorine can exhibit +3 oxidation state due to promotion of

Each of the three singly occupied orbitals overlap with p-orbital of each of three fluorine atoms to form ClF₃ (trigonal bipyramidal geometry, T-shaped molecule).
No (i-orbitals are present in the valency shell of fluorine and thus excitation is not possible. Fluorine, therefore, does not exhibit positive oxidation state and so the formation of FCl₃ is not possible.

Question 45. Out of H₂O and H₂S, which one has higher bond angle and why?
Solution: Oxygen is more electronegative than sulphur therefore bond pair of electrons of O – H bond will be closer to oxygen. As a result, there will be more bond pair-bond pair repulsion between bond pairs of two O – H bonds.

Question 46. SF₆ is known but SCl₆ is not. Why?
Solution: Due to small size, six fluorine atoms can be accommodated around sulphur atom while chlorine atoms being larger in size are difficult to accommodate.
The other reason is that the fluorine being highly electronegative and oxidising in nature is capable of unpairing the paired orbitals of the values shell of sulphur atom and thereby showing the highest, oxidation state of +6 while chlorine is not able to do this.

Question 47. On reaction with Cl₂, phosphorus forms two types of halides ‘A’ and ‘B’. Halide A is yellowish-white powder but halide ‘B’ is colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products.
Solution: When dry chlorine is passed over white phosphorus heated gently in a retort, first phosphorus trichloride is formed. PCl₃ further reacts with Cl₂ and forms phosphorus pentachloride PCl₅.

Question 48. In the ring test of NO₃ ion.Fe²⁺ion reduces nitrate ion to nitric oxide, which
combines with Fe²⁺ (aq.) ion to form brown complex. Write the reactions involved in the formation of brown ring.
Solution:

Question 49. Explain why does the stability of oxoacids of chlorine increase in the order given below:
HClO < HClO₂ < HClO₃ < HClO₄
Solution: Oxygen is more electronegative than chlorine, therefore dispersal of negative charge present on chlorine increases from CIO to ClO₄ ion because number of oxygen atoms attached to chlorine increases. Therefore, stability of ions will increase in the order given below:
ClO<ClO₂<ClO₃<ClO₅
Thus due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order HClO < HClO₂ < HClO₅ < HClO₄

Question 50. Explain why ozone is thermodynamically less stable than oxygen.
Solution: Ozone is thermodynamically unstable with respect to oxygen because it

Question 51. P₄O₆ reacts with water according to equation P₄O₆ + 6H₂O
Calculate the volume of 0.1 M NaOH solution required to neutralize the acid formed by dissolving 1.1 g of P₄O₆ in H₂O.
Solution:

Question 52. White phosphorus reacts with chlorine and the product hydrolysis in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water.
Solution:

Question 53. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
Solution: The three oxyacids of nitrogen are:

Question 54. Nitric acid forms an oxide of nitrogen on reaction with P₄O₁₀. Write the reaction involved. Also write the resonating structures of the oxide of nitrogen formed.
Solution:

Reactivity: White phosphorus is very reactive in comparison to red phosphorus. This is due to angular strain in white phosphorus on account of bond angles (60°).

Question 55. Phosphorus has three allotropic fonns —(i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus on the basis of their structure and reactivity.
Solution: Yellow or white phosphorus: It is a discrete tetrahedral molecule. The molecular formula is P4. The four phosphorus atoms lie at the comers of a regular tetrahedron. Each P atom is linked to each of the three other atoms by single covalent bonds i.e., six P – P bonds are present and each P atom has a lone pair.

Question 56. Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.
Solution: Dilute and concentrated nitric acid give different oxidation products on reaction with metals. For example, zinc with very dilute HNO₃ (6%) forms NH₄NO₃, with dilute HNO₃(20%) forms N₂O and with cone. HNO₃ (70%) forms NO₂.

Question 57. PCl₅ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH₃ solution. Write the reactions involved to explain what happens.
Solution: PCl₅ reacts with silver to form white silver salt (AgCl). This dissolves in aqueous ammonia to form soluble complex.

Question 58. Phosphorus forms a number of oxoacids. Out of these oxoacids phosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.
Solution: Phosphinic acid is hypophosphorus acid (H₃PO₂) which acts as a reducing agent due to possession of P – H bonds.

Matching Column Type Questions

Question 59. Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option.

Solution: (a) (A -> 1), (B -> 3), (C -> 4), (D -» 2)

Question 60. Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.

Solution: (b) (A —> 4), (B->1), (C-> 2), (D ->3)

Question 61. Match the items of Column I and Column II and mark the correct option.


Solution: (a) (A -> 4), (B -» 3), (C -> 1), (D -» 2)
(a) H₂SO₄ is used in storage batteries
(b) CCl₃NO₂ is known as tear gas.
(c) Cl₂ has highest electron gain enthalpy.
(d) Sulphur is a member of chalcogen i.e., one producing elements.

Question 62. Match the species given in Column I with the shape given in Column II and mark the correct option.

Solution: (b) (A ->3), (B -> 4), (C —> 2), (D —> 1)

Question 63. Match the items of Column I and Column II and mark the correct option.

Solution:

Assertion and Reason Type Questions

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the following choices:
(a) Both Assertion and Reason are correct statements and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct statements but Reason is not the correct explanation of Assertion.
(c) Assertion is correct but Beason is wrong.
(d) Assertion is wrong but Reason is correct.
(e) Both Assertion and Reason are wrong.

Question 64. Assertion (A): N₂ is less reactive than P₄.
Reason (R): Nitrogen has more electron gain enthalpy than phosphorus.
Solution: (c) N₂ is less reactive due to high bond dissociation energy. Its electron gain enthalpy is less than phosphorus.

Question 65. Assertion (A): HNO₃ makes from passive.
Reason (R): HNO₃ forms a protective layer of ferric nitrate on the surface of iron.
Solution: (c) Passivity is attained by formation of a thin film of oxide on iron.

Question 66. Assertion (A): HI cannot be prepared by the reaction of KI with concentrated
H₂SO₄.
Reason (R): HI has lowest H – X bond strength among halogen acids.
Solution: (b) Both statements are correct but are independent of each other.

Question 67. Assertion (A): Both rhombic and monoclinic sulphur exist as S₈ but oxygen exists as Oπ.
Reason (R): Oxygen forms pπ-pπ multiple bond due to small size and small bond length but pπ-pπ bonding is not possible in sulphur.
Solution: (a) S exists as S₈ but oxygen exists as O₂ because oxygen forms pπ-pπ multiple bonds which is not possible in S.

Question 68. Assertion (A): NaCl reacts with concentrated H₂SO₄ to give colourless fumes with pungent smell. But on adding MnO₂ the fumes become greenish yellow.
Reason (R): MnO₂ oxidises HCl to chlorine gas which is greenish yellow.
Solution: (a) Colourless fumes of HCl become greenish yellow because MnO₂ oxidises HCl to chlorine gas.

Question 69. Assertion (A): SF₆ cannot be hydrolysed but SF₄ can be.
Reason (R): Six F atoms in SF₆ prevent the attack of H₂O on sulphur atom of SF₆.
Solution: (a) SF₆ is sterically protected due to presence of six F atoms around S which prevents the attack of H₂O on SF₆.

Long Answer Type Questions

Question 70. An amorphous solid “A” bums in air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO₄ solution and reduces Fe³⁺ to Fe⁺². Identify the solid “A” and the gas “B” and write the reactions involved.
Solution: A is sulphur (Sg) while B is sulphur dioxide (SO₂).

Question 71. On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid C. Identify A, B and C and also write reactions involved and draw the structures of B and C.
Solution: The gas A is NO₂  . The reactions are explained as under:

Question 72. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H₂  ) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved. –
Solution:

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