80 Conceptual Questions of the P-Block Elements Class 12

In this particular blog post, you will be able to access a total of 80 significant and relevant questions pertaining to the P-Block Elements. This insightful compilation of questions is tailored to assist you in your upcoming Class 12 Board Exam, giving you an advantageous edge in your academic journey.

Note: No conceptual question will come in exam except from these 80 conceptual questions in P-block elements.

1. Why is N₂ less reactive at room temperature?

Ans: N₂ is less reactive at room temperature because of very high bond dissociation enthalpy of N≡N.

2. N cannot extend its coordination number beyond 4, why?

Ans: ₇N: 1s² 2s² 2p³

As N has four valence orbitals, i.e., one 2s and three 2p, so it can form maximum of 4 bonds, So, the maximum covalency or coordination number of N is 4.

It is also clear that the d-orbital is absent for 2nd shell, so it cannot extend its coordination number beyond 4.

3. R₃P=O is possible while R₃N=O does not exist.

Or PCl₅ exist while NCl₅ does not exist, why?

Or N does not form pentahalides, why?

Ans: It is so because of the absence of d-orbitals N can not form 5 bonds but P has d-orbitals, so it can form 5 bonds.

4. Single N-N bond is weaker than P-P bond, why?

Or Single O-O bond is weaker than S-S bond, why?

Or Single F-F bond is weaker than Cl-Cl bond, why?

Ans: It is because of the interelectronic repulsion between two N atoms, the N-N bond is weaker.

(Same reason for O-O and F-F)

5. N exists as N₂ while P exists as P₄, why?

Or Catenation tendency is weaker in N as compared to P, why?

Or Catenation of O is weaker than in S, why?

Ans: Catenation depends upon strength of single bond. Since, P-P bond is stronger than N-N, so the catenation tendency of P is more than N.

(Same reason for catenation of O and S)

6. NH₃ is more basic than PH₃, why?

Or NH₃ has a higher proton affinity than PH₃, why?

Ans: Due to smaller size, higher electronegativity of N, the N-H bond is stronger than P-H bond. That’s why NH₃ is more basic (or has higher proton affinity) than PH₃.

7. NCl₃ is an endothermic while NF₃ is exothermic compound, why?

Ans: Due to smaller size of F than Cl, the bond dissociation enthalpy of F-F is lower than Cl-Cl as F-F bond is weaker bond while N-Cl bond is weaker so bond dissociation enthalpy of N-F is greater than N-Cl.

N≡N + Cl-Cl → NCl₃

(Cl-Cl bond is strong while N-Cl bond is weak)

N≡N + F-F → NF₃

(F-F bond is weak while N-F bond is strong)

So, the energy released during NCl₃ formation is less than the energy needed to break N₂ and Cl₂ bonds while energy released during NF₃ formation is more than the energy needed to break N₂ and F₂ bonds.

Check out: Unique Guide to Organic Conversions (Cheat Sheet included)

8. Out of white phosphorous and red phosphorous, which one is more reactive and why?

Ans: In white P, the PPP bond angle is 60° and various P4 molecules have van der waals forces between them while in red P, the PPP bond angle is greater than 60° and various P4 tetrahedra are joined together by strong covalent bond.

So, white phosphorous is more reactive than red phosphorous.

9. Are all 5 bonds in PCl5 equivalent? Justify your answer.

Ans: No. All 5 bonds are not equivalent in PCl5.

The two axial bonds are repelled by three bond pairs while three equatorial bonds are repelled by two bond pairs. The bond pair-bond pair repulsion is greater in case of axial bonds, so the axial bonds are longer than equatorial bonds.

10. Give chemical reaction in support of the statement that all bonds in PCl₅ are not equivalent.

Ans: On heating PCl₅, the two axial bonds are broken as they are less stable due to greater bond pair-bond pair repulsion.

PCl₅ → PCl₃ + Cl₂

11. PH₃ has lower boiling point than NH₃, why?

Ans: Due to high electronegativity and smaller size of nitrogen atom, it can form H-bonding but phosphorous can not. That’s why NH₃ has higher boiling point than PH₃.

12. Why does NO₂ dimerise?

Ans: As it is an odd electron species, it is less stable. So, to become more stable, it dimerises to form N₂O₄. NO₂ is dark brown in colour while N₂O₄ is colourless, and the process of dimerisation is exothermic.

13. NH₃ is a good complexing agent, why?

Or NH₃ acts as a ligand, why?

Ans: Due to the presence of lone pair of electron on nitrogen atom, it acts as a ligand and form complex compounds with transition metal ions, e.g., [Ag(NH₃)₂]Cl and [Cu(NH₃)₄]SO₄.

14. Mention the conditions to increase the yield of NH3?

Ans: The reaction for the production of ammonia is-

N2 + 3H2 ⇌ 2NH3 ΔH=-92.4 kJ mol^-1

Conditions required to increase the yield of ammonia are-

a. High pressure of about 200 atm

b. Temperature around 700 K

c. Iron catalyst with K₂O and Al₂O₃.

15. Why are pentahalides more covalent than trihalides?

Ans: According to Fajan’s rule, greater the charge, greater will be the covalency. So, pentahalides (PX₅) having greater charge, i.e., +5, are more covalent as compared to trihalide (PX₃) having +3 charge.

16. What is the basicity of H₃PO₂, H₃PO₃ and H₃PO₄?

Ans: In the structures of the given acids, we have to check the hydrogen atom which is bonded with an electronegative atom, i.e., oxygen, so that hydrogen will be acidic. The structures are as follows-

Now, check the hydrogen bonded with oxygen.

In phosphinic acid (H3PO2), there is 1 hydrogen bonded with oxygen, so basicity is 1.

In phosphorous acid (H3PO3), there are 2 hydrogen bonded with oxygen, so basicity is 2.

In phosphoric acid (H3PO4), there are 3 hydrogen bonded with oxygen, so basicity is 3.

17. Account for the reducing behaviour of H₃PO₂.

Or H₃PO₂ is better reducing agent than H₃PO₃, explain.

Ans: Reducing behaviour of these acids depends on the number of P-H bonds. More the P-H bonds, more will be the reducing strength. As H₃PO₂ has two P-H bond, H₃PO₃ has one P-H bond and H₃PO₄ has no P-H bond, so the reducing character is as- H₃PO₂>H₃PO₃>H₃PO₄.

18. Which has higher bond angle: NO₂⁺ or NO₂^–.

Ans: NO₂⁺ has higher bond angle because in NO₂^–, there is a lone pair which decreases the bond angle.

19. Can it be proved that PH₃ is basic in nature, and in what way?

Ans: Due to presence of a lone pair on P atom, it acts as a lewis base and reacts with acid like HI to form salt.

PH₃ + HI → PH₄⁺I^–

20. The HNH bond angle is higher than HPH, HAsH and HSbH, why?

Or Why is the bond angle greater in NH₃ than other hydrides of Group 15 elements?

Ans: From N to P to As to Sb, the electronegativity decreases, so the electrons in bond pairs lie away from central atom. We can say that, repulsion between the bond pairs goes on decreasing, so the bond angle also keeps on decreasing.

Bond angle is as: NH₃ 107.8°, PH₃ 93.6°, AsH₃ 91.8°, SbH₃ 91.3°.

21. Explain why NH₃ is basic, while BiH₃ is feebly basic?

Or Basic character of hydrides of Group 15 elements decreases down the group, why?

Ans: Due to high electronegativity, the electron density on N is greater than than on less electronegative Bi, so the tendency to donate its lone pair decreases from N to Bi. Thus, the basic strength decreases from NH₃ to BiH₃.

22. PCl₅ is ionic in nature, in the solid state, why?

Ans: In the solid state, PCl₅ exists as [PCl₄]⁺[PCl₆]^–.

2PCl₅ → [PCl₄]⁺+ [PCl₆]^–

So, the PCl₅ conducts electricity in molten state.

23. Of Bi(V) and Sb(V) which may be a stronger oxidising agent and why?

Or Bismuth is a strong oxidising agent in pentavalent state, why?

Or Bi (V) is a strong oxidising agent, why?

Ans: The stability of +5 state decreases down the group due to inert pair effect, so that Bi (V) is unstable and it convert into stable Bi (III) state. So, it is a strong oxidising agent.

In Sb (V), the inert pair effect is less than in Bi (V), so that Bi (V) is stronger oxidising agent than Sb (V).

24. Bond angle in PH₄⁺ is higher than that in PH₃, why?

Or Bond angle in NH₄⁺ is higher than that in NH₃, why?

Ans: In PH₃, there are 3 bond pairs and 1 lone pair, while in PH₄⁺, there are 4 bond pairs. As the lone pair-bond pair (lp-bp) repulsion is greater than bond pair-bond pair (bp-bp) repulsion, so the bond angle in PH₄⁺ is greater than PH₃.

In NH₃, there are 3 bond pairs and 1 lone pair, while in NH₄⁺, there are 4 bond pairs. As the lone pair-bond pair (lp-bp) repulsion is greater than bond pair-bond pair (bp-bp) repulsion, so the bond angle in NH₄⁺ is greater than NH₃.

25. What happens when ammonia react with a solution of Cu²⁺?

Ans: Ammonia reacts with Cu(II) solution to form a deep blue coloured complex compound.

e.g., Cu²⁺ + NH₃ → [Cu(NH₃)₄]²⁺

26. What happens when white phosphorous is heated with aqueous NaOH in an inert atmosphere of CO₂?

Ans: White phosphorous reacts with NaOH to form phosphine gas (PH₃).

P₄ + NaOH + H₂O → PH₃ + NaH₂PO₂

27. What happens when PCl5 is heated?

Ans: On heating PCl₅, the two longer and weak P-Cl bonds break to form PCl₃ and Cl₂.

PCl₅ + Heat → PCl₃ + Cl₂

28. Why does PCl₃ fumes in presence of moisture?

Ans: PCl₃ reacts with moisture, to give fumes of HCl.

PCl₃ + H₂O → H₃PO₃ + HCl

29. In HNO₃, the N-O bond is shorter than the N-OH bond, why?

Ans: In HNO₃, the N-O bond has partial double bond character, while the N-OH bond is pure single covalent bond. That’s why N-O bond is shorter than N-OH bond.

30. Nitric oxide becomes brown, when released in air, why?

Ans: Nitric oxide reacts with atmospheric O₂ to form NO₂, which is brown in colour.

2NO + O₂ → 2NO₂

31. O exists as O₂, while S exist as S₈, why?

Or Catenation power of O is weaker than that of S, why?

Ans: O-O bond is weaker than S-S bond. It is because of smaller size of oxygen than sulphur, there exist a repulsion force between lone pair of oxygen, and the catenation power depends upon the strength of single bond.

32. Oxygen is a gas while sulphur is a solid, why?

Ans: Oxygen exists as O₂ while sulphur exists as S₈. The extent of Vander Waals bonding is greater in S₈ because of large surface area, so that S₈ is solid while O₂ is a gas.

33. The electron gain enthalpy with negative sign for oxygen is less than that of sulphur, why?

Ans: Due to smaller size of oxygen, the electron-electron repulsion is greater in oxygen than that in sulphur. So, less energy is released in O than S.

34. H₂S is more acidic than H₂O, why?

Or Why H₂S is acidic while H₂O is neutral?

Ans: Due to larger size of sulphur than oxygen, the S-H bond is weaker than O-H bond. So, H₂S can give H⁺ ions easily in aqueous medium.

35. SF₆ is known while SH₆ is not known, why?

Or SF₆ is known while SCl₆ is not known, why?

Ans: F is very strong oxidising agent and is most electronegative, so it can oxidise S to its +6 oxidation state, while H or Cl being a weaker oxidising agent, cannot oxidise S to its +6 oxidation state.

Hence, SF₆ is known while SH₆ or SCl₆ is not known.

36. Sulphur in vapour state exhibits paramagnetic behaviour, why?

Or Which form of sulphur is paramagnetic and why?

Ans: In vapour state, sulphur exists as S₂, which has 2 unpaired electrons in anti-bonding π* orbital, so it exhibits paramagnetic behaviour.

37. Why SF₄ undergo hydrolysis but SF₆ does not?

Or SF₆ is inert towards hydrolysis, why?

Or SF₆ is less reactive than SF₄, why?

Ans: In SF₆, the S atom is protected by 6 F atoms, so water molecules cannot attack on sulphur, while in SF₄, the S atom is not strictly protected by 4 F atoms, so water molecules can attack easily.

38. H₂O is a liquid while H₂S is a gas, why?

Or H₂O has higher boiling point than H₂S, why?

Ans: As oxygen is more electronegative than sulphur, so oxygen can form hydrogen bonding while sulphur cannot. So H₂O has higher boiling point and is in liquid state.

39. H₂S is less acidic than H₂Te, why?

Ans: From O to Te, atomic size increases, so bond strength decreases, thus acidic character increases.

H-Te bond length is larger than H-S bond length, so H₂Te is more acidic than H₂S.

40. In solution of H₂SO₄ in water, the 2nd dissociation constant Ka₂ is less than Ka₁, why?

Ans: It is because HSO₄^– has less tendency to donate H⁺ as compared to H₂SO₄.

41. Write the conditions to increase the yield of H₂SO₄ by contact process.

Ans: In the preparation of H₂SO₄, the reaction 2SO₂ + O₂ ⇌ 2SO₃ ΔH = -196.6 kJ/mol, is exothermic and reversible. Low temperature around 720 K, high pressure of 2 bar and V₂O₅ is used as a catalyst to increase the yield.

42. How is O₃ estimated quantitatively?

Ans: When O₃ is treated with excess of KI solution, I₂ is liberated.

2I- + H₂O + O₃ → 2OH- + I₂ + O₂

The I₂ liberated is titrated against a standard solution of sodium thiosulphate.

2Na₂S₂O₃ + I₂ → Na₂S₄O₆ + 2NaI

43. H₂S acts only as reducing agent, but SO₂ acts both as reducing agent as well as oxidising agent, why?

Ans: The range of oxidation state of sulphur is from -2 to +6. In H₂S, S is in -2 state, so it can oxidise only to higher state and hence acts as a reducing agent.

While in SO₂, S is in +4 oxidation state, so it can act both as an oxidising as well as a reducing agent, as it can increase its state till +6 and also decrease its state till -2.

44. Bond dissociation energy of F₂ is less than that of Cl₂, explain.

Or F-F bond is weaker than Cl-Cl bond, why?

Ans: In F₂, due to smaller size of F, there is large repulsion between the lone pairs and bond pair, while in Cl₂, this repulsion is smaller due to large size of Cl. So, Cl-Cl bond is stronger than F-F bond.

45. The negative electron gain enthalpy is less for F than that for Cl, why?

Ans: Due to small size of F, there is large electron-electron repulsion which is very small in case of Cl. So, Cl can accept electron easily but F cannot.

46. Halogens have most negative electron gain enthalpy in the respective period of periodic table, why?

Ans: Halogens have smallest size and highest effective nuclear charge, due to which they readily accept an electron.

47. F does not show any positive oxidation state, why?

Or F shows only -1 oxidation state, why?

Ans: As F is most electronegative element, so it cannot show any positive oxidation state like +1, +3, +5 or +7. Also, it does not contain d-subshell, so it cannot expand its octet and does not form more than one bond. Hence, only -1 oxidation state is possible for fluorine.

48. Explain why F forms only one oxoacid, HOF.

Ans: Due to the absence of d-orbitals, F can not show +3, +5 or +7 states, so that F does not form HOFO, HOFO₂, HOFO₃.

49. Amongst halogens, F₂ is strongest oxidising agent, why?

Or F₂ is stronger oxidising agent than Cl₂, why?

Ans: The E° value of F is most positive, as it has weak F-F bond so that it reduces most readily and acts as strong oxidising agent.

50. Why interhalogens are more reactive than halogens?

Or ICl is more reactive than Cl₂, why?

Ans: It is because of less effective overlapping between orbitals of different halogens and polar nature the bond X-X’ has low bond dissociation enthalpy than X-X bond. Thus, X-X’ compounds (interhalogens) are more reactive than XX compounds (halogens).

51. F₂ is exceptionally more reactive than its interhalogens, why?

Ans: Because of its very low bond dissociation enthalpy of F-F bond.

52. Arrange HClO₄, HClO₃, HClO₂ and HClO in order of (i) acidic strength (ii) oxidising power. Give reasons.

Ans: (i) HClO₄ > HClO₃ > HClO₂ > HClO

This order is due to the stability of oxoanion, formed after loosing H+ ion. Greater the stability of oxoanion, stronger will be the acid.

(ii) HClO₄ < HClO₃ < HClO₂ < HClO

As the stability of oxoanion increases, its tendency to donate oxygen decreases, so the oxidising power is the reverse of stability of oxoanions.

53. Arrange HClO, HBrO, HIO in order of decreasing acidic strength with reason.

Ans: HOCl > HOBr > HOI

This is due to decrease in electronegativity from Cl to I.

54. Arrange in increasing acid strength HF, HCl, HBr and HI.

Or F is more electronegative than I, yet HF has lower acid strength than HI, explain.

Ans: From F to I, the size increases, so the bond dissociation energy decreases from H-F to H-I, hence the acid strength increases in this order HF < HCl < HBr < HI.

55. Bleaching of flowers by Cl₂ is permanent while that by SO₂ is temporary, why?

Ans: Cl₂ bleaches coloured material (flowers) by oxidation and does not reduce again. While SO₂ bleaches flowers by reduction and in air it again oxidised, so that its bleaching is temporary.

56. F never acts as the central atom in interhalogen compounds, explain.

Ans: Due to very small size of F and absence of d-orbitals F cannot show +3, +5, +7 states and hence does not play the role of central atom in interhalogen compounds.

57. ClF₃ exists but FCl₃ does not, why?

Ans: Cl has d-orbitals and can show +3 oxidation state in ClF₃ but F cannot show +3 state because of highest electronegativity and absence of d-orbitals, that’s why FCl₃ does not exist.

58. Why HF is stored in wax coated glass bottles?

Or HF is not stored in glass bottles, why?

Ans: Because HF does not react with wax but reacts with glass (SiO₂).

6HF + SiO₂ → H₂SiF₆ + 2H₂O

59. I₂ is more soluble in KI than water, why?

Ans: Because I₂ reacts with KI to form KI₃.

KI + I₂ → KI₃

60. ClF₃ has T-shaped structure and not trigonal planar, why?

Ans: Because ClF₃ has 2 lone pairs and 3 bond pairs, so the structure is T-shaped.

61. Why are halogens coloured?

Ans: Halogens absorb light in the visible region as a result of which their electrons excited to higher energy levels, while the remaining light is transmitted and this transmitted light is responsible for colour.

62. Arrange in increasing order of bond dissociation enthalpy: F₂, Cl₂, Br₂, I₂. Give reason.

Ans: I₂ < F₂ < Br₂ < Cl₂

As the size increases bond dissociation enthalpy decreases from Cl₂ to I₂. But in case of F₂ due to small size there is electron-electron repulsion so it has low bond dissociation enthalpy.

63. What inspired N. Bartlett for carrying out reaction between Xe and PtF₆?

Or What prompted Bartlett to the discovery of noble gas compounds?

Ans: Because the ionisation enthalpy of O₂ (1175 kJ/mol) and that of Xe (1170 kJ/mol) are almost similar, so the reacted Xe with PtF₆ in place of O₂.

Xe + PtF₆ → Xe⁺PtF₆^–

64. The majority of noble gas compounds are of Xe, why?

Or Among noble gases, only Xe forms compounds, why?

Ans: Except Rn which is radioactive, only Xe has lowest ionisation enthalpy amongst noble gases so that it reacts to form compounds with oxygen and fluorine.

65. No chemical compound of He is known, why?

Or No distinct compound of He is known, why?

Ans: Because of very high ionisation enthalpy and absence of d-orbitals He does not form chemical compounds.

66. Noble gases are mostly inert, explain.

Ans: This is because of completely filled electrons configuration, high ionisation enthalpy and positive electron gain enthalpy.

67. Xe does not fluorides like XeF₃ and XeF₅, why?

Ans: Xe has fully filled orbitals, so promotion of 1, 2 or 3 electrons give rise to 2, 4 or 6 half filled orbitals respectively. That’s why Xe can form even number of bonds only, and hence does not form XeF₃ and XeF₅.

68. Why He is used in diving apparatus?

Ans: Because of its very low solubility in blood, a mixture of O₂ and He is used in diving apparatus.

69. Why has it been difficult to study the chemistry of radon (Rn) element?

Ans: Because it is a radioactive element with half life of 3.82 days.

70. Does the hydrolysis of XeF₆ lead to a redox reaction?

Or Out of XeF₂, XeF₄ and XeF₆ whose hydrolysis is not a redox reaction.

Ans: Hydrolysis of XeF₂ and XeF₄ are redox reaction while hydrolysis of XeF₆ is not a redox reaction.

XeF₂ + H₂O → Xe + HF + O₂

XeF₄ + H₂O → Xe + XeO₃ + HF + O₂

XeF₆ + 3H₂O → XeO₃ + 6HF

71. Elements of group 16 generally show lower value of first ionisation enthalpy as compared to the elements of group 15.

Ans: Due to extra stability of half filled electronic configuration of group 15, large amount of energy is required to remove electron as compared to group 16.

72. The two O-O bond lengths are same in O₃ molecule, why?

Or Explain why two S-O bonds are of equal length in SO₂.

Ans: The two O-O bonds have partial double bond character due to resonance, that’s why two O-O bonds are equal in length.

Similarly for the case of sulphur dioxide.

73. BiCl₃ is less covalent than PCl₃, why?

Ans: Acoording to Fajan’s rule, smaller the size of cation, greater will be tha covalent character. As P is smaller than Bi, so PCl₃ is more covalent than BiCl₃.

74. O₃ acts as a powerful oxidising agent, why?

Ans: O₃ is very unstable compound and decomposes readily to give atomic oxygen and hence oxidises the other.

O₃ → O₂ + O

75. F forms largest number of interhalogen compounds, why?

Ans: Because F has smallest size and it is most electronegative amongst halogens. So, F forms largest number of interhalogen compounds.

76. The acidic strength increases in this order: PH₃<H₂S<HCl, why?

Ans: Acidic strength increases with increase in electronegativity, i.e., P < S< Cl.

77. Why do noble gases have very low boiling point?

Ans: Noble gases have very weak Vander Waals forces, i.e., dispersion forces among molecules so that they have very low boiling point.

78. What happens when Cl₂ is passed through hot concentrated solution of NaOH?

Ans: Chlorine reacts with hot concentrated NaOH, to form sodium chloride and sodium chlorate are formed.

Cl₂ + NaOH → NaCl NaClO₃ + H₂O

79. Why is N-O bond in NO₂^– shorter than N-O bond in NO₃^–?

Ans: In NO₂^–, the partial double bond occurs in two N-O bonds while in NO₃^–, the one double bond is for three N-O bonds. That’s why NO₂^– has small N-O bond than in NO₃^–.

80. Why is F₂ most reactive of all the four halogens?

Ans: Because of high electronegativity, low F-F bond dissociation enthalpy and great tendency to accept electron.

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