Chapter 7: The p-Block Element Class 12 Chemistry NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 7 – The p-Block Element (In-text Questions)

Question 7.1: Why are pentahalides more covalent than trihalides?

Answer: Pentahalides are more covalent because the +5 oxidation state of group 15 elements promotes electron sharing rather than electron loss. Hence, pentahalides are more covalent than trihalides.

Question 7.2: Why is BiH₃ the strongest reducing agent among the hydrides of group 15 elements?

Answer: As we move down group 15, the E–H bond weakens due to increased atomic size. Bi–H bonds are weakest, so BiH₃ readily releases hydrogen, making it the strongest reducing agent.

Question 7.3: Why is N₂ less reactive at room temperature?

Answer: N₂ has a strong triple bond between nitrogen atoms that requires high energy to break. Hence, it is inert at room temperature.

Question 7.4: What are the conditions required to maximise the yield of ammonia?

Answer: High pressure (~200 atm), moderate temperature (~700 K), iron catalyst with molybdenum promoter, and continuous removal of ammonia favour maximum yield in the Haber process.

Question 7.5: How does ammonia react with a solution of Cu²⁺?

Answer: Ammonia reacts with Cu²⁺ ions to form a deep blue complex:
Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺

Question 7.6: What is the covalency of nitrogen in N₂O₅?

Answer: Each nitrogen atom forms four covalent bonds in N₂O₅. Hence, its covalency is 4.

Question 7.7: Why is the bond angle in PH₄⁺ higher than in PH₃?

Answer: PH₃ has a lone pair causing lone pair–bond pair repulsion, reducing bond angle. PH₄⁺ has no lone pair, leading to a perfect tetrahedral angle of 109.5°.

Question 7.8: What happens when white phosphorus is heated with concentrated NaOH in the presence of CO₂?

Answer: White phosphorus reacts to form phosphine (PH₃) and sodium hypophosphite:
P₄ + 4NaOH + 2H₂O → 2PH₃ + 2NaH₂PO₂

Question 7.9: What happens when PCl₅ is heated?

Answer: PCl₅ decomposes into PCl₃ and Cl₂:
PCl₅ → PCl₃ + Cl₂

Question 7.10: Write a balanced equation for the hydrolysis of PCl₅ in heavy water.

Answer: PCl₅ + 4D₂O → D₃PO₄ + 5DCl

Question 7.11: What is the basicity of H₃PO₄?

Answer: The basicity of H₃PO₄ is 3 as it can donate three protons (H⁺).

Question 7.12: What happens when H₃PO₃ is heated?

Answer: H₃PO₃ undergoes disproportionation to form phosphine (PH₃) and H₃PO₄.

Question 7.13: List the important sources of sulphur.

Answer: Sulphur sources include:
• Sulphates: Gypsum, Epsom salt, Baryte
• Sulphides: Galena, Zinc blende, Copper pyrites, Iron pyrites
• Hydrogen sulphide gas and organic matter.

Question 7.14: Write the order of thermal stability of the hydrides of Group 16 elements.

Answer: Order: H₂O > H₂S > H₂Se > H₂Te > H₂Po

Question 7.15: Why is H₂O a liquid while H₂S is a gas at room temperature?

Answer: Due to strong hydrogen bonding in H₂O molecules, water exists as a liquid; H₂S lacks hydrogen bonding and exists as a gas.

Question 7.16: Which of the following does not react with oxygen directly: Zn, Ti, Pt, Fe?

Answer: Platinum (Pt) does not react directly with oxygen due to its noble character.

Question 7.17: Complete the following reactions:

Answer: (i) C₂H₂ + 5/2O₂ → 2CO₂ + H₂O
(ii) 4Al + 3O₂ → 2Al₂O₃

Question 7.18: Why does ozone (O₃) act as a strong oxidising agent?

Answer: Ozone decomposes to give nascent oxygen [O], which is highly reactive:
O₃ → O₂ + [O]

Question 7.19: How is ozone quantitatively estimated?

Answer: Ozone oxidises iodide to iodine in buffered KI solution, and the liberated iodine is titrated against sodium thiosulphate using starch as indicator.

Question 7.20: What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?

Answer: SO₂ reduces Fe³⁺ to Fe²⁺:
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄²⁻ + 4H⁺

Question 7.21: Comment on the nature of the two S–O bonds in SO₂. Are they equal?

Answer: Due to resonance, both S–O bonds in SO₂ are identical and have partial double bond character.

Question 7.22: How is the presence of SO₂ detected?

Answer: Tests:
• Pungent smell
• Turns acidified potassium dichromate from orange to green
• Decolorises acidified KMnO₄ solution

Question 7.23: Mention three uses of sulphuric acid (H₂SO₄).

Answer: • Manufacture of fertilizers
• Production of explosives
• Petroleum refining

Question 7.24: Write the conditions for maximum yield of H₂SO₄ in the Contact process.

Answer: • Catalyst: V₂O₅
• Temperature: ~720 K
• Pressure: ~2 atm
• Excess O₂ supply

Question 7.25: Why is Ka₂ much less than Ka₁ for H₂SO₄?

Answer: First proton ionises completely (strong acid), but the second proton from HSO₄⁻ ionises partially, hence Ka₂ << Ka₁.

Question 7.26: Compare the oxidising powers of F₂ and Cl₂ using bond energy, electron affinity, and hydration enthalpy.

Answer: F₂ is a stronger oxidising agent than Cl₂ because of:
• Lower bond dissociation energy
• Higher hydration enthalpy of F⁻
• Slightly lower electron affinity compensated by other factors

Question 7.27: Give two examples showing fluorine’s anomalous behaviour.

Answer: • Highest electronegativity, ionisation enthalpy, and oxidising power
• Shows only –1 oxidation state; no d-orbitals

Question 7.28: Sea is the greatest source of some halogens. Comment.

Answer: Sea water contains salts like NaCl, MgCl₂, KBr, and iodine in seaweeds, making it the largest natural source of Cl, Br, and I.

Question 7.29: Give the reason for bleaching action of chlorine.

Answer: Chlorine produces nascent oxygen which oxidises coloured substances to colourless compounds:
Cl₂ + H₂O → HCl + [O]

Question 7.30: Name two poisonous gases prepared from chlorine.

Answer: • Phosgene (COCl₂)
• Tear gas (CCl₃NO₂)

Question 7.31: Why is ICl more reactive than I₂?

Answer: ICl bond is weaker than I–I bond, making ICl more reactive.

Question 7.32: Why is helium used in diving equipment?

Answer: Helium is inert and poorly soluble in blood, preventing decompression sickness in divers.

Question 7.33: Balance the equation: XeF₆ + H₂O → XeO₂F₂ + HF

Answer: XeF₆ + 2H₂O → XeO₂F₂ + 4HF

Question 7.34: Why is it difficult to study radon chemistry?

Answer: Radon is radioactive with a very short half-life, making chemical studies difficult.

Question 7.35: How are XeO₃ and XeOF₄ prepared?

Answer: • XeF₆ + 3H₂O → XeO₃ + 6HF
• XeF₆ + H₂O → XeOF₄ + 2HF

Question 7.36: Arrange the following:

Answer: (i) Bond Dissociation: I₂ < F₂ < Br₂ < Cl₂
(ii) Acid Strength: HF < HCl < HBr < HI
(iii) Basic Strength: BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃

Question 7.37: Which one does not exist?

Answer: NeF₂ does not exist.

Question 7.38: Give the formula and structure of noble gas species isostructural with:

Answer: (i) XeF₄ (square planar like ICl₄⁻)
(ii) XeF₂ (linear like IBr₂⁻)
(iii) XeO₃ (pyramidal like BrO₃⁻)

Question 7.39: Why do noble gases have comparatively large atomic sizes?

Answer: Noble gases’ atomic size is measured by van der Waals radii, which are larger than covalent radii.

Question 7.40: List the uses of neon and argon gases.

Answer: • Neon: Used in advertising signs and glow lamps
• Argon: Used in arc welding, filling light bulbs, and laboratory work