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Class 12 · Tamil Nadu State Board

Samacheer Kalvi 12th Chemistry Chapter 3 Exercise Solutions

Tamil Nadu Class 12 Chemistry Chapter 3 p-Block Elements II exercise solutions with nitrogen, oxygen, halogens, noble gases, oxoacids, interhalogens, and equation practice.

Updated 17 June 20268 min read

Chapter 3 — p-Block Elements II Exercise Solutions

This chapter has many facts and reactions. Use tables for trends; group equations by element.

Key Concepts Quick Recap

ConceptKey Fact
Group 15 elementsN, P, As, Sb, Bi — valence config ns²np³
Group 16 elementsO, S, Se, Te, Po — ns²np⁴
Group 17 (Halogens)F, Cl, Br, I, At — ns²np⁵; strongest oxidising agents
Group 18 (Noble gases)He, Ne, Ar, Kr, Xe, Rn — completely filled shells
Dinitrogen stabilityTriple bond (N≡N) — bond energy 941 kJ/mol
Ozone (O₃)Allotrope of oxygen; angular; strong oxidising agent
Interhalogen compoundsFormed between two different halogens — e.g. ClF₃, BrF₅
Fluorine anomalyHighest electronegativity, no d orbitals, only −1 oxidation state
Xenon compoundsXeF₂, XeF₄, XeF₆ — formed with F₂ under UV/heat
Haber processN₂ + 3H₂ ⇌ 2NH₃ at 450 °C, 200 atm, Fe catalyst
Contact process2SO₂ + O₂ ⇌ 2SO₃ using V₂O₅ catalyst

Important Areas

GroupFocus Topics
Group 15Ammonia, nitric acid, oxoacids of P, allotropes of P
Group 16Oxygen allotropes, ozone, sulphuric acid (contact process)
Group 17Reactivity order, oxoacids of Cl, interhalogens
Group 18Xenon fluorides, uses of noble gases

I. Choose the Best Answer (MCQs)

1. Which process is used for industrial manufacture of ammonia? (a) Ostwald process (b) Haber process (c) Contact process (d) Solvay process

Answer(b) Haber process N₂ + 3H₂ ⇌ 2NH₃ at ~450 °C, 200 atm, Fe catalyst with Al₂O₃ and K₂O promoters.

2. The oxidation state of nitrogen in N₂O₅ is: (a) +3 (b) +4 (c) +5 (d) +2

Answer(c) +5 2x + 5(−2) = 0 → x = +5. N₂O₅ is the anhydride of nitric acid (HNO₃).

3. Which halogen has the highest electronegativity? (a) Cl (b) Br (c) I (d) F

Answer(d) F Fluorine has the highest electronegativity (4.0 Pauling scale). It has no d orbitals and only shows the −1 oxidation state.

4. The shape of XeF₄ is: (a) Tetrahedral (b) Square planar (c) Pyramidal (d) See-saw

Answer(b) Square planar XeF₄ has 6 electron pairs around Xe (4 bond + 2 lone) → octahedral geometry → square planar shape.

5. Ozone is used as a disinfectant because it is: (a) A reducing agent (b) A strong oxidising agent (c) Paramagnetic (d) A noble gas

Answer(b) A strong oxidising agent O₃ → O₂ + [O]. The nascent oxygen released destroys bacteria.

6. The bleaching action of SO₂ is due to: (a) Oxidation (b) Reduction (c) Hydration (d) Substitution

Answer(b) Reduction SO₂ bleaches by reduction — adding hydrogen to the coloured substance. This bleaching is temporary (colour returns on exposure to air).

7. Which noble gas is used in filling airships (and is non-flammable)? (a) He (b) Ne (c) Ar (d) Kr

Answer(a) He Helium is lighter than air, non-flammable, and safe — used in airships and balloons. Hydrogen is flammable and dangerous.

II. Short Answer Questions

Q1. Why does nitrogen form multiple oxides?
Answer

Nitrogen shows multiple oxidation states (−3, 0, +1, +2, +3, +4, +5) due to its small size and high electronegativity. The oxides are:

OxideOxidation StateCommon Name
N₂O+1Laughing gas (anaesthetic)
NO+2Colourless, toxic
N₂O₃+3Blue solid
NO₂+4Brown, poisonous, acidic
N₂O₅+5Anhydride of HNO₃

Q2. Compare the oxidising power of F₂ and Cl₂.
Answer
  • F₂ is a stronger oxidising agent than Cl₂.
  • Fluorine has higher electronegativity (4.0 vs 3.0 for Cl).
  • F−F bond energy is low (154 kJ/mol) — easy to break, releasing highly reactive F atoms.
  • Fluorine has no d orbitals — it always acts as an oxidant, never a reductant.
  • Cl can show +1 to +7 oxidation states using d orbitals, making it a weaker oxidant than F₂.

Q3. Why does fluorine not exhibit positive oxidation states?
Answer

Fluorine is the most electronegative element (4.0) and has no vacant d orbitals in the valence shell. Other halogens (Cl, Br, I) use d orbitals to expand their octet, allowing them to show +1, +3, +5, +7 oxidation states. Since fluorine cannot do this, it only shows 0 (in F₂) and −1.


Q4. What are interhalogen compounds? Give two examples and their shapes.
Answer

Interhalogen compounds are compounds formed between two different halogens. General formula: XYₙ (n = 1, 3, 5, 7) where X is the larger, less electronegative halogen.

CompoundShapeExample
XY (n=1)LinearClF, BrCl, ICl
XY₃ (n=3)T-shapedClF₃, BrF₃
XY₅ (n=5)Square pyramidalBrF₅, IF₅
XY₇ (n=7)Pentagonal bipyramidalIF₇

Properties: More reactive than parent halogens; used as fluorinating agents.


Q5. Write the structure and basicity of H₃PO₃ and H₃PO₄.
Answer
  • H₃PO₃ (Phosphorous acid): One P−H bond (non-ionisable) + two P−OH bonds → bibasic (basicity = 2). It is a reducing agent (direct P−H bond).
  • H₃PO₄ (Phosphoric acid): Three P−OH bonds, no direct P−H → tribasic (basicity = 3). Not a reducing agent.

Rule: Only P−OH bonds contribute to basicity. Direct P−H bonds do not ionise.


Worked Examples

Example 1 — Haber process conditions and yield

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ/mol Conditions: 450 °C, 200 atm, Fe catalyst (promoters: Al₂O₃, K₂O) - High pressure favours NH₃ (fewer moles on product side). - Moderate temperature balances rate and yield. - Catalyst increases rate without changing equilibrium position. - ~15% yield per pass; unconverted gases are recycled.

Example 2 — Contact process for H₂SO₄

Step 1: S + O₂ → SO₂ Step 2: 2SO₂ + O₂ ⇌ 2SO₃ (V₂O₅, 450 °C, 1–2 atm) Step 3: SO₃ + H₂SO₄ → H₂S₂O₇ (oleum / fuming H₂SO₄) Step 4: H₂S₂O₇ + H₂O → 2H₂SO₄ Note: SO₃ is NOT dissolved directly in water — it is explosive.

Example 3 — Halogen displacement reactions

Reactivity: F₂ > Cl₂ > Br₂ > I₂ A more reactive halogen displaces a less reactive one from its salt: Cl₂ + 2KBr → 2KCl + Br₂ (Cl₂ displaces Br⁻) Br₂ + 2KI → 2KBr + I₂ (Br₂ displaces I⁻) F₂ + 2NaCl → 2NaF + Cl₂ (F₂ displaces Cl⁻) I₂ cannot displace any other halide (least reactive).

Example 4 — Xenon fluorides

Xe + F₂ →(UV light, 2:1) XeF₂ (linear) Xe + 2F₂ →(400 °C, 1 atm) XeF₄ (square planar) Xe + 3F₂ →(400 °C, 60 atm) XeF₆ (distorted octahedral)

Example 5 — Oxoacids of chlorine: oxidation states and acid strength

CompoundFormulaCl Oxidation StateAcid Strength
Hypochlorous acidHOCl+1Weakest
Chlorous acidHClO₂+3↑
Chloric acidHClO₃+5↑
Perchloric acidHClO₄+7Strongest

Higher oxidation state of Cl → more O atoms → greater electron withdrawal from O−H → stronger acid.

Example 6 — Ozone vs oxygen

PropertyO₂O₃
ColourColourlessPale blue
SmellOdourlessPungent
StabilityStableDecomposes to O₂
Magnetic natureParamagneticDiamagnetic
Bond angle180° (linear)117° (angular)
Reaction with HgNoHg turns dull (tailing)
Bond length121 pm128 pm

Common Mistakes

  • Basicity of H₃PO₃ is 2, not 3 — only P−OH counts, not P−H.
  • Wrong halogen reactivity order — always F > Cl > Br > I.
  • Saying noble gases cannot form any compounds — Xe does form XeF₂, XeF₄, XeF₆.
  • Mixing contact process (H₂SO₄) with Ostwald process (HNO₃).
  • Forgetting that SO₂ bleaches by reduction (temporary), while Cl₂ bleaches by oxidation (permanent).

Exam Checklist

  • Formula of each product is correct.
  • Oxidation states are verified (sum = charge).
  • Trend order (reactivity, acidity, electronegativity) is in correct direction.
  • Reason is included alongside the fact.
  • Industrial process conditions (temperature, pressure, catalyst) are stated.
  • Equations are balanced.

Frequently Asked Questions

Oxoacids, halogen trends, interhalogen compounds, noble gas compounds, and preparation/properties of important compounds are scoring areas.
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