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
| Concept | Key Fact |
|---|
| Group 15 elements | N, P, As, Sb, Bi — valence config ns²np³ |
| Group 16 elements | O, 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 stability | Triple bond (N≡N) — bond energy 941 kJ/mol |
| Ozone (O₃) | Allotrope of oxygen; angular; strong oxidising agent |
| Interhalogen compounds | Formed between two different halogens — e.g. ClF₃, BrF₅ |
| Fluorine anomaly | Highest electronegativity, no d orbitals, only −1 oxidation state |
| Xenon compounds | XeF₂, XeF₄, XeF₆ — formed with F₂ under UV/heat |
| Haber process | N₂ + 3H₂ ⇌ 2NH₃ at 450 °C, 200 atm, Fe catalyst |
| Contact process | 2SO₂ + O₂ ⇌ 2SO₃ using V₂O₅ catalyst |
Important Areas
| Group | Focus Topics |
|---|
| Group 15 | Ammonia, nitric acid, oxoacids of P, allotropes of P |
| Group 16 | Oxygen allotropes, ozone, sulphuric acid (contact process) |
| Group 17 | Reactivity order, oxoacids of Cl, interhalogens |
| Group 18 | Xenon 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:
| Oxide | Oxidation State | Common Name |
|---|
| N₂O | +1 | Laughing gas (anaesthetic) |
| NO | +2 | Colourless, toxic |
| N₂O₃ | +3 | Blue solid |
| NO₂ | +4 | Brown, poisonous, acidic |
| N₂O₅ | +5 | Anhydride 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.
| Compound | Shape | Example |
|---|
| XY (n=1) | Linear | ClF, BrCl, ICl |
| XY₃ (n=3) | T-shaped | ClF₃, BrF₃ |
| XY₅ (n=5) | Square pyramidal | BrF₅, IF₅ |
| XY₇ (n=7) | Pentagonal bipyramidal | IF₇ |
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
| Compound | Formula | Cl Oxidation State | Acid Strength |
|---|
| Hypochlorous acid | HOCl | +1 | Weakest |
| Chlorous acid | HClO₂ | +3 | ↑ |
| Chloric acid | HClO₃ | +5 | ↑ |
| Perchloric acid | HClO₄ | +7 | Strongest |
Higher oxidation state of Cl → more O atoms → greater electron withdrawal from O−H → stronger acid.
Example 6 — Ozone vs oxygen
| Property | O₂ | O₃ |
|---|
| Colour | Colourless | Pale blue |
| Smell | Odourless | Pungent |
| Stability | Stable | Decomposes to O₂ |
| Magnetic nature | Paramagnetic | Diamagnetic |
| Bond angle | 180° (linear) | 117° (angular) |
| Reaction with Hg | No | Hg turns dull (tailing) |
| Bond length | 121 pm | 128 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.
