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Playbook

Chemical Bonding

11 q · 0% HARD. Ionic vs covalent prediction from electronegativity difference, oxidation-state assignment in a compound (rule sequence: H = +1, O = −2, group I = +1, sum to zero), valency from electronic config, molecular formula → atom count. Pure rule application, every q tractable.

questions in the bank
11
tagged HARD
0%
subtopic(s)
3
worked examples
2

When you’ll see it

An ionic vs covalent prediction, an oxidation-state assignment (e.g. V in V₂O₅), a valency-from-electronic-configuration question, or a molecular formula → atom count.

How this chapter is tested

11 q in 10 years · ZERO HARD. Pure rule-application chapter. Ionic vs Covalent (5 q) decided by electronegativity (EN) difference: ΔEN > 1.7 → ionic (NaCl ΔEN ≈ 2.1, MgO ΔEN ≈ 2.2); ΔEN < 1.7 → covalent (H₂O ΔEN ≈ 1.4, NH₃ ΔEN ≈ 0.9); ΔEN < 0.4 → mostly non-polar covalent (H₂, N₂, CH₄). Ionic compounds: high melting + boiling point, conduct electricity in molten/aqueous state, crystalline solids, soluble in water. Covalent: lower MP/BP, poor conductors, soluble in non-polar solvents.

Oxidation-state assignment (4 q) uses a rule sequence: (a) free elements = 0 (Cu, O₂, N₂). (b) H = +1 except in metal hydrides (NaH, H = −1). (c) O = −2 except in peroxides (H₂O₂, O = −1) and OF₂ (O = +2). (d) Group I = +1, Group II = +2. (e) F always = −1; other halogens = −1 except with O or F. (f) Sum of ox-states = molecule's net charge. For V₂O₅: 2V + 5(−2) = 0 → 2V = 10 → V = +5. For Mn in KMnO₄: 1 + Mn + 4(−2) = 0 → Mn = +7.

Bond Counting (2 q): atoms count in a molecular formula (H₂SO₄ has 2 H, 1 S, 4 O). Anion valency: O²⁻ = −2, S²⁻ = −2, N³⁻ = −3, P³⁻ = −3, F⁻/Cl⁻/Br⁻/I⁻ = −1. Cation valency: Na⁺ = +1, Ca²⁺ = +2, Al³⁺ = +3, Fe²⁺/Fe³⁺ variable.

The sub-skills

The rules and habits that decide whether you get a question right.

  • Ionic vs covalent from EN

    ΔEN > 1.7: ionic. ΔEN 0.4–1.7: polar covalent. ΔEN < 0.4: non-polar covalent. EN values from Pauling: F 4.0, O 3.5, N 3.0, Cl 3.0, C 2.5, H 2.1, Na 0.9, K 0.8. Memorise top 5 + alkali metals.

  • Oxidation-state assignment sequence

    1. Free element = 0. 2. H = +1 (−1 in metal hydrides). 3. O = −2 (−1 in peroxides, +2 in OF₂). 4. Group I = +1, Group II = +2. 5. F = −1. 6. Sum = molecule's charge. Apply in order; solve for the unknown.

  • Property prediction by bond type

    Ionic: high MP/BP (~800–1500 °C), conduct in molten/aqueous, crystalline, water-soluble. Covalent: low MP/BP, poor conductors, often gas/liquid at RT, soluble in organic solvents. Metallic: malleable, ductile, conduct solid+molten.

  • Atom counting + valency match

    Molecular formula gives explicit atom counts. Valency check: total +ve = total −ve. AlCl₃ valency check: Al³⁺ × 1 = +3; Cl⁻ × 3 = −3; balanced. Empirical formula from valency ratios (NPP).

2 worked examples from the bank

Real past-year questions illustrating the playbook. Click to reveal options + solution.

Example 1Chemical BondingMODERATE
Which one of the following nitrogen oxides may dimerize?

[Q82 · Apr · 2025]

Example 2Chemical BondingMODERATE
Which one among the following oxides has the highest melting point ?

[Q107 · Apr · 2024]

Traps to expect

Distractor shapes specific to this chapter. The page-wide Traps section covers the bank-level patterns.

  • All metal-non-metal bonds are ionic

    Mostly true but not always. AlCl₃ has significant covalent character (small Al³⁺ polarises Cl⁻ heavily — Fajans' rules). HgCl₂ is covalent despite Hg being metallic. Borderline cases (BeCl₂, AlCl₃, FeCl₃) act covalent.

  • Forgetting variable oxidation states

    Fe = +2 or +3 depending on compound. Cu = +1 or +2. Mn = +2/+4/+6/+7. Don't assume a default. Cl = −1 most often but +1 in HOCl, +5 in HClO₃, +7 in HClO₄.

Drill every chemical bonding question

11 questions from the bank, scoped to 3 bundled subtopics.

Drill the 11 questions

Related playbooks

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