Interleaved — Phase 3

Chemistry interleaved practice

printable — key stays hidden on paper

Instructions: Solve each problem in order. These questions deliberately mix different topics — before answering, identify which concept/method each requires. Show reasoning. Total: 50 marks. No calculators needed except where stated.

1. (5 marks) Arrange the following in order of increasing acid strength and justify using oxidation-state / electronegativity reasoning: $\text{HClO}, \quad \text{HClO}_2, \quad \text{HClO}_3, \quad \text{HClO}_4$

2. (6 marks) Calculate the mass of $\text{CaO}$ (in g) that must be added to soften 1000 L of temporarily hard water containing $\text{Ca(HCO}_3)_2$ at a concentration of $0.001\ \text{mol L}^{-1}$ (Clark's process). Write the balanced equation. [Atomic masses: Ca = 40, O = 16]

3. (4 marks) Explain why $\text{H}_3\text{PO}_3$ is dibasic whereas $\text{H}_3\text{PO}_4$ is tribasic, despite both having three H atoms. Draw both structures.

4. (5 marks) Diborane $\text{B}_2\text{H}_6$ has only $12$ valence electrons but $8$ bonds. Explain the 3c–2e bonding in the bridge, state how many terminal vs bridging H atoms exist, and account for the total electron count.

5. (4 marks) A student claims " $\text{Li}$ resembles $\text{Mg}$ more than it resembles $\text{Na}$ ." Give three experimental facts supporting this diagonal relationship.

6. (6 marks) Hydrogen peroxide is both an oxidising and reducing agent. Write balanced ionic half-reactions / equations showing $\text{H}_2\text{O}_2$ acting as: (a) an oxidising agent in acidic medium (with $\text{Fe}^{2+}$ ), (b) a reducing agent in acidic medium (with $\text{KMnO}_4$ ).

7. (3 marks) The observed atomic radii of $\text{Zr}$ (160 pm) and $\text{Hf}$ (159 pm) are nearly identical, though Hf lies one period below Zr. Name the phenomenon and explain its cause in one or two sentences.

8. (5 marks) Predict the shape and hybridisation of the central atom in each, using VSEPR: (a) $\text{XeF}_4$ (b) $\text{XeO}_3$ (c) $\text{XeF}_2$

9. (6 marks) In the industrial synthesis of $\text{HNO}_3$ (Ostwald process), $\text{NH}_3$ is oxidised. Write the three sequential balanced equations and calculate the theoretical mass of $\text{HNO}_3$ (in kg) obtainable from 17 kg of $\text{NH}_3$ , assuming 100% conversion. [N=14, H=1, O=16]

10. (6 marks) Identify the odd one out in each group and justify in one line: (a) $\text{LiH}$ , $\text{NaH}$ , $\text{B}_2\text{H}_6$ , $\text{CaH}_2$ (b) diamond, graphite, fullerene, quartz (c) protium, deuterium, tritium, positronium

Answer keyMark scheme & solutions

1. (Tests 3.2.10 — Oxoacids of halogens acidity trend) Order: $\text{HClO} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4$ . As the number of terminal (non-hydroxyl) O atoms increases, the oxidation state of Cl rises (+1, +3, +5, +7). More electronegative O atoms pull electron density from the O–H bond, stabilising the conjugate base by delocalising the negative charge. Hence $\text{HClO}_4$ (strongest) → $\text{HClO}$ (weakest). Why this method: Recognise it's an oxoacid trend, not a HX trend — the deciding factor is O count / oxidation state, not atomic size.

2. (Tests 3.1.5 — hardness/softening; stoichiometry) Clark's reaction: $\text{Ca(HCO}_3)_2 + \text{Ca(OH)}_2 \rightarrow 2\text{CaCO}_3\downarrow + 2\text{H}_2\text{O}$ . But CaO is added → forms $\text{Ca(OH)}_2$ . 1 mol $\text{Ca(HCO}_3)_2$ needs 1 mol $\text{Ca(OH)}_2$ = 1 mol CaO. Moles hardness = $0.001 \times 1000 = 1\ \text{mol}$ . Mass CaO $= 1 \times 56 = \mathbf{56\ g}$ . Why: Trap — must use CaO molar mass (56), not Ca(OH)₂; 1:1 stoichiometry for temporary hardness.

3. (Tests 3.2.6 — oxoacid basicity) Basicity = number of ionisable P–OH hydrogens.

$\text{H}_3\text{PO}_3$ (phosphorous acid): one P=O, two P–OH, and one P–H (non-ionisable) → dibasic.
$\text{H}_3\text{PO}_4$ (phosphoric acid): one P=O, three P–OH → tribasic. Structures: H₃PO₃ has H directly bonded to P; H₃PO₄ has all three H via O. Why: Count P–OH bonds, not total H — the P–H bond doesn't ionise.

4. (Tests 3.2.1 — diborane 3c–2e bond) $\text{B}_2\text{H}_6$ : 4 terminal H (normal 2c–2e) + 2 bridging H. Each B–H–B bridge is a three-centre two-electron (banana) bond. Electron count: 2 B (3 e each) + 6 H (1 e each) = 12 valence e = 6 pairs. Four terminal B–H bonds use 8 e; remaining 4 e form the two 3c–2e bridge bonds (2 e each). Bridging B–H bonds are longer/weaker than terminal ones. Why: Electron-deficient molecule — must invoke 3c–2e, not ordinary covalent bonds.

5. (Tests 3.1.7 — Li–Mg diagonal relationship) Any three:

$\text{Li}$ and $\text{Mg}$ form nitrides ( $\text{Li}_3\text{N}$ , $\text{Mg}_3\text{N}_2$ ) directly with N₂; other alkali metals don't.
Their carbonates ( $\text{Li}_2\text{CO}_3$ , $\text{MgCO}_3$ ) decompose on heating to oxides; Na₂CO₃ is stable.
Both form covalent, LiCl/MgCl₂ soluble in organic solvents (deliquescent, hydrated).
Both hydroxides are weakly basic and decompose on heating. Why: Diagonal relationship due to similar charge/radius ratio.

6. (Tests 3.1.6 — H₂O₂ redox) (a) Oxidising (acidic), with Fe²⁺: $\text{H}_2\text{O}_2 + 2\text{Fe}^{2+} + 2\text{H}^+ \rightarrow 2\text{Fe}^{3+} + 2\text{H}_2\text{O}$ (b) Reducing (acidic), with KMnO₄: $5\text{H}_2\text{O}_2 + 2\text{MnO}_4^- + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}$ Here O in H₂O₂ (−1) → 0, releasing O₂. Why: Identify the partner's role — MnO₄⁻ is a strong oxidiser, forcing H₂O₂ to reduce.

7. (Tests 3.3.3 — lanthanide contraction) Lanthanide contraction: poor shielding of the nucleus by the 4f electrons causes a steady decrease in size across the lanthanides, so the expected size increase from Zr→Hf is cancelled, giving nearly equal radii. Why: Post-lanthanide anomaly, not simple periodicity.

8. (Tests 3.2.11 — Xe compound structure/VSEPR) (a) $\text{XeF}_4$ : 6 e-pairs (4 bond + 2 lp), $sp^3d^2$ , square planar. (b) $\text{XeO}_3$ : 4 e-domains (3 bond + 1 lp), $sp^3$ , pyramidal. (c) $\text{XeF}_2$ : 5 e-pairs (2 bond + 3 lp), $sp^3d$ , linear. Why: Count lone pairs on Xe to fix geometry from electron domains.

9. (Tests 3.2.5 — Ostwald process, stoichiometry) Equations: $4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}$ $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ $3\text{NO}_2 + \text{H}_2\text{O} \rightarrow 2\text{HNO}_3 + \text{NO}$ Overall (ignoring recycled NO): $1\ \text{NH}_3 \rightarrow 1\ \text{HNO}_3$ (1:1 N balance). Moles NH₃ = $17000/17 = 1000\ \text{mol}$ → 1000 mol HNO₃. Mass HNO₃ $= 1000 \times 63 = 63000\ \text{g} = \mathbf{63\ kg}$ . Why: Track the N atom through all steps — 1:1 overall despite multiple stages.

10. (Tests 3.1.4 / 3.2.3 / 3.1.2 — hydrides, allotropes, isotopes) (a) $\text{B}_2\text{H}_6$ — covalent (electron-deficient) hydride; others are ionic (saline) hydrides. (b) quartz ( $\text{SiO}_2$ ) — a silicate, not a carbon allotrope; others are allotropes of carbon. (c) positronium — not an isotope of hydrogen (electron–positron system); protium/deuterium/tritium are. Why: Each part tests classifying by bonding/element identity, forcing recall of definitions.

[
  {"claim":"CaO needed to soften water = 56 g",
   "code":"mol_hardness=0.001*1000\nmol_CaO=mol_hardness*1\nmass=mol_CaO*(40+16)\nresult = (mass==56)"},
  {"claim":"HNO3 obtainable from 17 kg NH3 = 63 kg (1:1 mol)",
   "code":"mol_NH3=17000/17\nmol_HNO3=mol_NH3*1\nmass_g=mol_HNO3*(1+14+48)\nresult = (mass_g/1000==63)"},
  {"claim":"XeF4 has 6 electron pairs (4 bonding + 2 lone) -> sp3d2",
   "code":"bonding=4\nlone=(8-4)//2\ntotal=bonding+lone\nresult = (total==6 and lone==2)"}
]