OCR A Level Chemistry A: Bonding and structure (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

Questions in this set all come from section 2.2.2, Bonding and structure, from OCR’s A Level Chemistry A specification (H432, Version 3.1 dated May 2026; first assessment 2017), statements (a) to (o). H432 is the full A level with no tiers, and Module 2 is assessed in all three written components.

Before you start, read the study guide and the revision notes. Use the course hub or the checklist for the whole course to plan what comes next.

Questions

1. Covalent bonding.

(a) Define the term covalent bond. [2] (b) Define the term dative covalent bond. [1] (c) The average bond enthalpy of C=O is larger than that of C–O. State what this tells you about the two bonds. [1]

2. Strontium burns in nitrogen to form strontium nitride, Sr₃N₂. Draw a dot-and-cross diagram for strontium nitride, showing outer electrons only. [3]

3. Strontium bromide, SrBr₂, is an ionic compound.

(a) Describe the structure of solid strontium bromide. [2] (b) Explain why strontium bromide has a high melting point. [2] (c) Explain why strontium bromide conducts electricity when dissolved in water but not as a solid. [2]

4. Aluminium hydride, AlH₃, accepts a hydride ion, H⁻, to form the tetrahydridoaluminate ion, AlH₄⁻.

(a) Draw a dot-and-cross diagram for AlH₄⁻ and identify the dative covalent bond. [2] (b) Predict the shape of AlH₄⁻ and its bond angle. Show how you decided. [2] (c) Explain why all four Al–H bonds in AlH₄⁻ are identical. [1]

5. Use electron pair repulsion to answer this question.

(a) Predict the shape and bond angle of each species: AsF₆⁻, the methyl cation, CH₃⁺, and the methyl anion, CH₃⁻. [3] (b) Explain why the bond angle in CH₃⁻ is smaller than in CH₄. [2]

6. Use these Pauling electronegativity values: C 2.55, O 3.44, S 2.58, Cl 3.16.

(a) Define electronegativity. [1] (b) Carbonyl chloride, COCl₂, contains one C=O bond and two C–Cl bonds. State which type of bond is more polar, giving a reason. [1] (c) Explain why COCl₂ is a polar molecule but sulfur trioxide, SO₃, is not. [4]

7. Use these boiling points: CH₄ −162 °C, SiH₄ −112 °C, GeH₄ −88 °C.

(a) Describe and explain the trend in boiling point from CH₄ to GeH₄. [3] (b) Explain why there are no permanent dipole–dipole interactions between these molecules. [1]

8. A sealed plastic bottle holds 250 g of liquid water at 0 °C. Use these densities: liquid water 1.00 g cm⁻³, ice 0.917 g cm⁻³.

(a) Calculate the volume of the ice when all the water freezes, and the percentage increase in volume. Give both answers to 3 significant figures. [3] (b) Explain, in terms of hydrogen bonding, why ice has a lower density than liquid water. [3]

9. Ammonia, NH₃, forms hydrogen bonds.

(a) Draw a diagram to show one hydrogen bond between two ammonia molecules. Include relevant dipoles and lone pairs. [3] (b) Explain why ammonia is very soluble in water. [2]

10. Use these boiling points of the hydrogen halides: HF 19 °C, HCl −85 °C, HBr −67 °C, HI −35 °C, and these Pauling values: H 2.20, F 3.98, Cl 3.16, Br 2.96, I 2.66.

Explain the trend in boiling point from HCl to HI, and why HF has the highest boiling point. Refer to the types of intermolecular force present. [7]

11. Potassium iodide, KI, melts at 681 °C. Iodine, I₂, melts at 114 °C.

(a) Describe the structure of each solid. [2] (b) Explain the difference in melting points. [2] (c) Potassium iodide dissolves in water. Iodine dissolves well in hexane. Explain each observation. [2] (d) State and explain whether each substance conducts electricity when molten. [2]

Answers

1. (a) Strong electrostatic attraction [1] between a shared pair of electrons and the nuclei of the bonded atoms [1]. (b) A shared pair of electrons in which both electrons come from the same atom [1]. (c) The C=O bond is stronger than the C–O bond [1]. Examiner insight: a definition that says only “sharing electrons” is incomplete; state the attraction to both nuclei as well.

2. Three [Sr]²⁺ ions, each with an empty outer shell (or a full shell of eight crosses if the shell below is shown) [1]. Two [N]³⁻ ions, each with eight outer electrons: five dots and three crosses [1]. Square brackets with the charges 2+ and 3− outside, in the ratio 3 : 2 [1]. Examiner insight: shared pairs between Sr and N would show covalent bonding; keep every ion separate, in its own brackets.

3. (a) A giant ionic lattice [1] of Sr²⁺ and Br⁻ ions held by strong electrostatic attraction acting in all directions [1]. (b) Strong electrostatic attractions hold every Sr²⁺ to neighbouring Br⁻ ions [1], and breaking down the whole lattice takes a large amount of energy [1]. (c) In the solid the Sr²⁺ and Br⁻ ions are held in place [1]. In solution the ions are mobile and carry charge [1]. Examiner insight: mentioning moving electrons or intermolecular forces in an ionic compound is chemically wrong and can undermine an otherwise correct answer.

4. (a) Four shared pairs round Al and no lone pair; three pairs have one electron from Al and one from H, and the fourth has both from the hydride ion [1]. The bond formed from the H⁻ pair into the space on Al is the dative bond, H→Al [1]. Whole ion in brackets with a − charge. (b) 3 + 4 + 1 = 8 electrons: four bonding pairs, zero lone pairs [1]. Tetrahedral, 109.5° [1]. (c) Once formed, a dative bond is the same as any other covalent bond between those atoms [1]. Examiner insight: a bare shape does not show how you decided; state how many bonding and lone pairs there are.

5. (a) CH₃⁺: 4 + 3 − 1 = 6 electrons, three bonding pairs and no lone pairs, trigonal planar, 120° [1]. AsF₆⁻: 5 + 6 + 1 = 12 electrons, six bonding pairs, octahedral, 90° [1]. CH₃⁻: 4 + 3 + 1 = 8 electrons, three bonding pairs and one lone pair, pyramidal, 107° [1]. (b) The lone pair on carbon repels more than the bonding pairs do [1], pushing the three C–H bonds closer together, so the angle falls from 109.5° to about 107° [1]. Examiner insight: in (a), give the shape and the angle for every species; a correct shape with a wrong angle is incomplete.

6. (a) An atom’s ability to attract the bonding electrons (the shared pair) in a covalent bond [1]. (b) C=O, because the electronegativity difference (0.89) is larger than for C–Cl (0.61) [1]. (c) COCl₂ has three bonding regions round C, so it is trigonal planar [1]. The C=O and C–Cl dipoles are different in size, so they do not cancel and there is an overall dipole [1]. In SO₃ the S=O bonds are polar (difference 0.86) [1], but the three identical bonds sit at 120° in a symmetrical trigonal planar molecule, so their dipoles cancel and there is no overall dipole [1]. Examiner insight: when comparing polarity, deal with both bond polarity and shape for each molecule, not just one of them.

7. (a) Boiling point increases from CH₄ to GeH₄ [1]. The number of electrons increases (10, 18, 36), so the induced dipole–dipole (London) forces get stronger [1]. More energy is needed to overcome them [1]. (b) All three are symmetrical tetrahedral molecules, so any bond dipoles cancel and each molecule is non-polar [1]. Examiner insight: explain London forces through the number of electrons; “heavier molecules” does not explain the trend.

8. (a) V = m ÷ ρ = 250 ÷ 0.917 [1] = 273 cm³ [1]. Increase = 272.63 − 250 = 22.63 cm³; percentage = 22.63 ÷ 250 × 100 = 9.05% [1]. (b) In ice each H₂O molecule forms hydrogen bonds to four others [1] in a tetrahedral arrangement, holding the molecules in an open lattice with large spaces [1]. When ice melts, some hydrogen bonds break and the molecules move closer together, so water is denser [1]. Examiner insight: carry unrounded values through the calculation and round only at the end; rounding 272.6 to 273 before subtracting gives 9.20%, which does not match.

9. (a) Two NH₃ molecules with δ− on N and δ+ on H [1]. A lone pair drawn on the N of one molecule [1]. A dashed line from that lone pair to an H on the other molecule, roughly in line with the N–H bond [1]. (b) Ammonia forms hydrogen bonds with water molecules [1]: the N lone pair to an H of water, and an O lone pair of water to an H of ammonia [1]. Examiner insight: draw the hydrogen bond from the lone pair on N to the H; a line to the N atom in general misses the role of the lone pair.

10. All the hydrogen halides have London forces between molecules [1]. From HCl to HI the number of electrons increases (18, 36, 54), so the London forces get stronger [1], and more energy is needed to separate the molecules [1]. All are polar and have permanent dipole–dipole interactions, but these get weaker from HCl to HI as the electronegativity difference falls (0.96, 0.76, 0.46), so the rise in London forces controls the trend [1]. HF has hydrogen bonds because F is highly electronegative (difference 1.78), making H strongly δ+ [1], and the H is attracted to a lone pair on F of a neighbouring molecule [1]. Hydrogen bonds are stronger than the other intermolecular forces, so HF has the highest boiling point despite having the fewest electrons (10) [1]. Examiner insight: in an extended answer, link each force to its effect on boiling point and say which force controls the trend; a bare list of forces explains little.

11. (a) KI: a giant ionic lattice of K⁺ and I⁻ ions [1]. I₂: a simple molecular lattice of I₂ molecules held by London forces [1]. (b) Melting KI means overcoming strong attractions between oppositely charged ions throughout the lattice [1]. Melting I₂ only overcomes weak London forces between molecules; the I–I covalent bonds do not break [1]. (c) Polar water molecules attract and surround the K⁺ and I⁻ ions, pulling them out of the lattice [1]. I₂ and hexane are both non-polar; London forces form between them, similar to those broken [1]. (d) Molten KI conducts because its ions are mobile [1]. Molten iodine does not conduct because it has no ions or delocalised electrons [1]. Examiner insight: writing that covalent bonds break when iodine melts is a common error; say clearly that only intermolecular forces are overcome.

Where marks are usually lost

• Leaving lone pairs off the central atom, which also leads to the wrong shape.
• Writing the charge inside the square brackets of an ion.
• Treating a double bond as two regions when counting pairs for shape.
• Giving the electron-pair arrangement (tetrahedral) instead of the shape (pyramidal or non-linear).
• Deciding on polarity from the bonds alone and ignoring the symmetry of the shape.
• Comparing London forces by mass rather than by number of electrons.
• Drawing hydrogen bonds to hydrogens bonded to carbon, or leaving out the lone pair.
• Writing that iodine’s I–I bonds break on melting.
• Rounding intermediate values in density calculations.

Next steps

• Revision notes and quick self-test
• All free 10-minute diagnostics
• Book a free trial class
• Bonding and structure study guide
• Every OCR A Level Chemistry page, on the course hub
• Checklist to print and tick

Official syllabus

OCR, A Level Chemistry A specification (H432), Version 3.1 dated May 2026, first assessment 2017: section 2.2.2 Bonding and structure.