OCR A Level Chemistry A: Bonding and structure (H432)

This guide teaches 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). It covers every lettered statement, (a) to (o), with the additional guidance. H432 is the full A level with no tiers, and Module 2 content is assessed in all three written components (01, 02 and 03).

Alongside this guide, use the condensed notes plus the practice set with worked answers. Every page for this course is gathered on the course hub, and each statement can be ticked off on the OCR chemistry checklist. Electron configurations are covered in the OCR electron structure guide. A short diagnostic test will show your gaps before you begin.

What this unit covers

2.2.2	What you must be able to do	Links
(a)	Ionic bonding as attraction between ions; dot-and-cross diagrams
(b)	Giant ionic lattices, e.g. NaCl
(c)	Melting and boiling points, solubility and conductivity of ionic compounds	HSW1
(d)	Covalent bond as attraction between a shared pair and both nuclei
(e)	Dot-and-cross diagrams: single, multiple and dative bonds, up to six pairs
(f)	Average bond enthalpy as a measure of bond strength	3.2.1 f
(g)	Shapes and bond angles, up to six pairs; 3-D diagrams	M4.1, M4.2, HSW1,2
(h)	Linear, non-linear, trigonal planar, pyramidal, tetrahedral, octahedral
(i)	Electronegativity; Pauling values	HSW1,2
(j)	Polar bonds, permanent dipoles, polar molecules
(k)	Permanent dipole–dipole and induced dipole–dipole (London) forces	HSW1,2
(l)	Hydrogen bonding, including the role of lone pairs
(m)	Density of ice; high melting and boiling points of water	HSW1
(n)	Simple molecular lattices, e.g. I₂ and ice
(o)	Properties of simple molecular substances

The specification links no PAG to this section.

Ionic bonding: (a) to (c)

An ionic bond is the electrostatic attraction between positive and negative ions. Metal atoms lose electrons to form cations; non-metal atoms gain them to form anions.

In a dot-and-cross diagram, enclose each ion in square brackets and write its charge at the top right, outside the brackets. Use dots for one element’s electrons and crosses for the other’s. Show outer shells only unless asked.

Worked example 1: barium fluoride, BaF₂

Ba (Group 2) has two outer electrons; F (Group 17) has seven. One Ba gives one electron to each of two F atoms, so Ba : F = 1 : 2. Draw [Ba]²⁺ with an empty outer shell, and two [F]⁻ ions each with eight outer electrons: seven dots and one cross.

Giant ionic lattice. In solid NaCl each Na⁺ is surrounded by six Cl⁻ ions and each Cl⁻ by six Na⁺. The oppositely charged ions attract strongly in all directions, building one huge three-dimensional lattice. There are no molecules.

Property	Ionic compounds	Why
Melting and boiling points	High	Much energy is needed to overcome the strong attractions between oppositely charged ions throughout the lattice. Higher charges and smaller ions attract more strongly.
Solubility	Many dissolve in water	Polar water molecules attract the ions (δ− O towards cations, δ+ H towards anions), pull them from the lattice and surround them. Some, such as calcium carbonate, are insoluble because the lattice attractions are too strong.
Conductivity as a solid	None	Ions are fixed in the lattice.
Conductivity when molten or aqueous	Conducts	Ions are mobile and carry charge.

Covalent bonding: (d) to (f)

A covalent bond is the strong electrostatic attraction between a shared pair of electrons and the nuclei of the bonded atoms.

In a dot-and-cross diagram, put shared pairs where the outer shells overlap and show every lone pair.

• Single bond: one shared pair. Double or triple bond: two or three shared pairs.
• Dative covalent (coordinate) bond: a shared pair whose two electrons are both supplied by one atom. The donor needs a lone pair; the acceptor needs space for it. In a displayed formula, draw it as an arrow from donor to acceptor.

Worked example 2: sulfur dichloride, SCl₂. S shares one electron with each Cl. Around S: two bonding pairs and two lone pairs. Each Cl: one bonding pair, three lone pairs.

Worked example 3: carbonyl sulfide, OCS. C shares two of its electrons with O and the other two with S, giving O=C=S. C has four bonding pairs and no lone pairs; O and S each keep two lone pairs.

Worked example 4: H₃N→BCl₃. Boron in BCl₃ has only three bonding pairs (six electrons), so it can accept a pair. Nitrogen in NH₃ donates its lone pair into that space, forming N→B. Boron now has four bonding pairs. Once formed, a dative bond is identical to any other covalent bond between the same atoms.

From Period 3 onwards, the central atom can hold more than eight electrons: SeF₆ has six bonding pairs round Se.

Average bond enthalpy measures covalent bond strength: the larger the value, the stronger the bond. Carbon–carbon values rise from C–C to C=C to C≡C. No definition or calculation is needed here; that comes in 3.2.1.

Shapes of molecules and ions: (g) and (h)

Electron pairs round a central atom repel and move as far apart as possible. Lone pairs are closer to the central atom and repel more. Two lone pairs repel most, then a lone pair and a bonding pair, and two bonding pairs least. Each lone pair closes the bond angle by about 2.5°.

Pairs	Bonding	Lone	Shape	Angle	Example
2	2	0	Linear	180°	BeCl₂
3	3	0	Trigonal planar	120°	BCl₃
4	4	0	Tetrahedral	109.5°	CH₄
4	3	1	Pyramidal	107°	NH₃
4	2	2	Non-linear	104.5°	H₂O
6	6	0	Octahedral	90°	SeF₆

A double or triple bond counts as one bonding region, so OCS, with two double bonds, is linear (180°). With six pairs of which two are lone pairs, as in XeF₄, the lone pairs sit opposite each other and the shape is square planar (90°).

Method

1. Take the central atom’s outer electrons.
2. Add one per singly bonded atom.
3. Add one per negative charge; subtract one per positive charge.
4. Halve it for the number of pairs. Subtract the bonded atoms to get the lone pairs.
5. Name the shape from the positions of the atoms, then give the angle.

Worked example 5: SCl₂. 6 + 2 = 8 electrons, 4 pairs: 2 bonding, 2 lone. Non-linear, 104.5°, the same arrangement as in water.

Worked example 6: SnCl₆²⁻. Tin is in Group 14: 4 + 6 + 2 = 12 electrons, 6 pairs, all bonding. Octahedral, 90°.

3-D diagrams. A solid line is in the plane of the paper, a solid wedge comes towards you and a dashed wedge goes away. A tetrahedral molecule has two lines, one wedge and one dashed wedge. Show lone pairs on pyramidal and non-linear species.

Electronegativity and polarity: (i) and (j)

Electronegativity is an atom’s ability to attract the bonding electrons in a covalent bond. Values rise towards fluorine: across a period and up a group. Use these Pauling values:

H	C	N	O	F	Si	S	Cl	Ba
2.20	2.55	3.04	3.44	3.98	1.90	2.58	3.16	0.89

Identical atoms (difference 0) give a non-polar bond, a small difference gives a polar covalent bond and a large difference means mostly ionic bonding. C–H differs by 0.35, C–F by 1.43 and Ba–F by 3.09, which fits BaF₂ being ionic.

A polar bond forms between atoms of different electronegativity. The more electronegative atom pulls the shared pair towards itself and becomes δ−; the other becomes δ+. This is a permanent dipole. A polar molecule needs polar bonds whose dipoles do not cancel because of the shape. The specification’s example: H₂O and CO₂ both have polar bonds, but linear CO₂ has no overall dipole, while non-linear H₂O does.

Worked example 7: SiH₄ and SiH₃Cl. Both are tetrahedral. In SiH₄ four identical Si–H dipoles point symmetrically and cancel: non-polar. In SiH₃Cl one bond is the much more polar Si–Cl (difference 1.26, against 0.30 for Si–H), so the dipoles do not cancel: polar.

Intermolecular forces: (k) and (l)

Intermolecular forces act between molecules and are much weaker than covalent bonds within them.

Induced dipole–dipole interactions (London forces). Moving electrons create an uneven charge at any instant: an instantaneous dipole. It induces a dipole in a neighbour, and the two attract. London forces act between all molecules and get stronger with more electrons. Use these boiling points: Ne −246 °C, Ar −186 °C, Kr −153 °C, Xe −108 °C. Down the group each atom has more electrons, so the forces and boiling points increase.

Permanent dipole–dipole interactions. The δ+ end of one polar molecule attracts the δ− end of another, in addition to London forces. Both types can be called van der Waals’ forces.

Worked example 8: H₂S and SiH₄. Each has 18 electrons, so London forces are similar. Non-linear H₂S is polar; symmetrical SiH₄ is not. H₂S boils at −60 °C, 52 °C above SiH₄ (−112 °C), because of its permanent dipole–dipole interactions.

Hydrogen bonding happens between a lone pair on N, O or F in one molecule and the H of an –NH, –OH or HF group in another. N, O and F are so electronegative, and H so small, that the H is strongly δ+ and can get close to the lone pair. In a diagram, label δ+ and δ−, draw the lone pair, and draw the hydrogen bond as a dashed line from the lone pair to the H, with O–H···O close to 180°. Hydrogen bonds are the strongest intermolecular force.

Anomalous properties of water: (m)

Ice is less dense than water. In ice each molecule hydrogen bonds to four others, tetrahedrally, holding them in an open lattice with lots of space. On melting, some hydrogen bonds break and the molecules move closer. Use ice 0.917 g cm⁻³ and water 1.00 g cm⁻³: water expands by about 9% when it freezes, and ice floats.

High melting and boiling points. Use these boiling points: H₂S −60 °C, H₂Se −41 °C, H₂O 100 °C. From H₂S to H₂Se the rise comes from more electrons and stronger London forces. On that trend water would boil well below −60 °C. It actually boils 160 °C above H₂S because hydrogen bonds must also be broken.

Simple molecular lattices: (n) and (o)

A simple molecular lattice is a regular arrangement of small covalent molecules held by intermolecular forces: I₂ molecules held by London forces in solid iodine, and H₂O molecules held by hydrogen bonds in ice.

Property	Simple molecular	Why
Melting and boiling points	Low	Only weak intermolecular forces are overcome; covalent bonds do not break.
Solubility	Non-polar substances dissolve in non-polar solvents (iodine in hexane); substances that hydrogen bond, such as simple alcohols, dissolve in water	Similar intermolecular forces form between solute and solvent.
Conductivity	None in any state	No ions or delocalised electrons.

Worked example 9: methanol and ethane. Both have 18 electrons. Methanol has an –OH group and hydrogen bonds; ethane has only London forces. Methanol boils at 65 °C, ethane at −89 °C: a 154 °C gap from the hydrogen bonds.

Common errors

• Writing “molecules” or “intermolecular forces” for an ionic lattice.
• Claiming that boiling iodine or water splits its covalent bonds.
• Missing lone pairs, or writing charges inside the brackets.
• Counting a double bond as two regions when predicting shape.
• Calling NH₃ tetrahedral: name the shape from the atoms.
• Drawing a hydrogen bond to an H bonded to carbon.
• Explaining London forces by mass rather than number of electrons.

Next steps

Revise with the revision notes, then test yourself on the practice set, or take a 10-minute diagnostic.

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.