Describe ionic, covalent (including dative) and metallic bonding, predict molecular shapes and bond angles using VSEPR, account for bond polarity and overall polarity, and relate intermolecular forces (van der Waals, hydrogen bonding) to physical properties

What this dot point is asking

SEAB wants you to describe the three primary bonding types (ionic, covalent including dative, and metallic), use VSEPR theory to predict shapes and bond angles, judge whether a molecule is polar overall, and connect intermolecular forces to measurable properties such as boiling point and solubility. Shape prediction and the explanation of anomalous boiling points are perennial exam staples.

The answer

The three bonding types

Ionic bonding is the electrostatic attraction between oppositely charged ions formed by electron transfer (typically metal to non-metal). It gives giant ionic lattices: high melting points, brittle, conduct when molten or aqueous.

Covalent bonding is a shared pair of electrons between two atoms. A dative (coordinate) bond is a covalent bond where both electrons come from the same atom, as in $\text{NH}_4^+$ or the donor bond in $\text{H}_3\text{O}^+$.

Metallic bonding is the attraction between a lattice of positive ions and a sea of delocalised valence electrons. It explains electrical conductivity, malleability, and high melting points.

VSEPR: predicting shape

Valence Shell Electron Pair Repulsion theory says electron pairs around a central atom arrange themselves to minimise repulsion. The repulsion order is:

Count the electron pairs (regions of electron density), choose the geometry, then account for any lone pairs:

Each lone pair compresses bond angles by roughly $2.5$ degrees because of its greater repulsion.

Bond polarity and overall polarity

A bond is polar if the two atoms differ in electronegativity, giving partial charges. A molecule is polar overall only if the bond dipoles do not cancel by symmetry:

• $\text{CO}_2$ has two polar C=O bonds but is linear, so the dipoles cancel; it is non-polar.
• $\text{H}_2\text{O}$ is bent, so the dipoles do not cancel; it is polar.

Intermolecular forces

Three types, increasing in strength:

1. Instantaneous dipole-induced dipole (dispersion) forces. Present in all molecules; strengthen with more electrons (larger $M_r$) and larger surface contact.
2. Permanent dipole-dipole forces between polar molecules.
3. Hydrogen bonding, the strongest, occurs when H is bonded to N, O or F and there is a lone pair on a neighbouring electronegative atom.

These control boiling point, melting point, viscosity, and solubility. Like dissolves like: polar and hydrogen-bonding solutes dissolve in polar solvents.

The boiling-point anomalies

Down a group the hydrides should boil higher (more electrons, stronger dispersion forces), and they mostly do. But $\text{H}_2\text{O}$, $\text{HF}$ and $\text{NH}_3$ boil anomalously high because of hydrogen bonding. Water is the standout: each molecule has two H atoms and two lone pairs, so it forms an extensive hydrogen-bonded network.

Examples in context

Example 1. Comparing boiling points of isomers. A Paper 2 question compares the boiling points of butan-1-ol and diethyl ether, both $C_4H_{10}O$. Butan-1-ol has an O-H group and forms hydrogen bonds between molecules, so it boils much higher (about 118 degrees Celsius) than diethyl ether (about 35 degrees Celsius), which can only accept hydrogen bonds (no O-H) and relies mainly on dipole-dipole and dispersion forces. This shows that the functional group, not the molecular formula, sets the dominant intermolecular force.

Example 2. Explaining solubility. Ethanol is fully miscible with water because its O-H group hydrogen bonds with water, while hexane is immiscible because it is non-polar and can only form dispersion forces, which cannot disrupt water's hydrogen-bonded network. This like-dissolves-like reasoning is exactly what SEAB expects in solubility explanations.

Try this

Q1. Predict the shape and bond angle of (a) $\text{BF}_3$ and (b) $\text{PCl}_3$. [2+2 marks]

• Cue. (a) Trigonal planar, $120$ degrees, no lone pairs. (b) Trigonal pyramidal, about $107$ degrees, one lone pair on P.

Q2. State and explain whether $\text{CCl}_4$ is polar. [2 marks]

• Cue. Tetrahedral and symmetric; the four polar C-Cl dipoles cancel, so $\text{CCl}_4$ is non-polar.

Q3. Place $\text{CH}_4$, $\text{NH}_3$ and $\text{H}_2\text{O}$ in order of increasing boiling point and explain. [3 marks]

• Cue. $\text{CH}_4 < \text{NH}_3 < \text{H}_2\text{O}$. $\text{CH}_4$ has only dispersion forces; $\text{NH}_3$ and $\text{H}_2\text{O}$ hydrogen bond, and water forms more hydrogen bonds per molecule (two H and two lone pairs).