Chemical Bonding: Question 3

Worked solution

Part (a): Shape and bond angle of ammonia

Nitrogen has $5$ outer-shell electrons. In $\text{NH}_3$, $3$ of these electrons pair up with the $3$ hydrogen electrons to form $3$ bonding pairs, leaving nitrogen’s remaining $2$ electrons as $1$ lone pair.

$\text{Total electron pairs around N} = 3 \text{ (bonding)} + 1 \text{ (lone)} = 4$

By VSEPR theory, $4$ electron pairs arrange themselves as far apart as possible, i.e. towards the corners of a tetrahedron, to minimise repulsion. However, the shape of a molecule is described by the positions of the atoms only, not the lone pair. With $3$ bonding pairs and $1$ lone pair, the molecular shape is pyramidal (trigonal pyramidal), and the H–N–H bond angle is $\boxed{107^\circ}$ (slightly less than the perfect tetrahedral angle of $109.5^\circ$, for the reason explored in part (c)).

Part (b): Shape and bond angles of PCl₅

Phosphorus forms $5$ bonding pairs to the $5$ chlorine atoms and has no lone pairs remaining, so all $5$ electron pairs are bonding pairs. VSEPR theory places $5$ electron pairs as far apart as possible in a trigonal bipyramidal arrangement: $3$ chlorine atoms lie in a triangular “equatorial” plane around phosphorus, and $2$ further chlorine atoms lie above and below this plane in “axial” positions.

Because the axial and equatorial positions are not geometrically equivalent, two different bond angles exist in the same molecule:

• $\boxed{90^\circ}$ between an axial Cl atom and any equatorial Cl atom.
• $\boxed{120^\circ}$ between any two of the three equatorial Cl atoms.

This is different from a tetrahedral or octahedral species, where every position is equivalent and only one bond angle exists.

Part (c): Why NH₃’s bond angle is smaller than CH₄’s

Both $\text{NH}_3$ and $\text{CH}_4$ have $4$ electron pairs around the central atom, so both start from the same underlying tetrahedral arrangement of electron pairs (ideal angle $109.5^\circ$). The difference is in what those $4$ pairs consist of:

• In $\text{CH}_4$, all $4$ pairs are bonding pairs (to $4$ H atoms). All repulsions are bonding pair–bonding pair, which are relatively weak and equal in every direction, so the tetrahedron stays perfectly regular at $109.5^\circ$.
• In $\text{NH}_3$, $3$ pairs are bonding but $1$ is a lone pair. A lone pair is held closer to the central nucleus (it is not shared with another atom, so its electron density is more concentrated near N) and so repels neighbouring electron pairs more strongly than a bonding pair does.

The repulsion strength order is: $\text{lone pair–lone pair} > \text{lone pair–bonding pair} > \text{bonding pair–bonding pair}$

Because $\text{NH}_3$‘s lone pair pushes the three N–H bonding pairs closer together, the H–N–H angle is compressed slightly below the ideal tetrahedral value, from $109.5^\circ$ down to $107^\circ$.

Final answers

• (a) $4$ electron pairs ($3$ bonding $+ 1$ lone); shape $=$ pyramidal; bond angle $= \boxed{107^\circ}$.
• (b) Shape $=$ trigonal bipyramidal; bond angles $= \boxed{90^\circ}$ (axial–equatorial) and $\boxed{120^\circ}$ (equatorial–equatorial), because the axial and equatorial positions are not equivalent.
• (c) $\text{NH}_3$‘s lone pair repels more strongly than a bonding pair, compressing the angle from $109.5^\circ$ (in $\text{CH}_4$, all bonding pairs) to $107^\circ$.