AS Chemistry 9701: Hydrolysis Equations for AlCl3 and SiCl4

Aluminium chloride and silicon tetrachloride are both chlorides of Period 3 elements, but each reacts differently when added to an excess of water.

(a) Aluminium chloride reacts with water to form a precipitate of aluminium hydroxide and hydrogen chloride. Construct the balanced symbol equation, including state symbols, for this reaction. [2]

(b) Silicon tetrachloride reacts with water to form solid silicon(IV) oxide and hydrogen chloride. Construct the balanced symbol equation, including state symbols, for this reaction. [2]

(c) Carbon tetrachloride, $\text{CCl}_4$ , is also a simple molecular tetrachloride, yet unlike $\text{SiCl}_4$ , it does not hydrolyse in water at all. Suggest, in terms of atomic structure, why $\text{SiCl}_4$ hydrolyses readily but $\text{CCl}_4$ does not. [3]

(d) State, with a brief reason, which of the two solutions formed in (a) and (b), from $\text{AlCl}_3$ or from $\text{SiCl}_4$ , has the lower pH. [1]

Part (a): Aluminium chloride with water

Aluminium chloride hydrolyses to form a white precipitate of aluminium hydroxide, releasing hydrogen chloride: $\text{AlCl}_3\text{(s)} + 3\text{H}_2\text{O(l)} \rightarrow \text{Al(OH)}_3\text{(s)} + 3\text{HCl(aq)}$

Checking the balance: $1$ Al atom, $3$ Cl atoms, $3$ O atoms, and $6$ H atoms on each side ( $3\times2$ on the left; $3$ in $\text{Al(OH)}_3$ $+\ 3$ in $3\text{HCl}$ on the right).

Part (b): Silicon tetrachloride with water

Silicon tetrachloride hydrolyses completely, forming solid silicon(IV) oxide and hydrogen chloride: $\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(aq)}$

Checking the balance: $1$ Si atom, $4$ Cl atoms, $2$ O atoms on each side, and $4$ H atoms ( $2\times2$ on the left; $4$ in $4\text{HCl}$ on the right).

Part (c): Why $\text{SiCl}_4$ hydrolyses but $\text{CCl}_4$ does not

Both $\text{SiCl}_4$ and $\text{CCl}_4$ are simple molecular tetrachlorides with similar shapes (tetrahedral) and polar bonds, so on electronegativity grounds alone both might be expected to hydrolyse. The key difference lies in the available orbitals of the central atom.

Silicon is in Period 3, so its valence shell ( $n=3$ ) includes empty $3d$ orbitals that are low enough in energy to be accessible. A water molecule’s oxygen atom can use one of its lone pairs to attack silicon by donating into one of these empty $3d$ orbitals, forming an intermediate that then loses $\text{HCl}$ . This gives hydrolysis a viable, low-energy mechanistic pathway.

Carbon is in Period 2, so its valence shell ( $n=2$ ) contains only $2s$ and $2p$ sub-shells. There is no $2d$ sub-shell at all. With no accessible empty orbital for water’s lone pair to attack, there is no pathway by which a water molecule can begin the hydrolysis reaction, so $\text{CCl}_4$ remains completely unreactive towards water even though the reaction would release energy overall.

Part (d): Comparing the pH of the two solutions

$\text{SiCl}_4$ reacts with water immediately and vigorously (it fumes visibly in moist air), releasing $4$ mol of HCl for every mole of $\text{SiCl}_4$ and hydrolysing essentially completely. $\text{AlCl}_3$ also hydrolyses, releasing $3$ mol of HCl per mole, but less vigorously.

Because $\text{SiCl}_4$ ‘s hydrolysis is more complete and releases a higher concentration of $\text{H}^+$ (via HCl) relative to the amount of chloride dissolved, the $\text{SiCl}_4$ solution has the lower pH (more strongly acidic) of the two.

Final answers

(a) $\text{AlCl}_3\text{(s)} + 3\text{H}_2\text{O(l)} \rightarrow \text{Al(OH)}_3\text{(s)} + 3\text{HCl(aq)}$
(b) $\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(aq)}$
(c) Silicon has accessible empty $3d$ orbitals that water’s oxygen lone pair can attack; carbon (n=2) has no accessible empty orbital, so it has no pathway for hydrolysis.
(d) $\text{SiCl}_4$ ‘s solution has the lower pH. Its hydrolysis is more complete and vigorous than $\text{AlCl}_3$ ‘s.