Which statement correctly explains why $\mathrm{Al}{}^{3+}$ is NOT detected as a sulphide when $\mathrm{H}{}_2\mathrm{S}$ is passed through the ammoniacal solution (group IV step) containing it?

1. A$\mathrm{Al}{}^{3+}$ is oxidised to $\mathrm{Al}{}^{4+}$ by $\mathrm{H}{}_2\mathrm{S}$
2. B$\mathrm{Al}{}_2\mathrm{S}{}_3$ is completely hydrolysed, so white $\mathrm{Al}(\mathrm{OH}){}_3$ is obtained, with release of $\mathrm{H}{}_2\mathrm{S}$
3. C$\mathrm{Al}{}_2\mathrm{S}{}_3$ is a yellow precipitate, so it cannot be mistaken for $\mathrm{Al}(\mathrm{OH}){}_3$
4. D$\mathrm{Al}{}_2\mathrm{S}{}_3$ is amphoteric and dissolves in ammonia