Aluminium hydroxide, formed as a white precipitate with a little NaOH, dissolves in excess NaOH. On adding solid $\mathrm{NH}{}_4\mathrm{Cl}$ to this clear solution and boiling, a white gelatinous precipitate reappears because

1. A$\mathrm{NH}{}_4{}^{+}$ removes $\mathrm{OH}{}^{-}$ as $\mathrm{NH}{}_3$, so $[\mathrm{Al}(\mathrm{OH}){}_4]{}^{-}$ reverts to $\mathrm{Al}(\mathrm{OH}){}_3$
2. B$\mathrm{NH}{}_4\mathrm{Cl}$ reacts with $\mathrm{Al}$ to give $\mathrm{AlCl}{}_3$ which precipitates
3. C$\mathrm{NH}{}_4\mathrm{Cl}$ oxidises $\mathrm{Al}{}^{3+}$ to a higher state
4. Dboiling decomposes $[\mathrm{Al}(\mathrm{OH}){}_4]{}^{-}$ to $\mathrm{Al}{}_2\mathrm{O}{}_3$ only