The residue of group III hydroxides is treated with NaOH and $\mathrm{H}{}_2\mathrm{O}{}_2$ and boiled. The mixture is filtered, and the yellow filtrate (containing $[\mathrm{Al}(\mathrm{OH}){}_4]{}^{-}$ and $\mathrm{CrO}{}_4{}^{2-}$) is boiled with solid $\mathrm{NH}{}_4\mathrm{Cl}$. The residue left on the first filter paper, and the precipitate that appears after adding $\mathrm{NH}{}_4\mathrm{Cl}$, respectively are

1. Agreen $\mathrm{Cr}(\mathrm{OH}){}_3$; white $\mathrm{Al}(\mathrm{OH}){}_3$
2. Bwhite $\mathrm{Al}(\mathrm{OH}){}_3$; brown $\mathrm{Fe}(\mathrm{OH}){}_3$
3. Cbrown $\mathrm{Fe}(\mathrm{OH}){}_3$; white $\mathrm{Al}(\mathrm{OH}){}_3$
4. Dbrown $\mathrm{Fe}(\mathrm{OH}){}_3$; yellow $\mathrm{PbCrO}{}_4$