The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions, is:
| List-I | List-II |
|---|---|
| P. Passing \( \mathrm{H_2S} \) in the presence of \( \mathrm{NH_4OH} \) | 1. \( \mathrm{Cu^{2+}} \) |
| Q. \( \mathrm{(NH_4)_2CO_3} \) in the presence of \( \mathrm{NH_4OH} \) | 2. \( \mathrm{Al^{3+}} \) |
| R. \( \mathrm{NH_4OH} \) in the presence of \( \mathrm{NH_4Cl} \) | 3. \( \mathrm{Mn^{2+}} \) |
| S. Passing \( \mathrm{H_2S} \) in the presence of dilute \( \mathrm{HCl} \) | 4. \( \mathrm{Ba^{2+}} \) |
| 5. \( \mathrm{Mg^{2+}} \) |
The reaction of \( \mathrm{K_3[Fe(CN)_6]} \) with freshly prepared \( \mathrm{FeSO_4} \) solution produces a dark blue precipitate called Turnbull’s blue. Reaction of \( \mathrm{K_4[Fe(CN)_6]} \) with the \( \mathrm{FeSO_4} \) solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the \( \mathrm{FeSO_4} \) solution with \( \mathrm{NaNO_3} \), followed by a slow addition of concentrated \( \mathrm{H_2SO_4} \) through the side of the test tube produces a brown ring.
Precipitate X is:
