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 K₃[Fe(CN)₆] with freshly prepared FeSO₄ solution produces a dark blue precipitate called Turnbull's blue. Reaction of K₄[Fe(CN)₆] with the FeSO₄ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO₄ solution with NaNO₃, followed by a slow addition of concentrated H₂SO₄ through the side of the test tube produces a brown ring.

Precipitate X is

The reaction of K₃[Fe(CN)₆] with freshly prepared FeSO₄ solution produces a dark blue precipitate called Turnbull's blue. Reaction of K₄[Fe(CN)₆] with the FeSO₄ solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO₄ solution with NaNO₃, followed by a slow addition of concentrated H₂SO₄ through the side of the test tube produces a brown ring.

Among the following, the brown ring is due to the formation of