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

An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarised in the scheme given below:

SCHEME:

M1, Q and R, respectively, are