Identify each group reagent from classical salt analysis

In qualitative inorganic analysis:

• Passing $\mathrm{H_2S}$ in acidic medium precipitates Group II cations as sulfides.
• Passing $\mathrm{H_2S}$ in ammoniacal/basic medium precipitates Group IV cations as sulfides.
• $\mathrm{NH_4OH}$ in presence of $\mathrm{NH_4Cl}$ precipitates Group III cations as hydroxides.
• $\mathrm{(NH_4)_2CO_3}$ in presence of $\mathrm{NH_4OH}$ precipitates Group V cations as carbonates.

Match each reagent with the correct metal ion

(P) Passing $\mathrm{H_2S}$ in the presence of $\mathrm{NH_4OH}$

This is Group IV reagent.

Group IV cations include ions like $\mathrm{Zn^{2+}}$, $\mathrm{Mn^{2+}}$, $\mathrm{Ni^{2+}}$, $\mathrm{Co^{2+}}$.

From the list, the suitable ion is: $\mathrm{Mn^{2+}} \Rightarrow (3)$

So, $P \to 3$

(Q) $\mathrm{(NH_4)_2CO_3}$ in the presence of $\mathrm{NH_4OH}$

This is Group V reagent.

Group V cations such as $\mathrm{Ba^{2+}}$, $\mathrm{Sr^{2+}}$, $\mathrm{Ca^{2+}}$ precipitate as carbonates.

From the list: $\mathrm{Ba^{2+}} \Rightarrow (4)$

So, $Q \to 4$

(R) $\mathrm{NH_4OH}$ in the presence of $\mathrm{NH_4Cl}$

This is Group III reagent.

Group III cations like $\mathrm{Fe^{3+}}$, $\mathrm{Al^{3+}}$, $\mathrm{Cr^{3+}}$ precipitate as hydroxides.

From the list: $\mathrm{Al^{3+}} \Rightarrow (2)$

So, $R \to 2$

(S) Passing $\mathrm{H_2S}$ in the presence of dilute $\mathrm{HCl}$

This is Group II reagent.

Group II cations include $\mathrm{Cu^{2+}}$, $\mathrm{Cd^{2+}}$, $\mathrm{Pb^{2+}}$, $\mathrm{Hg^{2+}}$, etc.

From the list: $\mathrm{Cu^{2+}} \Rightarrow (1)$

So, $S \to 1$

Final matching

$P \to 3, \quad Q \to 4, \quad R \to 2, \quad S \to 1$

Compare with options

This corresponds to Option A.

Comparison with stored correct answer

Stored correct answer = A

Derived answer = A

Hence, they agree.