Matching Qualitative Analysis Group Reagents with Metal Cations

To determine the correct match between the group reagents (List-I) and the metal cations (List-II), we analyze the principles of qualitative inorganic analysis based on group classification and solubility product constants ($K_{sp}$):

1. (S) Passing $\text{H}_2\text{S}$ in the presence of dilute $\text{HCl}$ (Group II Reagent): The presence of dilute $\text{HCl}$ (a strong acid) suppresses the ionization of $\text{H}_2\text{S}$ due to the common ion effect ($[\text{H}^+]$), keeping the sulfide ion concentration $[\text{S}^{2-}]$ sufficiently low so that only cations with very low $K_{sp}$ values precipitate as sulfides.

   • $\text{Cu}^{2+}$ belongs to Group II and precipitates as cupric sulfide ($\text{CuS}$).
   • Therefore, $\text{S} \rightarrow 1$.

2. (R) $\text{NH}_4\text{OH}$ in the presence of $\text{NH}_4\text{Cl}$ (Group III Reagent): The common ion effect from $\text{NH}_4\text{Cl}$ suppresses the dissociation of $\text{NH}_4\text{OH}$, providing a controlled concentration of hydroxide ions ($[\text{OH}^-]$) sufficient only to precipitate the less soluble metal hydroxides.

   • $\text{Al}^{3+}$ belongs to Group III and precipitates as aluminum hydroxide ($\text{Al(OH)}_3$).
   • Therefore, $\text{R} \rightarrow 2$.

3. (P) Passing $\text{H}_2\text{S}$ in the presence of $\text{NH}_4\text{OH}$ (Group IV Reagent): The basic medium provided by $\text{NH}_4\text{OH}$ facilitates higher dissociation of $\text{H}_2\text{S}$, resulting in a high concentration of $[\text{S}^{2-}]$, which allows the precipitation of metal sulfides having moderately higher $K_{sp}$ values.

   • $\text{Mn}^{2+}$ belongs to Group IV and precipitates as manganese(II) sulfide ($\text{MnS}$).
   • Therefore, $\text{P} \rightarrow 3$.

4. (Q) $(\text{NH}_4)_2\text{CO}_3$ in the presence of $\text{NH}_4\text{OH}$ (Group V Reagent): This reagent mixture precipitates the alkaline earth metal cations as their insoluble carbonates.

   • $\text{Ba}^{2+}$ belongs to Group V and precipitates as barium carbonate ($\text{BaCO}_3$).
   • Therefore, $\text{Q} \rightarrow 4$.

(Note: $\text{Mg}^{2+}$ belongs to Group VI and is precipitated using $(\text{NH}_4)_2\text{HPO}_4$ in the presence of $\text{NH}_4\text{OH}$ and $\text{NH}_4\text{Cl}$.)

Combining the matching results: $\text{P} \rightarrow 3; \quad \text{Q} \rightarrow 4; \quad \text{R} \rightarrow 2; \quad \text{S} \rightarrow 1$

This corresponds to Option (A).