Compound 'A' in acidic medium does not give a precipitate with $H_2S$ but in $NH_4OH$ medium gives a precipitate. Compound 'A' is:

This question is based on the principles of Qualitative Analysis of cations and the Solubility Product ($K_{sp}$).

1. Reagent Function: Hydrogen sulphide ($H_2S$) is used to precipitate metal sulphides. The concentration of sulphide ions ($S^{2-}$) depends on the pH of the medium.
2. Acidic Medium: In the presence of acid (like $HCl$), the dissociation of $H_2S$ is suppressed due to the Common Ion Effect. The sulphide ion concentration is low, sufficient only to precipitate Group II cations (like $Cu^{2+}, Pb^{2+}, Sn^{2+}$) which have very low $K_{sp}$ values. $SnCl_2$ would precipitate as $SnS$ here, so it is incorrect.
3. Basic Medium ($NH_4OH$): In the presence of a base, the $H^+$ ions from $H_2S$ are removed, shifting the equilibrium forward and increasing the sulphide ion concentration. This high $[S^{2-}]$ exceeds the $K_{sp}$ of Group IV cations (like $Zn^{2+}, Mn^{2+}, Ni^{2+}$), causing them to precipitate.
4. Analysis of Options: $SnCl_2$: Precipitates in acidic medium (Group II). $ZnCl_2$: Does not precipitate in acidic medium (high $K_{sp}$ of $ZnS$) but precipitates as Zinc Sulphide ($ZnS$, white) in alkaline/ammoniacal medium .

• $FeCl_3$ and $AlCl_3$: Group III cations usually precipitate as hydroxides with $NH_4OH$ directly, but the specific context of $H_2S$ acidic vs. basic differentiation points to the Group II vs. Group IV sulphide distinction, where Zinc is the classic example.