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$H_2S$ gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because -
A
Presence of HCl decreases the sulphide ion concentration
B
Sulphides of group IV cations are unstable in HCl
C
Solubility product of group II sulphides is more than that of group IV sulphides
D
Presence of HCl increases the sulphide ion concentration
Detailed Solution
$H_2S$ is a weak acid: $H_2S \rightleftharpoons 2H^+ + S^{2-}$
HCl is a strong acid and supplies a large concentration of $H^+$ ions, a common ion.
By the common ion effect the ionisation of $H_2S$ is suppressed, so the concentration of $S^{2-}$ ions becomes very low.
The sulphides of group II cations (such as CuS, CdS, PbS) have very low solubility products, so even this low $[S^{2-}]$ makes the ionic product exceed $K_{sp}$ and they are precipitated.
The sulphides of group IV cations (such as ZnS, MnS, NiS) have much higher solubility products, so the ionic product does not exceed $K_{sp}$ and they are not precipitated in acidic medium.
So the reason is that the presence of HCl decreases the sulphide ion concentration.
HCl is a strong acid and supplies a large concentration of $H^+$ ions, a common ion.
By the common ion effect the ionisation of $H_2S$ is suppressed, so the concentration of $S^{2-}$ ions becomes very low.
The sulphides of group II cations (such as CuS, CdS, PbS) have very low solubility products, so even this low $[S^{2-}]$ makes the ionic product exceed $K_{sp}$ and they are precipitated.
The sulphides of group IV cations (such as ZnS, MnS, NiS) have much higher solubility products, so the ionic product does not exceed $K_{sp}$ and they are not precipitated in acidic medium.
So the reason is that the presence of HCl decreases the sulphide ion concentration.