The cation that will not be precipitated by $$H_2S$$ in the presence of dil HCl is:

First, let us recall the basic idea used in salt-analysis. The precipitation of metal sulphides is brought about by passing hydrogen sulphide gas $$H_2S$$ through the solution containing the cation. The required sulphide ion concentration $$[S^{2-}]$$ is obtained from the dissociation

$$H_2S \; \rightleftharpoons \; 2H^+ \;+\; S^{2-}$$

The equilibrium expression is

$$K_a = \dfrac{[H^+]^2\,[S^{2-}]}{[H_2S]}.$$

Now, if we add dilute hydrochloric acid, we increase $$[H^+]$$. According to Le-Chatelier’s principle the equilibrium shifts to the left, thereby decreasing $$[S^{2-}]$$. Only those metal ions whose sulphides have extremely small solubility-product constants $$K_{sp}$$ can still get the necessary $$S^{2-}$$ and precipitate in this acidic medium. These ions constitute Group II of qualitative analysis.

The two sub-groups of Group II are

Group II A : $$Cu^{2+},\, Cd^{2+},\, Pb^{2+},\, Bi^{3+},\, Hg^{2+}\ ($$ and similar $$)$$

Group II B : $$As^{3+},\, As^{5+},\, Sb^{3+},\, Sb^{5+},\, Sn^{2+},\, Sn^{4+}.$$

All of these have very low $$K_{sp}$$ values for their sulphides, so they are precipitated by $$H_2S$$ even in the presence of dilute $$HCl$$.

Let us now examine the cations given in the options:

$$\begin{aligned} \text{Option A } & : Co^{2+} \\ \text{Option B } & : As^{3+} \\ \text{Option C } & : Pb^{2+} \\ \text{Option D } & : Cu^{2+} \end{aligned}$$

$$Co^{2+}$$ belongs to Group IV. The sulphide $$CoS$$ has a comparatively large solubility-product constant. In the acidic medium produced by dilute $$HCl$$, the concentration $$[S^{2-}]$$ becomes far smaller than that required to exceed $$K_{sp}(CoS)$$. Hence the ionic product

$$I.P = [Co^{2+}]\,[S^{2-}]$$

never reaches $$K_{sp}(CoS)$$ and no precipitate forms.

In contrast, $$As^{3+},\, Pb^{2+}$$ and $$Cu^{2+}$$ all have sulphides with such tiny $$K_{sp}$$ values that even the suppressed $$[S^{2-}]$$ in acidic medium is enough to satisfy

$$[M^{n+}]\,[S^{2-}] \ge K_{sp}(MS)$$

and their sulphides $$As_2S_3,\, PbS,$$ and $$CuS$$ respectively do precipitate.

Therefore, among the four ions only $$Co^{2+}$$ fails to precipitate with $$H_2S$$ in the presence of dilute $$HCl$$.

Hence, the correct answer is Option A.