In a quantitative analysis of second group in the lab, $H_2S$ gas is passed in an acidic medium for ppt. When $Cu^{2+}$ and $Cd^{2+}$ react with $KCN$, then in which of the following conditions will a precipitate not be formed due to relative stability:

When KCN is added to a mixture of $Cu^{2+}$ and $Cd^{2+}$, both form cyanide complexes. $Cu^{2+}$ is reduced to $Cu^+$ by the $CN^-$ ions and forms the highly stable complex $K_3[Cu(CN)_4]$ (potassium tetracyanidocuprate(I)). $Cd^{2+}$ forms a relatively less stable complex, $K_2[Cd(CN)_4]$ (potassium tetracyanidocadmate(II)).

$2Cu^{2+} + 10CN^- \rightarrow 2[Cu(CN)_4]^{3-} + (CN)_2 \uparrow$ $Cd^{2+} + 4CN^- \rightarrow [Cd(CN)_4]^{2-}$

When $H_2S$ gas is passed through this solution, the highly stable $K_3[Cu(CN)_4]$ does not dissociate to give enough $Cu^+$ ions to form a precipitate of $Cu_2S$. However, $K_2[Cd(CN)_4]$ is less stable and dissociates to give $Cd^{2+}$ ions, which react with $H_2S$ to form a yellow precipitate of $CdS$. Thus, Cu does not precipitate due to the higher stability of $K_3[Cu(CN)_4]$ relative to $K_2[Cd(CN)_4]$.