JEE Main 2026 — d and f Block Elements Question with Solution

The first row transition metal (M) that does not liberate $H_2$ gas from dilute $HCl$ is Copper ($Cu$), as it has a positive reduction potential ($E_{C{u}^{2+}/Cu}^{\circ }=+0.34\text{ V}$).
When $1\text{ mol}$ of $CuS{O}_4$ reacts with excess aqueous $KCN$, it first forms $Cu(CN{)}_2$, which is unstable and decomposes to $CuCN$ and cyanogen gas $(CN{)}_2$.
$2CuS{O}_4+4KCN\to 2Cu(CN{)}_2+2K_{2}S{O}_4$
$2Cu(CN{)}_2\to 2CuCN+(CN{)}_2$
The $CuCN$ then dissolves in excess $KCN$ to form a highly stable soluble complex, potassium tetracyanocuprate(I):
$CuCN+3KCN\to K_3[Cu(CN{)}_4]$
The complex $[Cu(CN{)}_4{]}^{3-}$ is so stable that its dissociation constant is extremely low. When $H_{2}S$ gas is passed through this solution, the concentration of $C{u}^{+}$ ions is insufficient to exceed the solubility product of $C{u}_{2}S$.
Therefore, no precipitate of metal sulphide is formed.
The amount of $MS$ (or $M_{2}S$) formed is $0\text{ mol}$.