A first row transition metal (M) does not liberate H_ 2 gas from...

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 ($E^\circ_{Cu^{2+}/Cu} = +0.34\text{ V}$).
When $1\text{ mol}$ of $CuSO_4$ reacts with excess aqueous $KCN$, it first forms $Cu(CN)_2$, which is unstable and decomposes to $CuCN$ and cyanogen gas $(CN)_2$.
$2CuSO_4 + 4KCN \rightarrow 2Cu(CN)_2 + 2K_2SO_4$
$2Cu(CN)_2 \rightarrow 2CuCN + (CN)_2$
The $CuCN$ then dissolves in excess $KCN$ to form a highly stable soluble complex, potassium tetracyanocuprate(I):
$CuCN + 3KCN \rightarrow K_3[Cu(CN)_4]$
The complex $[Cu(CN)_4]^{3-}$ is so stable that its dissociation constant is extremely low. When $H_2S$ gas is passed through this solution, the concentration of $Cu^+$ ions is insufficient to exceed the solubility product of $Cu_2S$.
Therefore, no precipitate of metal sulphide is formed.
The amount of $MS$ (or $M_2S$) formed is $0\text{ mol}$.