A first row transition metal (M) does not lib | vedclass

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(D) The metal $M$ that does not liberate $H_{2}$ gas from dilute $HCl$ is $Cu$ (Copper), as it has a positive standard reduction potential $(E^{\circ}

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(D) The metal $M$ that does not liberate $H_{2}$ gas from dilute $HCl$ is $Cu$ (Copper), as it has a positive standard reduction potential $(E^{\circ}_{Cu^{2+}/Cu} = +0.34 \ V)$.When $CuSO_{4}$ is treated with excess $KCN$, it forms a very stable complex $[Cu(CN)_{4}]^{3-}$.The reaction is: $2CuSO_{4} + 10KCN \rightarrow 2K_{3}[Cu(CN)_{4}] + (CN)_{2} + 2K_{2}SO_{4}$.Because the complex $[Cu(CN)_{4}]^{3-}$ is extremely stable (a 'perfect' complex), the concentration of free $Cu^{2+}$ ions in the solution is negligible.Therefore, when $H_{2}S_{(g)}$ is passed through this solution, the ionic product does not exceed the solubility product $(K_{sp})$ of $CuS$, and no precipitate of $CuS$ is formed.Thus, the amount of $MS$ formed is $0 \ mol$.

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