JEE Chemistry 2026 Question with Solution

Given: A first row transition metal $M$ does not liberate $\mathrm{H_2}$ from dilute HCl. $1$ mol of aqueous $\mathrm{MSO_4}$ is treated with excess aqueous KCN and then $\mathrm{H_2S(g)}$ is passed through the solution.

Find: The amount of $\mathrm{MS}$ formed.

The metal is Cu, because copper does not displace hydrogen from dilute acids.

With excess KCN, copper forms a stable cyanide complex:

On passing $\mathrm{H_2S}$, copper sulphide precipitates. Each mole of $\mathrm{Cu^{2+}}$ gives $1$ mole of $\mathrm{CuS}$.

Therefore, from $1$ mol of $\mathrm{MSO_4}$, the amount of $\mathrm{MS}$ formed is $1$ mol. Hence, the correct option is A.

Given: The first-row transition metal does not liberate $\mathrm{H_2}$ with dilute HCl.

Find: Moles of sulphide formed after treatment with KCN and $\mathrm{H_2S}$.

Step 1: Identify the metal. Among first-row transition metals, Cu is known for not liberating hydrogen gas from dilute HCl.

Step 2: Copper initially forms a cyanide complex in excess KCN:

Step 3: According to the given solution, copper still forms an insoluble sulphide on passing $\mathrm{H_2S}$.

Since the metal ion is divalent in $\mathrm{MSO_4}$, one mole of the salt contains $1$ mole of metal ions, and this gives $1$ mole of $\mathrm{MS}$.

Therefore, the amount of metal sulphide formed is $1$ mol.