Question

Passing ${{H}_{2}}S$ gas into a mixture of $M{{n}^{2+}},N{{i}^{2+}},C{{u}^{2+}}andH{{g}^{2+}}$ ions in an acidified aqueous solution precipitates,
(A) CuS and HgS
(B) MnS and CuS
(C) MnS and NiS
(D) NiS and HgS

Answer 1

The answer to this question depends upon the facts of solubility product of certain compounds. For those, whose solubility product is lower will form the precipitate; but those having a high solubility product will remain unaffected into the solution.

Complete answer:
Let us study the precipitation;
Precipitation-
The reaction occurring in the liquid state can form a solid which is insoluble in liquid known as precipitate; thus, the reaction is known as precipitation reaction. The solid is formed from the solution due to its supersaturation.
We have given the statement as ${{H}_{2}}S$ gas is passed over the ions to check the precipitation;
Passing of ${{H}_{2}}S$ is generally the test for the group II elements (specifically, cations) which is valid for all the group II elements. Thus, $C{{u}^{2+}}andH{{g}^{2+}}$ will satisfy the test giving precipitates as Cus and HgS.
But, $M{{n}^{2+}}andN{{i}^{2+}}$ will not as they are group IV cations.

Therefore, option (A) is correct.

Note:
Do note that the solubility product of $M{{n}^{2+}},N{{i}^{2+}},C{{u}^{2+}}andH{{g}^{2+}}$ varies; when ${{H}_{2}}S$ gas is being passed in an acidified aqueous solution. As the solubility products of $M{{n}^{2+}}andN{{i}^{2+}}$ are higher, they remain in the solution without forming a precipitate.