Question

Which of the following is not precipitated as sulphide by passing ${H_2}S$ in the presence of conc. $HCL$ ?
(A) Copper
(B) Arsenic
(C) Cadmium
(D) Lead

Answer 1

Hint: In the above question we have to do that, in the given elements which element is not precipitated as sulphide be passing hydrogen sulphide where in the presence of concentrated hydrochloric acid in which many factors are affecting where we have to focus on the reaction made between each element and where is the solubility product is most higher.

Complete Step by Step Solution:
As we know, the solubility product of cadmium is highest in comparison to the sulphides of Group 2.

From which we can say that, it only get precipitated when we do his dilution whereas in the
presence of concentrated HCL it will never get precipitated.

Now we can say that,
${K_{sp}} > Group{(II)_{sp}}(\operatorname{su} lphides)$
As in the above equation K stands for Cadmium and (sp) stands for solubility product.
As when we pass Copper, Arsenic and Lead in ${H_2}S$ as presence of conc. $HCL$ then we get all of them precipitated as their sulphides.

From which we can say that Cadmium is the correct answer here.
Hence, the correct option is (C).

Note: As per talking about the term practical organic chemistry from where this question is putted from there are various methods of purification of practical organic compounds in which the major ones are: Fractional Crystallisation, Sublimation, Chromatography, Distillation, differential extraction these are the major methods on the basis of their physical and chemical properties.