Question

Which of the following basic radicals will not be precipitated by \[{H_2}S\;\] gas in the presence of \[N{H_3}\] ​?
(A) $M{n^{ + 2}}$
(B) $N{i^{ + 2}}$
(C) $C{d^{ + 2}}$
(D) $C{a^{ + 2}}$

Answer 1

Hint:\[N{H_3}\] is a base. \[{H_2}S\;\] in presence of a base is used to precipitate group $4$ cations. Precipitation in group $4$ is different from group $2$ sulphides by the concentration of sulphide ion. Ionic products should exceed the solubility product to cause precipitation.

Complete Step by Step Solution:
 The base reacts with partially dissociated ${H^ + }$ from ${H_2}S$, thus the concentration of sulphide ion increases which is required to cause precipitation of group $4$ sulphides. The concentration required to precipitate group $4$is greater than the concentration of sulphides required for group $2$. Thus, in the basic condition group $2$ are also precipitated. Just the acid condition prevents the precipitation of the later group.

The group $2$includes two groups namely copper group and arsenic group. copper group includes $H{g^{ + 2}},P{b^{ + 2}},B{i^{ + 3}},C{u^{ + 2}},C{d^{ + 2}}$. Arsenic group include $S{b^{ + 3}}, S{b^{ + 5}},A{s^{ + 3}},A{s^{ + 5}},S{n^{ + 2}},S{n^{ + 4}}$. Thus, precipitation of $C{d^{ + 2}}$takes place, hence option C is incorrect.
$A{l^{ + 3}}$, $C{r^{ + 3}}$,$M{n^{ + 3}}$,$F{e^{ + 3}}$is precipitated in group $4$thus, option A and B is incorrect.
$C{a^{ + 2}}$is precipitated neither in group $2$nor group $4$. Hence $CaS$is not obtained on passing \[{H_2}S\;\]gas. Thus, the correct option is D.
Ferrous sulphate does not get precipitated rather ferric sulphate is always precipitated. Ferrous ions are oxidised by heating with concentrated $HN{O_3}$.

Note: Positively charged ions are called basic radicals. A cation is the positive half of the ionic compound. In a base, the negative compound is hydroxide and the positive part is the cation. Thus, these cations come from base, that's why they are called basic radicals.