Question

The reagents, $N{{H}_{4}}Cl$ and aqueous $N{{H}_{3}}$ will precipitate
(A) $C{{a}^{2+}}$
(B) $A{{l}^{3+}}$
(C) $B{{i}^{3+}}$
(D) $M{{g}^{2+}}$
(E) $Z{{n}^{2+}}$

Answer 1

Hint: The question involves the qualitative test of cations which includes group II, III and zinc (II) ions by precipitation method. The precipitation is the process in which a dissolved substance is transformed into an insoluble solid to separate a solid and solution. The solid formed upon precipitation is known as precipitate and the reagent leading to the process of precipitation is called as precipitant.

Complete Step by Step Solution:
Inorganic precipitants: These reagents are known to form slightly soluble salts or hydrous oxides with the solution. Precipitate formation using inorganic precipitant is one of the important tools for the detection of type of cation in a salt.

Among the given options, due to low solubility, group III elements i.e., $A{{l}^{3+}}$ and $B{{i}^{3+}}$ will be the cations which precipitate when $N{{H}_{4}}Cl$ and aqueous $N{{H}_{3}}$are used as reagents in the form of their hydroxides as follows:

Ammonium ion:
Ammonium ion when reacts with aqueous $N{{H}_{3}}$ in presence of $N{{H}_{4}}Cl$ to produce a white gelatinous precipitate of $Al{{(OH)}_{3}}$.
$A{{l}^{3+}}+3N{{H}_{3}}(aq)\xrightarrow{N{{H}_{4}}Cl}Al{{(OH)}_{3}}(s)$

Bismuth ion:
Aqueous ammonia reacts with the bismuth (III) ion in presence of $N{{H}_{4}}Cl$to form white precipitate of bismuth hydroxide.
$B{{i}^{3+}}(aq)+3N{{H}_{3}}(aq)\xrightarrow{N{{H}_{4}}Cl}Bi{{(OH)}_{3}}(s)$

Thus, the ions which precipitate in the presence of aqueous $N{{H}_{3}}$ and $N{{H}_{4}}Cl$ are $A{{l}^{3+}}$ and $B{{i}^{3+}}$. So, option (B) and (C) are the correct answer.

Note: It is important to note that group II cations are precipitated in the presence of ${{H}_{2}}S$ and $HCl$ as their sulphides whereas the zinc (II) is also confirmed when it is dissolved in the acidic solution of hydrogen sulphide to form bluish white precipitate.