Question

An inorganic salt solution on treatment with $\text{ HCl }$will not give a white precipitate of which metal ions?
A) $\text{ Hg }_{2}^{2+}\text{ }$
B) $\text{ H}{{\text{g}}^{\text{2+ }}}$
C) $\text{ Z}{{\text{n}}^{\text{2+}}}\text{ }$
D) $\text{ A}{{\text{l}}^{\text{3+}}}\text{ }$

Answer 1

The metals M reacts with the hydrochloric acid. The general reaction of the metal with the hydrochloric acid forms a metal salt and liberates the hydrogen gas. The general reaction is as shown below,
$\text{ M(s) + nHCl (aq) }\to \text{ }{{\text{M}}^{\text{n+}}}\text{C}{{\text{l}}_{\text{n}}}\text{ + }{{\text{H}}_{\text{2}}}\text{ }\uparrow \text{ }$

Complete answer:
The $\text{ Hg (I) }$or the mercury in the $\text{ +1 }$state reacts with the halogen (such as chloride ) and it precipitates as the $\text{ Hg (I) }$chlorides. The precipitate has a chemical formula$\text{ H}{{\text{g}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}\text{ }$. The reaction between the $\text{ Hg (I) }$and the chloride ion is as shown below,
$\text{ Hg}_{2}^{2+}(aq)\text{ + 2C}{{\text{l}}^{-}}(aq)\text{ }\rightleftharpoons \text{ }\begin{matrix} \text{H}{{\text{g}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}\text{ (}\downarrow \text{)} \\\ \text{(White)} \\\ \end{matrix}\text{ }$
The $\text{ H}{{\text{g}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}\text{ }$is a white precipitate. This precipitate when reacted with the ammonia, the forms a mixture of a $\text{ HgN}{{\text{H}}_{\text{2}}}\text{Cl }$and it is evenly distributed black$\text{ Hg }$. Thus, $\text{ Hg }$ the $\text{ +1 }$oxidation state forms a white precipitate.
The slight excess of the chloride in the solution increases the solubility of precipitate. The excess of chloride forms a chloro complex which dissolves the precipitate. The $\text{ Hg }_{2}^{2+}\text{ }$when reacts with the excess of $\text{ C}{{\text{l}}^{-}}\text{ }$ forms a $\text{ HgCl}_{4}^{2-}\text{ }$ complex. Here, the mercury exists in the $\text{ +1 }$oxidation state. The reaction is as follows,
$\text{ H}{{\text{g}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}\text{ + 2C}{{\text{l}}^{-}}\text{ }\rightleftharpoons \text{ HgCl}_{4}^{2-}\text{ (soluble) + Hg }$
Thus, $\text{ H}{{\text{g}}^{\text{2+}}}\text{ }$ forms a soluble complex with chloride from the hydrochloric acid.
Zinc in the $\text{ +2 }$oxidation state reacts with the hydrochloric acid. When a metal reacts with the acid, it bubbles out the hydrogen gas and forms a metal salt. The reaction of the $\text{ Zn (II) }$ with the hydrochloric acid is as shown below,
$\text{ Zn + 2HCl }\to \text{ ZnC}{{\text{l}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{ (}\uparrow \text{) }$
This is a single replacement reaction, where the zinc metal is displaced by the hydrogen which is provided by the hydrochloric acid. The $\text{ ZnC}{{\text{l}}_{\text{2}}}\text{ }$ is a soluble salt.
The aluminium reacts with the hydrochloric acid and produces the aluminium chloride $\text{ AlC}{{\text{l}}_{\text{3}}}\text{ }$ and liberates the hydrogen gas. The reaction between the aluminium and the hydrochloric acid is as shown below,
$\text{ 2Al(s) + 6HCl(aq) }\to \text{ 2AlC}{{\text{l}}_{3}}\text{(aq) + 3}{{\text{H}}_{\text{2}}}\text{ (}\uparrow \text{) }$
The aluminium chloride is a soluble salt of aluminium.
Thus, $\text{ Hg }_{2}^{2+}\text{ }$forms a white precipitate.

Hence, (B), (C), and (D) are the correct options.

Note:
Note that, the property of the $\text{ Hg }_{2}^{2+}\text{ }$ to form a precipitate with the halogen is used in the qualitative analysis of mercury from the mixture. The group I cations $\text{ }\left( \text{A}{{\text{g}}^{+}}\text{ , Hg }_{2}^{2+}\text{ , P}{{\text{b}}^{\text{2+}}} \right)\text{ }$ . The group I cations form the precipitate with the hydrochloric acid. The reactions are as follows,
$\begin{aligned} & \text{A}{{\text{g}}^{\text{+}}}\text{ + C}{{\text{l}}^{-}}(aq)\text{ }\rightleftharpoons \text{ AgCl}(\text{s})\text{ (white) } \\\ & \text{Hg}_{2}^{2+}\text{ + 2C}{{\text{l}}^{-}}(aq)\text{ }\rightleftharpoons \text{ H}{{\text{g}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}(\text{s})\text{ (white)} \\\ & \text{P}{{\text{b}}^{\text{2+}}}(aq)\text{ + 2C}{{\text{l}}^{-}}(aq)\text{ }\rightleftharpoons \text{ PbC}{{\text{l}}_{\text{2}}}(\text{s})\text{ (white) } \\\ \end{aligned}$