Question

\[{{H}}{{{g}}_2}^{2 + }\] when reacts with ${{{H}}_2}{{S}}$, black precipitate A formed which when reacts with ${{N}}{{{a}}_2}{{S}}$ followed by filtration leaving behind black precipitate B. The filtrate with ${{{H}}^ + }$ gives black precipitate C. A, B and C are:
A. ${{H}}{{{g}}_2}{{S, Hg, HgS}}$
B. ${{Hg + HgS, HgS, Hg}}$
C. ${{Hg + HgS, Hg, HgS}}$
D. ${{H}}{{{g}}_2}{{S, HgS, Hg}}$

Answer 1

Hint:Qualitative analysis is a process to identify an unknown substance. This identifies the cations and anions present in the unknown sample. It can also identify the chemical properties of the ions present in it. Cations are positively charged and anions are negatively charged.

Complete step by step solution:
Certain metal ions can be identified from a mixture using qualitative analysis.
All salts are ionic compounds and they contain both cations and anions. We can identify the cations and anions in the salt by adding certain reagents, i.e. chemicals.
There are some preliminary observations like color and nature of the substance. The color tells us about the possible cations present in the substance.
The given cation is insoluble in water. It precipitates in water.
\[{{H}}{{{g}}^{2 + }}\] comes under group $2$ cations while \[{{H}}{{{g}}_2}^{2 + }\] comes under group $1$ cations. When \[{{H}}{{{g}}_2}^{2 + }\] reacts with ${{{H}}_2}{{S}}$, ${{Hg + HgS}}$ is formed. Thus A is ${{Hg + HgS}}$. The reaction is given below:
${{H}}{{{g}}^{2 + }}_{\left( {{{aq}}} \right)} + {{{H}}_2}{{{S}}_{\left( {{{aq}}} \right)}} \rightleftharpoons {{H}}{{{g}}_{\left( {{s}} \right)}} + {{Hg}}{{{S}}_{\left( {{s}} \right)}} + 2{{H}}_{\left( {{{aq}}} \right)}^ + $
When ${{Hg + HgS}}$ is reacted with ${{N}}{{{a}}_2}{{S}}$, ${{N}}{{{a}}_2}{{{S}}_2}$ and \[{{Hg}}\] are formed. \[{{Hg}}\] is responsible for the black precipitate. When it is filtered, the filtrate contains ${{{H}}^ + }$ which precipitates due to the presence of ${{HgS}}$.
Usually the precipitates are taken using either filtration or centrifugation.
Thus the products A, B and C are ${{Hg + HgS, Hg, HgS}}$ respectively.

Hence the correct option is C.

Note:
When \[{{H}}{{{g}}^{2 + }}\] reacts with ${{{H}}_2}{{S}}$, only ${{HgS}}$ is formed. Also, when this ${{HgS}}$ is reacted with ${{N}}{{{a}}_2}{{S}}$, a complex, $\left[ {{{N}}{{{a}}_2}\left( {{{Hg}}{{{S}}_2}} \right)} \right]$ is formed. No other compounds are formed from this reaction. \[{{H}}{{{g}}^{2 + }}\] is insoluble in acid.
Qualitative analysis is useful for separating certain metal ions by selective precipitation and how these metals react with certain reagents.