Question

Passing ${{H}_{2}}S$ gas into a mixture of $M{{n}^{2+}},N{{i}^{2+}},C{{u}^{2+}},\text{ and }H{{g}^{2+}}$ ions in an acidified solution precipitates:
(A) $CuS$ and $HgS$
(B) $MnS$ and $CuS$
(C) $MnS$ and $NiS$
(D) $NiS$ and $HgS$

Answer 1

Hint: We have the cations divided into groups for detection. ${{H}_{2}}S$ gas is passed to the mixture of the elements for the formation of sulfides of the cations that are present in the mixture that belongs to the second group.

Complete step by step solution:
Quantitative analysis of cation is the technique that is used to detect the cation of the compound by stepwise addition of specific reagents and solutions.
The cations are classified because each group of cations is precipitated at every step. Each cation behaves to a common test of reagent that is different from another cation.
There are five groups in which the cations are precipitated:
(i)- Group I cations- These are silver ion ($A{{g}^{+}}$ ), mercury ion ($H{{g}^{2+}}$ ) and lead ion ($P{{b}^{2+}}$ ). These are insoluble chlorides and precipitate out at the first step.
(ii)- Group II cations- These are mercury ion ($H{{g}^{2+}}$ ), lead ion ($P{{b}^{2+}}$ ), copper ion ($C{{u}^{2+}}$ ), bismuth ion ($B{{i}^{3+}}$ ), cadmium ion ($C{{d}^{2+}}$ ), arsenic ion ($A{{s}^{3+}}$ ), tin ion ($S{{n}^{4+}}$ ) and antimony ion ($S{{b}^{3+}}$ ). In acidic medium they form in soluble sulfides.
(iii)- Group III cations- These are aluminium ion ($A{{l}^{3+}}$ ), ferric ion ($F{{e}^{3+}}$ ), cobalt ion ($C{{o}^{2+}}$ ), nickel ion ($N{{i}^{2+}}$ ), chromium ion ($C{{r}^{3+}}$ ), zinc ion ($Z{{n}^{2+}}$ ), and manganese ion ($M{{n}^{2+}}$ ). Their sulphide and hydroxide are insoluble in an alkaline medium.
(iv)- Group IV cations- These are calcium ion ($C{{a}^{2+}}$ ), strontium ion ($S{{r}^{2+}}$ ), and barium ion ($B{{a}^{2+}}$ ). There bicarbonates precipitate out.
(v)- Group V cations- These are magnesium ion ($M{{g}^{2+}}$ ), sodium ion ($N{{a}^{+}}$ ), potassium ion (${{K}^{+}}$), and ammonium ion ($N{{H}_{4}}^{+}$ ).
So, ${{H}_{2}}S$ gas is passed to the mixture of the elements for the formation of sulfides of the cations that are present in the mixture that belongs to the second group. Now among the following ions present in the question only copper ion and mercury ion is the second group ion. Hence they will precipitate at $CuS$ and $HgS$.

Therefore, the correct answer is an option (A) $CuS$ and $HgS$.

Note: At each step, there are specific reagents added that precipitates only the cations that belong to that category. Sulfides are only formed when the group cation is in an acidic solution.