Question

A coloured precipitate is obtained when ${{\text{H}}_2}{\text{S}}$ is passed through an aqueous solution of salt in the presence of ammonium hydroxide. The precipitate dissolves in dilute ${\text{HCl}}$ and reacts with ${\text{NaOH}}$ to give white precipitate which on standing turns out into a brown/black mass. The brown/black mass on fusion with ${\text{KN}}{{\text{O}}_3}$ and ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$ gives green mass. The cation of the salt is:
A.${\text{C}}{{\text{o}}^{2 + }}$
B.${\text{M}}{{\text{g}}^{2 + }}$
C.${\text{N}}{{\text{i}}^{2 + }}$
D.${\text{M}}{{\text{n}}^{2 + }}$

Answer 1

To answer this question it is required to have knowledge about how to write the chemical equation and the role of each reactant in the reaction. We have to proceed in a stepwise manner in order to find the product. The positively charged species written on the left hand side will be cation.

Complete step by step solution:
Let us start writing the equation along with the sequence given in the question. It is given that the cation formed coloured species with ${{\text{H}}_2}{\text{S}}$. Reaction with ${{\text{H}}_2}{\text{S}}$ will produce sulphite of respective ions. The cobalt and nickel forms black colour sulphite, whereas magnesium forms white colour sulphite. Manganese ion is the only one which forms buff colour precipitates of manganese sulphide. But to confirm this we need to check other properties as well. The precipitates are found to be dissolved in hydrochloric acid and sulphite ions are replaced by chlorine ions. The reactions are as follow:
${\text{M}}{{\text{n}}^{2 + }} + {{\text{H}}_2}{\text{S}} \to {\text{MnS }} + {\text{HCl}} \to {\text{MnC}}{{\text{l}}_2}$
The above formed manganese chloride reacts with ${\text{NaOH}}$ as follow:
${\text{MnC}}{{\text{l}}_2} + {\text{NaOH}} \to {\text{Mn}}{{\text{O}}_2}$
These manganese oxide forms turn black or brown on standing. The final reaction with ${\text{KN}}{{\text{O}}_3}$ and ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$ will be:
${\text{Mn}}{{\text{O}}_2} + 2{\text{KN}}{{\text{O}}_3} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} \to {\text{N}}{{\text{a}}_2}{\text{Mn}}{{\text{O}}_4} + 2{\text{KN}}{{\text{O}}_2} + 2{\text{NaOH }} + {\text{C}}{{\text{O}}_2}$
After fusion the solid mass of ${\text{N}}{{\text{a}}_2}{\text{Mn}}{{\text{O}}_4}$ will form.

Note: It is always necessary to check all the properties. We should not come to an answer if only one of the properties gets satisfied because the similar results can be shown by other options as well. The green coloured mass formed is known as sodium manganate. It is a deep green colour solid and is expensive as well as difficult to prepare.