Chemistry - Salt Analysis Question with Solution | TestHub

When a solution of a$\mathrm{H}{\mathrm{g}}^{2+}$is treated with KI a red precipitate of$\left(\mathrm{H}\mathrm{g}{\mathrm{I}}_2\right)$is obtained which dissolves in excess KI to give a colourless solution$(\mathrm{H}\mathrm{g}{\mathrm{I}}_4{)}^{2-}$. When the solution of$\mathrm{H}{\mathrm{g}}^{2+}$is treated with a solution of cobalt (II) thiocyanate a deep blue crystalline precipitate of$\mathrm{H}\mathrm{g}(\mathrm{S}\mathrm{C}\mathrm{N}{)}_2$is formed. The reactions are as follows
$\mathrm{H}{\mathrm{g}}^{2+}+2{\mathrm{I}}^{–}\to \mathrm{H}\mathrm{g}{\mathrm{I}}_2$
$\underset{\left(\mathrm{e}\mathrm{x}\mathrm{c}\mathrm{e}\mathrm{s}\mathrm{s}\right)}{\mathrm{H}\mathrm{g}{\mathrm{I}}_2}+2{\mathrm{I}}^{–}\to \underset{\mathrm{S}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{b}\mathrm{l}\mathrm{e}}{[\mathrm{H}\mathrm{g}{\mathrm{I}}_4{]}^{2–}}$
${\text{Hg}}^{2+}+\text{Co}{(\text{SCN})}_2\to \underset{\text{Blue} \text{crystal} \text{line} \text{ppt}.}{\text{Hg}{(\text{SCN})}_2}$

Here other ion$\mathrm{P}{\mathrm{b}}^{2+}$,$\mathrm{H}{\mathrm{g}}^{2+}$and$\mathrm{C}{\mathrm{o}}^{2+}$do not give precipitates formed in these reactions.