An aqueous solution of colorless metal sulfate M, gives a white precipitate with ${\mathrm{NH}}_4\mathrm{OH}$. This was soluble in excess of ${\mathrm{NH}}_4\mathrm{OH}$. On passing ${\mathrm{H}}_2\mathrm{S}$ through this solution a white precipitate is formed. The metal M in the salt is :

$\begin{array}{l}{\mathrm{CoCl}}_2+2{\mathrm{NaHCO}}_3⟶\mathrm{Co}{\left({\mathrm{HCO}}_3\right)}_2+2\mathrm{NaCl}\\ \mathrm{Co}{\left({\mathrm{HCO}}_3\right)}_2+4{\mathrm{NaHCO}}_3\\ ⟶{\mathrm{Na}}_4\left[\mathrm{Co}{\left({\mathrm{CO}}_3\right)}_3\right]+3{\mathrm{H}}_2\mathrm{O}+3{\mathrm{CO}}_2\\ {\mathrm{Bn}}_2+{\mathrm{H}}_2\mathrm{O}⟶2\mathrm{HBr}+\left[\mathrm{O}\right]\\ 2{\mathrm{Na}}_4\left[\mathrm{Co}{\left({\mathrm{CO}}_3\right)}_3+{\mathrm{H}}_2\mathrm{O}+\left[\mathrm{O}\right]\right.\\ ⟶2{\mathrm{Na}}_3\left[\mathrm{Co}{\left({\mathrm{CO}}_3\right)}_3\right]\left(\text{ Green }\right)+2\mathrm{NaOH}\\ \text{ (Sodium cobalticarbonate) }\end{array}$

Hence,  the correct option is D