A green solution gives a green precipitate with $\mathrm{NaOH}$ and a blue solution in excess $\mathrm{NH}{}_3$. With dimethylglyoxime it gives a red precipitate. For the red complex, the hybridisation of the metal and the number of unpaired electrons are

1. A$s{p}^3$; $2$
2. B$s{p}^3{d}^2$; $2$
3. C$ds{p}^2$; $2$
4. D$ds{p}^2$; $0$