The coordination number and the oxidation state of the element E in the complex $\left[\mathrm{E}(\mathrm{en}{)}_2\left({\mathrm{C}}_2{\mathrm{O}}_4\right)\right]{\mathrm{NO}}_2$, where en is ethylenediamine, are respectively

Both en and ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$ are bidentate ligands. Hence, the coordination number ofE is 2 x 2 + 2 = 6. Its oxidation state is +III since ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$ and ${\mathrm{NO}}_2^{-}$ carry a total of three negative charge.