Write the complex clearly

The compound is: $[E(en)_2(C_2O_4)]NO_2$

Here:

• $en$ = ethylenediamine, a neutral bidentate ligand
• $C_2O_4^{2-}$ = oxalate, a bidentate ligand
• $NO_2^-$ is the counter ion outside the coordination sphere

Find the charge on the complex ion

Since $NO_2^-$ is outside the bracket, the complex ion must have charge $+1$: $[E(en)_2(C_2O_4)]^+$

Find the oxidation state of $E$

Let the oxidation state of $E$ be $x$.

Contributions of ligands:

• Each $en$ is neutral, so $2(en)$ contributes $0$
• One oxalate ligand contributes $-2$

Total charge on the complex is $+1$, so: $x + 0 + (-2) = +1$ $x - 2 = 1$ $x = 3$

So, the oxidation state of $E$ is $+3$.

Find the coordination number

Coordination number is the number of donor atoms attached to the metal.

• Each $en$ is bidentate $\,\Rightarrow\,$ donates $2$ atoms
• Two $en$ ligands donate: $2 \times 2 = 4$
• Oxalate $(C_2O_4^{2-})$ is also bidentate $\,\Rightarrow\,$ donates $2$ atoms

Therefore, total coordination number: $4 + 2 = 6$

Match with the options

• Coordination number = $6$
• Oxidation state = $3$

Hence the correct option is: D: 6 and 3