The coordination number and the oxidation state of the element '$$E$$' in the complex $$[E(en)_2 (C_2 O_4)] NO_2$$ (where (en) is ethylene diamine) are, respectively,

The complex is $$[E(en)_2(C_2O_4)]NO_2$$, where
 • (en) = ethane-1,2-diamine, a neutral bidentate ligand.
 • $$C_2O_4^{2-}$$ = oxalate ion, a bidentate ligand with charge $$-2$$.
 • $$NO_2^-$$ (nitrite) is the counter-anion outside the coordination sphere.

Coordination number of $$E$$
Each ligand contributes a number of donor atoms equal to its denticity.
 • Two (en) ligands: $$2 \times 2 = 4$$ donor atoms.
 • One oxalate ligand: $$1 \times 2 = 2$$ donor atoms.
Total donor atoms = $$4 + 2 = 6$$. Hence the coordination number of $$E$$ is $$6$$.

Oxidation state of $$E$$
Let the oxidation state of $$E$$ be $$x$$. The overall charge on the complex ion $$[E(en)_2(C_2O_4)]^{\,q}$$ must balance the $$NO_2^-$$ outside:

Charge balance: $$q + (-1) = 0 \;\; \Rightarrow \;\; q = +1$$.

Inside the coordination sphere:
 • (en) is neutral: charge $$0$$.
 • $$C_2O_4^{2-}$$ carries charge $$-2$$.

So, $$x + 0 + (-2) = +1 \;\; \Rightarrow \;\; x = +3$$.

Therefore, the coordination number is $$6$$ and the oxidation state is $$+3$$.

Option D which is: 6 and 3