Question

The coordination number and the oxidation state of element $E$ in the complex $\left[E(e n)_{2}\left( C _{2} O _{4}\right)\right] NO _{2}($ where $(e n)$ is ethylenediamine) are, respectively

• A. $6$ and $3$
• B. $6$ and $2$
• C. $4$ and $2$
• D. $4$ and $3$

Answer 1

Answer: (A)

In the given complex $\left[E(e n)_{2}\left( C _{2} O _{4}\right)\right] NO _{2}$ ethylenediamine is a bidentate ligand and $\left( C _{2} O _{4}^{2-}\right)$ oxalate ion is also bidentate ligand. Therefore, coordination number of the complex is $6$ i.e., it is an octahedral complex.

Oxidation number of $E$ in the given complex is $x+2 \times 0+1 \times(-2)=+1 \Rightarrow x=+3$