Question

Coordination number and oxidation state of $Cr$ in $K_3\left[Cr\right(C_2{O}_4{)}_3]$ are, respectively :

• A. $3$ and $+3$
• B. $3$ and $0$
• C. $6$ and $+3$
• D. $4$ and $+2$

Answer 1

Answer: (C)

$6$ and $+3$

As oxalate is a bidentate ligand, coordination number of $Cr$ is $6$.
Since the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$
Let the oxidation number of $Cr$ be $x$.
Therefore, $3+x+3\times (-2)=0$
or, $x=+3$