Question

What is the charge on the complex $\left[Cr\right(C_2{O}_4{)}_2\left(H_{2}O{)}_2\right]$ formed by $Cr\left(III\right)$?

• A. $+3$
• B. $+1$
• C. $+2$
• D. $-1$

Answer 1

The correct option is D $-1$

$H_{2}O$ is neutral, chromium is in $+3$ oxidation state and oxalato anion has $-2$ charge; due to two $C_2{O}_4^{2-}$. It will be $-4$ and net charge on the complex is algebraic sum of the charge on metal ion and ligands present in the coordniation sphere. So $3+2\times (-2)=-1$. So, the formulae of complex $\left[Cr\right(C_2{O}_4{)}_2(H_{2}O{)}_2{]}^{-}$.