Question

In $\left[Cr\right(O_2\left)\right(N{H}_3{)}_4{H}_{2}O]C{l}_2$, oxidation number of Cr is $+3$, then oxygen will be in the form of:

• A. dioxo
• B. peroxo
• C. superoxo
• D. oxo

Answer 1

The correct option is C superoxo

Oxidation number of $Cr$ is +3 as given. Oxidation number of $N{H}_3$ and $H_{2}O$ is 0 as it is neutral ligand and of $Cl$ is -1. Now we can find oxidation number of $O$ as

$(+3)+2\left(x\right)+4\left(0\right)+0+2(-1)=0$

$2\left(x\right)=-1$

As we find $O_2$ has -1 oxidation number so it is superoxo.