Question

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

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

Answer 1

Answer: (C)

superoxo

Let the oxidation number of $O_2$ be $x$.
Since the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$.
$\left[\underset{+3}{Cr}\underset{+x}{\left(O_2\right)}\underset{+0}{{\left({NH}_3\right)}_4}\underset{+0}{\left(H_{2}O\right)}\right]\underset{-1}{{Cl}_2}$
Therefore, $x+3+0+0-2=0;x=-1$
So, $\left(O_2\right)$ exists as $O_2^{-}$ (superoxide ion).