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  4. In [Cr(O2)(NH3)4H2O]Cl2 , oxidation number of Cr is +3, then oxygen will be in the form

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Q. In $[Cr(O_2)(NH_3)_4H_2O]Cl_2$ , oxidation number of Cr is +3, then oxygen will be in the form

NTA AbhyasNTA Abhyas 2020Redox Reactions

Solution:

The complex is $[Cr(O_2)(NH_3)_4H_2O]Cl_2$ , let oxidation number of oxygen is x

So $+3+2x+0\times 4+0-2=0$

$2x=-1$

$x=-\frac{1}{2}$ .

Hence, oxygen is in superoxo form $\left(\right.O_{2}^{-}\left.\right)$ .