Q. In $\left[\right.Cr\left(\right.O_{2}\left.\right)\left(\right.NH_{3}\left(\left.\right)_{4}H_{2}O\left]\right.Cl_{2}$ , oxidation number of Cr is +3, then oxygen will be in the form

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dioxo

peroxo

superoxo

oxo

Solution:

The complex is $\left[\right.Cr\left(\right.O_{2}\left.\right)\left(\right.NH_{3}\left(\left.\right)_{4}H_{2}O\left]\right.Cl_{2}$ , 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 $\left(\right.O_{2}^{-}\left.\right)$ .

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