The oxidation state of $\mathrm{Cr}$ in $\mathrm{CrO}{}_5$ is:

Solution & Explanation

\textbf{Answer:} (c) $+6$ \textbf{Solution:} $\mathrm{CrO}{}_5$ is chromium peroxide, and its correct structure is $\mathrm{CrO}(\mathrm{O}{}_2){}_2$: it contains one oxo oxygen ($\mathrm{O}{}^{2-}$) and two peroxo linkages ($\mathrm{O}{}_2{}^{2-}$), not five normal oxide ions. Charge balance: let the oxidation state of $\mathrm{Cr}$ be $x$. One oxo oxygen contributes $-2$, and each of the two peroxide groups contributes $-2$ (so $2\times (-2)=-4$). $x+(-2)+(-4)=0\Rightarrow x=+6$. If one wrongly treated all five oxygens as $-2$, one would get the impossible value $+10/+12$. Recognising the peroxo $\mathrm{O}{}_2{}^{2-}$ linkages gives the correct oxidation state $\mathrm{Cr}$ $=+6$.