Consider the following compounds: $ \mathrm{KO}_2, \mathrm{H}_2 \mathrm{O}_2 ext { and } \mathrm{H

Question

Consider the following compounds: $ \mathrm{KO}_2, \mathrm{H}_2 \mathrm{O}_2 	ext { and } \mathrm{H}_2 \mathrm{SO}_4 $ The oxidation states of the underlined elements in them are, respectively,

Professor PC Thomas Classes · Answer

(c) : $\mathrm{KO}_2$ ionises as $\mathrm{K}^{+}$and $\mathrm{O}_2^{-}$ions as it is a superoxide.
Thus, oxidation state of K is +1 .
$\mathrm{H}_2 \mathrm{O}_2$ ionises as $\mathrm{H}^{+}$and $\mathrm{O}_2^{2-}$ as it is a peroxide.
Oxidation state of each O in $\mathrm{O}_2^{2-}$ is -1 .
In $\mathrm{H}_2 \mathrm{SO}_4$, let the oxidation state of S be $x$; hence
$
\begin{aligned}
& 2(+1)+x+4(-2)=0 \
& 2+x-8=0 \
& x=+6
\end{aligned}
$