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Q.
The oxidation state of the most electronegative element in the products of the reaction. $ BaO_2 $ with dil, $ H_2SO_4 $ are
IIT JEEIIT JEE 1991The s-Block Elements
Solution:
Reaction between $BaO _{2}$ and $H _{2} SO _{4}$ is as follows:
$
\begin{array}{r}
BaO _{2}+ H _{2} SO _{4} \rightarrow H _{2} O _{2}+ BaSO _{4} \\
-1 -2
\end{array}
$
The most electronegative element is oxygen in both the products and its oxidation state is $-1$ and $-2$ respectively, in $H _{2} O _{2}$ and $BaSO _{4}$.