Question

The oxidation states of the most electronegative element in the products of the reaction between $ BaO_{2} $ and $ H_{2}SO_{4} $ are:

• A. $ 0 $ and $ -1 $
• B. $ -1 $ and $ -2 $
• C. $ -2 $ and $ -3 $
• D. $ -2 $ and $ 0 $

Answer 1

Answer: (B)

$BaO _{2}+ H _{2} SO _{4} \rightarrow BaSO ^{-2}{ }_{4}+ H _{2} O ^{-1}{ }_{2}$
Oxygen being the most electronegative element in the reaction has the oxidation states of $-1$ (in peroxide, $ie , H _{2} O _{2}$ ) and $-2$ in $BaSO _{4}$.