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(B) The reaction is: $BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$.The products are $BaSO_4$ and $H_2O_2$.The most electronegative element in these pr

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$-1$ and $-2$

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(B) The reaction is: $BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$.The products are $BaSO_4$ and $H_2O_2$.The most electronegative element in these products is oxygen $(O)$.In $BaSO_4$,the oxidation state of $O$ is calculated as: $(+2) + (+6) + 4(x) = 0$,which gives $4x = -8$,so $x = -2$.In $H_2O_2$,the oxidation state of $O$ is calculated as: $2(+1) + 2(x) = 0$,which gives $2x = -2$,so $x = -1$.Thus,the oxidation states of oxygen are $-2$ and $-1$.

What are the oxidation states of the most electronegative element in the products of the reaction between $BaO_2$ and $H_2SO_4$?

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