The oxidation states of the most electronegat | vedclass

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(B) The reaction of $BaO_2$ with dilute $H_2SO_4$ is given by: $BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$ In the products,the most electronegative

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$-1$ and $-2$

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(B) The reaction of $BaO_2$ with dilute $H_2SO_4$ is given by: $BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2$ In the products,the most electronegative element is oxygen. In $H_2O_2$,oxygen is in the peroxide state,so its oxidation state is $-1$. In $BaSO_4$,oxygen is in the oxide state,so its oxidation state is $-2$. Therefore,the oxidation states are $-1$ and $-2$.

The oxidation states of the most electronegative element in the products of the reaction of $BaO_2$ with dilute $H_2SO_4$ are

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