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(C) To determine the oxidation number of oxygen in each compound:$1$. In $BaO_2$ (barium peroxide),oxygen is in the peroxide form,so its oxidation num

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$H_2O_2, O_2F_2$

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(C) To determine the oxidation number of oxygen in each compound:$1$. In $BaO_2$ (barium peroxide),oxygen is in the peroxide form,so its oxidation number is $-1$.$2$. In $H_2O_2$ (hydrogen peroxide),oxygen is in the peroxide form,so its oxidation number is $-1$.$3$. In $CO_2$ (carbon dioxide),oxygen is in the oxide form,so its oxidation number is $-2$.$4$. In $SO_3$ (sulfur trioxide),oxygen is in the oxide form,so its oxidation number is $-2$.$5$. In $H_2O_2$,the oxidation number of oxygen is $-1$.$6$. In $O_2F_2$ (dioxygen difluoride),fluorine is more electronegative than oxygen,so fluorine has an oxidation number of $-1$. Calculating for oxygen: $2x + 2(-1) = 0$,which gives $2x = +2$,so $x = +1$.Since $H_2O_2$ has an oxidation number of $-1$ and $O_2F_2$ has an oxidation number of $+1$,this pair has different oxidation numbers for oxygen.

Which of the following pairs will have different oxidation numbers for oxygen?

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