Peroxide linkage is present in:

H_{2} O_{2}

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B a O_{2}

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O F_{2}

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H_{2} S_{2} O_{8}

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Solution

The correct options are
A $H_{2} O_{2}$
B $B a O_{2}$
D $H_{2} S_{2} O_{8}$
In peroxide linkage, oxygen has -1 oxidation state so we calculate the oxidation of oxygen to know about the presence of peroxide linkage.

In $H_{2} O_{2},$ as oxidation state of hydrogen is +1 so the oxidation state of oxygen is -1. So it has peroxide linkage.

In $B a O_{2},$ as oxidation state of barium is +2 so the oxidation state of oxygen is -1. So it has peroxide linkage.

In $O F_{2},$ as oxidation state of fluorine is -1 as it is more electronegative than oxygen so the oxidation state of oxygen is +2. So it has no peroxide linkage.

In $H_{2} S_{2} O_{8},$ the highest possible oxidation state of S is +6 and two oxygen must be attached to each other through peroxide linkage. So it has peroxide linkage.

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