For each compound, the oxidation state of the metal is given: S =+6 in K2S2O8 (K =+1), Cr =+6 in CrO5, Pb =+2 (two atoms) and +4 (one atom) in Pb3O4, and Mn =+7 in Mn2O7. Taking O as -2 unless forced otherwise, select the compounds that MUST contain at least one peroxo (O-O) link.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3bn049

For each compound, the oxidation state of the metal is given: S $= + 6$ in $K X_{2} S X_{2} O X_{8}$ (K $= + 1$ ), Cr $= + 6$ in $CrO X_{5}$ , Pb $= + 2$ (two atoms) and $+ 4$ (one atom) in $Pb X_{3} O X_{4}$ , and Mn $= + 7$ in $Mn X_{2} O X_{7}$ . Taking O as $- 2$ unless forced otherwise, select the compounds that MUST contain at least one peroxo ( $O - O$ ) link.

If all O were

- 2

the oxygen total would be

- 2 n

K X_{2} S X_{2} O X_{8}

: positive

= 2 + 12 = 14

but 8 O would need

16

, so peroxo O must be present.

CrO X_{5}

6

against

10

, so peroxo links exist.

Pb X_{3} O X_{4}

4 + 4 = 8

equals

2 \times 4

, so normal oxide.

Mn X_{2} O X_{7}

14 = 2 \times 7

, normal.