Each peroxo link O-O has two O atoms at -1, while all other O atoms are -2. Given the oxidation states S =+6 in H2S2O8 and in H2SO5, Cr =+6 in CrO5, Na =+1 in Na2O2 and H =+1, deduce the number of O-O links in one formula unit of each and report the TOTAL for all four compounds together.

JEE Advanced levelNumerical answerDifficulty 3 of 3bn046

Each peroxo link $O - O$ has two O atoms at $- 1$ , while all other O atoms are $- 2$ . Given the oxidation states S $= + 6$ in $H X_{2} S X_{2} O X_{8}$ and in $H X_{2} SO X_{5}$ , Cr $= + 6$ in $CrO X_{5}$ , Na $= + 1$ in $Na X_{2} O X_{2}$ and H $= + 1$ , deduce the number of $O - O$ links in one formula unit of each and report the TOTAL for all four compounds together.

Numerical answer type.

Let

p

be the number of peroxo O atoms.

H X_{2} S X_{2} O X_{8}

: positive charge

= 2 + 12 = 14

, O total

= - p - 2 (8 - p) = p - 16

, so

p = 2

(1 link).

H X_{2} SO X_{5}

8 = 10 - p

p = 2

(1 link).

CrO X_{5}

6 = 10 - p

p = 4

(2 links).

Na X_{2} O X_{2}

2 = 4 - p

p = 2

(1 link). Total

= 1 + 1 + 2 + 1 = 5