Arrange the oxidation numbers of O by INCREASING value in: (i) Na2O, (ii) H2O2, (iii) KO2, (iv) O2F2 (structure F-O-O-F, O-O bond shared equally, F is -1), (v) OF2.

(i) < (ii) < (iii) < (v) < (iv)

JEE Advanced levelSingle correct optionDifficulty 3 of 3bn160

Arrange the oxidation numbers of O by INCREASING value in: (i) $Na X_{2} O$ , (ii) $H X_{2} O X_{2}$ , (iii) $KO X_{2}$ , (iv) $O X_{2} F X_{2}$ (structure $F - O - O - F$ , O-O bond shared equally, F is $- 1$ ), (v) $OF X_{2}$ .

Na X_{2} O

- 2

H X_{2} O X_{2}

- 1

KO X_{2}

- 1/2

O X_{2} F X_{2}

2 x - 2 = 0

gives

+ 1

OF X_{2}

+ 2

. The order is (i) < (ii) < (iii) < (iv) < (v).