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JEE Main levelSingle correct optionDifficulty 2 of 3pn119

Arrange the following in increasing order of the oxidation state of oxygen: $OF X_{2}$ , $H X_{2} O X_{2}$ , $O X_{2} F X_{2}$ , $KO X_{2}$ .

A $KO X_{2} < H X_{2} O X_{2} < OF X_{2} < O X_{2} F X_{2}$
B $H X_{2} O X_{2} < O X_{2} F X_{2} < KO X_{2} < OF X_{2}$
C $H X_{2} O X_{2} < KO X_{2} < O X_{2} F X_{2} < OF X_{2}$
D $OF X_{2} < O X_{2} F X_{2} < KO X_{2} < H X_{2} O X_{2}$

Answer: C

Oxidation states of O:

H X_{2} O X_{2}

is

- 1

;

KO X_{2}

(superoxide) is

- 1/2

;

O X_{2} F X_{2}

is

+ 1

;

OF X_{2}

is

+ 2

(F is more electronegative). The increasing order is H2O2 < KO2 < O2F2 < OF2.

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