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JEE Advanced levelSingle correct optionDifficulty 3 of 3rx036

Arrange the following in decreasing order of the average oxidation number of oxygen (potassium is $+ 1$ , ozonide is $O X_{3} X^{-}$ ): $K X_{2} O$ , $K X_{2} O X_{2}$ , $KO X_{2}$ , $KO X_{3}$ .

A $KO X_{3} > KO X_{2} > K X_{2} O X_{2} > K X_{2} O$
B $K X_{2} O > K X_{2} O X_{2} > KO X_{2} > KO X_{3}$
C $KO X_{2} > KO X_{3} > K X_{2} O X_{2} > K X_{2} O$
D $K X_{2} O X_{2} > KO X_{2} > KO X_{3} > K X_{2} O$

Answer: A

Average values:

KO X_{3}

is

- 1/3

,

KO X_{2}

is

- 1/2

,

K X_{2} O X_{2}

is

- 1

,

K X_{2} O

is

- 2

. A less negative value is higher, giving KO3 first and K2O last.

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