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JEE Advanced levelSingle correct optionDifficulty 3 of 3bn048

A binary compound of potassium and oxygen contains $44.8%$ K by mass (K $= 39$ , O $= 16$ ). Which option correctly gives its formula, the average oxidation number of O and its name?

A $KO X_{3}$ , $- \frac{1}{3}$ , potassium ozonide
B $KO X_{3}$ , $- 1$ , potassium trioxide
C $K X_{2} O X_{2}$ , $- 1$ , potassium peroxide
D $KO X_{2}$ , $- \frac{1}{2}$ , potassium superoxide

Answer: A

Mass ratio: K

= 44.8/39 = 1.149

mol and O

= 55.2/16 = 3.45

mol, ratio

1 : 3

, so the formula is

KO X_{3}

. Then

1 + 3 x = 0

gives

x = - \frac{1}{3}

; it contains the ozonide ion

O X_{3} X^{-}

. (

KO X_{2}

would contain

54.9%

K.)

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