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Average oxidation number of iodine in ${KI}_{3}$

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$+ 1 / 3$

$- 1 / 3$

$+ 3$

$- 1$

answer is B.

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Detailed Solution

Let us assume $X$ be the oxidation state of iodine in ${KI}_{3}$ . Potassium has one

Electron in its valence shell, so it exists in +1 oxidation state.

So, $1 + 3 X = 0$

$3 X = - 1$

$X = \frac{- 1}{3}$

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