In which of the following compounds, the oxidation state $I$ atom is highest?

K I_{3}

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K I O_{4}

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K I O_{3}

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I F_{5}

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Solution

The correct option is B $K I O_{4}$
(a) $K I_{3}$
Oxidation state of $K = + 1$
Let, the oxidation state of $I$ be $x$
$1 + 3 x = 0 3 x = - 1 x = - \frac{1}{3}$
(b) $K I O_{4}$
Oxidation state of $K = + 1$
Oxidation state of $O = - 2$
Let, the oxidation state of $I$ be $x$
$+ 1 + x + (- 2 \times 4) = 0 1 + x - 8 = 0 x = + 7$
(c) $K I O_{3}$
Oxidation state of $K = + 1$
Oxidation state of $O = - 2$
Let, the oxidation state of $I$ be $x$
$+ 1 + x + (- 2 \times 3) = 0 1 + x - 6 = 0 x = + 5$
(d) $I F_{5}$
Oxidation state of $F = - 1$
Let, the oxidation state of $I$ be $x$
$x + (- 1 \times 5) = 0$
$I = + 5$

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