Determine the oxidation number of underlined atoms in following.

$3 {I -}_{2} + 6 N a O H \to N a I - O_{3} + 5 N a I - + 3 H_{2} O$

0, +5, -1

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0, +4, -2

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0, +3, -1

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None of these

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Solution

The correct option is C 0, +5, -1
The oxidation number of underlined atoms $3 {I -}_{2} + 6 N a O H \to N a I - O_{3} + 5 N a I - + 3 H_{2} O$ are $0, + 5, - 1$ respectively. Let X be the oxidation number of iodine in $3 {I -}_{2}$ .
$2 X = 0$
$X = 0$
Let X be the oxidation number of iodine in $N a I - O_{3}$ .
$1 + X + 3 (- 2) = 0$
$X = + 5$
Let X be the oxidation number of iodine in $N a I -$ .
$1 + X = 0$
$X = - 1$

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