Find oxidation number of Sulphur in $K A l {(S O_{4})}_{2} . 12 H_{2} O$ .

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Solution

Let the oxidation number of sulphur be $x$ .
The oxidation number of $K, A l, O, H$ are $+ 1, + 3, - 2, + 1$ respectively.
$∴ 1 + 3 + 2 x + 4 \times 2 \times (- 2) + 12 \times 2 \times (+ 1) + 12 (- 2) = 0$
$∴ 4 + 2 x - 16 + 24 - 24 = 0$
$∴ 2 x - 12 = 0$
$∴ 2 x - 12 = 0$
$∴ x = + 6$
Therefore, the oxidation number of sulphur is +6.

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