What will be the oxidation number of underlined element : $K A l (S - - O_{4})_{2} .12 H_{2} O$

+ 2

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+ 6

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- 2

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+ 3

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Solution

The correct option is B $+ 6$
For $K A l (S - - O_{4})_{2} . 12 H_{2} O$
Let, the oxidation state of $S$ be $x$ in $K A l (S - - O_{4})_{2} .12 H_{2} O$
Oxidation number of $K = + 1$
Oxidation number of $A l = + 3$
Oxidation number of $O = - 2$
Oxidation number of $H = + 1$
$+ 1 + 3 + 2 x + 3 (- 2) + 12 (2 \times 1 - 2) = 0$
$x = + 6$

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K A l (S - - O_{4})_{2} .12 H_{2} O

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