Assign oxidation number to the underlined elements in each of the following species:
KAl(SO4)2.12 H2O
Let the oxidation number(O.N.) of S be ‘x’
We know that,
O.N. of K = + 1
O.N. of H = + 1
O.N. of O = -2
O.N. of Al = + 3
Therefore, we have,
1(+ 1) + 1(+ 3) + 2(x) + 8(-2) + 24(+ 1) + 12(-2) = 0
1 + 3 + 2x-16 + 24-24 = 0
X = + 6
The O.N. of S is + 6.
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