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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