Assign oxidation number to the underlined elements in each of the following species:

NaHSO4


Let the oxidation number (O.N.) of S be ‘x’


We know that,


O.N. of Na = + 1


O.N. of H = + 1


O.N. of SO4 = -2


Therefore, we have,


1(+1) + 1(+ 1)1(x) + 4(2) = 0


1 + 1 + x-8 = 0


X = + 6


The O.N. of S is + 6.


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