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.