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.