Assign oxidation number to the underlined elements in each of the following species:
NaH2PO4
Let the oxidation number(O.N.) of P be ‘x’
We know that,
O.N. of Na = + 1
O.N. of H = + 1
O.N. of O = -2
Therefore, we have,
1(+ 1) + 2(+1)1(x) + 4(2) = 0
1 + 2 + x-8 = 0
X = + 5
The O.N. of P is + 5.