Question

Assign oxidation number to the underlined elements
(a) NaH2PO4
(b) NaHSO4​

Answer 1

Answer:

a) NaH2po4- 2

b) NaHSO4 - 1

Answer 2

Answer:

O.N. of P is + 5

O.N. of S is + 6

Explanation:

Given :

Compound :

NaH₂PO₄

We have to calculate O.N. of P

Let O.N. of P be x

We know :

O.N. of Na , H and O are + 1 , + 1 and - 2 respectively.

= > 1 + 2 + x + ( - 8 ) = 0  [ Net charge on compound is zero ]

= > 3 - 8 + x = 0

= > x = + 5

Therefore , O.N. of P in NaH₂PO₄ is + 5 .

2 .We have NaHSO₄

We are asked to calculate O.N. of S :

Let O.N. of s be y

1 + 1 + y + ( -8 ) = 0

= > 2 - 8 + y = 0

= > y = + 6 .

Therefore , O.N. of P in  NaHSO₄  is + 6 .