Question

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

(a) NaH₂PO₄ (b) NaHSO₄ (c) H₄P₂O₇ (d) K₂MnO₄ 

(e) CaO₂ (f) NaBH₄ (g) H₂S₂O₇ (h) KAl(SO₄)₂.12 H₂O 

Answer 1

(a) Na₂PO₄

Let the oxidation number of P be x. 

We know that, 

Oxidation number of Na = +1 

Oxidation number of H = +1 

Oxidation number of O = –2 

Na⁺¹H⁺¹₂P^xO^-2₄

Then, we have 

Hence, the oxidation number of P is +5. 

(b) Na⁺¹H⁺¹S^xO^-2₄

Then, we have 

Hence, the oxidation number of S is + 6. 

(c) H⁺¹₄P^x₂O^-2₇

Then, we have 

Hence, the oxidation number of P is + 5. 

(d) K⁺¹₂Mn^xO^-2₄

Then, we have 

Hence, the oxidation number of Mn is + 6. 

(e) Ca⁺²O^x₂

Then, we have 

Hence, the oxidation number of O is – 1. 

(f) Na⁺¹B^xH^-1₄

Then, we have 

Hence, the oxidation number of B is + 3

(g) H⁺²S^x₂O^-72

Then, we have 

Hence, the oxidation number of S is + 6. 

(h) K⁺¹Al³⁺(S^xO^2-₄).12H⁺¹₂O^-2

Then, we have 

Or, We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have 

Hence, the oxidation number of S is + 6.