Question

Calculate the oxidation state of the underlined atoms. 

(a) H₂S₂O₇ 

(b)  NaHSO₄

(c) CH₃COOH

(d) Cu(NH₃)₄ SO₄

(e) NH⁺₄

Answer 1

(a) O.N. of S in H₂S₂O₇ 

Let the oxidation number of S be x. 

We know that, 

Oxidation number of H = +1 

Oxidation number of O=-2 

Then, we have, 

2(+1) + 2(x) + 7(-2) = 0 

2 + 2x - 14 = 0 

2x = 12 

X = +6 

(b) O.N. of S in NaHSO₄ 

Let the oxidation number of S be x. 

We know that, 

Oxidation number of Na = +1 

Oxidation number of H = +1 

Oxidation number of O= -2 

Then, we have. 

(+1) + 1(+1) + (x) + 4(-2) = 0 

1 + 1 + x = 8

x = +6 

(c) O.N. of C in CH₃COOH, Oxidation number of 

hydrogen = + 1 

Let the oxidation number of carbon be x. 

Then we have. 

1(x) + 3(+1) + 1(x) + 2(−2) + 1(+1) = 0 

4 + 2x - 4 = 0 

x = 0 

(d) O.N. of N in Cu in Cu(NH₃)₄ SO₄ 

Let the oxidation number of Cu is x. 

+ (4 × O.N. of NH3) + O.N. of S + 4 × (O.N. of O)) = 0 

x + 0 + 6 + (-8) = 0 

x - 2 = 0 

x = +2

(e) O.N. of N in NH⁺₄ 

Let the oxidation number of N is x. 

x + (4 × O.N. of H) − + 1 

x + {4 × (+1)} = +1 

x + 4 = +1 

x = - 3