Calculate the oxidation number of underlined atoms. a. H2SO4 b. HNO3

a. H₂SO₄ :

Oxidation number of H = +1 

Oxidation number of O = -2

H₂SO₄ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms of H₂SO₄ = 0

∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0

∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0

∴ Oxidation number of S + 2 – 8 = 0

∴ Oxidation number of S in H₂SO₄ = +6

b. HNO₃ :

Oxidation number of H = +1 

Oxidation number of O = -2

HNO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms of HNO₃ = 0

∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0

∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0 

∴ Oxidation number of N + 1 – 6 = 0 

∴ Oxidation number of N in HNO₃ = +5

c. H₃PO₃ :

Oxidation number of O = -2 

Oxidation number of H = +1 

H₃PO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0 

∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0 

∴ Oxidation number of P + 3 – 6 = 0 

Oxidation number of P is H₃PO₃ = +3

d. K₂C₂O₄ :

Oxidation number of K = +1 

Oxidation number of O = -2

K₂C₂O₄ is a neutral molecule.

∴ Sum of the oxidation number of all atoms = 0 

∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0 

∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0 

∴ 2 × (Oxidation number of C) + 2 – 8 = 0 

∴ 2 × (Oxidation number of C) = + 6

∴ Oxidation number of C = \(+\frac{6}{2}\)

∴ Oxidation number of C in K₂C₂O₄ = +3

e. H₂S₄O₆ :

Oxidation number of H = +1 

Oxidation number of O = -2

H₂S₄O₆ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0

∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0 

∴ 4 × (Oxidation number of S) + 2 – 12 = 0 

∴ 4 × (Oxidation number of S) = + 10

∴ Oxidation number of S = \(+\frac{10}{4}\)

∴ Oxidation number of S in H₂S₄O₆ = +2.5

f. Cr₂O₇^2- :

Oxidation of O = -2

Cr₂O₇^2- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 2 

∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2 

∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2 

∴ 2 × (Oxidation number of Cr) – 14 = – 2 

∴ 2 × (Oxidation number of Cr) = – 2 + 14 

∴ Oxidation number of Cr = \(+\frac{12}{2}\) 

∴ Oxidation number of Cr in Cr₂O₇^2- = +6

g. NaH₂PO₄ :

Oxidation number of Na = +1 

Oxidation number of H = +1 

Oxidation number of O = -2

NaH₂PO₄ is a neutral molecule

Sum of the oxidation numbers of all atoms = 0 

(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0 (+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0

(Oxidation number of P) + 3 – 8 = 0

Oxidation number of P in NaH₂PO₄ = +5