1.

Calculate the oxidation number of underlined atoms.a. H2SO4b. HNO3c. H3PO3d. K2C2O4e. H2S4O6f. Cr2O72-g. NaH2PO4

Answer»

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. HNO:

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. Cr2O72- :

Oxidation of O = -2

Cr2O72- 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 Cr2O72- = +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



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