Question

Find the oxidation numbers of the underlined species in the following compounds or ions :

i. PF₆^-

ii. NaIO₃

iii. NaHCO₃

iv. ClF₃

v. SbF₆^-

vi. NaBH₄

vii. H₂PtCl₆

viii. H₅P₃O₁₀

ix. V₂O₇^4-

x.  CuSO₄

xi. BiO₃^-

xii. CH₃OH

Answer 1

i. PF₆^-

Oxidation number of F = -1

\(PF_6^-\) is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1 

∴ (Oxidation number of P) + 6 × (Oxidation number of F) = – 1 

∴ Oxidation number of P + 6 × (-1) = -1 

∴ Oxidation number of P – 6 = – 1

Oxidation number of P in \(PF_6^-\) = +5

ii. NaIO₃

Oxidation number of Na = +1 

Oxidation number of O = -2 

NaIO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 (Oxidation number of Na) + (Oxidation number of I) + 3 × (Oxidation number of O) = 0

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

Oxidation number of I + 1 – 6 = 0

Oxidation number of I in NaIO₃ = +5

iii. NaHCO₃

Oxidation number of Na = +1 

Oxidation number of H = +1 

Oxidation number of O = -2

NaHCO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

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

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

∴ Oxidation number of C + 2 – 6 = 0

∴ Oxidation number of C in NaHCO₃ = +4

iv. ClF₃

Oxidation number of F = -1

ClF₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ (Oxidation number of Cl) + 3 × (Oxidation number of F) = 0 

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

∴ Oxidation number of Cl in ClF₃ = +3

v. SbF₆^-

Oxidation number of F = -1

\(SbF_6^-\) is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1

∴ (Oxidation number of Sb) + 6 × (Oxidation number of F) = – 1

∴ Oxidation number of Sb + 6 × (-1) = -1

∴ Oxidation number of Sb in \(SbF_6^-\) = +5

vi. NaBH₄

Oxidation number of Na =+1 

Oxidation number of H = -1 (for Hydride) 

NaBH₄ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ (Oxidation number of Na) + (Oxidation number of B) + 4 × (Oxidation number of H) = 0 

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

∴ Oxidation number of B + 1 – 4 = 0

∴ Oxidation number of B in NaBH₄ = +3

vii. H₂PtCl₆

Oxidation number of H = +1 

Oxidation number of Cl = -1 

H₂PtCl₆ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ 2 × (Oxidation number of H) + (Oxidation number of Pt) + 6 × (Oxidation number of Cl) = 0

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

(Oxidation number of Pt) + 2 – 6 = 0

∴ Oxidation number of Pt in H₂PtCl₆ = +4

viii. H₅P₃O₁₀

Oxidation number of H = +1 

Oxidation number of O = -2

H₅P₃O₁₀ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 5 × (Oxidation number of H) + 3 × (Oxidation number of P) +10 × (Oxidation number of O) = 0

∴ 5 × (+1) + 3 × (Oxidation number of P) + 10 × (-2) = 0

∴ 3 × (Oxidation number of P) + 5 – 20 = 0

Oxidation number of P = +\(\frac{15}{3}\)

∴ Oxidation number of P in H₅P₃O₁₀ = +5

ix. V₂O₇^4-

Oxidation number of O = -2

\(V_2O_7^{4-}\) is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 4

∴ 2 × (Oxidation number of V) + 7 × (Oxidation number of O) = – 4

∴ 2 × (Oxidation number of V) + 7 × (-2) = – 4

∴ 2 × (Oxidation number of V) = – 4 + 14

∴ Oxidation number of V = \(+\frac{10}{2}\)

∴ Oxidation number of V in \(V_2O_7^{4-}\) = +5

x.  CuSO₄

Oxidation number of Cu = +2 

Oxidation number of O = -2

CuSO₄ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

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

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

∴ Oxidation number of S + 2 – 8 = 0

∴ Oxidation number of S in CuSO₄ = +6

xi. BiO₃^-

Oxidation number of O = -2

\(BiO_3^-\) is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1 

∴ (Oxidation number of Bi) + 3 × (Oxidation number of O) = – 1 

∴ Oxidation number of Bi + 3 × (-2) = – 1 

∴ Oxidation number of Bi = – 1 + 6

∴ Oxidation number of Bi in \(BiO_3^-\) = +5

xii. CH₃OH

Oxidation number of H = +1 

Oxidation number of O = -2

CH₃OH is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

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

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

∴ Oxidation number of C + 2 = 0

∴ Oxidation number of C in CH₃OH = -2