Assign oxidation number to each atom in the following species.

a. Cr(OH)₄^-

Oxidation number of O = -2 

Oxidation number of H = +1

Cr(OH)₄^- is an ionic species.

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

∴ Oxidation number of Cr + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = – 1 

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

∴ Oxidation number of Cr – 8 + 4 = –1 

∴ Oxidation number of Cr – 4 = –1

∴ Oxidation number of Cr = –1 + 4 

∴ Oxidation number of Cr in Cr(OH)₄^- = +3

b. Na₂S₂O₃

Oxidation number of Na = +1 

Oxidation number of O = -2

Na₂S₂O₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 2 × (Oxidation number of Na) + 2 × (Oxidation number of S) + 3 × (Oxidation number of O) = 0

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

∴ 2 × (Oxidation number of S) + 2 – 6 = 0 

∴ 2 × (Oxidation number of S) = + 4

∴ Oxidation number of S = +\(\frac{4}{2}\)

∴ Oxidation number of S in Na₂S₂O₃ = +2

c. H₃BO₃

Oxidation number of H = +1 

Oxidation number of O = -2

H₃BO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

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

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

∴ Oxidation number of B + 3 – 6 = 0

∴ Oxidation number of B in H₃BO₃ = +3