Question

Calculate the oxidation number of the underlined elements in the following compounds:

(i) K₂CrO₄

(ii) K₂Cr₂O₇

(iii) Cr₂O₂Cl₂

(iv) Cr₂(SO₄)₃

(v) MnO₄^-

Answer 1

(i) Let x be the oxidation no. of chromium in K₂CrO₄

O.N. of K = +1

O.N. of O = – 2

\(\therefore\) 2 x (+1) + \(\mathrm{x}\) + 4 x (– 2) = 6

or \(\mathrm{x}\) – 6 = 0 or \(\mathrm{x}\) = +6

Hence, oxidation no. of Cr in K₂CrO₄ = 6.

(ii) Let \(\mathrm{x}\) be the oxidation no. of Cr in K₂Cr₂O₇

Oxidation no. of K = +1

Oxidation no. of O = – 2

\(\therefore\) 2 × (+ 1) + 2\(\mathrm{x}\) + 7 × (– 2) = 0

or  2\(\mathrm{x}\) - 12 = 0

or \(\mathrm{x}\) = +6

Hence, oxidation no. of Cr in K₂Cr₂O₇

= +6

(iii) Let \(\mathrm{x}\) be the oxidation no. of Cr in CrO₂CI₂.

Oxidation no. of CI = – 1

\(\therefore\) \(\mathrm{x}\) + 2 x (-2) + 2 x (-1) = 0

or \(\mathrm{x}\) - 6 = 0

or \(\mathrm{x}\) = +6

Hence oxidation no. Cr in CrO₂CI₂

= + 6.

(iv) Let \(\mathrm{x}\) be the oxidation no. of Cr in Cr₂(SO₄)₃.

Oxidation no. of SO₄^2– = – 2

\(\therefore\) 2\(\mathrm{x}\) + 3 × (– 2) = 0

or 2\(\mathrm{x}\) = +6

or \(\mathrm{x}\) = +3

Hence oxidation no. of Cr in Cr₂(SO₄)₃ = +3

(v) Let, \(\mathrm{x}\) be the oxidation no. of Mn in Mno₄^-

Oxidation of O = – 2

\(\therefore\) \(\mathrm{x}\) + 4 x (-2) = -1

\(\mathrm{x}\) - 8 = -1

\(\mathrm{x}\) = -1 + 8 = +7