Question

Calculate the oxidation number of the underlined element in each species:

(i) VO₂+

(ii) UO₂²⁺

(iii) Ba₂XeO₆

(iv) K₄P₂O₇

(v) K₂S

Answer 1

(i) VO₂⁺

Let, oxidation number of V = x

\(\therefore\) x + 2(-2) = 1

x - 4 = 1

x = 1 + 4 = +5

(ii) UO₂²⁺

Let, oxidation number of U = x

\(\therefore\) x + 2 (– 2) = 2

x – 4 = 2

x = 2 + 4 = +6

(iii) Ba₂XeO₆

Let, oxidation number Xe = x

\(\therefore\) 2(2) + x + 6(– 2) = 0

4 + x – 12 = 0

x – 8 = 0

x = +8

(iv) K₄P₂O₇

Let, oxidation number of P = x

4(1) + 2(x) + 7 (– 2) = 0

4 + 2x – 14 = 0

2x – 10 = 0

2x = 10

X = \(\frac{10}{2}\) = +5

(v) K₂S

Let, oxidation number of S = x

\(\therefore\) 2(1) + x = 0

x = – 2