Experimentally it was found that a metal oxide has formula M _ 0.98 O . Metal M, is present as M ^ 2 + and M ^ 3 + in its oxide. Fraction of the metal which exists as M ^ 3 + would be:

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Correct answer: D. 4 . 08%

Consider one mole of the oxide, then

Moles of M = 0.98; Moles of O^2-= 1

Let 'x' moles of M are in +3 oxidation state, then Moles of M²⁺ = 0.98-x

On doing charge balancing we get

(0.98 - x)2 + 3x = 2

Or 1.96 - 2x + 3x - 2 = 0

Or x = 0.04

∴ % of{M}^{3 +} = \dfrac{0 .04}{0 .98} \times 100 = 4 .08 %

Experimentally it was found that a metal oxide has formula M _ 0.98...

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