Experimentally it was found that a metal oxide has a formula of M0.98 . Metal M is present as M^+2 and M^+3 in its oxide. Fraction of the metal which exists as M^+3 would be ?

Options -----

(A) 4.08 percentage

(B) 6.05 percentage

(C) 5.08 percentage

(D) 7.01 percentage

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Solution

M0.98O
Let M2+ be x so that M3+ will be 0.98-x. Total charge on the compound must be zero so that

+2(x) + 3(0.98-x) - 2 = 0 (charge on oxygen atom = -2)
2x + 2.94 - 3x - 2 = 0
x = 0.94

Amount of M3+ present = 0.98 - 0.94 = 0.04

Fraction of M3+ present = 0.04/0.98

= 0.0408

so option A is correct

Hope you understand the query 😀

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