Question

Experimentally it was found that a metal oxide has formula
M₀.₉₈O. Metal M, present as M²⁺ and M³⁺ in its oxide. Fraction
of the metal which exists as M³⁺ would be :
(a) 7.01% (b) 4.08% (c) 6.05% (d) 5.08%

Answer 1

answer : option (b) 4.08 %

Experimentally it was found that a metal oxide has formula M₀.₉₈O .

Let fraction of M²⁺ is x and then fraction of M³⁺ is (0.98 - x).

here, 2x + 3(0.98 - x) = 2 [ because o.n of O = -2 and then o.n of M is +2]

⇒2x + 2.94 - 3x = 2

⇒-x = -0.94

⇒x = 0.94

and hence, fraction of M³⁺ is 0.98 - 0.94 = 0.04

now percentage of M³⁺ = fraction of M³⁺/total M × 100

= 0.04/0.98 × 100

= 4/0.98

= 4.08 %

hence option (b) is correct choice.

Answer 2

Fraction  of the metal which exists as M³⁺ is 4.08%

Explanation:

Oxidation number of the metal in a compound denotes its charge that makes the whole compound as neutral.

Given that metal oxide has formula M₀.₉₈O.

Metal M is present as M²⁺

Now, if x ions are present in +3 oxidation state, then

$3x+(0.98-x)\times2=2$

$\Rightarrow x=0.04$

Therefore the percentage of M in +3 oxidation state = $\frac{0.04}{0.98} \times 100=4.08\%$

Hence, the correct answer is option (b)