Experimentally it was found that a metal oxide has formula M_0.98O . Metal M is present as M^2+ and M^3+ in its oxide. Fraction of the metal which exi - Chemistry MCQ JEE Main 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

Let the number of M^{3 +} ions be x .Total number of metal ions = 0.98 .Then, number of M^{2 +} ions = 0.98 - x .The oxide ion is O^{2 -}, so total negative charge = 2 .For electrical neutrality: 3 (x) + 2 (0.98 - x) = 2 3 x + 1.96 - 2 x = 2 x = 2 - 1.96 = 0.04 Fraction as M^{3 +} = \frac{x}{Total M} \times 100 = \frac{0.04}{0.98} \times 100 = 4.08% .

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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:

4.08%

6.05%

5.08%

7.01%

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