A metal oxide is found to have the non-stoichiometric formula M_ 0.98 O. If the metal M exists in both M^ 2+ and M^ 3+ states, calculate the percentage of the metal that exists as M^ 3+ .

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EEJ MAIN Chemistry QUESTION #7184

Question 1

A metal oxide is found to have the non-stoichiometric formula $M_{0.98}O$. If the metal $M$ exists in both $M^{2+}$ and $M^{3+}$ states, calculate the percentage of the metal that exists as $M^{3+}$.

$7.01\%$
$4.08\%$✔️
$6.05\%$
$5.08\%$

Correct Answer Explanation

Applying the principle of charge neutrality for $M_{0.98}O$:

Let the number of $M^{3+}$ ions be $x$. Then $M^{2+}$ ions are $(0.98 - x)$.
Total positive charge = Total negative charge ($2$ for one $O^{2-}$).
$3x + 2(0.98 - x) = 2 \implies 3x + 1.96 - 2x = 2 \implies x = 0.04$.
Fraction of $M^{3+} = 0.04 / 0.98 \approx 0.0408$ or $4.08\%$.

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