Solid State - Result Question 17
####17. Experimentally it was found that a metal oxide has formula $M _{0.98} O$. Metal $M$, present as $M^{2+}$ and $M^{3+}$ in its oxide. Fraction of the metal which exists as $M^{3+}$ would be
(a) $7.01 \text{\%}$
(b) $4.08 \text{\%}$
(c) $6.05 \text{\%}$
(d) $5.08 \text{\%}$
(2013 Main)
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Answer:
Correct Answer: 17. (b)
Solution:
- From the valency of $M^{2+}$ and $M^{3+}$, it is clear that three $M^{2+}$ ions will be replaced by $M^{3+}$ causing a loss of one $M^{3+}$ ion. Total loss of them from one molecule of $M O=1-0.98=0.02$
Total $M^{3+}$ present in one molecule of
$\begin{aligned} M O & =2 \times 0.02 \\ \text { That } M^{2+} \text { and } M^{3+} & =0.98\end{aligned}$
$$ M O=2 \times 0.02=0.04 $$
Thus, $\quad \text{\%}$ of $M^{3+}=\frac{0.04 \times 100}{0.98}=4.08 \text{\%}$
