Question

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:

• A. 5.08%
• B. 7.01%
• C. 4.08%
• D. 6.05%

Answer 1

Answer: (C)

4.08%

Since the oxidation state of oxygen is $-2$. So, for $M_{0.98}0$ to be neutral, the total oxidation state of $M_{0.98}$ has to be $+2$.

Let the fraction of $M^{3+}$ be $x$.

Then fraction of $M^{2+}$ will be $(0.98-x)$.

Now for the compound to be neutal,

$3x+2(0.98-x)=23x+1.96-2x=2x=2-1.96x=.04$

So, fraction of $M^{3+}$ will be 4%.

So, the correct answer is option $C$.