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. $4.08\%$
• B. $6.05\%$
• C. $5.08\%$
• D. $7.01\%$

Answer 1

The correct option is A $4.08\%$

The oxidation state of oxygen is -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 neutral,

3x+2(0.98−x) = 2
3x+1.96−2x =2
x = 2−1.96x
x = 0.04
So, the fraction of $M^{3+}$ will be 4%.