Question

The formula of a metal oxide is M_0.96 O₁. The fractions of metal that exists as M³⁺and M²⁺ ions in that oxide are respectively

• A. 0.083, 0.916
• B. 0.916, 0.083
• C. 0.88,0.12
• D. 0.12, 0.88

Answer 1

The Correct Option is A

To solve the problem, we need to find the fractions of metal existing as $M^{3+}$ and $M^{2+}$ in the oxide with the formula $M_{0.96}O_1$.

1. Understand the oxidation states and formula:
Let the number of moles of $M^{3+}$ be $x$, and that of $M^{2+}$ be $(0.96 - x)$. The total negative charge from 1 mole of oxygen atoms (as $O^{2-}$) is -2. For the compound to be neutral, the total positive charge must also be +2.

2. Set up the equation for charge neutrality:
$3x + 2(0.96 - x) = 2$
Simplifying:
$3x + 1.92 - 2x = 2$
$x + 1.92 = 2$
$x = 2 - 1.92 = 0.08$

3. Find the fractions:
Fraction of $M^{3+} = \frac{0.08}{0.96} = 0.083$
Fraction of $M^{2+} = \frac{0.88}{0.96} = 0.916$

Final Answer:
The fractions of metal that exist as $M^{3+}$ and $M^{2+}$ are 0.083 and 0.916 respectively.