Question

A metal exists as an oxide with formula M_0.96​O. Metal M can exist as M⁺² and M⁺³ in its oxide M0.96​O. The percentage of M⁺³ in the oxide is nearly

• A. 9.6%
• B. 8.3%
• C. 4.6%
• D. 5%

Answer 1

The Correct Option is B

The given oxide formula is \( M_{0.96}O \). Let the amount of metal M present as \( M^{+2} \) be \( x \) and the amount of metal M present as \( M^{+3} \) be \( y \). The total number of metal atoms in the oxide is \( 0.96 \), so: \[ x + y = 0.96. \] The charge balance equation for the oxide is: \[ 2x + 3y = 2 \quad \text{(since the oxide is neutral, and oxygen has a charge of -2)}. \] Now, solve the system of equations: 1. \( x + y = 0.96 \) 2. \( 2x + 3y = 2 \) From equation (1), solve for \( x \): \[ x = 0.96 - y. \] Substitute this into equation (2): \[ 2(0.96 - y) + 3y = 2, \] \[ 1.92 - 2y + 3y = 2, \] \[ 1.92 + y = 2, \] \[ y = 2 - 1.92 = 0.08. \] So, the percentage of \( M^{+3} \) in the oxide is: \[ \frac{y}{0.96} \times 100 = \frac{0.08}{0.96} \times 100 \approx 8.3\%. \]

Thus, the correct answer is (B): \( 8.3\% \).