The formula of a metal oxide is M_{0.96} O_1. | vedclass

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Question

(A) Let $x$ be the number of $M^{3+}$ ions and $y$ be the number of $M^{2+}$ ions.Total number of metal ions is $x + y = 0.96$.Therefore, $y = 0.96 -

Vedclass · Accepted Answer

$0.083, 0.916$

Vedclass · Answer

(A) Let $x$ be the number of $M^{3+}$ ions and $y$ be the number of $M^{2+}$ ions.Total number of metal ions is $x + y = 0.96$.Therefore, $y = 0.96 - x$.Since the oxide is electrically neutral, the total positive charge must equal the total negative charge ($-2$ for $O^{2-}$).$3x + 2y = 2$.Substituting $y = 0.96 - x$ into the equation:$3x + 2(0.96 - x) = 2$.$3x + 1.92 - 2x = 2$.$x = 0.08$.Thus, $y = 0.96 - 0.08 = 0.88$.The fraction of $M^{3+} = \frac{0.08}{0.96} = 0.083$.The fraction of $M^{2+} = \frac{0.88}{0.96} = 0.917$.

The formula of a metal oxide is $M_{0.96} O_1$. The fractions of metal that exist as $M^{3+}$ and $M^{2+}$ ions in that oxide are respectively

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