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 $M_{0.96}O$. The percentage of $M^{3+}$ in the oxide is nearly

Updated On: May 15, 2024
  • 0.083
  • 0.046
  • 0.05
  • 0.096
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The Correct Option is A

Solution and Explanation

Let, the number of $M ^{2+}$ ions $= x$ Then, the number of $M ^{3+}$ ions will be $0.96- x$ We know, the overall charge in the metal oxide is zero. So, $x(2)+(0.96-x)(3)+1(-2)=0$ $\Rightarrow 2 x+2.88-3 x=2$ $\Rightarrow- x =-0.88$ $\Rightarrow x =0.88$ $\therefore$ Number of $M ^{3+}$ ions $=0.96-0.88=0.08$ $\therefore$ Percentage of $M ^{3+}$ ions $=0.08 / 0.96 \times 100=8.33 \%$
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