In a metal deficient oxide sample,M_X Y_2 O_4 | vedclass

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(D) The total charge of the oxide $M_X Y_2 O_4$ must be zero.Given that $Y$ is in $+3$ oxidation state and $O$ is in $-2$ oxidation state,the total ne

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$0.75$

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(D) The total charge of the oxide $M_X Y_2 O_4$ must be zero.Given that $Y$ is in $+3$ oxidation state and $O$ is in $-2$ oxidation state,the total negative charge is $4 	imes (-2) = -8$.The total positive charge from $Y$ is $2 	imes (+3) = +6$.Therefore,the total positive charge from $M$ must be $8 - 6 = +2$.Let the total number of $M$ ions be $X$. The fraction of $M^{2+}$ is $\frac{1}{3}$,so the number of $M^{2+}$ ions is $\frac{X}{3}$ and the number of $M^{3+}$ ions is $\frac{2X}{3}$.The total charge contributed by $M$ is $(\frac{X}{3} 	imes 2) + (\frac{2X}{3} 	imes 3) = \frac{2X}{3} + 2X = \frac{8X}{3}$.Equating the total positive charge from $M$ to $+2$,we get $\frac{8X}{3} = 2$.Solving for $X$,we get $X = \frac{6}{8} = 0.75$.

In a metal deficient oxide sample,$M_X Y_2 O_4$ ($M$ and $Y$ are metals),$M$ is present in both $+2$ and $+3$ oxidation states and $Y$ is in $+3$ oxidation state. If the fraction of $M^{2+}$ ions present in $M$ is $\frac{1}{3}$,the value of $X$ is. . . . . . .

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