A compound formed by Mg,Al,and O is found to | vedclass

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(C) Let the number of oxide ions $(O^{2-})$ in the $CCP$ lattice be $N = 4$.Number of octahedral voids $(OV)$ = $N = 4$.Number of tetrahedral voids $(

Vedclass · Accepted Answer

$MgAl_2O_4$

Vedclass · Answer

(C) Let the number of oxide ions $(O^{2-})$ in the $CCP$ lattice be $N = 4$.Number of octahedral voids $(OV)$ = $N = 4$.Number of tetrahedral voids $(TV)$ = $2N = 8$.Given that $Mg^{2+}$ ions occupy $\frac{1}{8}^{th}$ of the tetrahedral voids:Number of $Mg^{2+} = \frac{1}{8} 	imes 8 = 1$.Given that $Al^{3+}$ ions occupy $\frac{1}{2}$ of the octahedral voids:Number of $Al^{3+} = \frac{1}{2} 	imes 4 = 2$.Thus,the ratio of $Mg : Al : O$ is $1 : 2 : 4$.The formula of the compound is $MgAl_2O_4$.

$A$ compound formed by $Mg$,$Al$,and $O$ is found to have a cubic close-packed $(CCP)$ array of oxide ions,in which $Mg^{2+}$ ions occupy $\frac{1}{8}^{th}$ of the tetrahedral voids and $Al^{3+}$ ions occupy $\frac{1}{2}$ of the octahedral voids. The formula for the compound is:

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