The cubic unit cell structure of a compound c | vedclass

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(A) From the given unit cell,$M$ atoms are at the face centers and $X$ atoms are at the edge centers.$(A)$ Number of $M$ atoms $(Z_M)$ = $6 	imes \fr

Vedclass · Accepted Answer

$A, B$

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(A) From the given unit cell,$M$ atoms are at the face centers and $X$ atoms are at the edge centers.$(A)$ Number of $M$ atoms $(Z_M)$ = $6 	imes \frac{1}{2} = 3$. Number of $X$ atoms $(Z_X)$ = $12 	imes \frac{1}{4} = 3$. The ratio $Z_M : Z_X = 1:1$. Thus,the empirical formula is $MX$.$(B)$ Both $M$ and $X$ occupy sites with the same coordination number,so they have the same coordination geometry.$(C)$ The distance between a face-centered atom $M$ and an edge-centered atom $X$ is the bond length $d = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2 + 0^2} = \frac{a}{\sqrt{2}} = 0.707a$. The statement is incorrect.$(D)$ For this structure,the contact occurs between $M$ and $X$. The distance $d = r_M + r_X = \frac{a}{\sqrt{2}}$. Given $r_M Therefore,only statement $(A)$ is correct. However,based on standard multiple-choice patterns,if $(A)$ is the only correct statement,the provided options might be flawed. Re-evaluating the structure: if $M$ is at body center and $X$ at corners,it is $CsCl$ type. The provided image shows $M$ at face centers and $X$ at edge centers,which is not a standard lattice. Given the options,$(A)$ is the most plausible correct statement.

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