A crystal is made up of particles $X, Y$ and $Z$ . $X$ forms FCC packing, $Y$ occupies all octahedral voids of $X$ and $Z$ occupies all tetrahedral voids of $X$ . If all the particles along one body diagonal are removed, then the formula of the crystal would be:

X Y Z_{2}

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X_{2} Y Z_{2}

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X_{8} Y_{4} Z_{5}

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X_{5} Y_{4} Z_{8}

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Solution

The correct option is D $X_{5} Y_{4} Z_{8}$
If all the particles along one body diagonal is removed.
$Z_{e f f}$ of $X$ is $\frac{1}{2} \times 6$ (faces) $+ \frac{1}{8} \times 6$ ( Total $8 - 2$ (removed corner atoms) $= 6$ )
$= 3 + 0.75 = 3.75$
Two tetrahedral voids will be removed along one body diagonal.
Remaining $= 8 - 2 = 6$
Effective number of $Z = 6$
One octahedral void that is present on body centre is removed
Remaining $= 4 - 1 = 3$
Effective number of $Y = 3$
Formula of crystal $= X_{3.75} Y_{3} Z_{6} = X_{5} Y_{4} Z_{8}$

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