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


A
XYZ2
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B
X2YZ2
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C
X8Y4Z5
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D
X5Y4Z8
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Solution

The correct option is C X5Y4Z8
Since X forms FCC, the number of X particles = 4
The number of octahedral voids = number of particles in the unit cell.
The number of tetrahedral voids = twice the number of particles in the unit cell.
Number of Y particles = 4
Number Z particles = 8
After removing the particles along a body diagonal, two corner particles from the lattice are removed (X), one octahedral void is removed (Y - the body centre) and two tetrahedral voids are removed (Z - each at a distance of 1/4th the length of the body diagonal from the corner particles)
So, we get:

of X atoms in unit cell = 4 - 18=154
Y atoms in unit cell = 4 - 1 = 3
No. of Z atoms in unit cell = 8 - 2 = 6
Hence, the formula will be X15/4Y3Z6
On simplifying this, we get - X15Y12Z24= X5Y4Z8

Chemistry

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