  Question

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  B
X2YZ2  C
X8Y4Z5  D
X5Y4Z8  Solution

The correct option is C X5Y4Z8Since 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= X5Y4Z8Chemistry

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