The number of octahedral and tetrahedral sites in a cubical close packed array of N spheres respectively is

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N and 2N

N/2 and N

2N and N

4N and 2N

answer is A.

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Detailed Solution

Number of octahedral sites = number of spheres in packing Number of tetrahedral voids =2× number of spheres in packing Number of tetrahedral voids =2×octahedral voids So, the answer is N and 2N.NOTE : For fcc arrangement THV'S are located along body diagonal and on each body diagonal 2 THV'S are present.Therefore Number of THV'S per unit cell =4×2×1=8Hence in fcc , Number of THV'S is twice to the number of particles.⇒OH'S are located at edge centre and body centre . Number of octahedral per unit cell =12×14 + 1×1 = 4Hence in fcc , Number of OHV'S is equal to the number of particles.

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