1
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Question

# Calcium crystallizes in a cubic unit cell with density 3.2 g/cc. Edge length of the unit cell is 437 pm.The number of nearest neighbours of a Ca atom is ______ .

A
4
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B
6
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C
8
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D
12
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Solution

## The correct option is D 12The density of the crystal can be found out by applying the formula as follows:d=M×zN0×a3where d= density of the crystal,M= atomic or molecular weight of the species,z= number of atoms in the unit cell,N0 = Avogadro number,a= edge length.We have to find out the value of z in order to determine the type of unit cell.From the above equation, we have z=d×N0×a3MNow substituting the different values in the above equation,z=3.22×6.023×1023×(437×10−10)340z=4.04≈4.0Since z=4, the type of unit cell is face centred cubic (FCC).Hence, the number of nearest neighbours (coordination number) of a Ca atom is 12.

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