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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 12
The density of the crystal can be found out by applying the formula as follows:
d=M×zN0×a3

where 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×a3M

Now substituting the different values in the above equation,
z=3.22×6.023×1023×(437×1010)340

z=4.044.0

Since 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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