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. $4$
• B. $6$
• C. $8$
• D. $12$

Answer 1

The correct option is D $12$

The density of the crystal can be found out by applying the formula as follows:
$d=\frac{M\times z}{N_0\times a^3}$

where $d=$ density of the crystal,
$M=$ atomic or molecular weight of the species,
$z=$ number of atoms in the unit cell,
$N_0$ = 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 $z=d\times N_0\times \frac{a^3}{M}$

Now substituting the different values in the above equation,
$z=\frac{3.22\times 6.023\times {10}^{23}\times (437\times {10}^{-10}{)}^3}{40}$

$z=4.04\approx 4.0$

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 $12$.