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

Answer: (B)

$6$

Density $\left(d\right)=\frac{nM}{V{N}_A}=\frac{n\times 40}{\left(\right(437\times {10}^{-10}{)}^3\times 6\times {10}^{23})}=3.2$ g/cc
$⟹n=4$
Therefore Calcium crystallizes in FCC structure and number of nearest neighbours of Ca atom is 6.