Question

$N{H}_{4}Cl$ crystallizes in a body-centred cubic type lattice with a unit cell edge length of $387$ pm. The distance (in pm) between the oppositely charged ions in the lattice is:

• A. $335.10$
• B. $83.77$
• C. $274.46$
• D. $137.23$

Answer 1

The correct option is A $335.10$

For bcc structure $d=\frac{\sqrt{3}}{2}a=\frac{\sqrt{3}}{2}\times 387=335.10$ pm.

Here d is the distance between the oppositely charged ions in the lattice and a is the edge length.

Hence, the correct option is $\text{A}$