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

The explanation for the correct option:

Option(C) $335.15pm$

• In a BCC lattice, the distance between two oppositely charged ions $=\frac{\sqrt{3}}{2}a$ ,(where a = edge length of unit cell)

$=\frac{\sqrt{3}}{2}\times 387=335.15pm$

• BCC lattice has atoms at each corner of the cube and an atom at the center of the structure.
• The total number of atoms present per unit cell = 2 atoms

Hence, ​Option(C) $335.15pm$ is the distance between the oppositely charged ions in the lattice.