Question

$ N{H_4}Cl $ crystallizes in a body-centred cubic type lattice with a unit cell edge length of $ 387{\text{ }}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

Hint :A sort of atom arrangement found in nature is known as body-centered cubic (BCC). A body-centered cubic unit cell structure is made up of atoms organised in a cube with one atom in the centre and one atom in each corner sharing an atom.

Complete Step By Step Answer:
A unit cell is the simplest repeating unit in a crystal. Each unit cell is defined in terms of lattice points, which are the points in space where the particles in a crystal are free to vibrate.
The simplest repeating unit in a body-centered cubic structure is the body-centered cubic unit cell. On the eight corners of the unit cell, there are eight identical particles once more. This time, however, there is a ninth identical particle in the centre of the unit cell's body.
Now, coming to the question;
The bcc structure
 $ d = \dfrac{{\sqrt 3 }}{2}a \\
  d = \dfrac{{\sqrt 3 }}{2} \times 387 \\
  d = 335.10pm $
Therefore, the distance (in pm) between the oppositely charged ions in the lattice is: $ 335.10pm $
So, the correct answer is (A) $ 335.10 $ .

Additional Information:
Metals with a bcc structure are typically harder and less malleable than metals with a close-packed structure, such as gold. The planes of atoms must slip over each other when the metal is deformed, and this is more difficult in the bcc structure. Other key methods for hardening materials, such as the introduction of impurities or imperfections that make slippage more difficult, should be noted.

Note :
The atoms in the bcc configuration cannot pack as tightly as they may in the fcc or hcp layouts. The bcc structure is a high-temperature version of metals that are tightly packed at low temperatures.