Question

AB crystallizes in a bcc lattice with an edge length equal to 387 pm. The distance between two oppositely charged ions in the lattice is:
A.335 pm
B.250 pm
C.200 pm
D.300 pm

Answer 1

Hint: In a bcc lattice or body centred unit cell, there is one additional particle present at the centre within the body of the unit cell in addition to the particles at the corners of the unit cell.
-The number of atoms present per unit cell in a bcc lattice is 2.
-The distance between the two oppositely charged ions is the nearest neighbour distance.

Complete step by step answer:
Given that AB crystallizes in a bcc lattice with an edge length equal to 387 pm.
We need to find out the distance between the two oppositely charged ions in the crystal lattice.
We know that if ‘a’ is the edge length of the unit cell of a bcc lattice and ‘d’ is the nearest neighbour distance, then the relationship between the nearest neighbour distance ‘d’ and the edge length ‘a’ of the unit cell is given by the expression:
${\text{d}} = \dfrac{{\sqrt 3 }}{2}{\text{a}}$
According to the question, ‘a’ is equal to 387 pm.
So, substitute the value of ‘a’ in the above expression. Thus, we will have:
$
  {\text{d}} = \dfrac{{\sqrt 3 }}{2} \times 387{\text{pm}} \\
   \Rightarrow {\text{d}} = 335.14{\text{pm}} \\
   \Rightarrow {\text{d}} \approx 335{\text{pm}} \\
 $
Therefore, the nearest neighbour distance is approximately 335 pm. This means the distance between the two oppositely charged ions in the lattice is approximately 335 pm.

So, the correct option is A.

Note:
If ‘r’ is the atomic radius, then r will be half of the nearest neighbour distance and then we have:
\[{\text{r}} = \dfrac{{\sqrt 3 }}{4}{\text{a}}\]
The nearest neighbour distance for a simple cubic unit cell is ${\text{d}} = {\text{a}}$ and for a face centred cubic unit cell is ${\text{d}} = \dfrac{{\text{a}}}{{\sqrt 2 }}$ .