Question

AB crystalizes in a bcc lattice. If the distance between two oppositely charged ions in the lattice is 335 pm, then the edge length of it (in pm) is

• A. 376.8
• B. 366.8
• C. 386.8
• D. 396.8 \bigskip

Hint:

For a bcc lattice: \[ \text{Nearest neighbour distance} =\frac{\sqrt3\,a}{2} \] and \[ a=\frac{2d}{\sqrt3} \] where \(d\) is the distance between nearest neighbouring ions.

Answer 1

The Correct Option is C

Concept: In a body-centered cubic (bcc) ionic lattice, the body-centered ion and corner ion touch each other along the body diagonal. The length of the body diagonal of a cube is: \[ \sqrt{3}a \] where \(a\) is the edge length. Since the body-centered ion lies midway along the body diagonal, \[ \text{Nearest neighbour distance} =\frac{\sqrt{3}a}{2} \]

Step 1: Write the relation between edge length and nearest neighbour distance.
Given distance between oppositely charged ions: \[ d=335\ \text{pm} \] For a bcc structure, \[ d=\frac{\sqrt3\,a}{2} \]

Step 2: Substitute the given value.
\[ 335=\frac{\sqrt3\,a}{2} \] Multiplying both sides by 2, \[ 670=\sqrt3\,a \] \[ a=\frac{670}{\sqrt3} \]

Step 3: Calculate the numerical value.
\[ a=\frac{670}{1.732} \] \[ a\approx 386.8\ \text{pm} \]

Step 4: Identify the correct option.
\[ \boxed{a=386.8\ \text{pm}} \] Hence the correct answer is Option (C).