Question

CsBr has BCC structure. The length of its one side is $4.3\text{ }\overset{\circ }{\mathop{\text{A}}}\,$. The minimum distance between $C{{s}^{+}}$ and $B{{r}^{-}}$ ion will be:
(a)- $\text{0}\text{.897 }\overset{\circ }{\mathop{\text{A}}}\,$
(b)- $\text{3}\text{.72 }\overset{\circ }{\mathop{\text{A}}}\,$
(c)- $1.749\text{ }\overset{\circ }{\mathop{\text{A}}}\,$
(d)- None of the above

Answer 1

Hint: To solve this problem we can use the formula $d=\dfrac{\sqrt{3}}{2}a$. So, the d will be the distance between the nearest neighbor element, and a will be the edge length of the cubic crystal.

Complete answer:
The given compound is CsBr in which there are two ions, i.e., $C{{s}^{+}}$ and $B{{r}^{-}}$. We are given that the CsBr compound crystallizes in the BCC structure. The BCC structure means body-centered cubic structure. BCC means in the cubic structure the atoms are present at the corners of the cube as well as one atom is present at the center of the cube.
We are also given that the length of its one side is $4.3\text{ }\overset{\circ }{\mathop{\text{A}}}\,$, this means that the value $4.3\text{ }\overset{\circ }{\mathop{\text{A}}}\,$is the edge length of the cubic crystal and this is denoted by the d or the distance between the nearest neighbor atoms.
So, here we can use the formula for BCC (body-centered cubic) structure will be:
$d=\dfrac{\sqrt{3}}{2}a$
Hence the d will be the distance between the nearest neighbor element, and a will be the edge length of the cubic crystal.
Now putting the value of the edge length in the above formula, we get:
$d=\dfrac{\sqrt{3}}{2}\text{ x 4}\text{.3}$
$d=3.72\text{ }\overset{\circ }{\mathop{\text{A}}}\,$
So, the nearest distance will be $3.72\text{ }\overset{\circ }{\mathop{\text{A}}}\,$.

Therefore, the correct answer is an option (b).

Note:
 When the given structure is a simple cubic crystal, then the formula will be: d = a. When the given structure is FCC or face-centered cubic crystal, then the formula will be: $d=\dfrac{a}{\sqrt{2}}$.