Question

CsBr  has a bcc structure with edge length of \[4.3A^o\] The shortest inter-ionic distance(in pm) between cation and anion is:

A. $3.72A^o$

B. $1.86A^o$

C. $7.44A^o$

D. $4.30A^o$

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 step by step 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 edge length is \[43pm\]  this means that the value 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:

${{r}_{C{{s}^{+}}}}+{{r}_{B{{r}^{-}}}}=\dfrac{\sqrt{3}}{2}\times a$

Thus distance between \[C{{s}^{+}}\] and \[B{{r}^{-}}\]$=\dfrac{\sqrt{3}}{2}\times a=\dfrac{\sqrt{3}}{2}\times 4.3=3.72pm$

So, the nearest distance between cation and anion will be $3.72A^o$

Therefore, option (A) is correct.

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: $d=\dfrac{a}{\sqrt{2}}.$