Solid State - Result Question 15

####15. $CsCl$ crystallises in body centred cubic lattice. If ’ $a$ ’ its edge length, then which of the following expressions is correct?

(a) $r _{Cs^{+}}+r _{Cl^{-}}=3 a$

(b) $r _{Cs^{+}}+r _{Cl^{-}}=\frac{3 a}{2}$

(c) $r _{Cs^{+}}+r _{Cl^{-}}=\frac{\sqrt{3}}{2} a$

(d) $r _{Cs^{+}}+r _{Cl^{-}}=\sqrt{3} a$

(2014 Main)

Show Answer

Answer:

Correct Answer: 15. (c)

Solution:

  1. In $CsCl, Cl^{-}$lies at corners of simple cube and $Cs^{+}$at the body centre. Hence, along the body diagonal, $Cs^{+}$and $Cl^{-}$touch each other so $r _{Cs^{+}}+r _{Cl^{-}}=2 r$

Calculation of $r$

In $\triangle E D F$,

Body centred cubic unit cell

In $\triangle A F D$,

$$ F D=b=\sqrt{a^{2}+a^{2}}=\sqrt{2} a $$

$$ \begin{aligned} & c^{2}=a^{2}+b^{2}=a^{2}+(\sqrt{2} a)^{2}=a^{2}+2 a^{2} \\ & c^{2}=3 a^{2} \Rightarrow \quad c=\sqrt{3} a \end{aligned} $$

As $\triangle A F D$ is an equilateral triangle.

$$ \begin{array}{rlrl} & \therefore & \sqrt{3} a & =4 r \\ \Rightarrow & r & =\frac{\sqrt{3} a}{4} \\ & \text { Hence, } & r _{Cs^{+}}+r _{Cl^{-}} & =2 r=2 \times \frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{2} a \end{array} $$



NCERT Chapter Video Solution

Dual Pane