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) $r _{Cs^{+}}+r _{Cl^{-}}=\dfrac{3 a}{2}$
(c) $r _{Cs^{+}}+r _{Cl^{-}}=\dfrac{\sqrt{3}}{2} a$
(d) $r _{Cs^{+}}+r _{Cl^{-}}=\sqrt{3} a$
(2014 Main)
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Answer:
Correct Answer: 15. (c)
Solution:
- 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
$ F D=b=\sqrt{a^{2}+a^{2}}=\sqrt{2} a $
In $\triangle A F D$,
$ \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 c=\sqrt{3} a \end{aligned} $
As $\triangle A F D$ is an equilateral triangle.
$\therefore \sqrt{3} a =4 r \quad$ $[\therefore C = 3r + r + r ]$
$\Rightarrow r =\dfrac{\sqrt{3} a}{4} $
Hence, $ r _{Cs^{+}}+r _{Cl^{-}} =2 r=2 \times \dfrac{\sqrt{3}}{4} a=\dfrac{\sqrt{3}}{2} a$
BETA
