Question

Select the correct statements among the following.

• A. Nearest neighbour distance in $NaCl=\frac{a}{2}$
• B. Nearest neighbour distance in $Ca{F}_2=\frac{\sqrt{3}a}{4}$
• C. Nearest neighbour distance in $N{a}_{2}O=\frac{\sqrt{3}a}{4}$
• D. Nearest neighbour distance in $CsCl=\frac{\sqrt{3}a}{2}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Nearest neighbour distance in $NaCl=\frac{a}{2}$
B Nearest neighbour distance in $Ca{F}_2=\frac{\sqrt{3}a}{4}$
C Nearest neighbour distance in $N{a}_{2}O=\frac{\sqrt{3}a}{4}$
D Nearest neighbour distance in $CsCl=\frac{\sqrt{3}a}{2}$

(A) $C{l}^{-}$ is present at the corner and the face centres and $N{a}^{+}$ is present at the edges and the body centre. Edge length is 'a' and hence distance between $N{a}^{+}$ and $C{l}^{-}$ is $\frac{a}{2}$.


(B) In $Ca{F}_2$, $Ca$ is present at corner of the cube and $F$ is present at tetrahedral void.
$\therefore$ distance between them is $\frac{\text{body diagonal}}{4}=\frac{\sqrt{3}a}{4}$

(C) $N{a}_{2}O$ has antifluorite structure. Hence $O^{2-}$ occupies at the corner and $N{a}^{+}$ occupies all the tetrahedral voids.
$\therefore$ distance $=\frac{\text{body diagonal}}{4}=\frac{\sqrt{3}a}{4}$


(D) In $CsCl$, $Cl$ forms lattice and is present at corner of the cube and $Cs$ is present at the centre of the cube.
$\therefore$ distance $=\frac{\text{body diagonal}}{2}=\frac{\sqrt{3}a}{2}$