Question

If x is the length of body diagonal, then the distance between two nearest cations in rock salt structure is:

• A. $\frac{X}{\sqrt{6}}$
• B. $\frac{X}{\sqrt{5}}$
• C. $\frac{X}{\sqrt{3}}$
• D. $\frac{X}{\sqrt{2}}$

Answer 1

Answer: (A)

$\frac{X}{\sqrt{6}}$

In $NaCl$ structure, $C{l}^{-}$ ions form FCC and $N{a}^{+}$ are present at body center and edge center.

Now, if $a$ is the edge length of the cube, then body diagonal is given by

$d=\sqrt{3}a$

Given, $d=x$

$\Rightarrow x=\sqrt{3}a$.......(i)

Since, atoms at face centere and corners are touching each other, so distance between two nearest cation $=\frac{\sqrt{2}a}{2}$........(ii)

(i) and (ii)

$\Rightarrow \frac{x}{\sqrt{3}}=\sqrt{2}{d}^{\prime }$

$\Rightarrow d^{\prime }=\frac{x}{\sqrt{6}}$