Question

The distance between $N{a}^{+}$ and $C{l}^{-}$ ions in $NaCl$ with a density $3.165$ g $c{m}^{-3}$ is:

• A. $300$ pm
• B. $248.5$ pm
• C. $248.5$ cm
• D. $24.85$ pm

Answer 1

The correct option is B $248.5$ pm

Theoretical density of crystal, $\rho =\frac{nM}{N_o{a}^3}g/c{m}^3$

In $NaCl$ crystal, there are $4$ formula units per unit cell. So, $n=4$

$\Rightarrow a^3=\frac{nM}{N_o\rho }=\frac{4\times 58.5}{6.022\times {10}^{23}\times 3.165}$

$\Rightarrow a=(123{)}^{\left(1/3\right)}\times ({10}^{-24}{)}^{1/3}cm$

$\Rightarrow a=4.97\times {10}^{-8}cm=497pm$

Now, in $NaCl$, $C{l}^{-}$ ions are present at corners and at face centers and $N{a}^{+}$ ions are present at edge centers and body centers.

So, distance between $N{a}^{+}$ and $C{l}^{-}=\frac{a}{2}=\frac{497}{2}=248.5pm$