Question

The radii, $rN{a^ + }\; = {\text{ }}95\;pm$ and $rC{l^ - }\; = 181\;pm$ in NaCl (rock salt) structure. What is the shortest distance (in pm) between $N{a^ + }$ ions?
A.778.3
B.276.2
C.195.7
D.390.3

Answer 1

NaCl (rock salt) forms fcc structure and since the radius of the cation and anion are given in the question, so we take ‘a’ as the edge length of the atom. Hence, the edge length ‘a’ will be equal to $2\left( {rN{a^ + } + rC{l^ - }} \right)$. The shortest distance between two $N{a^ + }$ ions will be $\dfrac{a}{{\sqrt 2 a}}$.
Complete step by step answer:
Radius of cation, $rN{a^ + }\; = {\text{ }}95{\text{ }}pm$
Radius of anion, $rC{l^ - } = 181{\text{ }}pm$
Let ‘a’ be the length of each side of the cube.
NaCl forms fcc structure, Na individually forms fcc structure and Cl also individually forms fcc structure. So, in fcc structure the nearest distance between 2 atoms is $\dfrac{a}{{\sqrt 2 }}$ .
We know that,

$$\left( {rN{a^ + } + rC{l^ - }} \right) = \dfrac{a}{2} \\\ \Rightarrow 95 + 181 = \dfrac{a}{2} \\\ \Rightarrow 2 \times 276 = a \\\ \Rightarrow a = 552 \\\$$

The shortest distance between $N{a^ + }$ and $C{l^ - }$ is $\dfrac{a}{2} = 276{\text{ }}pm$
So, Shortest distance between 2 $N{a^ + }$ ions is

\sqrt {\left( {2762 + 2762} \right)} \\\ = \sqrt {\left( {152352} \right)} \\\ { = 390.32pm} \end{array}$$ So, the shortest distance between $$N{a^ + }$$ ions is 390.32 pm **Therefore, the correct answer is option (D).** **Note:** The rock salt also called NaCl which is an ionic compound. It occurs naturally as white cubic crystals and it is extracted from the mineral form halite or evaporation of seawater. The structure of NaCl is formed by repeating the face centred cubic unit cell. It has an organized structure and has a $$1:1$$ ratio of $$Na:Cl$$ . Since NaCl are one to one ratio as a compound, the coordination numbers of Na and Cl are equal.