Question

A solid AB has a NaCl structure. If the radius of the cation ${A^ + }$ is 170 pm, then the maximum possible radius of the anion. ${B^ - }$ is:
A. 397.4 pm
B. 347.9 pm
C.210.9 pm
D. 410.6 pm

Answer 1

To solve this problem, we must first understand the type of unit cell present in the structure of NaCl. It is a solid crystal that has a cubic structure that can be represented as a face-centered cubic array of anions with an interpenetrating fcc cation lattice (or vice-versa). This unit cell of NaCl looks the same whether we start with cations or anions at the corners. Each of these ions is 6-coordinate and has a local octahedral geometry.
Formula used:
$\dfrac{{{I_{{A^ + }}}}}{{{I_{{B^ - }}}}} = 0.414$

Complete step by step answer:
NaCl is an ionic crystalline compound. While the formation of the crystal structure of sodium chloride, it crystallizes into a packing structure known as a face-centered cubic. This packing structure forms a single unit in the multi – molecule lattice structure of sodium chloride. In this type of unit – cell, i.e. face-centered cubic, the following characteristics are observed:
1. In face-centered cubic structure, there are atoms on all the vertices of the cube
Atoms are also present on all the sides of the cube
The vertices have $\dfrac{1}{8}$ th of an atom on every vertex due to geometric constraints
Similarly, all faces have the only $\dfrac{1}{2}$ of an atom on every face due to geometric constraints
Sodium atoms area at corners and the chlorine atoms are at the center of each face.
Now for 6 coordination number the
$\dfrac{{{I_{{A^ + }}}}}{{{I_{{B^ - }}}}} = 0.414$ . Where ${I_{{A^ + }}}$ and ${I_{{B^ - }}}$ are the radius of ${A^ + }$ and ${B^ - }$ respectively.
Now put the given data as follows and find out the radius of ${B^ - }$ as follows,

$$\dfrac{{{I_{{A^ + }}}}}{{{I_{{B^ - }}}}} = 0.414 \\\ \dfrac{{170}}{{{I_{{B^ - }}}}} = 0.414 \\\ {I_{{B^ - }}} = \dfrac{{170}}{{0.414}} \\\ {I_{{B^ - }}} = 410.6 \\\$$

So, the correct answer is D.

Note:
From the picture shown above, we can infer that there 8 portions of the $\dfrac{1}{8}$ th of an atom on the 8 vertices of the cube, while all the 6 sides of the cube have $\dfrac{1}{2}$ atoms each. Hence the total number of atoms present in the face-centered cubic structure can be calculated to be:
Total number of atoms = $\dfrac{1}{8}$ (8 vertices) + $\dfrac{1}{2}$ (6 faces) = 1 + 3 = 4 atoms
Hence, there are 4 atoms present in each unit cell of a face-centered cubic structure.
Hence, the number of NaCl units present in a unit cell of NaCl is 4