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

Hint: 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