Question

Cadmium oxide (CdO) has NaCl type structure with density equal to $ 8.27{\text{ g c}}{{\text{m}}^{ - 3}} $ . If the ionic radius of $ {O^{2 - }} $ is $ 1.24{\text{ }}{{\text{A}}^ \circ } $ , determine the ionic radius of $ C{d^{2 + }} $ ?
(A) $ 1.5 $
(B) $ 1.1 $
(C) $ 1.9 $
(D) $ 1.6 $

Answer 1

Hint: NaCl has FCC structure. FCC is also called as face centered cubic structure. In FCC, atoms are arranged at the corners and center of each cube face. To solve this question, we will use the density formula and formula connecting radius of cation, anion and edge length. We shall substitute the values in the equation given to find the edge length and use it to calculate the ionic radius.
Formula used
The formulae used in this equation are as follows,
 $ d = \dfrac{{{\text{Molar mass }} \times {\text{ 2}}}}{{{{\text{N}}_a}{\text{ }} \times {\text{ }}{{\text{a}}^3}}} $
 $ {r^ + } + {r^ - } = \dfrac{a}{2} $

Complete step by step solution:
The density of the cadmium oxide crystal lattice is given as $ 8.27{\text{ g c}}{{\text{m}}^{ - 3}} $ . Let a be the edge length of the unit cell of FCC lattice. The Avogadro number is represented by $ {{\text{N}}_a} $ . Z is the coordination number and for FCC lattice, the value of Z is 4. The molar mass of cadmium oxide is 128.
The edge length can be calculated using the following formula,
 $ d = \dfrac{{{\text{Molar mass }} \times {\text{ 2}}}}{{{{\text{N}}_a}{\text{ }} \times {\text{ }}{{\text{a}}^3}}} $
 $ {a^3} = \dfrac{{4 \times 128}}{{8.27 \times 6.02 \times {{10}^{23}}}} $
 $ a = 4.68 \times {10^{ - 10}}m = 4.68{\text{ }}{{\text{A}}^ \circ } $
Hence the edge length of cadmium oxide crystal lattice is $ 4.68{\text{ }}{{\text{A}}^ \circ } $ .For NaCl type structure,
 $ {r^ + } + {r^ - } = \dfrac{a}{2} $
In cadmium oxide, $ C{d^{2 + }} $ is a cation and $ {O^{2 - }} $ is an anion.
Hence, the ionic radius of $ C{d^{2 + }} $ can be found out by the following method,
 $ \begin{array}{l}{r^ + } + {r^ - } = \dfrac{a}{2}\\\therefore {r^ + } = \dfrac{{4.68}}{2} - 1.24 = 1.1{\text{ }}{{\text{A}}^ \circ }\end{array} $
The ionic radius of $ C{d^{2 + }} $ is $ 1.1{\text{ }}{{\text{A}}^ \circ } $ .
Therefore, the correct answer is option B.

Note:
Coordination number is the number of the particles that each particle in a crystalline solid contacts. In other words, it is the number of the nearest neighboring atoms or the ions surrounding an atom or an ion.