Question

Ferrous oxide has a cubic structure with an edge length of the unit cell equal to 5.0 A˚. Assuming the density of ferrous oxide to be $3.84\dfrac{g}{{c{m^3}}}$ , the number of \[F{e^{2 + }}\] and \[{O^{2 - }}\] ions present in each unit cell will be respectively, (use \[{N_A} = {\text{ }}6{\text{ }} \times {\text{ }}{10^{23}}\])
A) 4 and 4
B) 2 and 2
C) 1 and 1
D) 3 and 4

Answer 1

Hint: In the ferrous oxide structure, both ions ( \[F{e^{2 + }}\] and \[{O^{2 - }}\] ) are arranged in such a way that they show a cubic arrangement. Since the edge length of the unit cell and the density of ferrous oxide is given in the question, and we know the expression of density, $d = \dfrac{{nM}}{{V{N_A}}}$. Substituting the values, we will calculate the number of \[F{e^{2 + }}\] and \[{O^{2 - }}\] ions present in each unit cell.

Complete step by step answer:
Given in the question is that ferrous oxide has cubic structure with all sides equal.
The expression for density (d) is as follows,
 ​
$
  d = \dfrac{{nM}}{{V{N_A}}} \\
   \Rightarrow n = \dfrac{{dV{N_A}}}{M} \\
 $
Here, ‘a’ is the edge length of the unit cell, M is the molecular mass and V is the volume of a unit cell.

Given in the question are,
$A = 5{A^0} = 5 \times {10^{ - 8}}cm$,
$d = 3.84\dfrac{g}{{c{m^3}}}$
Let the units of ferrous oxide in a unit cell =n
Molecular weight of Ferrous Oxide (FeO) = 56 + 16 = 72 $\dfrac{g}{{mol}}$
Volume of one unit = cube of length of corner $ = {\left( {5 \times {{10}^{ - 8}}} \right)^3}$
Now, substituting all the values in the expression of density,
\[
  n = \dfrac{{dV{N_A}}}{M} \\
   \Rightarrow n = \dfrac{{3.84 \times {{\left( {5 \times {{10}^{ - 8}}} \right)}^3} \times 6.023 \times {{10}^{23}}}}{{72}} \\
   \Rightarrow n = 4 \\
 \]
So, the number of 4 \[F{e^{2 + }}\] and 4 \[{O^{2 - }}\] ions present in each unit cell will be respectively.

Therefore, the correct answer is option (A).

Note: The term \[{N_A}\] is the Avogadro number which corresponds to \[6{\text{ }} \times {\text{ }}{10^{23}}\] . We know that one mole of a substance contains \[{N_A}\] i.e. \[6{\text{ }} \times {\text{ }}{10^{23}}\] number of molecules or ions or atoms.