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Question: Ferrous oxide has a cubic structure with an edge length of the unit cell equal to 5.0 A˚. Assuming t...

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 3.84gcm33.84\dfrac{g}{{c{m^3}}} , the number of Fe2+F{e^{2 + }} and O2{O^{2 - }} ions present in each unit cell will be respectively, (use NA= 6 × 1023{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

Explanation

Solution

In the ferrous oxide structure, both ions ( Fe2+F{e^{2 + }} and O2{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=nMVNAd = \dfrac{{nM}}{{V{N_A}}}. Substituting the values, we will calculate the number of Fe2+F{e^{2 + }} and O2{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=nMVNA n=dVNAM  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=5A0=5×108cmA = 5{A^0} = 5 \times {10^{ - 8}}cm,
d=3.84gcm3d = 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 gmol\dfrac{g}{{mol}}
Volume of one unit = cube of length of corner =(5×108)3 = {\left( {5 \times {{10}^{ - 8}}} \right)^3}
Now, substituting all the values in the expression of density,

n=dVNAM n=3.84×(5×108)3×6.023×102372 n=4 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 Fe2+F{e^{2 + }} and 4 O2{O^{2 - }} ions present in each unit cell will be respectively.

Therefore, the correct answer is option (A).

Note: The term NA{N_A} is the Avogadro number which corresponds to 6 × 10236{\text{ }} \times {\text{ }}{10^{23}} . We know that one mole of a substance contains NA{N_A} i.e. 6 × 10236{\text{ }} \times {\text{ }}{10^{23}} number of molecules or ions or atoms.