Question

The effective radius of an iron atom is 1.42 $\text{A}^{\circ}$. It has FCC structure. Calculate its density in ${\text{gc}}{{\text{m}}^{{\text{ - 3}}}}$ if it is given that atomic mass of \[Fe = 56\]
A.$2.47$
B.$5.74$
C.$8.24$
D.None of these

Answer 1

Hint: To answer this question, you should recall the concept of the unit cell. The unit cells are systematically arranged in such a way that fills the space without overlapping. A crystal lattice can be defined as the 3D arrangement of atoms, molecules or ions inside a crystal. Each unit of this crystal lattice is known as a unit cell.

Formula used:
\[d = \dfrac{{Z \times M}}{{{N_{A}} \times V}}\]
where $d$ is density, $Z$ is atoms occupying space, $M$ is mass, \[{N_{A}}\] is Avogadro number and $V$ is volume.
\[a = 2\sqrt 2 r\]
where $r$=radius= 1.42 $\text{A}^{\circ}$ and $a$=length of the side of the unit cell.

Complete step by step solution:
These crystal lattices can be broadly classified into primitive cubic, body-centred cubic (BCC) or face-centred cubic (FCC).
In the case of the FCC unit cell, the edge length can be calculated using the relation \[a = 2\sqrt 2 r\]
Substituting the values, we can calculate the value:
$a = 4.02 \times {10^{ - 8}}\;{\text{cm}}$
Therefore, the volume of the unit cell is \[{(4.02 \times {10^{ - 8}})^3}{\text{c}}{{\text{m}}^{\text{3}}}\]
Now using the formula of density, we get:
\[d = \dfrac{{4 \times 56}}{{6.023 \times {{10}^{23}} \times {{(4.02 \times {{10}^{ - 8}})}^3}}}\]
After solving this we get the answer as \[d = 5.74\;{\text{gc}}{{\text{m}}^{{\text{ - 3}}}}\]

Hence, the correct option is B.

Note: Crystalline solids show regular and repeating pattern arrangement of constituent particles resulting in two types of interstitial voids in a 3D structure:
Tetrahedral voids: In case of a cubic close-packed structure, the second layer of spheres is present over the triangular voids of the first layer. This results in each sphere touching the three spheres of the first layer. When we join the centre of these four spheres, we get a tetrahedron and the space left over by joining the centre of these spheres forms a tetrahedral void.
Octahedral voids: Adjacent to tetrahedral voids you can find octahedral voids. When the triangular voids of the first layer coincide with the triangular voids of the layer above or below it, we get a void that is formed by enclosing six spheres. This vacant space which is formed by the combination of the initial formed triangular voids of the first layer and that of the second layer is known as Octahedral Voids.