Question

Iron exhibits bcc structure at room temperature. Above ${900^ \circ }{\text{C}}$, it transforms to an fcc structure. The ratio of density of iron at room temperature to that at ${900^ \circ }{\text{C}}$ (assuming molar mass and atomic radii of iron remains constant with temperature) is:
A. $\dfrac{{3\sqrt 3 }}{{4\sqrt 2 }}$
B. $\dfrac{{\sqrt 3 }}{{\sqrt 2 }}$
C. $\dfrac{1}{2}$
D. $\dfrac{{4\sqrt 3 }}{{3\sqrt 2 }}$

Answer 1

Hint:Atoms or groups of atoms forming a building block of the smallest acceptable size of the whole volume of crystal is defined as a unit cell. For the cubic crystal, we have simple cubic, face-centered and body-centered.

Complete step by step answer:

FCC has an atom at each corner of the cube and one atom at the intersection of the diagonals of each of the six faces of the cube, but none at the center of the cube.
The number of atoms per unit cell in bcc, ${{\text{Z}}_1} = 2$
The edge length, ${{\text{a}}_1}{\text{ = }}\dfrac{4}{{\sqrt 3 }}{\text{r}}$, where ${\text{r}}$ is the radius of atom.
The number of atoms per unit cell in fcc, ${{\text{Z}}_2} = 4$
The edge length, ${{\text{a}}_2} = \dfrac{4}{{\sqrt 2 }}{\text{r}}$
Density of iron can be determined by the formula:
${\text{d}} = \dfrac{{{\text{ZM}}}}{{{{\text{a}}^3}{{\text{N}}_{\text{A}}}}}$, where ${\text{M}}$ is the molar mass.
${{\text{N}}_{\text{A}}}$ is the Avogadro number.
Consider the bcc unit cell. The density of iron in bcc, ${{\text{d}}_1} = \dfrac{{{{\text{Z}}_1}{\text{M}}}}{{{{\text{a}}_1}^3{{\text{N}}_{\text{A}}}}}$
Substituting the values, we get
${{\text{d}}_1} = \dfrac{{{\text{2M}}}}{{{{\left( {\dfrac{4}{{\sqrt 3 }}{\text{r}}} \right)}^3}{{\text{N}}_{\text{A}}}}} \to \left( 1 \right)$
Consider fcc unit cell and the density of iron in fcc, ${{\text{d}}_2} = \dfrac{{{{\text{Z}}_2}{\text{M}}}}{{{{\text{a}}_2}^3{{\text{N}}_{\text{A}}}}}$
Substituting the values, we get
${{\text{d}}_2} = \dfrac{{{\text{4M}}}}{{{{\left( {\dfrac{4}{{\sqrt 2 }}{\text{r}}} \right)}^3}{{\text{N}}_{\text{A}}}}} \to \left( 2 \right)$
Now we have to calculate the ratio of iron at room temperature and at ${900^ \circ }{\text{C}}$.
i.e. $\dfrac{{{{\text{d}}_1}}}{{{{\text{d}}_2}}} = \dfrac{{\dfrac{{2{\text{M}}}}{{{{\left( {\dfrac{4}{{\sqrt 3 }}{\text{r}}} \right)}^3}{{\text{N}}_{\text{A}}}}}}}{{\dfrac{{{\text{4M}}}}{{{{\left( {\dfrac{4}{{\sqrt 2 }}{\text{r}}} \right)}^3}{{\text{N}}_{\text{A}}}}}}}$
Taking reciprocal, we get
\[\dfrac{{{{\text{d}}_1}}}{{{{\text{d}}_2}}} = \dfrac{{2{\text{M}}}}{{{{\left( {\dfrac{4}{{\sqrt 3 }}{\text{r}}} \right)}^3}{{\text{N}}_{\text{A}}}}} \times \dfrac{{{{\left( {\dfrac{4}{{\sqrt 2 }}{\text{r}}} \right)}^3}{{\text{N}}_{\text{A}}}}}{{{\text{4M}}}} = \dfrac{{{{\left( {\dfrac{4}{{\sqrt 2 }}{\text{r}}} \right)}^3}{{\text{N}}_{\text{A}}}}}{{{{\left( {\dfrac{4}{{\sqrt 3 }}{\text{r}}} \right)}^3}{{\text{N}}_{\text{A}}}}} \times \dfrac{{2{\text{M}}}}{{{\text{4M}}}}\]
Canceling the common terms, we get
\[\dfrac{{{{\text{d}}_1}}}{{{{\text{d}}_2}}} = \dfrac{{{{\left( {\dfrac{4}{{\sqrt 2 }}} \right)}^3}}}{{{{\left( {\dfrac{4}{{\sqrt 3 }}} \right)}^3}}} \times \dfrac{2}{{\text{4}}} = {\left( {\dfrac{4}{{\sqrt 2 }}} \right)^3} \times {\left( {\dfrac{{\sqrt 3 }}{4}} \right)^3} \times \dfrac{1}{2} = \dfrac{{{{\sqrt 3 }^3}}}{{{{\sqrt 2 }^3}}} \times \dfrac{1}{2}\]
On simplification,
$\dfrac{{{{\text{d}}_1}}}{{{{\text{d}}_2}}} = \dfrac{{3\sqrt 3 }}{{2\sqrt 2 }} \times \dfrac{1}{2} = \dfrac{{3\sqrt 3 }}{{4\sqrt 2 }}$
Hence the ratio of density is $\dfrac{{3\sqrt 3 }}{{4\sqrt 2 }}$
So the correct option is A.

Note:
A bcc unit cell has one atom in the center of the cube and one atom at each of the corners. The radius is dependent on the lattice parameter. From the density of the unit cell, we can calculate the atomic packing fraction or packing efficiency. It is the ratio of volume of atoms per unit cell to the total volume of the unit cell.