Question

When heated above ${{916}^{\circ }}C$, iron changes its bcc crystalline form to fcc without the change in the radius of atoms. The ratio of density of the crystal before heating and after heating is:
(a)- 1.069
(b)- 0.918
(c)- 0.725
(d)- 1.231

Answer 1

Hint: The bcc crystal structure means body-centered cubic structure and fcc means face-centered cubic structure. The density of the crystal can be calculated by using the formula:
$d=\dfrac{\text{Z x M}}{{{(a)}^{3}}\text{ x }{{\text{N}}_{a}}}$
Where Z is the number of atoms, M is the atomic mass of the element, a is the edge length of the cubic cell, and ${{N}_{a}}$ is the Avogadro’s number.

Complete answer:
We have to take the ratio of the density of the crystal. The formula that can be used is:
$d=\dfrac{\text{Z x M}}{{{(a)}^{3}}\text{ x }{{\text{N}}_{a}}}$
Where Z is the number of atoms, M is the atomic mass of the element, a is the edge length of the cubic cell, and ${{N}_{a}}$ is the Avogadro’s number.
Before ${{916}^{\circ }}C$ the crystal of iron is body-centered cubic crystal (BCC), so the Z will be 2, and M will be 56. Since, there is no change in the radius of the atom, the relation between the edge length and radius of the cubic structure is:
$r=\dfrac{\sqrt{3}}{4}a$
$a=\dfrac{4}{\sqrt{3}}r$
Keeping this as density one, we can write:
${{d}_{1}}=\dfrac{2\text{ x 56}}{{{\left( \dfrac{4}{\sqrt{3}}r \right)}^{3}}\text{ x }{{\text{N}}_{a}}}$
After ${{916}^{\circ }}C$ the crystal of iron is face-centered cubic crystal(FCC), so the Z will be 4, and M will be 56. Since, there is no change in the radius of the atom, the relation between the edge length and radius of the cubic structure is:
$r=\dfrac{a}{2\sqrt{2}}$
$a=2\sqrt{2}r$
Keeping this as density second, we can write:
${{d}_{2}}=\dfrac{\text{4 x 56}}{{{\left( 2\sqrt{2}r \right)}^{3}}\text{ x }{{\text{N}}_{a}}}$
Now, we can take the ratio of this, we get:
$\dfrac{{{d}_{1}}}{{{d}_{2}}}=\dfrac{\dfrac{2\text{ x 56}}{{{\left( \dfrac{4}{\sqrt{3}}r \right)}^{3}}\text{ x }{{\text{N}}_{a}}}}{\dfrac{\text{4 x 56}}{{{\left( 2\sqrt{2}r \right)}^{3}}\text{ x }{{\text{N}}_{a}}}}=0.918$
So, the ratio is 0.918.

Therefore, the correct answer is an option (b).

Note:
If the density was greater than 1, means the density of the iron has increased above the temperature. Here the value is less than 1, so after ${{916}^{\circ }}C$ the density of iron has decreased.