Question

Naturally gold is face centered cube and has a density of $19.3{\text{g}}.{\text{c}}{{\text{m}}^{ - 3}}$. Find the atomic radius of gold $\left( {{\text{Au = 197g}}{\text{.mo}}{{\text{l}}^{ - 1}}} \right)$.

Answer 1

Hint: The fundamental structural unit of the structure of crystal is termed as the unit cell. There are four basic arrangements. They are simple, body-centered, hexagonal close-packed and face centered cubic structures. Along face diagonals, the atoms are touched with each other.

Complete step by step solution:
The main properties of unit cell are number of atoms per unit cell, effective number of atoms per unit cell, coordination number and atomic packing factor.
Face-centered cubic structure is also known as cubic close packing. It is the most efficient packing of hard spheres of any lattice. All angles in fcc are equal to ${90^ \circ }$.
It is given that the density of gold, ${\rho _{\text{G}}} = 19.3{\text{g}}.{\text{c}}{{\text{m}}^{ - 3}}$ and atomic mass of gold, ${{\text{M}}_{\text{G}}} = 197{\text{g}}.{\text{mo}}{{\text{l}}^{ - 1}}$
The number of atoms in face-centered cubic unit cell, ${\text{z = 4}}$
And we know that the Avogadro number, ${{\text{N}}_{\text{A}}} = 6.022 \times {10^{23}}$
Now according to the question, density can be calculated by the given formula:
${\rho _{\text{G}}} = \dfrac{{{\text{zM}}}}{{{{\text{N}}_{\text{A}}}{{\text{a}}^3}}}$
Rearranging the equation, we get
${{\text{a}}^3} = \dfrac{{{\text{zM}}}}{{{\rho _{\text{G}}}{{\text{N}}_{\text{A}}}}} = \dfrac{{4 \times 197}}{{19.3 \times 6.022 \times {{10}^{23}}}}$
On simplification, we get
${{\text{a}}^3} = \dfrac{{788}}{{1.16 \times {{10}^{25}}}} = 6.78 \times {10^{ - 23}}$
Taking cube root, we get the value of the length of the cubic unit cell, ${\text{a}}$.
i.e. ${\text{a}} = \sqrt[3]{{6.78 \times {{10}^{ - 23}}}} = 4.08 \times {10^{ - 8}}{\text{cm}}$
When we consider fcc structure, atomic radius can be calculated by the given formula:
Atomic radius, ${\text{r}} = \dfrac{{\text{a}}}{{\sqrt 8 }}$
Substituting the value of ${\text{a}}$, we get
${\text{r}} = \dfrac{{4.08 \times {{10}^{ - 8}}}}{{\sqrt 8 }} = 1.44 \times {10^{ - 8}}{\text{cm}} = 144 \times {10^{ - 10}}{\text{m}}$
$1 \times {10^{ - 10}}{\text{m}} = 1{\text{pm}}$

So, atomic radius of gold, ${\text{r = 144pm}}$.

Note: The effective number of atoms per unit cell denotes the number of atoms which belong to a unit cell. The density of a material can be predicted if the atomic mass, atomic radius, and the crystal geometry. Fcc, bcc and hcp are common metallic crystal structures. Atomic packing factor and coordination numbers are the same for these three structures.