Question

The density of copper metal is 8.95 g$\text{c}{{\text{m}}^{-3}}$, if the radius of copper atom is 127.8 pm is the copper r unit cell a simple cubic, a body centered cubic or a face centered cubic structure?( At mass of Cu = 63.54 g/mole and ${{\text{N}}_{a}}$ = 6.02 × ${{10}^{23}}$ $\text{mol}{{\text{e}}^{-1}}$)

Answer 1

Hint: We need to first find the densities of copper metal taking the different values of Z and r according to simple cube, bcc and fcc. For simple cube Z = 1 and a = 2r , for bcc Z = 2 and a = $\dfrac{\text{4r}}{\sqrt{3}}$ and for fcc Z = 4 and a = $\dfrac{\text{4r}}{\sqrt{2}}$ and then the by applying the formula for density = $\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
, we can find their densities and after that we can easily know whether the copper unit cell is simple cube, bcc or fcc.

Complete step by step answer:
To know about the unit cell of copper, we have to find the densities of copper metal in the simple cube, bcc, and fcc.
So, for simple cubic, we know that Z=1 and a=2r
r= 127.8 pm (given)
then,
\[\begin{align}
& \Rightarrow a=2\times 127.8\\
 & \Rightarrow \text{ }=\text{ }255.6pm \\
 & \Rightarrow \text{ }=2.55\times {{10}^{-12}}m\text{ }(1m\text{ }={{10}^{12}}pm) \\
 & \Rightarrow \text{ }=2.55\times {{10}^{-8}}cm\text{ }(1m=100cm) \\
\end{align}\]
As we know that, density =$\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
Atomic mass of Cu=63.54 $\text{gmol}{{\text{e}}^{-1}}$ (given)
${{\text{N}}_{a}}$= $6.02\times ~{{10}^{23}}\text{mol}{{\text{e}}^{-1}}$ (given), then;
density = $\dfrac{1\times 63.5}{{{(2.55)}^{3}}\times {{10}^{-24}}\times 6.023\times {{10}^{23}}}$
=$\dfrac{1\times 63.54\times 10}{16.58\times 6.023}$
=$\dfrac{635.4}{99.86}$
=6.36 g$\text{c}{{\text{m}}^{-3}}$
So, the density of a simple cube is not the same as that of the density of copper metal given in the statement. So, the copper (Cu) unit cell is not a simple cube.
For body centered, we know that Z=2 and a=$\dfrac{\text{4r}}{\sqrt{3}}$
r= 127.8 pm (given)

Then,
 \[\begin{align}
 & \Rightarrow a=\dfrac{4\times 127.8}{1.732}\\
 & \Rightarrow \text{ } = \text{ }295.15pm\\
 & \Rightarrow \text{ } =295.15\times {{10}^{-12}}m\text{ }(1m\text{ }={{10}^{12}}pm) \\
 & \Rightarrow \text{ } = 2.95\times {{10}^{-8}}cm\text{ }(1m = 100cm) \\
\end{align}\]
As we know that, density =$\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$

Atomic mass of Cu = 63.54 $\text{gmol}{{\text{e}}^{-1}}$ (given)
${{\text{N}}_{a}}$ = $6.02\times ~{{10}^{23}}\text{mol}{{\text{e}}^{-1}}$(given), then
Density = $\dfrac{2\times 63.5}{{{(2.95)}^{3}}\times {{10}^{-24}}\times 6.023\times {{10}^{23}}}$
=$\dfrac{2\times 63.54\times 10}{25.67\times 6.023}$
=$\dfrac{1270.8}{154.61}$
 =8.21 $g \text{c}{{\text{m}}^{-3}}$

So, the density of body centered cube (bcc) is nearly equal to that of the density of copper metal given in the statement but not the same. So, the copper (Cu) unit cell is not a body centered cube.
For face centered, we know that Z=4 and a=$\dfrac{\text{4r}}{\sqrt{2}}$
r= 127.8 pm (given)
Then,
\[\begin{align}
 & \Rightarrow a=\dfrac{4\times 127.8}{1.414}\\
 & \Rightarrow \text{ }=\text{ 361}\text{.52}pm \\
 & \Rightarrow \text{ }=361.52\times {{10}^{-12}}m\text{ }(1m\text{ }={{10}^{12}}pm) \\
 & \Rightarrow \text{ }=3.61\times {{10}^{-8}}cm\text{ }(1m=100cm) \\
\end{align}\]
As we know that, density =$\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$
Atomic mass of Cu=63.54 $\text{gmol}{{\text{e}}^{-1}}$ (given)
${{\text{N}}_{a}}$=$6.02\times ~{{10}^{23}}\text{mol}{{\text{e}}^{-1}}$(given), then
Density = $\dfrac{4\times 63.5}{{{(3.61)}^{3}}\times {{10}^{-24}}\times 6.023\times {{10}^{23}}}$
 =$\dfrac{4\times 63.54\times 10}{47.04\times 6.023}$
=$\dfrac{2540}{282}$
=8.96 g$\text{c}{{\text{m}}^{-3}}$
The density of face centered is same as that of the density of copper metal given in the statement.
Hence, the copper (Cu) unit cell is face centered cubic(fcc).

Note: Don’t get confused in the a and r, a is the edge length (i.e. edge means the line segment where the two edges meet and the cube have 12 edges and the length of one edge to another is called as the edge length) and r is the radius of the cube. If we know, edge length and density, we can calculate radius from the density formula i.e. $\dfrac{\text{ZM}}{{{\text{a}}^{3}}{{\text{N}}_{\text{0}}}}$ and vice versa.