Question

Which of the following statement is not true about the hexagonal close packing?

• A. It has $74\%$ packing efficiency.
• B. Tetrahedral voids of the second layer are covered by the spheres of the third layer.
• C. In this arrangement spheres of the fourth layer are exactly aligned with those of the first layer.
• D. The coordination number is $12$.

Answer 1

The correct option is C In this arrangement spheres of the fourth layer are exactly aligned with those of the first layer.

$\text{Analyzing the option}$

Option (A):

Hexagonal packing has layers arranged in $ABAB$….. Pattern. In $hcp$ each sphere in a layer is surrounded by $6$ spheres in the same plane, $3$ spheres above it and $3$ spheres below it. Therefore, the coordination number for this arrangement is $12$.



Option (B):

Packing Efficiency:

$=\frac{\text{Volume occupied by atoms in a unit cell}}{\text{Total volume of the unit cell}}\times 100\%$

Packing efficiency:

$=\frac{Z\times \frac{4}{3}\pi r^3}{VolumeofHCPUnitcell}\times 100$

Where,

$r=$ radius on an atom
$Z=$ effective number of atoms in a unit cell
$a=$ edge length of a $HCP$ unit cell


Effective number of atoms in $HCP$ unit cell $\left(Z\right)$:

$=(\frac{1}{6}\times 12)+(\frac{1}{2}\times 2)+3=6$

$\frac{1}{6}$th contribution of one corner atom & there are $12$ corner atoms, half contribution of face centered atom & there are $2$ such atoms and $3$ atoms at body centre of $HCP$ unit cell.


Volume of $HCP$ unit cell:
$=\text{Area of the base}\left(A\right)\times \text{Height of the unit cell}\left(C\right)$


Area of the base is calculated by the area of six equilateral triangles.

Area of the base:

$=6\times \frac{\sqrt{3}}{4}{a}^2=6\times \frac{\sqrt{3}}{4}(2r{)}^2$


${}^{\prime }{h}^{\prime }$ can be calculated by applying Pythagoras theorem using Fig. $2$:

According to Pythagoras theorem applied on right angle triangle $4^{\prime }43$:

$h^2+x^2=a^2$ ...............$\left(i\right)$
where, $a=2r$

Using another small triangle $3^{\prime }{4}^{\prime }3$:
$Cos{30}^{\circ }=\frac{3^{\prime }3}{34^{\prime }}=\frac{r}{x}$

$\frac{r}{x}=\frac{\sqrt{3}}{2}$

$\Rightarrow x=\frac{2r}{\sqrt{3}}$

Putting the values of $a$ and $x$ in equation $\left(i\right)$:

$h^2+(2r{)}^2={\left(\frac{2r}{\sqrt{3}}\right)}^2$

$\Rightarrow h=2r\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$

From Fig.$1$:
$C=2h=4r\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$

Volume of $HCP$ unit cell $\left(V\right)$:

$=\text{Area of the base}\left(A\right)\times \text{Height of the unit cell}\left(C\right)$

$V=6\times \frac{\sqrt{3}}{4}(2r{)}^2\times 4r\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$

Packing efficiency:

$=\frac{Z\times \frac{4}{3}\pi r^3}{VolumeofHCPunitcell}\times 100$

$=\frac{6\times \frac{4}{3}\pi r^3}{6\times \frac{\sqrt{3}}{4}(2r{)}^2\times 4r\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}\times 100$

$=\frac{\pi }{3\sqrt{2}}\times 100=74.01\%\approx 74\%$


Option (C):

The voids surrounded by four spheres sitting at the corners of a regular tetrahedron are called tetrahedral voids.

Tetrahedral voids of the second layer $\left(B\right)$ are covered by the spheres of the third layer $\left(A\right)$.


Option (D):

Hexagonal packing has layers arranged in $ABAB$.. pattern. The spheres of the first layer are aligned with the third layer and not the fourth layer. The $1st$ layer and $4th$ layer are not exactly aligned.