The Solid State • Topic 2 of 3

Packing, Density & Voids

To understand how particles fill space we treat them as identical hard spheres and arrange them as compactly as possible — close packing. In a single layer (layer A) each sphere touches six neighbours, giving the maximum two-dimensional packing (hexagonal close packing of one layer).

Stacking layers leads to two important three-dimensional patterns:

Hexagonal close packing (hcp) — the third layer lies directly above the first, giving the repeating sequence ABAB... (e.g. Mg, Zn).
Cubic close packing (ccp), identical to face-centred cubic (fcc) — the third layer occupies new positions, giving the sequence ABCABC... (e.g. Cu, Ag, Au).

Both hcp and ccp give the same maximum coordination number of 12 (the number of nearest neighbours touching a given sphere) and the same packing efficiency of 74%. A bcc structure has coordination number 8 and a simple cubic structure has coordination number 6.

The empty spaces left between packed spheres are voids (interstitial sites). There are two kinds: a tetrahedral void is surrounded by 4 spheres (smaller, radius ratio $r/R=0.225$) and an octahedral void is surrounded by 6 spheres (larger, $r/R=0.414$). For $N$ close-packed spheres there are $2N$ tetrahedral voids and $N$ octahedral voids.

Packing efficiency is the percentage of total space actually occupied by spheres: $$\text{Packing efficiency}=\frac{z\times\tfrac{4}{3}\pi r^3}{a^3}\times100$$ Using the edge–radius relations:

Simple cubic: $a=2r$ gives 52.4%.
Body-centred cubic: $\sqrt{3}\,a=4r$ gives 68%.
Face-centred (ccp) / hcp: $\sqrt{2}\,a=4r$ gives 74% — the most efficient.

Finally, the density of a unit cell connects the microscopic structure to a measurable bulk property:

$$d=\frac{z\,M}{a^3\,N_A}$$ where $z$ = number of atoms (formula units) per unit cell, $M$ = molar mass, $a$ = edge length and $N_A=6.022\times10^{23}$ is Avogadro's number. When $a$ is in cm and $M$ in g/mol, $d$ comes out in $\text{g cm}^{-3}$. This single relation lets us find any one of $d,z,M,a$ when the other three are known, and is the workhorse of all numerical problems in this chapter.

Solved Examples

Worked Example

A metal crystallises in a face-centred cubic lattice with edge length $a=400\ \text{pm}$ and atomic radius $r$. Find $r$.

Solution

For fcc, atoms touch along the face diagonal: $\sqrt{2}\,a=4r$.
$r=\frac{\sqrt{2}\,a}{4}=\frac{1.414\times400}{4}$ pm.
$r=\frac{565.6}{4}=141.4\ \text{pm}$.

Answer: $r\approx141.4\ \text{pm}$.

Worked Example

Show that the packing efficiency of a simple cubic structure is 52.4%.

Solution

Simple cubic has $z=1$ and atoms touch along the edge: $a=2r$.
Efficiency $=\frac{1\times\frac{4}{3}\pi r^3}{a^3}=\frac{\frac{4}{3}\pi r^3}{(2r)^3}=\frac{\frac{4}{3}\pi r^3}{8r^3}$.
$=\frac{\pi}{6}=0.524$.
$\times100=52.4\%$.

Answer: 52.4%.

Worked Example

Show that the packing efficiency of a bcc structure is 68%.

Solution

For bcc, $z=2$ and atoms touch along the body diagonal: $\sqrt{3}\,a=4r$, so $r=\frac{\sqrt{3}}{4}a$.
Efficiency $=\frac{2\times\frac{4}{3}\pi r^3}{a^3}$. Substitute $r=\frac{\sqrt{3}}{4}a$.
$r^3=\frac{3\sqrt{3}}{64}a^3$, so numerator $=2\times\frac{4}{3}\pi\times\frac{3\sqrt{3}}{64}a^3=\frac{\sqrt{3}\pi}{8}a^3$.
Efficiency $=\frac{\sqrt{3}\pi}{8}=0.680=68\%$.

Answer: 68%.

Worked Example

An element (atomic mass $M=56\ \text{g/mol}$) crystallises in bcc with edge $a=2.86\times10^{-8}\ \text{cm}$. Find its density. ($N_A=6.022\times10^{23}$)

Solution

Use $d=\frac{zM}{a^3 N_A}$ with $z=2$ (bcc).
$a^3=(2.86\times10^{-8})^3=2.34\times10^{-23}\ \text{cm}^3$.
$d=\frac{2\times56}{(2.34\times10^{-23})(6.022\times10^{23})}=\frac{112}{14.09}$.
$d\approx7.95\ \text{g cm}^{-3}$.

Answer: $d\approx7.95\ \text{g cm}^{-3}$ (this is iron).

Worked Example

In a close-packed structure of 0.5 mol of spheres, how many tetrahedral and octahedral voids are present?

Solution

Number of spheres $N=0.5\times6.022\times10^{23}=3.011\times10^{23}$.
Octahedral voids $=N=3.011\times10^{23}$.
Tetrahedral voids $=2N=6.022\times10^{23}$.

Answer: $3.011\times10^{23}$ octahedral and $6.022\times10^{23}$ tetrahedral voids.

Worked Example

A face-centred cubic metal has density $d=8.92\ \text{g cm}^{-3}$ and molar mass $M=63.5\ \text{g/mol}$. Find the edge length $a$. ($N_A=6.022\times10^{23}$)

Solution

Rearrange $d=\frac{zM}{a^3 N_A}$ to $a^3=\frac{zM}{d\,N_A}$ with $z=4$.
$a^3=\frac{4\times63.5}{8.92\times6.022\times10^{23}}=\frac{254}{5.372\times10^{24}}$.
$a^3=4.728\times10^{-23}\ \text{cm}^3$, so $a=(4.728\times10^{-23})^{1/3}$.
$a\approx3.61\times10^{-8}\ \text{cm}=361\ \text{pm}$.

Answer: $a\approx361\ \text{pm}$ (this is copper).

Key Points

Close packing gives hcp (ABAB...) and ccp/fcc (ABCABC...), both with coordination number 12 and 74% packing efficiency.
For $N$ spheres there are $2N$ tetrahedral voids ($r/R=0.225$) and $N$ octahedral voids ($r/R=0.414$).
Packing efficiency: simple cubic 52.4% ($a=2r$), bcc 68% ($\sqrt{3}a=4r$), fcc/ccp 74% ($\sqrt{2}a=4r$).
Coordination numbers: simple cubic 6, bcc 8, fcc/hcp 12.
Density of a unit cell $d=\frac{zM}{a^3 N_A}$ links $z$, molar mass, edge length and $N_A$ to bulk density.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.The packing efficiency of a face-centred cubic (ccp) structure is:

Explanation: Both ccp/fcc and hcp pack spheres most efficiently, occupying 74% of space.

Q2.For $N$ close-packed spheres, the number of octahedral voids is:

Explanation: There are $N$ octahedral voids and $2N$ tetrahedral voids for $N$ spheres.

Q3.In a body-centred cubic lattice, atoms touch along the:

Explanation: In bcc the corner and body-centre atoms touch along the body diagonal, so $\sqrt{3}\,a=4r$.

Q4.The coordination number of an atom in a simple cubic lattice is:

Explanation: Each atom in a simple cubic lattice has 6 nearest neighbours, so the coordination number is 6.

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