Geometry: Area, Volume & Molecular Shape — Maths for General Chemistry

Shape, size and space

From beakers to bond angles

Chemistry happens in three dimensions, so a handful of geometry formulae come up constantly. The single most important idea is scaling: when you scale a shape by a factor of $k$ , lengths grow as $k$ , areas as $k^{2}$ , and volumes as $k^{3}$ . That cubic growth is why a crystal unit cell a few hundred picometres across fits millions of atoms per centimetre.

V_{sphere} = \frac{4}{3} π r^{3}, A_{sphere} = 4 π r^{2}, V_{cube} = s^{3}, V_{cyl} = π r^{2} h

Molecular shape is then determined by electron-pair repulsion (VSEPR): four bonding pairs arrange themselves tetrahedrally (109.5°), six pairs octahedrally (90°), and three pairs trigonally (120°). These angles are exact — they come from the geometry of the Platonic solids, not approximations.

For a crystallographer, the same formulae give unit-cell volume from the cell edge $a$ , atomic radius from the packing fraction, and density from mass divided by volume. For a catalyst engineer, the surface-area-to-volume ratio $(SA/V \propto 1/ r)$ explains why finely ground powders react far faster than large lumps.

Key formulas — shape at a glance

Sphere

V = \frac{4}{3} π r^{3}, A = 4 π r^{2}

Cube (unit cell)

V = a^{3}, A = 6 a^{2}

Cylinder (pellet / rod)

V = π r^{2} h, A = 2 π r (r + h)

SA / V ratio

For a sphere:

SA/V = \frac{3}{r}

. Halving

r

doubles the ratio.

Tetrahedral angle

θ = arccos (- \frac{1}{3}) = 109.4 7^{\circ}

Octahedral angle

90° between adjacent ligands, 180° between trans ligands.

Cubic unit cells — packing at a glance

Simple cubic (sc)

1 atom/cell (

8 \times \frac{1}{8}

). Touch along edge:

2 r = a

. Packing fraction 52.4 %. Coordination number 6. Rare in metals (Po only).

Body-centred cubic (bcc)

2 atoms/cell (

8 \times \frac{1}{8} + 1

). Touch along body diagonal:

4 r = a 3

. Packing fraction 68.0 %. CN = 8. e.g. Fe, W, Cr.

Face-centred cubic (fcc)

4 atoms/cell (

8 \times \frac{1}{8} + 6 \times \frac{1}{2}

). Touch along face diagonal:

4 r = a 2

. Packing fraction 74.0 %. CN = 12. e.g. Cu, Au, Ni.

Void space

% void = 100 % − packing %. sc: 47.6 %, bcc: 32.0 %, fcc: 26.0 %. These voids host interstitial atoms (C in Fe giving steel; H in Pd for hydrogen storage).

Common pitfalls

Wrong r–a relation for the cell type. Using $4 r = a 2$ (fcc) for a bcc cell — or vice versa — gives a radius off by a factor of ~1.22. Always identify the cell type first, then use the correct touch direction: edge (sc), body diagonal (bcc), face diagonal (fcc).
Forgetting fractional atom counts. A corner atom is shared by 8 unit cells ( $\frac{1}{8}$ each); a face-centre atom is shared by 2 cells ( $\frac{1}{2}$ each). Counting them as whole atoms inflates the density by 2–4×.
Not cubing lengths before computing density. The unit-cell volume is $a^{3}$ , not $a$ . Converting pm → cm must happen before cubing: 1 pm = 10⁻¹⁰ cm, so a value in pm gives $a^{3}$ in units of $1 0^{- 30} c m^{3}$ .
Confusing SA with volume. $4 π r^{2}$ is the sphere's surface area; $\frac{4}{3} π r^{3}$ is the volume. Plugging the wrong formula into the packing fraction gives nonsensical efficiencies above 100 %.

Where it shows up

Unit-cell volumes and density, atomic radius from packing geometry, surface-area-to-volume for catalysts and nanoparticles, and the bond angles that VSEPR predicts. Try the volume & area calculator and the molecular shapes tool to build intuition.

Going deeper

The body-centred cubic radius–edge relation $4 r = a 3$ follows directly from the 3D Pythagorean theorem applied to the body diagonal of a cube: the diagonal has length $a^{2} + a^{2} + a^{2} = a 3$ , and it passes through two corner atoms and one body-centre atom, spanning exactly four radii. Rearranging gives $r = \frac{a 3}{4}$ . The packing fraction for bcc is then $η = \frac{2 \times \frac{4}{3} π r ^{3}}{a ^{3}} = \frac{π 3}{8} \approx 68.0%$ . Knowing this, and the coordination numbers (sc 6, bcc 8, fcc 12), lets you infer the cell type from diffraction data alone.

Volume & area

Choose a shape and drag the size slider — volume (r³) outruns surface area (r²) as the object grows, so the SA/V ratio falls.

Shape:

V = \frac{4}{3} π r^{3} A = 4 π r^{2}

radius r2.0 units

volume33.51

surface area50.27

SA / V ratio1.50

As the object grows, volume (r³) outruns surface area (r²), so the SA/V ratio falls. That ratio governs catalyst activity, crystal dissolution rate and heat loss.

Worked example 1Unit-cell volume and density of copper

Copper adopts a face-centred cubic (fcc) structure with cell edge $a = 361.5 pm$ . There are 4 Cu atoms per unit cell (M = 63.55 g mol⁻¹). Calculate the density of copper.

Worked example 2Surface-area-to-volume ratio of a catalyst pellet

A spherical catalyst pellet has diameter 4.0 mm. What is its SA/V ratio in mm⁻¹? If the pellet is ground into spheres of diameter 0.10 mm (100 µm), how does SA/V change?

Molecular shapes

VSEPR predicts geometry by maximising the separation of electron pairs. Select a geometry and read off the ideal bond angle.

VSEPR geometry:

109.5°

ideal bond angle

geometryTetrahedral

examplesCH₄, NH₄⁺

VSEPR theory says electron pairs repel to the geometry that maximises their angular separation. The exact angles come straight from the geometry of regular solids.

Worked example 3Atomic radius from an fcc unit cell

In the face-centred cubic structure of gold (M = 197.0 g mol⁻¹, $ρ = 19.32 g c m^{- 3}$ ), the atoms touch along the face diagonal. Show that the atomic radius is $r = a 2 /4$ and calculate it from the known density.

Worked example 4Why NH₃ has a smaller angle than CH₄

CH₄ has a bond angle of exactly 109.47° but NH₃ has 107.3° and H₂O has 104.5°. Use VSEPR to explain the trend.

Worked examples

Pull together volume, area and molecular shape — try each before revealing the solution.

Worked example 5Sphere-packing efficiency in a face-centred cubic lattice

In an fcc lattice the atoms (radius $r$ ) touch along the face diagonal, so $4 r = a 2$ . Show that the packing efficiency — the fraction of the unit cell occupied by atoms — is $π 2 /6 \approx 74%$ .

Worked example 6Octahedral complex: number of 90° and 180° angles

In an octahedral complex such as $[Co (N H_{3})_{6}]^{3 +}$ , six ligands are arranged at the six faces of a cube around the metal. How many 90° L–M–L angles and how many 180° angles are there?

Worked example 7Molar volume of a gas at STP

Use the formula for a cylinder to estimate the volume occupied by 1 mol of an ideal gas at STP (0 °C, 1 atm) if it were compressed into a cylinder of radius 10 cm. Find the height $h$ .

Worked example 8bcc radius–edge relation and density of iron

Iron adopts a body-centred cubic (bcc) structure with cell edge $a = 286.7 pm$ . Show that atoms touch along the body diagonal ( $4 r = a 3$ ), find the atomic radius of iron, and calculate the density. $M (Fe) = 55.85 g mo l^{- 1}$ .

Worked example 9% packing efficiency of bcc vs fcc

Derive and compare the packing efficiencies of bcc ( $4 r = a 3$ , 2 atoms) and fcc ( $4 r = a 2$ , 4 atoms). Which is more efficiently packed, and what does the void space represent physically?

ChallengeIdentify the crystal structure from density and atomic radius

An element has molar mass $M = 107.87 g mo l^{- 1}$ (silver), density $ρ = 10.50 g c m^{- 3}$ , and atomic radius $r = 144 pm$ . Determine (a) the number of atoms per unit cell and (b) which cubic structure this corresponds to (sc, bcc, or fcc). Verify with the predicted r–a relation.

Check yourself

Four quick questions on shape, scaling and molecular geometry.

Question 1 of 4 · Score 0

The volume of a sphere of radius r is:

V_{s p h ere}

Choose an answer.