Solid State (Class 12): Unit Cells, Packing and the Density Numerical

The three cubic unit cells

Everything in this chapter starts with counting how many atoms genuinely belong to one unit cell. A corner atom is shared by eight cells, a face atom by two, an edge atom by four, and a body-centre atom by none. Counting carelessly here makes every later step wrong while looking perfectly reasonable.

The density numerical, worked properly

The formula is d = zM / (a³ N_A), where M is molar mass in g mol⁻¹, a is edge length in cm and N_A is 6.022 × 10²³. Three things go wrong far more often than the algebra:

1. Units. Edge length is almost always given in pm. 1 pm = 10⁻¹⁰ cm. Convert before cubing, never after — cubing a wrong unit hides the error completely.
2. The wrong z. Read the question for the lattice type. “Fcc”, “ccp” and “cubic close packed” are the same thing and all mean z = 4.
3. Rearranging under pressure. If the question asks for M or a, rearrange on paper first and substitute afterwards. Doing both at once is where most method marks are lost.

Voids, and why the numbers matter

In a close-packed structure with N spheres there are N octahedral voids and 2N tetrahedral voids. That single fact answers most stoichiometry questions in this chapter: if cations occupy all octahedral voids of an fcc anion lattice, the formula is AB; if they occupy half the tetrahedral voids, it is again AB; if all tetrahedral voids, AB₂. Work from the void count rather than trying to remember formulae.

Defects, briefly but correctly

Schottky defects remove a cation and an anion together, so density falls. Frenkel defects move a cation into an interstitial site, so density is unchanged. Students routinely reverse these two in one-mark questions, and the distinction is entirely about whether anything left the crystal.

What to practise

• Ten density numericals across all three cubic cells, converting pm to cm every time.
• Formula determination from void occupancy, at least five variations.
• One packing-efficiency derivation written out in full — it appears as a three-mark question.

FAQs

Why does a bcc lattice use √3 a = 4r?

Because in a bcc cell the atoms touch along the body diagonal, not along an edge. The body diagonal of a cube of side a is √3 a, and it passes through four radii — corner atom, whole central atom, corner atom. In fcc the contact is along the face diagonal, giving √2 a = 4r.

Is ccp the same as fcc?

Yes. Cubic close packed describes the packing pattern (ABCABC layers) and face-centred cubic describes the unit cell that results. Both mean z = 4 and 74% packing efficiency. Examiners use the terms interchangeably.

Do Schottky defects change the density?

Yes, they lower it, because equal numbers of cations and anions are missing from the crystal. Frenkel defects leave density unchanged, since the ion has only moved to an interstitial position and nothing has left.

How much of this chapter is numerical?

A large share of the marks. Theory questions on defects and packing appear, but the density and formula-determination numericals are the reliable scoring part and they repeat in structure year after year.