JEE Chemistry The Solid State Notes Set 02

Density of a Cubic Crystal

The mass of one unit cell is Z \(\times\) M/N_A, where Z is the number of formula units per cell and M the molar mass. Dividing by the cell volume a³ gives
\[ \rho = \dfrac{Z\,M}{N_A\,a^3} \]
Use a in cm (1 pm = 10⁻¹⁰ cm) to get \(\rho\) in g cm⁻³. The same relation can be used backwards to find Z, and hence the lattice type: Z = 1 simple cubic, 2 body-centred, 4 face-centred.
To find the number of unit cells in a given mass m, compute the moles m/M, then the formula units (or atoms), then divide by Z.

Radius Ratio and Coordination Number

In an ionic crystal the cation tries to touch as many anions as possible. The ratio r₊/r₋ decides how many anions can surround the cation without the anions pressing on each other.

The limiting values arise when the anions just touch each other and the cation just touches all of them. For the octahedral case, the cation fits the hole in the square of four anions: \( 2r_- = \sqrt{2}\,(r_+ + r_-) \), which gives r₊/r₋ = 0.414.

Distances in Common Cubic Structures

In diamond, the C-C bond length is \(\sqrt{3}\,a/4\) = 0.433 a. For a metal in FCC the nearest-neighbour distance is \(a/\sqrt{2}\) = 0.707 a, and in BCC it is \(\sqrt{3}\,a/2\) = 0.866 a.

Crystal Defects

A real crystal always contains defects. A point defect is a fault at a site of the lattice, and the number of point defects increases with temperature.
Stoichiometric defects keep the formula of the compound unchanged. They are of two kinds.

Non-stoichiometric defects change the ratio of the ions.
Metal excess: an anion site is empty and holds an electron (an F-centre, which gives the crystal colour, for example yellow NaCl heated in sodium vapour), or an extra cation sits in an interstitial site with an electron trapped near it (ZnO, which turns yellow on heating). Such crystals behave like n-type semiconductors.
Metal deficiency: a cation is missing and the charge is balanced by another cation in a higher oxidation state (Fe_0.95O, FeS, NiO). Such crystals behave like p-type semiconductors.
Doping an ionic crystal with a cation of higher charge (for example SrCl₂ in NaCl) creates cation vacancies equal to the number of dopant ions.

Worked Examples

Illustration 1. Aluminium has FCC structure with cell edge 405 pm. Find its density (Al = 27).
Answer:
\( \rho = \dfrac{4 \times 27}{6.022\times10^{23}\times(4.05\times10^{-8})^3} \).
\( (4.05\times10^{-8})^3 = 6.643\times10^{-23}\ \text{cm}^3 \), so \( \rho = 108/40.01 \) = 2.70 g cm⁻³.

Illustration 2. A cation has radius 0.6 \(\mathring{A}\) and the anion has radius 1.4 \(\mathring{A}\). What is the coordination number of the cation?
Answer:
r₊/r₋ = 0.6/1.4 = 0.43.
This lies between 0.414 and 0.732, so the coordination number is 6 (octahedral).

Illustration 3. A crystal of NaCl type has a = 564 pm. Find the Na⁺-Cl⁻ distance.
Answer:
In the NaCl structure the ions touch along the edge, so the distance is a/2.
Distance = 564/2 = 282 pm.