CSIR-NET Solid State and Materials Chemistry — Structures, Defects and Bands
Solid state chemistry is one of the few CSIR-NET topics where the same small set of ideas keeps returning in different clothes: a structure type, a defect equilibrium, a band picture, and one arithmetic relation linking the unit cell to a bulk property. If you can move confidently between those four, most questions on this chapter become short. This guide works through the quantitative core — density, defect concentration, band gap, and the spinel and perovskite site preferences — with every number actually computed.
The one equation that links the unit cell to the laboratory
Everything measurable about a crystal — its density, its formula, the number of formula units in the cell — hangs on a single relation.
Here ρ is the density (g cm⁻³), Z is the number of formula units per unit cell, M is the molar mass of one formula unit (g mol⁻¹), a is the cubic cell edge in cm, and NA = 6.022 × 10²³ mol⁻¹. The single most common slip is leaving a in pm or Å; convert first (1 pm = 10⁻¹⁰ cm, 1 Å = 10⁻⁸ cm). For a non-cubic cell replace a³ by the cell volume V.
Worked example 1 — density of rock salt from its lattice parameter
NaCl is face-centred cubic with Z = 4 formula units per cell and a = 564 pm. Take M(NaCl) = 22.990 + 35.45 = 58.44 g mol⁻¹.
Step 1 — convert the edge: a = 564 pm = 5.64 × 10⁻⁸ cm.
Step 2 — cube it: (5.64)³ = 5.64 × 5.64 = 31.8096; 31.8096 × 5.64 = 179.41.
So a³ = 179.41 × 10⁻²⁴ = 1.7941 × 10⁻²² cm³.
Step 3 — denominator: 1.7941 × 10⁻²² × 6.022 × 10²³ = 108.04.
Step 4 — numerator: Z × M = 4 × 58.44 = 233.76.
ρ = 233.76 / 108.04 = 2.16 g cm⁻³.
Cross-check: the measured density of NaCl is close to 2.16 g cm⁻³, so the assumed Z = 4 is confirmed. That reverse use — solving for Z from a measured density — is the commoner exam version of the question.
Structure types you are expected to recognise instantly
Most inorganic structures in the syllabus are described as a close-packed anion array with cations in some fraction of the holes. A cubic close-packed (ccp) array of N anions provides N octahedral and 2N tetrahedral holes; hexagonal close packing (hcp) provides the same counts. Which holes are filled, and how many, fixes the stoichiometry.
| Structure | Example | Anion array | Cation holes filled | Cation CN : anion CN |
|---|---|---|---|---|
| Rock salt | NaCl, MgO | ccp | all octahedral | 6 : 6 |
| Zinc blende | ZnS, GaAs | ccp | half the tetrahedral | 4 : 4 |
| Wurtzite | ZnS, ZnO | hcp | half the tetrahedral | 4 : 4 |
| Fluorite | CaF₂ | ccp of Ca²⁺ | all tetrahedral by F⁻ | 8 : 4 |
| Antifluorite | Na₂O, K₂S | ccp of O²⁻ | all tetrahedral by Na⁺ | 4 : 8 |
| Rutile | TiO₂ | distorted hcp | half the octahedral | 6 : 3 |
| Perovskite | CaTiO₃, SrTiO₃ | ccp of A + O together | B in ¼ of octahedral | B 6, A 12 |
| Spinel | MgAl₂O₄ | ccp of O²⁻ | ⅛ tetrahedral + ½ octahedral | A 4, B 6 |
The CsCl structure is the standard exception — it is not body-centred cubic in the Bravais sense, because the two ions are different. It is a simple cubic lattice of one ion with the other at the body centre, giving 8 : 8 coordination and Z = 1.
Point defects — the entropy argument
A perfect crystal is thermodynamically impossible above 0 K. Creating n defects costs enthalpy but gains configurational entropy, and minimising G = nΔH − TΔS gives an exponential defect population.
The factor of 2 in the denominator is there because one Schottky "defect" is a pair — a cation vacancy plus an anion vacancy — and ΔHs is the enthalpy for the pair. Dropping that 2 is the classic marks-losing error. N is the number of lattice sites and Ni the number of interstitial sites. Use R with ΔH in J mol⁻¹, or kB with ΔH in J per defect — never mix them.
Worked example 2. An ionic solid has ΔHs = 200 kJ mol⁻¹. What fraction of sites is vacant at 1000 K?
n/N = exp[ −200000 / (2 × 8.314 × 1000) ] = exp( −200000 / 16628 ) = exp(−12.03).
exp(−12) = 6.144 × 10⁻⁶ and exp(−0.03) = 0.9704, so
n/N ≈ 5.96 × 10⁻⁶, about 6 vacancies per million sites.
Sanity check on the physics. Repeat at 1500 K: 200000 / (2 × 8.314 × 1500) = 200000 / 24942 = 8.019, and exp(−8.019) = 3.3 × 10⁻⁴. The vacancy fraction has risen by a factor of about 55 for a 50% rise in temperature — defects multiply steeply with heating, which is exactly why ionic conductivity in these solids is so temperature sensitive. If your answer moves the other way, you have put T in the wrong place.
| Defect | What it is | Effect on density | Favoured when | Typical hosts |
|---|---|---|---|---|
| Schottky | Cation + anion vacancy pair | Decreases | r⁺/r⁻ close to 1, high CN | NaCl, KCl, CsCl |
| Frenkel | Ion displaced into an interstitial site | Unchanged | Large cation–anion size difference, low CN | AgCl, AgBr, ZnS |
| F-centre | Anion vacancy holding a trapped electron | Slightly decreases | Metal excess non-stoichiometry | NaCl heated in Na vapour |
| Metal deficient | Cation vacancies balanced by higher oxidation state | — | Variable-valence metal | Fe0.95O (p-type) |
Two consequences worth memorising: Frenkel defects leave the density unchanged because nothing leaves the crystal, and F-centres absorb visible light because the trapped electron has particle-in-a-box-like levels — which is why non-stoichiometric NaCl is yellow and KCl is violet.
From bonds to bands
Bring N atoms together and each atomic level splits into N closely spaced levels; when N is of the order of 10²³ these merge into a band. Conduction requires empty states immediately above filled ones.
| Class | Band picture | Band gap | σ with rising T |
|---|---|---|---|
| Metal | Partly filled band, or overlapping bands | None | Decreases (phonon scattering) |
| Semiconductor | Filled valence band, small gap to empty conduction band | Roughly ≤ 3 eV | Increases (more carriers) |
| Insulator | Same, but a large gap | Large | Effectively zero |
Doping Si (group 14) with P (group 15) puts a donor level just below the conduction band — an n-type semiconductor. Doping with B or Ga (group 13) creates an acceptor level just above the valence band — p-type, where the carriers are holes. Non-stoichiometry does the same job chemically: metal-excess ZnO is n-type, metal-deficient Fe0.95O is p-type.
Worked example 3 — band gap from the optical absorption edge. A semiconductor stops absorbing beyond 1100 nm. Find its band gap in eV.
E = hc/λ with h = 6.626 × 10⁻³⁴ J s, c = 2.998 × 10⁸ m s⁻¹, λ = 1100 × 10⁻⁹ m.
hc = 6.626 × 10⁻³⁴ × 2.998 × 10⁸ = 1.9865 × 10⁻²⁵ J m.
E = 1.9865 × 10⁻²⁵ / 1.100 × 10⁻⁶ = 1.806 × 10⁻¹⁹ J.
In electronvolts: 1.806 × 10⁻¹⁹ / 1.602 × 10⁻¹⁹ =
1.13 eV.
That is essentially silicon's gap of about 1.1 eV — a useful memory anchor. A handy shortcut for the exam hall: E(eV) ≈ 1240 / λ(nm), and 1240/1100 = 1.13.
Spinels — where crystal field theory decides the structure
A spinel AB₂O₄ has a ccp oxide array with A²⁺ nominally in tetrahedral (Td) holes and B³⁺ in octahedral (Oh) holes. In an inverse spinel half the B³⁺ swaps with the A²⁺, giving B(AB)O₄. Which arrangement wins is often decided by octahedral site preference energy, and Fe₃O₄ is the standard question.
Worked example 4 — why Fe₃O₄ is inverse. Write it as Fe²⁺(Fe³⁺)₂O₄. Both ions are high spin.
Fe³⁺ is d⁵. Octahedral high spin t2g³eg²: CFSE = 3(−0.4Δo) + 2(+0.6Δo) = −1.2 + 1.2 = 0. Tetrahedral high spin e²t₂³: CFSE = 2(−0.6Δt) + 3(+0.4Δt) = −1.2 + 1.2 = 0. Fe³⁺ has no site preference at all.
Fe²⁺ is d⁶. Octahedral high spin t2g⁴eg²: CFSE = 4(−0.4) + 2(+0.6) = −1.6 + 1.2 = −0.4Δo. Tetrahedral high spin e³t₂³: CFSE = 3(−0.6) + 3(+0.4) = −1.8 + 1.2 = −0.6Δt. Using Δt ≈ (4/9)Δo, that is −0.6 × 0.444 = −0.267Δo.
Octahedral site preference for Fe²⁺ = −0.400 − (−0.267) = −0.133 Δo, i.e. Fe²⁺ prefers the octahedral hole. Since Fe³⁺ is indifferent, Fe²⁺ takes an octahedral site and one Fe³⁺ is pushed into a tetrahedral one: the structure is inverse, Fe³⁺(Fe²⁺Fe³⁺)O₄. Its ferrimagnetism follows — the tetrahedral and octahedral sublattices oppose, and the leftover Fe²⁺ moment survives.
Perovskites and the tolerance factor
A cubic perovskite ABO₃ needs t close to 1; roughly 0.9–1.0 gives cubic, lower values give tilted orthorhombic variants, higher values push towards hexagonal packing.
Worked example 5 — SrTiO₃. Using Shannon radii r(Sr²⁺, CN 12) = 144 pm, r(Ti⁴⁺, CN 6) = 60.5 pm, r(O²⁻, CN 6) = 140 pm:
Numerator = 144 + 140 = 284 pm.
Denominator = 1.4142 × (60.5 + 140) = 1.4142 × 200.5 = 283.55 pm.
t = 284 / 283.55 = 1.00 — as close to ideal as the model allows, and
SrTiO₃ is indeed the textbook cubic perovskite.
Common mistakes in this chapter
- Leaving the cell edge in pm. The density formula demands centimetres; an unconverted edge is out by 10³⁰.
- Forgetting the 2 in exp(−ΔH/2RT). Both Schottky and Frenkel expressions carry it because the defect is a pair.
- Saying Frenkel defects lower the density. Nothing leaves the crystal, so mass and volume are unchanged.
- Using Δt = Δo in site-preference arguments. Δt ≈ (4/9)Δo; ignoring it reverses the spinel conclusion.
- Calling CsCl "body-centred cubic". The lattice is simple cubic with a two-ion basis; only in a monatomic solid would that arrangement be bcc.
- Assuming every d-metal oxide is stoichiometric. Fe0.95O and Zn1+xO are syllabus examples where it is exactly the deviation that matters.
How this topic appears across exams
| Exam | Typical demand from solid state |
|---|---|
| IIT-JAM | Unit cell counting, packing fraction, density and radius-ratio numericals |
| GATE Chemistry | Structure types, X-ray diffraction indexing, defect equilibria |
| CSIR-NET | Defect thermodynamics, band and doping arguments, spinel site preference, magnetic ordering |
| CUET-PG | Definition-level recall of structures, defects and semiconductor types |
The CSIR-NET paper is set in three parts — a general aptitude Part A and subject Parts B and C. For the number of questions, marks and any negative marking, always read the current official notification rather than any coaching summary, including this one.
Do the density arithmetic without slips. Every unit-cell density problem starts with M for the formula unit — NaCl, MgO, CaF₂, MgAl₂O₄, YBa₂Cu₃O₇ — and that is exactly what the Molar Mass & Composition tool returns, element by element, so the only thing left for you is the cube and the conversion.
Open the Molar Mass Calculator →Preparing for CSIR-NET, GATE, IIT-JAM or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and fully online for students anywhere in India — details at abcchemistry.in.