Chemistry · Lesson

Crystal structures, defects and solid-state bonding

Count shared sites in a specified ionic cell, relate cell contents to density, and conserve charge in two deliberately limited defect models.

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By the end, you can

Distinguish boundary-sharing weights, site occupancy and formula-unit count.
Calculate density with compatible volume and molar-mass units.
Check charge compensation under an explicitly selected defect route.
Separate ideal-cell results from actual material and phase evidence.

A crystal model repeats a structure through space

A periodic crystal model describes a repeat lattice together with the chemical contents associated with its points. Translating the chosen unit cell builds the extended structure. The cell is a bookkeeping region; an atom drawn on its boundary is shared with neighbouring cells.

A lattice point is different from an atom, and an atom count is different from a formula-unit count. Distinguish the primitive repeat from a larger conventional cell. The periodic examples here do not imply that all crystalline order is periodic or that every solid is crystalline.

Count occupants through the chosen cell boundary

Specify an ideal conventional cubic rock-salt-style arrangement with fully occupied Mg²⁺ at eight corners and six face centres, and O²⁻ at twelve edge centres and the cube centre. A corner occupant is shared by eight cells, a face occupant by two and an edge occupant by four. These sharing weights differ from fractional site occupancy.

The Mg count is 8×⅛+6×½=4; the O count is 12×¼+1=4. The chosen cell contains four MgO formula units and charge number 4×2+4×(−2)=0. Mg²⁺ and O²⁻ each have ten electrons, giving eighty electrons in this cell’s constituent inventory. The ion labels supply a formal ionic account, rather than a claim that isolated-ion electron densities fully describe every bond.

A stipulated primitive cell of the same periodic model has one-quarter of this conventional cell’s volume and one Mg/O pair: one formula unit, twenty electrons and the same density. A primitive lattice point is not one atom. The nonintegral contributions of boundary drawings do not mean that an individual atom has been divided chemically.

The contents and bonding determine what the repeat means

The same geometric arrangement can represent different chemical contents. Metallic, ionic, covalent-network and molecular solids describe different bonding patterns; a geometric picture alone does not assign a bonding category. Coordination, charge balance and the identities of occupied sites help connect structure to chemistry.

A unit-cell formula and molar mass can support a model density when cell volume and the number of formula units are known. Agreement with one bulk quantity does not by itself identify a unique crystal structure.

A density and a defect count need separate ledgers

For the previous ideal cell assign edge a=0.426 nm and formula-unit molar mass M=40.30 g mol⁻¹. These are teaching inputs rather than measured specimen values. Here Z=4 is the formula-unit count per chosen cell. With the Avogadro constant N_A=6.02214076×10²³ mol⁻¹, cell mass is ZM/N_A=2.67678898×10⁻²² g. Since 1 nm=10⁻⁷ cm, cell volume is a³=7.73087760×10⁻²³ cm³ and model density is 3.46246457 g cm⁻³.

Increasing a by 1.00% at unchanged contents multiplies volume by 1.01³ and density by 1/1.01³=0.970590148. The new density is 3.36063400 g cm⁻³. The graph isolates this inverse-volume relation; it does not include pores or identify a unique structure from a matching density.

In a separate ideal inventory with 1,000 Mg²⁺ and 1,000 O²⁻ sites, removing twelve occupants of each leaves 988 of each and twenty-four vacancies in total. The selected paired vacancy ledger remains neutral. Removing only twelve Mg²⁺ initially leaves residual charge number −24; twelve O²⁻ vacancies restore neutrality under the stipulated paired-vacancy route. This is a chosen Schottky-type account, rather than a calculated defect equilibrium.

For another invented fixed-valence ledger, use one hundred Sr²⁺ sites, one hundred B sites and an initial three hundred O²⁻ sites. Substituting eight Al³⁺ for eight Ti⁴⁺ lowers positive charge by eight. If oxygen vacancies are the only allowed compensation, four O²⁻ vacancies give Sr₁₀₀Ti₉₂Al₈O₂₉₆ and mean formula SrTi₀.₉₂Al₀.₀₈O₂.₉₆. The charge check is 200+368+24−592=0. This ledger neither proves that a material of this composition is stable nor assigns another compensation mechanism.

The same cell contents give a lower density at a larger cubic edge

Model density / g cm⁻³

Cubic edge ratio a/a₀

Assigned cell-density cases
— Fixed contents: ρ₀/(a/a₀)³

Underlying data

Cubic edge ratio a/a₀	Model density / g cm⁻³
1	3.462464567133027 · Assigned cell-density cases
1.01	3.3606339964078717 · Assigned cell-density cases

Original ideal-cell density model with four formula units, M=40.30 g mol⁻¹ and a₀=0.426 nm. Circles mark the assigned original cell and its 1.00% larger-edge transfer case; the dashed curve follows ρ(x)=ρ₀/x³ for x=a/a₀. All contents remain fixed. These are calculated values, not measured MgO densities or a structural identification.

A real crystal can differ from its ideal repeat

A vacancy is a missing occupant at a site relative to the chosen ideal model. Interstitials, substitutions, dislocations and boundaries represent other departures. Defects can affect transport, mechanical response and other properties; they are part of a material’s state rather than a universal sign of poor quality.

An average site occupancy describes the model’s population of a site. It does not alone locate every vacancy or distinguish every kind of disorder. Ionic defect accounts must also address charge compensation; creating one charged defect cannot silently leave a neutral bulk ledger unbalanced.

A property depends on more than the ideal lattice

Composition, phase, defect population, grain structure, orientation and processing history can all affect a measured property. A material may contain several phases or crystalline and noncrystalline regions. State which level of structure a claim concerns.

Use an ideal cell to learn a relation, then test whether that description fits the actual material and measurement. The original calculations below do not identify a commercial crystal, measured defect concentration or universally valid mechanical property.

Worked examples

Count the shared ionic sites

Use eight Mg corners/six Mg faces and twelve O edges/one O centre, with full occupancy.

Mg count is 8/8+6/2=4; O count is 12/4+1=4.
Four MgO units give charge number zero.
Each selected ion has ten electrons, giving 80 electrons per conventional cell.

Boundary sharing and partial site occupancy are different operations.

Convert the cubic volume before dividing

Use Z=4, M=40.30 g mol⁻¹ and a=0.426 nm.

Cell mass is 2.67678898×10⁻²² g.
Edge is 4.26×10⁻⁸ cm; cubic volume is 7.73087760×10⁻²³ cm³.
Their ratio gives ρ=3.46246457 g cm⁻³.

The density belongs to the assigned ideal-cell model.

Count vacancies separately from remaining ions

Remove twelve Mg²⁺ and twelve O²⁻ from 1,000 sites of each.

Each ion inventory becomes 988.
There are twelve cation and twelve anion vacancies: twenty-four total.
The charge sum 2×988−2×988 remains zero.

A selected paired-vacancy account does not determine an equilibrium defect concentration.

Compensate the assigned heterovalent substitution

Substitute eight Al³⁺ for Ti⁴⁺ in one hundred B sites; allow only O²⁻ vacancies.

The positive charge falls by eight.
Removing four O²⁻ supplies a relative charge change +8.
Counts Sr100/Ti92/Al8/O296 give mean formula SrTi₀.₉₂Al₀.₀₈O₂.₉₆ and zero net charge.

Neutrality of the invented ledger is different from evidence of a stable phase.

Try it yourself

Use a stipulated primitive cell of the same model with one-quarter conventional volume. Find Mg/O/formula-unit/electron counts and density change.
Show the answer and reasoning
One Mg²⁺, one O²⁻, one formula unit, twenty electrons; density factor 1.
Mass and volume both fall by four. The primitive cell is not necessarily a cube of one-quarter edge.
Increase the original cubic edge by 1.00% at fixed contents and molar mass. Find density, its ratio and what this proves about structure.
Show the answer and reasoning
ρ=3.36063400 g cm⁻³; ratio 0.970590148. A unique structure is not established.
Volume scales as a³, so density scales as a⁻³ under the stated conditions.
Remove twelve Mg²⁺ but initially no O²⁻ from the 1,000/1,000 neutral inventory. Find the charge number and O vacancies needed under the selected compensation route.
Show the answer and reasoning
Initial residual charge number −24; twelve additional O²⁻ vacancies give final zero.
Removing negative occupants compensates the relative loss of positive charge only in this chosen route.
Use eighteen Al³⁺ among the same hundred B sites and allow only O²⁻ vacancies. Find oxygen vacancies, occupied O and the average formula.
Show the answer and reasoning
Nine oxygen vacancies, 291 O occupants; SrTi₀.₈₂Al₀.₁₈O₂.₉₁; net charge zero.
Two substitutions each lowering positive charge by one are compensated by one missing O²⁻. No phase stability is inferred.

What to carry forward

Boundary sharing, site occupancy and formula-unit counts are different quantities.
Use the stated cell volume and mass to calculate density.
Defect accounts must close their charge and constituent inventories.
A neutral composition ledger does not prove that a phase forms.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.