← Explore exam analytics

Solid State · TOPIC LESSON

Crystal Lattices and Packing

Connect a repeating three-dimensional structure to particle counts, contact distances, density and composition.

JEE Main · JEE Advanced · NEET UG51 min study guide10 indexed sections

BY THE END, YOU CAN

  • Distinguish crystalline order, a lattice, a basis and a unit cell.
  • Classify solids by their structural particles and dominant bonding, with important exceptions.
  • Derive simple-cubic, body-centred-cubic and face-centred-cubic counts, contact relations and packing fractions.
  • Relate close-packed layers, interstitial-site counts and ionic formulas.
  • Solve density, radius, occupancy and elementary diffraction problems with consistent units.

BEFORE YOU START

  • Three-dimensional distances, sphere volume and basic geometry.
  • Moles, molar mass, density and unit conversions.
  • Ionic formulas, charge balance and intermolecular versus covalent bonding.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. A crystal structure is a repeating pattern with a basis

A lattice is an ideal set of points related by translations, with the same surroundings at every lattice point. A basis is the atom or group of atoms associated with each lattice point. Repeating the basis at all lattice points constructs a crystal structure. A lattice point is therefore not automatically one atom: a basis can contain several chemically different atoms.

A unit cell is a region that reproduces the periodic structure by translation. A primitive cell contains one lattice point, while a conventional cell may contain more than one because a larger cell makes the symmetry easier to see. Thus the familiar face-centred cubic conventional cell is a valid unit cell even though it is not primitive.

Crystalline solids possess long-range periodic order in the elementary structures considered here. Amorphous solids lack that long-range translational periodicity, although they can retain local short-range order. Glass is not accurately described as having no structure at all. The same chemical composition may form crystalline and amorphous materials under different preparation conditions.

A pure equilibrium crystal often has a well-defined melting temperature at a specified pressure, while a glass typically softens over a range and exhibits a glass transition. Real impurities, mixtures, polymorphs and decomposition complicate these observations. Similarly, a single crystal may have direction-dependent properties, whereas a randomly oriented polycrystalline sample can appear approximately isotropic on a larger scale.

02 / UNDERSTAND & APPLY

2. Classify by particles and bonding, not just hardness

Molecular solids contain identifiable molecular units held together mainly by intermolecular forces. Solid iodine contains I₂ molecules; dry ice contains CO₂ molecules; ordinary ice contains H₂O molecules joined by an extended hydrogen-bond network. Melting or subliming such a solid does not normally require breaking the principal covalent bonds within every molecule.

Nonpolar molecular units interact through dispersion forces, polar units also have permanent electrostatic interactions, and suitable donor–acceptor combinations form hydrogen bonds. Larger polarizable molecules can have substantial intermolecular attractions, so a nonpolar label alone does not establish an absolute melting-point ranking. Molecular shape and packing also matter.

Ionic solids contain cations and anions in an extended array rather than discrete neutral formula-unit molecules. They are often hard and brittle. In an ideal ordinary ionic crystal the ions are not freely mobile, so electrical conduction is poor; a melt or suitable solution can conduct through mobile ions. Some real solids are appreciable ionic conductors because of defects or special structures, so the elementary insulating rule has a scope.

Metallic solids have bonding involving delocalized electrons, supporting electrical and thermal conduction and often malleability. Metals do not all have very high melting points or the same hardness. Their band structure and atomic arrangement influence their properties. A metallic crystal is not held together by a lattice of isolated neutral molecules.

Covalent-network solids contain extended covalent connectivity. Diamond and silica are common three-dimensional examples. Graphite has strong covalent bonding within sheets but much weaker interactions between sheets, helping explain its softness in shear. It also conducts anisotropically through its electronic structure. Silicon is a semiconductor rather than a universal perfect insulator. These exceptions make bonding and structure more useful than a memorized hardness/conductivity slogan.

A molecular solid can contain covalent bonds inside its molecules without being a covalent-network solid. Conversely, a material may combine ionic, covalent and metallic contributions; the categories are idealized descriptions of dominant structural features. For an unfamiliar example, identify the repeating particles and ask what must be disrupted to separate them.

ORIGINAL WORKED EXAMPLE

Separate within-unit bonds from between-unit forces

Why are solid iodine and diamond placed in different solid categories even though both contain covalent bonds?

  1. Iodine has discrete I₂ molecules with an I–I covalent bond within each molecule.
  2. The crystal assembles those molecules mainly through intermolecular interactions.
  3. Diamond instead has an extended covalent network without separate small C molecules.

Answer Iodine is molecular; diamond is covalent-network. The relevant distinction is connectivity throughout the solid.

03 / UNDERSTAND & APPLY

3. Cell parameters and the limits of a shape mnemonic

A general cell is specified by lengths a, b and c and angles α between b and c, β between a and c, and γ between a and b. For a cubic cell, a=b=c and all three angles are 90°. Its volume is a³. A rectangular orthorhombic cell has volume abc; the product abc is not generally the volume of an oblique cell.

Common conventional metric patterns are tetragonal: a=b with a distinct c and all right angles; orthorhombic: independent lengths with all right angles; monoclinic in the unique-b setting: α=γ=90° with β unconstrained; and triclinic: no symmetry-imposed length equalities or right angles. Hexagonal axes have a=b, α=β=90° and γ=120°. Rhombohedral primitive axes have equal lengths and equal angles, generally not 90°.

There are fourteen three-dimensional Bravais lattice types. In the standard lattice-system grouping their counts are cubic 3, tetragonal 2, orthorhombic 4, monoclinic 2, triclinic 1, hexagonal 1 and rhombohedral 1. Primitive, body-centred, face-centred and base-centred describe different conventional centring patterns, but not every proposed centring in every shape creates a new distinct lattice type.

Precise crystallographic language distinguishes lattice systems from crystal systems. The seven crystal systems include trigonal, while rhombohedral is the name of a lattice system. Rhombohedral crystals belong to the trigonal crystal system, but the terms are not universally interchangeable. Elementary tables sometimes blur this distinction; use the convention actually specified and recognize what is being classified.

Cell dimensions provide useful constraints, but accidental metric equalities do not by themselves prove the full symmetry of an actual atomic structure. The basis can reduce the crystal symmetry even when the translation cell looks geometrically special. The worked packing models below specify both particle positions and the appropriate cubic geometry.

04 / UNDERSTAND & APPLY

4. Count the fraction actually belonging to one cell

For a cubic conventional cell, a particle centered at a corner is shared by eight neighboring cells and contributes one eighth. A face-centred particle is shared by two and contributes one half. An edge-centred particle is shared by four and contributes one quarter. A particle strictly inside the cell contributes one.

These fractions are a geometric accounting device, not a claim that physical atoms are cut into isolated pieces. Equivalent boundary choices give the same number density when the periodic structure is counted consistently. Count every chemical species separately before reducing the formula.

A simple cubic cell with atoms only at its eight corners contains one atom in total. A body-centred cubic cell adds one fully internal atom, giving two. A face-centred cubic cell adds six half-shared face atoms to the corner contribution, giving four. These counts apply to monatomic structures with one atom at each listed site.

The symbol Z in a compound density calculation denotes formula units per chosen cell, not necessarily total ions. A cell with four A⁺ ions and four B⁻ ions contains four AB formula units, not eight. For an elemental metal, each atom is the relevant formula unit, so the distinction is hidden.

N_cell=N_corner/8+N_edge/4+N_face/2+N_inside

Cubic-cell counting when the listed sites are fully occupied.

ORIGINAL WORKED EXAMPLE

Reduce a cell composition

A cubic cell has X at all corners and face centres and Y at every edge centre and the body centre. Find the formula and number of formula units.

  1. X count: 8/8+6/2=4.
  2. Y count: 12/4+1=4.
  3. The ratio X:Y is 1:1, giving XY.
  4. Four X and four Y constitute four XY formula units.

Answer Formula XY; Z=4.

05 / UNDERSTAND & APPLY

5. Find the line on which hard spheres touch

In simple cubic packing of identical touching spheres, nearest neighbours lie along a cell edge. The centre-to-centre distance is a=2r. Each sphere has six nearest neighbours, one along each of ±x, ±y and ±z. Atoms visible in one isolated cube are not the complete neighbour environment of a corner atom; neighboring cells continue the pattern.

In body-centred cubic packing, a corner atom touches the body-centred atom along a body diagonal. The entire body diagonal runs corner–centre–opposite corner and spans four radii. Its geometric length is √3a, so √3a=4r. Corner atoms do not touch each other along an edge in this model. Every atom has eight nearest neighbours.

In face-centred cubic packing, contact lies along a face diagonal. A corner atom touches the face-centred atom, which touches the opposite corner, giving four radii along √2a. Hence √2a=4r. Each atom has twelve nearest neighbours in the extended lattice. This is the same monatomic close-packed structure called cubic close packing.

Coordination number counts nearest neighbours of a particle, while Z counts particles or formula units per chosen cell. They are different quantities: FCC has coordination number 12 but Z=4. Also, a hard-sphere radius here is a model contact radius for a particular structure, not a universal element size independent of bonding and conditions.

SC: a=2r; BCC: √3a=4r; FCC: √2a=4r

These relations assume identical spheres touching along the appropriate edge or diagonal.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

SC BCC and FCC cell centres and contact linesThree projected cubes show simple cubic with eight corners, body-centred cubic with eight corners plus the body centre, and face-centred cubic with eight corners plus all six face centres. Highlighted lines are respectively a cell edge, a body diagonal and a face diagonal, giving a=2r, sqrt(3)a=4r and sqrt(2)a=4r for equal touching spheres. Conventional cell counts are 1,2,4 and nearest-neighbour counts 6,8,12. Markers indicate centres, not drawn sphere sizes.Cubic cells: locate the contact lineSimple cubica = 2r8 × ⅛ = 1 atom; neighbours = 6Body-centred cubic√3a = 4r8 × ⅛ + 1 = 2 atoms; neighbours = 8Face-centred cubic√2a = 4r8 × ⅛ + 6 × ½ = 4 atoms; neighbours = 12Markers show centres, not hard-sphere radii.Red: edge / body diagonal / face diagonal contact.Neighbour counts refer to the complete extended lattice.
Perspective centre-position schematics. Small markers are not space-filling atoms; grey cell edges are geometric guides rather than chemical bonds.

ORIGINAL WORKED EXAMPLE

Infer a radius from a BCC edge

A hypothetical monatomic BCC metal has cell edge 320 pm. Find its contact radius.

  1. Use the body-diagonal contact relation, not a=2r.
  2. r=√3a/4=√3(320)/4 pm.
  3. This gives 138.6 pm to four significant figures.

Answer r≈139 pm.

06 / UNDERSTAND & APPLY

6. Derive packing efficiency instead of mixing formulas

The packing fraction of the hard-sphere model is the volume occupied by the counted spheres divided by the cell volume. For a monatomic cubic cell it is Z(4πr³/3)/a³. Use the Z and contact relation belonging to the same structure. Combining FCC Z with a BCC radius relation produces a meaningless result.

For simple cubic packing, Z=1 and a=2r. The fraction is (4πr³/3)/(8r³)=π/6≈0.524. About 52.4% of the model cell volume lies within the spheres and 47.6% outside them.

For BCC, Z=2 and a=4r/√3. The fraction simplifies to π√3/8≈0.680. For FCC, Z=4 and a=2√2r, giving π/(3√2)≈0.740. These are ideal equal-sphere values; they should not be treated as measured void percentages of arbitrary ionic solids with unequal ions.

Packing fraction is dimensionless and independent of the sphere radius for a given ideal geometry. Scaling every length multiplies both sphere volume and cell volume by the same factor. A greater packing fraction also does not automatically mean a greater mass density across different elements, since density depends on mass per particle and the absolute cell size.

ORIGINAL WORKED EXAMPLE

A radius change at fixed structure

An ideal FCC hard-sphere solid is scaled so its radius and cell edge both increase by 5.00%. What happens to packing fraction and, at fixed atomic mass and Z, density?

  1. The packing fraction is unchanged because numerator and denominator both scale as length cubed.
  2. The cell volume increases by 1.05³=1.157625.
  3. The mass per cell remains fixed, so density is divided by 1.157625.

Answer Packing fraction remains 0.740; density becomes 0.8638 of its original value, a decrease of about 13.6%.

07 / UNDERSTAND & APPLY

7. Close-packed layers: ABAB and ABCABC

In a two-dimensional triangular close-packed layer, each sphere touches six neighbours. A second layer fits into one set of triangular hollows of the first; call the first layer A and the displaced second layer B. The second-layer spheres each touch three spheres beneath them.

The third layer has two close-packed choices. Placing it above the original A positions gives ABAB stacking, called hexagonal close packing. Choosing the other set of available hollows gives a new C position and ABCABC stacking, called cubic close packing. The difference is the stacking sequence, not the number of contacts within a single layer.

Both ideal HCP and CCP have twelve nearest neighbours per sphere: six in its own layer, three above and three below. Both have the same equal-sphere packing fraction π/(3√2). CCP is equivalent to FCC viewed along its close-packed stacking direction. HCP is not a Bravais lattice with one atom per lattice point; its usual description uses a hexagonal translation lattice with a two-atom basis.

The usual small hexagonal HCP cell has two atoms, while the larger full hexagonal-prism drawing often used in school counting contains six atoms. Those are different cell choices, so using six atoms with the smaller cell volume gives a wrong density. For ideal touching spheres the axial ratio c/a is √(8/3)≈1.633; real HCP metals can deviate from this ideal ratio.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

ABAB versus ABCABC layer registryA schematic layer-registry diagram shows HCP as A-B-A-B and CCP/FCC as A-B-C-A. A,B,C mark distinct lateral positions of two-dimensional triangular close-packed layers, not different elements. In HCP the third layer returns to the first registry; in CCP it occupies the third registry and the fourth returns to A. Both structures have twelve neighbours per sphere and ideal equal-sphere packing fraction approximately 0.740.Close-packed layer registriesA, B and C denote different lateral positions.HCPABABThird layer returns to A.CCP / FCCABCAThird layer occupies a new C registry.Registry diagram: offsets and vertical gaps are schematic.Actual close-packed layers are triangular arrays.Both: 6 neighbours in-layer + 3 above + 3 below = 12.Both ideal equal-sphere packing fractions ≈ 0.740.
Registry schematic rather than a space-filling side section: horizontal offsets and interlayer gaps are not quantitative. The actual individual layers are triangular two-dimensional arrays.

08 / UNDERSTAND & APPLY

8. Interstitial sites connect geometry to formulas

A tetrahedral interstitial site is surrounded by four host particles at tetrahedral corners; an octahedral site is surrounded by six. The names describe the local surrounding host geometry, not the shape of the empty space as a literal tetrahedron or octahedron. A hole is an available site until a guest particle occupies it.

For N host particles in an ideal close-packed array, there are N octahedral sites and 2N tetrahedral sites. In an FCC conventional host cell, four hosts therefore accompany four octahedral and eight tetrahedral sites. The octahedral sites occur at the body centre and edge centres: 1+12/4=4. The tetrahedral sites occur at the eight combinations of fractional coordinates 1/4 and 3/4.

If all octahedral sites of a close-packed anion array are occupied by cations, the cation:anion ratio is 1:1. If half the tetrahedral sites are occupied, the ratio is also 1:1, but the coordination geometry differs. Composition alone cannot distinguish rock-salt and zinc-blende arrangements.

In the rock-salt model, one ion type forms an FCC array and the other occupies every octahedral site. Each ion has six oppositely charged nearest neighbours and Z=4 for AB. In the zinc-blende model, half the tetrahedral sites of an FCC array are occupied; each ion has four opposite-ion neighbours and Z=4 for AB.

The CsCl-type structure instead has one species at cube corners and the other at the body centre, giving Z=1 and an 8:8 coordination pattern. Although its drawing resembles BCC, it is not a monatomic BCC Bravais array because the corner and centre species differ. In fluorite-type AB₂, A forms an FCC array and B occupies all eight tetrahedral sites, giving four AB₂ formula units. The A coordination is eight and B coordination four.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Tetrahedral and octahedral local host arrangementsA regular tetrahedron of four host centres surrounds a central tetrahedral site. A regular octahedron of six host centres surrounds a central octahedral site. The tetrahedron uses vertices proportional to (1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1); the octahedron uses the six signed Cartesian unit vectors. Projection can distort apparent angles. The orange centre marks a possible guest site, not a fifth or seventh host. In close packing N hosts provide 2N tetrahedral and N octahedral sites.Local geometry around an interstitial siteTetrahedral site4 surrounding host centresOctahedral site6 surrounding host centresOrange centre: available site, not an extra host atom.Markers are centre locations; sphere sizes are not to scale.For N close-packed hosts:2N tetrahedral sites; N octahedral sites.
Projected local centre geometries, not space-filling drawings. Lines are geometric guides, not bonds. Hole occupancy is separate from the existence of a site.

ORIGINAL WORKED EXAMPLE

Convert partial site occupancy to a formula

Anions X form a close-packed array. Cations M occupy one third of its tetrahedral sites and no other sites. Find the simplest M:X ratio.

  1. For N anions there are 2N tetrahedral sites.
  2. One-third occupancy gives 2N/3 cations.
  3. M:X=(2N/3):N=2:3.
  4. The formula is M₂X₃; an actual ionic compound must also satisfy charge balance.

Answer M₂X₃ from occupancy; charges must be compatible.

09 / UNDERSTAND & APPLY

9. Density counts mass per cell

A cell containing Z formula units of molar mass M has mass ZM/N_A. Divide by its volume to obtain density. For a cubic cell, ρ=ZM/(N_Aa³). M must refer to the whole formula unit counted by Z. For AB₂, using only the molar mass of A would omit both B atoms.

Unit conversion is often the dominant numerical trap. One picometre is 10⁻¹⁰ cm, so an edge of 400 pm is 4.00×10⁻⁸ cm. Cube the converted length: the volume is 6.40×10⁻²³ cm³. Do not cube 400 and then apply the linear conversion factor only once.

In ionic radius questions, identify the nearest opposite-ion distance and equate it to r₊+r₋ only under the stated touching-ion model. Rock salt has nearest opposite-ion distance a/2. CsCl-type has √3a/2. Zinc blende has √3a/4. Those are distances between different ion centres, not automatically twice either one radius.

An inferred Z near 1, 2 or 4 can identify a likely monatomic cubic structure when those are the allowed models. A noninteger result may instead reflect rounding, a wrong formula mass, vacancies, a different cell choice or an invalid structural assumption. Do not force every measured solid into SC, BCC or FCC.

ρ=ZM/(N_A V_cell); for cubic cells, V_cell=a³

M is the formula-unit molar mass consistent with Z.

ORIGINAL WORKED EXAMPLE

Compute a hypothetical FCC density

A monatomic FCC solid has M=60.0 g mol⁻¹ and a=400 pm. Calculate density using N_A=6.022×10²³ mol⁻¹.

  1. FCC has Z=4, so the cell mass is 4(60.0)/(6.022×10²³)=3.985×10⁻²² g.
  2. Convert the edge to 4.00×10⁻⁸ cm.
  3. The cell volume is (4.00×10⁻⁸)³=6.40×10⁻²³ cm³.
  4. Divide mass by volume: ρ=6.227 g cm⁻³.

Answer ρ=6.23 g cm⁻³.

ORIGINAL WORKED EXAMPLE

Use the correct opposite-ion line

A rock-salt-type AB crystal has edge 560 pm. If the anion radius is 170 pm in the touching-ion model, find the cation radius.

  1. The nearest opposite-ion distance is a/2=280 pm.
  2. Set r₊+r₋=280 pm.
  3. Subtract the given anion radius: r₊=280−170=110 pm.

Answer r₊=110 pm within the stated model.

10 / UNDERSTAND & APPLY

10. A measured spacing and an algebraic check

X-ray diffraction probes repeating spacings because waves scattered from regularly separated planes can interfere constructively. In the elementary Bragg description, nλ=2d sinθ, where d is the spacing of the relevant planes and θ is the angle between the incoming beam and those planes. Many instruments report 2θ, so halve that reported angle before using the formula.

For cubic cells, the interplanar spacing associated with Miller indices h,k,l obeys d_hkl=a/√(h²+k²+l²). This geometric relation does not mean every index set produces a visible reflection: the lattice and basis can cause systematic absences. A peak assignment requires more than guessing a nearest-neighbour distance.

Bragg’s equation is a path-difference condition. Its d is a plane spacing, while the contact relations earlier involved particular atom-to-atom distances. Those lengths may be related in a known structure but should never be substituted for one another without a geometric argument.

A final solution audit should state the structure, the cell choice, the counted unit, the contact assumption and the units. For composition problems, check charge balance separately from geometric occupancy. For property questions, explain the bonding mechanism and preserve known exceptions rather than treating a four-row classification table as an absolute law.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Plane spacing and the Bragg angleTwo horizontal lattice planes separated by d scatter parallel incident rays into parallel outgoing rays at equal angles theta to the planes. The lower path has extra length 2d sin(theta); constructive interference requires that to equal n wavelengths. Theta is measured from the planes and is half the commonly reported instrument scattering angle 2theta. The figure does not equate d with an atomic radius.Bragg angle is measured from the planesdincident raysscattered raysθθExtra path for the lower ray = 2d sinθConstructive condition: nλ = 2d sinθθ: beam-to-plane angle, not beam-to-normal.A reported instrument angle 2θ must be halved.d is plane spacing, not automatically an atomic radius.
Geometric path-difference model of diffraction; lines represent selected plane families and ray paths. Reflection drawing is a construction for coherent scattering, not literal mirror-like sheets.

ORIGINAL WORKED EXAMPLE

Remember the instrument angle

A first-order reflection occurs at 2θ=60.0° for X-rays of wavelength 150 pm. Find the plane spacing.

  1. The Bragg angle is θ=30.0°, not 60.0°.
  2. For n=1, d=λ/(2sinθ).
  3. d=150/[2(0.500)] pm=150 pm.

Answer d=150 pm for the assigned first-order reflection.

ORIGINAL PRACTICE · 1

A cubic cell has eight corner atoms and one body-centred atom. Find Z for a monatomic structure.

Reveal answer & reasoning

2.

The corner contribution is 8/8=1 and the internal atom contributes 1, giving 2.

ORIGINAL PRACTICE · 2

An FCC metal has contact radius 140 pm. Find a.

Reveal answer & reasoning

396 pm approximately.

a=2√2r=2√2(140)=395.98 pm. The contact line is a face diagonal.

ORIGINAL PRACTICE · 3

Does FCC coordination number 12 mean twelve atoms per conventional cell?

Reveal answer & reasoning

No; Z=4.

Coordination number counts neighbours of one atom; Z counts the fractional atoms belonging to the chosen cell.

ORIGINAL PRACTICE · 4

A close-packed array has 300 host particles. How many octahedral and tetrahedral sites exist in the bulk counting model?

Reveal answer & reasoning

300 octahedral and 600 tetrahedral sites.

For N close-packed hosts, the site counts are N and 2N. Surface effects are excluded.

ORIGINAL PRACTICE · 5

All tetrahedral sites in an FCC A array are occupied by B. Find composition and formula units per conventional cell.

Reveal answer & reasoning

AB₂, with Z=4.

The cell contains four A hosts and eight B guests, giving four AB₂ units.

ORIGINAL PRACTICE · 6

Which stacking is HCP: ABAB or ABCABC? Give its ideal coordination number.

Reveal answer & reasoning

ABAB; 12.

Each sphere contacts six in its own layer and three in each adjacent layer. ABCABC instead describes CCP/FCC.

ORIGINAL PRACTICE · 7

At fixed cell mass, the cubic edge doubles. What happens to density?

Reveal answer & reasoning

It becomes one eighth.

The volume scales as a³, so doubling a multiplies volume by eight while mass stays constant.

ORIGINAL PRACTICE · 8

For cubic a=450 pm, find d₁₁₀.

Reveal answer & reasoning

318 pm approximately.

d₁₁₀=450/√(1²+1²+0²)=450/√2=318.2 pm. This spacing calculation alone does not guarantee an allowed diffraction reflection.

KEEP THE BIG IDEAS

Quick revision

  • A crystal structure combines a translation lattice with a basis.
  • Count shared sites before calculating a formula or density.
  • SC, BCC and FCC touch along an edge, body diagonal and face diagonal respectively.
  • Equal-sphere packing fractions are about 52.4%, 68.0% and 74.0%.
  • HCP and CCP differ in stacking while sharing ideal coordination number and packing fraction.
  • Interstitial-site occupancy determines ratios but does not replace charge balance.
  • Density and diffraction use different geometrical quantities; label each carefully.
Back to the top ↑

READ WITH CONTEXT

References & scope

The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.

Explore this topic in context

Analytics use a separately reviewed, incomplete historical corpus. Opening a topic does not change its counts or denominator.

KEEP EXPLORING

Your next idea is waiting.

Back to Chemistry index →Browse all topic notes