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Solid State ChemistryBurdwan University · B.Sc. (NEP) · Inorganic
Crystal structures Chapter 4 · Important Crystal Structures

AB, AB₂ & Complex Crystal Structures

Dear student, this is the most "diagram-loving" chapter of the unit. Examiners ask you to draw and describe unit cells of NaCl, CsCl, fluorite, diamond and perovskite almost every year. Learn one fact-table per structure: how the ions are arranged, which voids are filled, the coordination number and the number of formula units — that is the whole game.

🧂 AB & AB₂ types🔷 Perovskite · Spinel💎 Diamond · Silicates
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AB Type Structures (1 : 1 compounds)

In an AB type ionic solid, equal numbers of cations (A) and anions (B) pack together. Which structure forms depends mainly on the radius ratio r⁺/r⁻ — bigger cations need bigger coordination numbers. Three structures you must know are given below.

1. Rock Salt Structure — Sodium Chloride (NaCl)

  • Arrangement: Cl⁻ ions form a face-centred cubic (fcc) lattice; Na⁺ ions occupy all the octahedral voids.
  • Coordination number: 6 : 6 — each Na⁺ is surrounded by 6 Cl⁻ and each Cl⁻ by 6 Na⁺.
  • Formula units per unit cell (Z): 4 (4 Na⁺ + 4 Cl⁻).
  • Radius ratio: r(Na⁺)/r(Cl⁻) ≈ 0.52 — fits the octahedral range 0.414–0.732.
  • Other examples: KCl, MgO, CaO, AgCl, LiF, MnO.
Rock Salt (NaCl) Structure
NaCl — 6:6 coordination Cl⁻ in fcc (ccp) Na⁺ in ALL octahedral voids (body-centre Na⁺ hidden in this face view)
Cl⁻ ions form an fcc lattice; Na⁺ ions fill every octahedral void — each ion is touched by 6 of the opposite kind (6:6).

2. Caesium Chloride Structure (CsCl)

  • Arrangement: Cl⁻ ions form a simple cubic lattice; one Cs⁺ ion sits at the body centre of the cube.
  • Coordination number: 8 : 8 — each Cs⁺ touches 8 Cl⁻ and vice versa.
  • Formula units per unit cell (Z): 1 (1 Cs⁺ + 1 Cl⁻).
  • Radius ratio: r(Cs⁺)/r(Cl⁻) ≈ 0.93 — above 0.732, so cubic (8-fold) coordination is stable.
  • Other examples: CsBr, CsI, TlCl, TlBr.
Caesium Chloride (CsCl) Structure
Cs⁺ CsCl — 8:8 coordination Cl⁻ at 8 corners (simple cubic array) Cs⁺ at body centre touches 8 Cl⁻
Cl⁻ ions sit at the corners of a cube with Cs⁺ at the body centre — each ion has 8 nearest neighbours of the opposite kind (8:8).
Exam tip: A very common 2-mark question: "Why does NaCl take the rock salt structure while CsCl takes the CsCl structure?" Answer: the Cs⁺ ion is much larger (r⁺/r⁻ ≈ 0.93), so 8 anions can fit around it; Na⁺ is smaller (r⁺/r⁻ ≈ 0.52), so only 6 anions fit comfortably. Always quote the radius-ratio ranges: 0.414–0.732 → octahedral (CN 6), above 0.732 → cubic (CN 8).

3. Zinc Blende vs Wurtzite (both ZnS)

Both are 4 : 4 coordinated structures of zinc sulphide. They differ only in the stacking of the sulphide ions. This comparison is a favourite 5-mark question.

PointZinc BlendeWurtzite
Packing of S²⁻ ionsCubic close packing (ccp / fcc)Hexagonal close packing (hcp)
Stacking sequenceABCABC…ABAB…
Zn²⁺ ions occupyHalf of the tetrahedral voidsHalf of the tetrahedral voids
Coordination number4 : 44 : 4
Formula units per cell42
ExamplesZnS (low temp.), CuCl, CdS, AgIZnS (high temp.), ZnO, CdS, BeO
Zinc Blende vs Wurtzite (ZnS)
Zinc blende — ABCABC A B C A B C S²⁻ in ccp, Zn²⁺ in ½ tetrahedral voids Wurtzite — ABAB A B A B A B S²⁻ in hcp, Zn²⁺ in ½ tetrahedral voids Both 4:4 — every Zn²⁺ sits in an S₄ tetrahedron around Zn²⁺
Zinc blende stacks sulphide layers ABCABC (ccp); wurtzite stacks them ABAB (hcp). In both, Zn²⁺ fills half the tetrahedral voids — 4:4 coordination.
Memory trick: "Blende is Best in Cubic" — Blende = BCC-like? No! Remember: zinc Blende = Cubic packing (ccp), Wurtzite = Hexagonal packing (hcp). Both fill half the tetrahedral voids, CN 4:4.
🔶

AB₂ Type Structures (1 : 2 compounds)

Here there are twice as many anions as cations. The cation usually has a higher coordination number than the anion.

1. Fluorite Structure — Calcium Fluoride (CaF₂)

  • Arrangement: Ca²⁺ ions form a ccp (fcc) lattice; F⁻ ions occupy all the tetrahedral voids.
  • Coordination number: 8 : 4 — each Ca²⁺ is surrounded by 8 F⁻, each F⁻ by 4 Ca²⁺.
  • Formula units per unit cell: 4 (4 Ca²⁺ + 8 F⁻).
  • Other examples: SrF₂, BaF₂, CdF₂, UO₂, ThO₂.

2. Antifluorite Structure — Sodium Oxide (Na₂O)

  • Arrangement: Exactly the reverse of fluorite — O²⁻ ions form the ccp lattice and Na⁺ ions occupy all the tetrahedral voids.
  • Coordination number: 4 : 8 — each Na⁺ is surrounded by 4 O²⁻ and each O²⁻ by 8 Na⁺.
  • Other examples: K₂O, Li₂O, Na₂S.
Fluorite (CaF₂) and Antifluorite (Na₂O)
CaF₂ — fluorite, 8:4 Ca²⁺ in ccp, F⁻ in ALL tetrahedral voids Na₂O — antifluorite (reversed) O²⁻ in ccp, Na⁺ in ALL tetrahedral voids
In fluorite, Ca²⁺ forms a ccp array with F⁻ in every tetrahedral void (8:4). Antifluorite swaps the roles: O²⁻ in ccp, Na⁺ in the tetrahedral voids.
Memory trick: "Anti = opposite." In fluorite the cation makes the close packing; in antifluorite the anion makes the close packing. Coordination numbers simply swap: 8:4 becomes 4:8.

3. Rutile Structure — Titanium Dioxide (TiO₂)

  • Arrangement: O²⁻ ions are approximately hexagonal close packed; Ti⁴⁺ ions occupy half of the octahedral voids.
  • Coordination number: 6 : 3 — each Ti⁴⁺ by 6 O²⁻, each O²⁻ by 3 Ti⁴⁺.
  • Other examples: MnO₂, SnO₂, PbO₂, GeO₂, MgF₂.

4. Cadmium Iodide (CdI₂) — Layer Structure

  • Arrangement: I⁻ ions are hexagonal close packed; Cd²⁺ ions fill the octahedral voids of alternate layers only.
  • The result is a layered (sandwich) structure: I–Cd–I sheets held to each other only by weak van der Waals forces, so the crystals are soft and cleave easily.
  • Coordination number: 6 : 3. Other examples: CdBr₂, FeCl₂, Mg(OH)₂.
Exam tip: Learn the "who packs, who fills" line for each structure — one line per structure is enough: NaCl = fcc Cl⁻ + all octahedral Na⁺; CsCl = simple cubic Cl⁻ + body-centre Cs⁺; CaF₂ = ccp Ca²⁺ + all tetrahedral F⁻; Na₂O = reverse; TiO₂ = hcp O²⁻ + half octahedral Ti⁴⁺; diamond = ccp C + half tetrahedral C. Write these six lines and you can answer any 5-mark "describe" question.
🔷

Perovskite Structure (ABX₃)

The mineral perovskite is CaTiO₃. The general formula is ABX₃, where A is a large cation, B a smaller cation and X usually oxygen (oxide perovskites) or a halide.

Ideal cubic structure

  • A²⁺ (Ca²⁺) — at the corners of the cube (coordination number 12).
  • B⁴⁺ (Ti⁴⁺) — at the body centre (coordination number 6, octahedral).
  • X²⁻ (O²⁻) — at the face centres (each B is surrounded by an octahedron of 6 X ions).
  • So the structure can be seen as a network of corner-sharing BX₆ octahedra with the large A cation sitting in the 12-coordinate cavity between them.
Perovskite Structure (CaTiO₃)
Ti CaTiO₃ — cubic perovskite Ca²⁺ corners (CN 12) Ti⁴⁺ body centre (CN 6) O²⁻ face centres → TiO₆ octahedron
Ca²⁺ at the cube corners, Ti⁴⁺ at the body centre, O²⁻ at the face centres — the six oxides form a TiO₆ octahedron around titanium.

Tolerance factor and structural distortion

Whether the ideal cubic structure actually forms depends on the relative sizes of A, B and X. This is measured by the Goldschmidt tolerance factor:

t = (rA + rX) / √2 (rB + rX)
  • t ≈ 0.9 – 1.0 → ideal cubic perovskite (e.g. SrTiO₃).
  • t < 0.9 → A cation too small; the BX₆ octahedra tilt and the structure distorts to orthorhombic or rhombohedral symmetry (e.g. CaTiO₃ itself is orthorhombic at room temperature).
  • t > 1.0 → A cation too large; distortion towards hexagonal structures.
Memory trick: Think of the tolerance factor as a "fitting test". t = 1 is a perfect fit (cubic). Too small an A ion (t < 0.9) makes the octahedra tilt like a collapsing tent; too big an A ion (t > 1) stretches the structure hexagonal.

Ilmenite structure (FeTiO₃)

  • Ilmenite is related to the corundum (α-Al₂O₃) structure: O²⁻ ions are approximately hcp and the cations fill two-thirds of the octahedral voids.
  • The difference from corundum is ordering: Fe²⁺ and Ti⁴⁺ occupy alternate layers of octahedral sites in an ordered fashion. It can be viewed as an ordered derivative of the corundum structure.

Applications

⚡ Ferroelectrics

BaTiO₃ (barium titanate) is a perovskite used in capacitors, transducers and memory devices because of its ferroelectricity.

☀️ Solar cells

Halide perovskites like CH₃NH₃PbI₃ are the active layer in low-cost, high-efficiency perovskite solar cells.

🧲 Multiferroics & catalysts

Perovskites such as LaMnO₃ are used in solid-oxide fuel-cell electrodes and oxidation catalysts.

⛏️ Ilmenite ore

FeTiO₃ is the main ore of titanium metal and TiO₂ pigment.

🧲

Spinel Structure (AB₂O₄)

The mineral spinel is MgAl₂O₄. General formula AB₂O₄: O²⁻ ions form a ccp lattice; the A²⁺ and B³⁺ cations distribute themselves between the tetrahedral and octahedral voids. How they distribute gives two types:

PointNormal SpinelInverse Spinel
General formulaA²⁺[B³⁺₂]O₄B³⁺[A²⁺B³⁺]O₄
Tetrahedral sites (1/8 filled)A²⁺ ionsHalf of the B³⁺ ions
Octahedral sites (1/2 filled)B³⁺ ionsA²⁺ ions + remaining B³⁺ ions
ExamplesMgAl₂O₄, ZnFe₂O₄, MnAl₂O₄Fe₃O₄ (magnetite), NiFe₂O₄, CoFe₂O₄
Spinel Structure (AB₂O₄)
O²⁻ in ccp + cation sites T T Oh Oh per 4 O²⁻: 8 T-holes + 4 Oh-holes blue T = tetrahedral, green Oh = octahedral Normal vs inverse spinel Normal — MgAl₂O₄ Mg²⁺ → 1/8 of T-sites Al³⁺ → 1/2 of Oh-sites Inverse — Fe₃O₄ Fe³⁺ → 1/8 of T-sites Fe²⁺ + Fe³⁺ → 1/2 of Oh-sites
Oxide ions are cubic close packed. In a normal spinel (MgAl₂O₄) the A cation takes 1/8 of tetrahedral holes and B takes 1/2 of octahedral holes; inverse spinels (Fe₃O₄) swap half the B ions into tetrahedral holes.

Degree of inversion: Real spinels often lie between the two extremes. The fraction of A²⁺ ions present in octahedral sites is called the inversion parameter — 0 for a perfect normal spinel and 1 for a perfect inverse spinel. Crystal field stabilisation energy (CFSE) decides the preference: ions with high octahedral CFSE (like Ni²⁺, Cr³⁺) prefer octahedral sites.

Cation distribution and magnetic implications

  • In magnetite, Fe₃O₄ (inverse spinel, Fe³⁺[Fe²⁺Fe³⁺]O₄), the magnetic moments of Fe³⁺ ions in tetrahedral sites align antiparallel to those in octahedral sites, so they cancel each other.
  • The net magnetic moment comes only from the Fe²⁺ ions in octahedral sites (4 unpaired electrons → 4 Bohr magnetons per formula unit).
  • This antiparallel-but-unequal arrangement is called ferrimagnetism — the reason magnetite is a natural magnet (lodestone).
Exam tip: For "explain the magnetism of Fe₃O₄", write three lines: (1) inverse spinel — Fe³⁺ in tetrahedral, Fe²⁺ + Fe³⁺ in octahedral; (2) tetrahedral and octahedral Fe³⁺ moments are antiparallel and cancel; (3) net moment = 4 BM from Fe²⁺ → ferrimagnetism. Full marks.
💎

Diamond Cubic Structure

  • Arrangement: Carbon atoms form a ccp lattice, and additional carbon atoms occupy half of the tetrahedral voids.
  • Coordination number: 4 — every carbon is tetrahedrally bonded to 4 others by strong covalent bonds.
  • Atoms per unit cell: 8 (4 from the ccp lattice + 4 from the filled tetrahedral voids).
  • Examples: Diamond (C), silicon (Si), germanium (Ge) — the basis of the semiconductor industry.
  • The strong directional covalent bonding in 3D gives diamond its extreme hardness and high melting point.
Diamond Cubic Structure
Diamond — each C tetrahedrally bonded (CN 4) grey = C in ccp black = C in ½ tetrahedral voids
Carbon atoms sit on a ccp lattice with half the tetrahedral voids also filled by carbon — every carbon is tetrahedrally bonded to 4 neighbours.

Diamond vs Zinc Blende vs Wurtzite — quick comparison

PointDiamondZinc BlendeWurtzite
Packing of main atomsccp of Cccp of S²⁻hcp of S²⁻
Atoms in tetrahedral voidsC (half of voids)Zn²⁺ (half of voids)Zn²⁺ (half of voids)
StackingABCABC…ABCABC…ABAB…
Coordination4 (all C)4 : 44 : 4
Atoms per unit cell84 ZnS units2 ZnS units
Bonding characterPure covalentCovalent + partial ionicCovalent + partial ionic
Memory trick: Diamond is just "zinc blende with both atoms the same" — ccp lattice + half tetrahedral voids filled, CN 4. If you know zinc blende, you know diamond; only the stacking (ABC vs AB) separates zinc blende from wurtzite.
🪨

Silicates — Structures Built from SiO₄ Tetrahedra

The basic building block of every silicate is the SiO₄ tetrahedron — one silicon atom at the centre, tetrahedrally bonded to four oxygen atoms. Silicates are classified by how many oxygens each tetrahedron shares with its neighbours. More sharing = more polymerised = higher Si : O ratio.

SiO₄ Tetrahedron — Basic Unit of Silicates
SiO₄⁴⁻ tetrahedron Si Corner-sharing linkage shared oxygen — Si–O–Si corner-sharing SiO₄ — basis of all silicates
Silicon sits at the centre of a tetrahedron of four oxygens. Tetrahedra link by sharing corner oxygens (Si–O–Si) to build every silicate structure.
TypeSharingBasic unit / formulaExamples
Ortho / Neso (isolated)No sharingSiO₄⁴⁻Olivine (Mg₂SiO₄), garnet, willemite (Zn₂SiO₄)
Pyro / Soro (paired)1 oxygen sharedSi₂O₇⁶⁻Thortveitite (Sc₂Si₂O₇)
Cyclo (rings)2 oxygens, closed ring(SiO₃)₃⁶⁻, (Si₆O₁₈)¹²⁻Beryl (Be₃Al₂Si₆O₁₈), tourmaline
Single chain — pyroxenes2 oxygens shared(SiO₃)ₙ²ⁿ⁻Diopside CaMg(SiO₃)₂, spodumene
Double chain — amphiboles2–3 oxygens shared(Si₄O₁₁)ₙ⁶ⁿ⁻Tremolite, asbestos minerals, hornblende
Sheet — phyllosilicates3 oxygens shared(Si₂O₅)ₙ²ⁿ⁻Talc, muscovite mica, clay minerals (kaolinite)
Framework — tectosilicatesAll 4 oxygens sharedSiO₂Quartz, feldspars, zeolites

Sheet silicates in daily life

🪶 Talc

Mg₃(Si₄O₁₀)(OH)₂ — softest mineral (hardness 1); sheets slide easily, used in talcum powder and lubricants.

✨ Mica (muscovite)

KAl₂(AlSi₃O₁₀)(OH)₂ — perfect basal cleavage into thin transparent sheets; electrical insulator used in electronics.

🏺 Clay minerals

Kaolinite Al₂Si₂O₅(OH)₄ — fine sheets that absorb water; basis of pottery, ceramics and paper coating.

🪟 Zeolites (framework)

Open 3D framework with cages and channels; used as molecular sieves, catalysts and water softeners (see below).

Isomorphous substitution and ion exchange

  • Isomorphous substitution means one ion replaces another of similar size in the crystal without changing the structure — e.g. Al³⁺ replaces Si⁴⁺ in the SiO₄ tetrahedra of sheet and framework silicates.
  • Since Al³⁺ has one less positive charge than Si⁴⁺, the framework gains a net negative charge.
  • This charge is balanced by loosely held cations (Na⁺, K⁺, Ca²⁺) sitting in the interlayer spaces or cavities.
  • These cations are not fixed — they can be exchanged for other cations from solution. This is the ion-exchange property of clays and zeolites.

Zeolites — industrial applications

  • Water softening (permutit process): sodium zeolite exchanges its Na⁺ for Ca²⁺ and Mg²⁺ in hard water: Na₂-zeolite + Ca²⁺ → Ca-zeolite + 2Na⁺.
  • Molecular sieves: uniform cage/channel sizes trap only molecules small enough to enter — used for drying gases and separating mixtures.
  • Catalysis: zeolites like ZSM-5 are solid-acid catalysts in petroleum cracking and petrochemical manufacture.
  • Detergents: zeolite A replaces phosphates as a builder, softening wash water.
Exam tip: The classic 5-mark question is "Classify silicates based on SiO₄ linkage with one example each." Draw the table above, add the Si : O ratio logic (isolated SiO₄ → SiO₂ framework), and finish with isomorphous substitution → ion exchange → zeolite uses. That single answer covers three syllabus points at once.

✎ PYQ Zone — Chapter 4

Every previous-year question from this chapter's topics, with year, paper, marks and a full exam-ready answer. Tap a question to open its detailed solution.

2020B.Sc. CC-115 marks🔁 also 2024

Q5 — “What are spinels? Using crystal field model, explain why Fe₃O₄ has an inverse spinel structure while Mn₃O₄ has a normal spinel structure. A complex of a transition metal ion with d⁶ electronic configuration is diamagnetic; is it an octahedral or a tetrahedral one?” ⭐⭐⭐⭐

Repeated: 2024 · B.Sc. CC-11 · Q2(a)(i) (“Using CFSE indicate spinels to be normal or inverse: Mn₃O₄, Fe₃O₄”, 2+2 marks)

📖 Detailed answer ▼

Spinels are oxides of formula AB₂O₄ built on a cubic close-packed array of O²⁻ ions. Per unit cell, 8 tetrahedral (A) holes and 16 octahedral (B) holes are filled by cations.

  • Normal spinel: A²⁺ in tetrahedral + 2B³⁺ in octahedral → AII[B₂III]O₄. Example: MgAl₂O₄, Mn₃O₄.
  • Inverse spinel: B³⁺ in tetrahedral + (A²⁺ + B³⁺) in octahedral → BIII[AIIBIII]O₄. Example: Fe₃O₄ (magnetite).

CFSE decides the sites (octahedral, high-spin values):

IonConfigurationCFSE (Oh)
Mn²⁺ / Fe³⁺d⁵ high-spin0
Fe²⁺d⁶ high-spin−0.4Δ₀
Mn³⁺d⁴ high-spin−0.6Δ₀
  • Fe₃O₄ = inverse spinel. The only ion with an octahedral preference is Fe²⁺ (−0.4Δ₀); Fe³⁺ (d⁵) has none. Fe²⁺ claims an octahedral site, pushing one Fe³⁺ into the tetrahedral site: FeIII[FeIIFeIII]O₄.
  • Mn₃O₄ = normal spinel. Mn³⁺ (d⁴, −0.6Δ₀, also Jahn–Teller stabilised in octahedral geometry) outranks Mn²⁺ (d⁵, zero preference). Both Mn³⁺ ions take octahedral sites and Mn²⁺ takes the tetrahedral site: MnII[Mn₂III]O₄.

The d⁶ half: a diamagnetic d⁶ ion has all six electrons paired → t2g⁶, which is possible only in a low-spin octahedral complex (strong field). Tetrahedral splitting Δt is always small (≈ 4/9 Δ₀), so tetrahedral complexes are always high-spin — d⁶ tetrahedral is e³t₂³ with 4 unpaired electrons, hence paramagnetic. Answer: it must be an octahedral complex.

2021B.Sc. CC-115 marks

Q3(a) — “What type of spinel structure do you expect for Co₃O₄ and NiCr₂O₄? Explain on the basis of CFT.” ⭐⭐⭐

📖 Detailed answer ▼

Rule: the ion with the larger octahedral-site preference energy claims the 16 octahedral B-sites; the other ion takes the 8 tetrahedral A-sites.

  • Co₃O₄ = NORMAL spinel: CoII[Co₂III]O₄. The deciding ion is Co³⁺: even in the weak oxide field it is low-spin d⁶ (t2g⁶) with an enormous octahedral CFSE of −2.4Δ₀ (plus pairing energy 2P) — far larger than the octahedral preference of Co²⁺ (high-spin d⁷, −0.8Δ₀). Both Co³⁺ ions therefore occupy octahedral sites and Co²⁺ goes tetrahedral. (Confirmed experimentally: Co₃O₄ contains diamagnetic low-spin Co³⁺ in octahedral holes; its magnetism comes from tetrahedral Co²⁺.)
  • NiCr₂O₄ = NORMAL spinel: NiII[Cr₂III]O₄. Cr³⁺ (d³) has the largest octahedral-site preference of the common 3d ions (CFSE −1.2Δ₀ — bigger than the site preference of Ni²⁺, d⁸), so both Cr³⁺ ions firmly take the octahedral sites and Ni²⁺ sits in the tetrahedral site.
Exam trap: both are normal — students often guess “inverse” for the chromite because Ni²⁺ also likes octahedral sites, but Cr³⁺’s octahedral preference is larger, so Cr³⁺ wins the octahedral holes.
2024B.Sc. CC-112+2 marks🔁 also 2020

Q2(a)(i) — “Using crystal field stabilization energy (CFSE) indicate spinels to be normal or inverse: Mn₃O₄, Fe₃O₄” ⭐⭐⭐⭐

Repeated: 2020 · B.Sc. CC-11 · Q5 (spinel half, 5 marks)

📖 Detailed answer ▼

CFSE in octahedral field (high-spin): d⁵ (Mn²⁺, Fe³⁺) = 0; d⁶ (Fe²⁺) = −0.4Δ₀; d⁴ (Mn³⁺) = −0.6Δ₀. The ion with the bigger octahedral CFSE takes the octahedral B-sites.

  • Mn₃O₄ = NORMAL: MnII[Mn₂III]O₄. Mn³⁺ (d⁴, −0.6Δ₀, Jahn–Teller active) strongly prefers octahedral; Mn²⁺ (d⁵, CFSE = 0) has no preference and goes tetrahedral.
  • Fe₃O₄ = INVERSE: FeIII[FeIIFeIII]O₄. Fe²⁺ (d⁶, −0.4Δ₀) is the only ion with an octahedral preference, so it claims an octahedral site and displaces one Fe³⁺ (d⁵, no preference) into the tetrahedral site.
One line: the ion with non-zero octahedral CFSE wins the octahedral holes — Mn³⁺ in Mn₃O₄ (normal), Fe²⁺ in Fe₃O₄ (inverse).
2019B.Sc. CC-92 marks

Q3(c)(iii) — “What are pyroxene and amphibole? Illustrate structurally.” ⭐⭐⭐

📖 Detailed answer ▼

Both are chain silicates — families of silicate minerals distinguished by how SiO₄ tetrahedra share corner oxygens.

  • Pyroxene — single chain. Each SiO₄ tetrahedron shares 2 of its 4 oxygens (2 bridging + 2 terminal), giving the repeating unit (SiO₃)ₙ²ⁿ⁻ (O:Si = 3:1). Examples: diopside CaMg(SiO₃)₂, jadeite NaAl(SiO₃)₂.
  • Amphibole — double chain. Two single chains cross-link: alternate tetrahedra share an extra oxygen, so each SiO₄ shares 2 or 3 oxygens, giving (Si₄O₁₁)ₙ⁶ⁿ⁻ (O:Si = 2.75:1). The cavity in the double chain holds OH⁻ groups. Examples: tremolite Ca₂Mg₅(Si₄O₁₁)₂(OH)₂, asbestos minerals.

Structural sketch: draw each SiO₄ as a triangle; join triangles corner-to-corner in a single row for pyroxene. For amphibole, draw two such rows side by side and link them through shared corners — the result is a ladder-like double chain. Cations (Ca²⁺, Mg²⁺) sit between the chains balancing the charge.

Memory hook: Pyroxene = one chain, (SiO₃); Amphibole = double chain, (Si₄O₁₁). “Amphi” = both/two.