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State Board · Class 12 · Chemistry

Lesson 9: Coordination Compounds

7 concepts · 56 practice questions with answers · Free to read

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Ligands, coordination terms and Werner's theory

From daily life

Like fingers gripping a ball: a metal ion at the centre, ligands holding it all around.

Simple definition

A complex is a central metal ion plus ligands joined by coordinate bonds. The coordination number is the number of donor atoms attached directly to the metal.

WernerWerner's experiments: CoCl₃·nNH₃ + excess AgNO₃FormulaAgCl (mol)IonsColour[Co(NH₃)₆]Cl₃34yellow[Co(NH₃)₅Cl]Cl₂23purple[Co(NH₃)₄Cl₂]Cl12green[Co(NH₃)₃Cl₃]00—Primary valence (ionisable) = Cl⁻; secondary = coordination number 6Inside [ ]: coordination sphere (non-ionisable); outside: ionisation sphereLigands: monodentate Cl⁻, NH₃ · bidentate en, ox²⁻ · hexadentate EDTA⁴⁻Ambidentate: NO₂⁻ / ONO⁻, SCN⁻ / NCS⁻

In the figure : Werner's experiments and ligands.

Rule 1: Types of ligand — Monodentate: Cl⁻, NH₃. Bidentate: en, ox²⁻. Hexadentate: EDTA⁴⁻. Ambidentate: NO₂⁻ (N or O), SCN⁻ (S or N).

  • Easy : H₂O is monodentate.
  • Medium : en binds through two N atoms.
  • Exam-type : EDTA grips at six points.

Rule 2: Werner's theory — Primary valence = ionisable (equal to oxidation number); secondary valence = coordination number, directional, non-ionisable.

  • Easy : CoCl₃·6NH₃ + AgNO₃ ⇒ 3 AgCl.
  • Medium : CoCl₃·5NH₃ ⇒ 2 AgCl.
  • Exam-type : CoCl₃·3NH₃ ⇒ 0 AgCl.

Rule 3: Coordination terms — Coordination sphere = inside [ ]; coordination polyhedron = shape of ligands; counter ion = outside [ ].

  • Easy : In [Co(NH₃)₆]Cl₃ Co³⁺ is the centre.
  • Medium : Coordination number 6.
  • Exam-type : 3 Cl⁻ counter ions.

Calculation: 0.01 mol of [Co(NH₃)₅Cl]Cl₂ + excess AgNO₃

  • Step 1: Ionisable Cl⁻ = 2 (outside the bracket).
  • Step 2: AgCl = 0.01 × 2 = 0.02 mol.
  • Step 3: Mass = 0.02 × 143.5 = 2.87 g.

CET level (MCQ)

  • CET 1) AgCl (mol) from [Co(NH₃)₄Cl₂]Cl? 1) 1 2) 2 3) 3 4) 0 — Answer: 1).
  • CET 2) Bidentate ligand? 1) en 2) Cl⁻ 3) NH₃ 4) H₂O — Answer: 1).

Real-life use: Haemoglobin, chlorophyll, drugs, pigments.

Remember

Ligands inside [ ] do not ionise; AgNO₃ precipitates only outside Cl⁻.

Common mistake

Counting inside-bracket Cl⁻ as AgCl.

CET trap

AgCl comes only from Cl⁻ outside [ ].

Shortcut

Memory clue: "only what is outside the bracket meets silver".

Spot the difference

TermMeaning
Ligandelectron-pair donor
Coordination no.donor atoms attached
Counter ionoutside [ ]

Practice questions

Easy
  1. 1What is a complex?
  2. 2Types of ligand.
  3. 3Werner's two valences.
Medium
  1. 4CoCl₃·6NH₃ ⇒ how much AgCl?
  2. 5Example of ambidentate ligand.
  3. 6What is coordination number?
Challenging
  1. 7Mass of AgCl from 0.01 mol [Co(NH₃)₅Cl]Cl₂.
  2. 8Denticity of EDTA?
Show answers (try them yourself first)
  1. 1Metal + ligands.
  2. 2Mono-, bi-, hexa-, ambidentate.
  3. 3Primary and secondary.
  4. 43 mol.
  5. 5NO₂⁻.
  6. 6Number of donor atoms.
  7. 72.87 g.
  8. 8Six.

Classification and IUPAC nomenclature

From daily life

Like writing an address in a fixed order, a complex name lists ligands first and the metal last.

Simple definition

Naming: cation first; ligands alphabetically; then the metal with Roman-numeral oxidation state.

NamingIUPAC naming of complexes1Cation first, then anion2Ligands alphabetically; metal last3di, tri, tetra; bis, tris for complex names4Anionic ligands –o: chlorido, cyanido; neutral: ammine, aqua, carbonyl5Roman numeral oxidation state; –ate for anionic complexes[Co(NH₃)₆]Cl₃hexaamminecobalt(III) chlorideK₄[Fe(CN)₆]potassium hexacyanidoferrate(II)[Pt(NH₃)₂Cl₂]diamminedichloridoplatinum(II)K₃[Fe(C₂O₄)₃]potassium trioxalatoferrate(III)[Cu(NH₃)₄]SO₄tetraamminecopper(II) sulfateAnionic complexes use Latin stems: ferrate, cuprate, argentate, aurate

In the figure : Naming rules.

Rule 1: Ligand names — Anionic: chlorido, cyanido, hydroxido, oxalato; neutral: ammine (NH₃), aqua (H₂O), carbonyl (CO).

  • Easy : NH₃ ⇒ ammine (two m).
  • Medium : Cl⁻ ⇒ chlorido.
  • Exam-type : CO ⇒ carbonyl.

Rule 2: Order and prefixes — Alphabetical order ignoring prefixes; di, tri, tetra; bis, tris for complex names like en.

  • Easy : diammine…dichlorido…: a before c.
  • Medium : tris(ethane-1,2-diamine).
  • Exam-type : tetra = 4, hexa = 6.

Rule 3: Anionic complex — The metal takes –ate: Fe ⇒ ferrate, Cu ⇒ cuprate, Ag ⇒ argentate, Au ⇒ aurate.

  • Easy : K₄[Fe(CN)₆] ⇒ potassium hexacyanidoferrate(II).
  • Medium : K₃[Fe(C₂O₄)₃] ⇒ potassium trioxalatoferrate(III).
  • Exam-type : Na[Ag(CN)₂] ⇒ sodium dicyanidoargentate(I).

Name: [Co(NH₃)₆]Cl₃ and [Pt(NH₃)₂Cl₂]

  • Step 1: Oxidation state of Co = +3 (3 Cl⁻ outside, NH₃ neutral).
  • Step 2: hexaammine + cobalt(III) + chloride.
  • Step 3: Pt: 2×0 + 2×(–1) + x = 0 ⇒ +2 ⇒ diamminedichloridoplatinum(II).

CET level (MCQ)

  • CET 1) Oxidation state of Fe in K₄[Fe(CN)₆]? 1) +2 2) +3 3) +4 4) 0 — Answer: 1).
  • CET 2) Ag in an anionic complex is named? 1) argentate 2) silverate 3) agate 4) argonate — Answer: 1).

Real-life use: Drug names, chemical catalogues, lab reports.

Remember

Cation first; ligands alphabetical; metal last; –ate for anionic complexes.

Common mistake

Wrong Roman numeral or ligand order.

CET trap

Oxidation number = overall charge – ligand charges.

Shortcut

Memory clue: "A–B–C first, metal last".

Spot the difference

LigandName
NH₃ammine
H₂Oaqua
Cl⁻chlorido
CN⁻cyanido
COcarbonyl

Practice questions

Easy
  1. 1Name [Co(NH₃)₆]Cl₃.
  2. 2Name K₄[Fe(CN)₆].
  3. 3Name [Pt(NH₃)₂Cl₂].
Medium
  1. 4Metal naming in anionic complexes?
  2. 5Name [Cu(NH₃)₄]SO₄.
  3. 6Write the formula of tetraamminecopper(II) sulfate.
Challenging
  1. 7When is bis used?
  2. 8Oxidation state of Fe in K₃[Fe(C₂O₄)₃].
Show answers (try them yourself first)
  1. 1hexaamminecobalt(III) chloride.
  2. 2potassium hexacyanidoferrate(II).
  3. 3diamminedichloridoplatinum(II).
  4. 4Latin stem + ate.
  5. 5tetraamminecopper(II) sulfate.
  6. 6[Cu(NH₃)₄]SO₄.
  7. 7For complex ligand names.
  8. 8+3.

Effective atomic number (EAN)

From daily life

A team is complete at a set number; the metal wants the electron count of the nearest noble gas.

Simple definition

EAN = Z of metal – oxidation number + 2 × coordination number. EAN equal to a noble-gas Z ⇒ stable.

EANEAN = Z – O.N. + 2 × (coordination number)ComplexZO.N.CNEANgas[Fe(CN)₆]⁴⁻262636Kr[Co(NH₃)₆]³⁺273636Kr[Ni(CO)₄]280436Kr[Fe(CO)₅]260536Kr[Cu(NH₃)₄]²⁺292435—[Cr(CO)₆]240636KrKr = 36; EAN = 36 ⇒ stable (metal carbonyls, cyanido complexes)Exceptions exist (e.g. [Cu(NH₃)₄]²⁺ = 35) — the rule is only a guide

In the figure : EAN table.

Rule 1: Method — Electrons in the ion = Z – O.N.; each ligand donates 2 e⁻.

  • Easy : [Fe(CN)₆]⁴⁻: 26 – 2 + 12 = 36.
  • Medium : [Co(NH₃)₆]³⁺: 27 – 3 + 12 = 36.
  • Exam-type : [Cu(NH₃)₄]²⁺: 29 – 2 + 8 = 35.

Rule 2: Metal carbonyls — Ni(CO)₄, Fe(CO)₅, Cr(CO)₆ — each EAN = 36, so stable.

  • Easy : Ni(CO)₄: 28 + 8 = 36.
  • Medium : Fe(CO)₅: 26 + 10 = 36.
  • Exam-type : Cr(CO)₆: 24 + 12 = 36.

Rule 3: Limitations — A guide only; many exceptions.

  • Easy : [Cu(NH₃)₄]²⁺ has EAN 35 yet is stable.
  • Medium : [Ni(NH₃)₆]²⁺ has EAN 38.
  • Exam-type : It ignores shape and charge effects.

Calculation: EAN of [Fe(CN)₆]⁴⁻

  • Step 1: Z of Fe = 26, O.N. = +2 ⇒ 24 e⁻.
  • Step 2: 6 CN⁻ × 2 = 12 e⁻.
  • Step 3: EAN = 24 + 12 = 36 (Kr).

CET level (MCQ)

  • CET 1) EAN of Ni(CO)₄? 1) 36 2) 28 3) 32 4) 40 — Answer: 1).
  • CET 2) EAN of [Co(NH₃)₆]³⁺? 1) 36 2) 27 3) 33 4) 30 — Answer: 1).

Real-life use: Metal carbonyl formation, catalysis.

Remember

EAN = Z – O.N. + 2×CN; 36 (Kr) ⇒ stable.

Common mistake

Forgetting to subtract O.N.

CET trap

Neutral ligands (CO, NH₃) also donate 2 e⁻.

Shortcut

Memory clue: "Z minus O.N., plus double CN".

Spot the difference

ComplexEAN
[Fe(CN)₆]⁴⁻36
[Co(NH₃)₆]³⁺36
Ni(CO)₄36
[Cu(NH₃)₄]²⁺35

Practice questions

Easy
  1. 1What is EAN?
  2. 2EAN of [Fe(CN)₆]⁴⁻.
  3. 3EAN of Fe(CO)₅.
Medium
  1. 4EAN of Cr(CO)₆.
  2. 5EAN of [Cu(NH₃)₄]²⁺.
  3. 6Limitation of the EAN rule.
Challenging
  1. 7EAN 36 matches which gas?
  2. 8How many e⁻ does each ligand donate?
Show answers (try them yourself first)
  1. 1Z – O.N. + 2×CN.
  2. 236.
  3. 326 + 10 = 36.
  4. 424 + 12 = 36.
  5. 529 – 2 + 8 = 35.
  6. 6Many exceptions.
  7. 7Krypton.
  8. 8Two.

Isomerism in complexes

From daily life

Twins look alike but behave differently — isomeric complexes.

Simple definition

Isomers share a formula but differ in structure or properties: structural (ionisation, hydrate, linkage, coordination) and stereoisomers (cis/trans, optical).

IsomerismIsomerism: cis / trans in [Pt(NH₃)₂Cl₂]ClClNH₃NH₃PtcisClNH₃NH₃ClPttranspolar, cisplatin (anticancer)non-polar, inactiveIonisation[Co(NH₃)₅Br]SO₄ ⇄ [Co(NH₃)₅SO₄]BrHydrate[Cr(H₂O)₆]Cl₃ ⇄ [Cr(H₂O)₅Cl]Cl₂·H₂OLinkage–NO₂ ⇄ –ONOCoordination[Co(NH₃)₆][Cr(CN)₆] ⇄ [Cr(NH₃)₆][Co(CN)₆]Optical: [Co(en)₃]³⁺, cis-[CoCl₂(en)₂]⁺ (non-superimposable mirror images)

In the figure : cis/trans and other isomerism.

Live figure

Isomerism in complexes

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Rule 1: Structural — Ionisation: [Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br; also hydrate, linkage (NO₂/ONO), coordination.

  • Easy : The first gives a precipitate with BaCl₂.
  • Medium : The second with AgNO₃.
  • Exam-type : –NO₂ (nitro) vs –ONO (nitrito).

Rule 2: Geometrical (cis/trans) — Square planar [Pt(NH₃)₂Cl₂] gives cis (polar) and trans (non-polar); tetrahedral complexes show none.

  • Easy : Cisplatin is used against cancer.
  • Medium : MA₂B₂ ⇒ 2 isomers.
  • Exam-type : Octahedral MA₄B₂ ⇒ 2.

Rule 3: Optical — Non-superimposable mirror images give d and l forms: [Co(en)₃]³⁺ and cis-[CoCl₂(en)₂]⁺.

  • Easy : The trans form is achiral.
  • Medium : They rotate plane-polarised light oppositely.
  • Exam-type : Chirality = no mirror symmetry.

Reasoning: isomers of [Pt(NH₃)₂Cl₂]

  • Step 1: Square planar, type MA₂B₂.
  • Step 2: Two Cl adjacent ⇒ cis; opposite ⇒ trans.
  • Step 3: Total 2 geometrical isomers.

CET level (MCQ)

  • CET 1) Anticancer drug? 1) cis-[Pt(NH₃)₂Cl₂] 2) trans-[Pt(NH₃)₂Cl₂] 3) [Ni(CO)₄] 4) K₄[Fe(CN)₆] — Answer: 1).
  • CET 2) Optically active? 1) [Co(en)₃]³⁺ 2) [Co(NH₃)₆]³⁺ 3) [Pt(NH₃)₂Cl₂] 4) [Ni(CO)₄] — Answer: 1).

Real-life use: Cancer therapy, catalyst selection.

Remember

Four structural + cis/trans + optical.

Common mistake

Assuming tetrahedral complexes show cis/trans.

CET trap

cis/trans only in square planar and octahedral.

Shortcut

Memory clue: "cis = same side, trans = across".

Spot the difference

TypeExample
Ionisation[Co(NH₃)₅Br]SO₄
Linkage–NO₂ / –ONO
Geometricalcis / trans
Optical[Co(en)₃]³⁺

Practice questions

Easy
  1. 1What is isomerism?
  2. 2Ionisation isomerism.
  3. 3Linkage isomerism.
Medium
  1. 4cis vs trans.
  2. 5Use of cisplatin.
  3. 6When does optical isomerism occur?
Challenging
  1. 7Why is [Co(en)₃]³⁺ chiral?
  2. 8Why no cis/trans in tetrahedral?
Show answers (try them yourself first)
  1. 1Same formula, different structure.
  2. 2Ion inside vs outside.
  3. 3Joining through N or O.
  4. 4Polar vs non-polar.
  5. 5Cancer treatment.
  6. 6Non-superimposable mirror image.
  7. 7Mirror image non-superimposable.
  8. 8All positions are equivalent.

Stability of complexes and the chelate effect

From daily life

A grip with two hands holds tighter than one — bidentate ligands give more stable complexes.

Simple definition

The stability constant β = K₁K₂…Kₙ; larger β ⇒ more stable. The chelate effect is the extra stability from ring-forming ligands.

StabilityStability: log β (larger ⇒ more stable)[Ag(NH₃)₂]⁺7.2[Cu(NH₃)₄]²⁺12.7[Cu(en)₂]²⁺20[Ag(CN)₂]⁻21.1[Fe(CN)₆]⁴⁻35[Fe(CN)₆]³⁻42Chelate effect: [Cu(en)₂]²⁺ (20.0) > [Cu(NH₃)₄]²⁺ (12.7)Factors: higher charge, smaller size, more basic ligand ⇒ more stableβ = K₁K₂…Kₙ; values approximate

In the figure : log β comparison (approximate).

Rule 1: Stability constant — M + 4L ⇌ ML₄: β = [ML₄]/([M][L]⁴); larger β ⇒ less dissociation.

  • Easy : [Ag(NH₃)₂]⁺ log β ≈ 7.2.
  • Medium : [Ag(CN)₂]⁻ ≈ 21.
  • Exam-type : Very stable ⇒ very large β.

Rule 2: Chelate effect — [Cu(en)₂]²⁺ (≈ 20) > [Cu(NH₃)₄]²⁺ (≈ 12.7); ring formation increases entropy.

  • Easy : EDTA gives very stable chelates.
  • Medium : Five- or six-membered rings are best.
  • Exam-type : Bidentate ⇒ two bonds from one ligand.

Rule 3: Factors affecting stability — Metal ion: higher charge, smaller size ⇒ stable. Ligand: more basic and chelating ⇒ stable.

  • Easy : Fe³⁺ complexes beat Fe²⁺ ones.
  • Medium : CN⁻ > NH₃ > H₂O.
  • Exam-type : EDTA is used to measure hardness.

Reasoning: why is [Cu(en)₂]²⁺ more stable?

  • Step 1: One en forms two bonds (a ring).
  • Step 2: Replacing NH₃ by en increases the number of free particles ⇒ entropy rises.
  • Step 3: ΔG is more negative ⇒ larger β.

CET level (MCQ)

  • CET 1) The chelate effect arises from? 1) ring formation 2) lower charge 3) bigger size 4) temperature — Answer: 1).
  • CET 2) Used to measure water hardness? 1) EDTA 2) en 3) CO 4) NH₃ — Answer: 1).

Real-life use: Water hardness, treating metal poisoning (EDTA).

Remember

β = K₁K₂…Kₙ; chelates are more stable.

Common mistake

Confusing β with a single K.

CET trap

β is the product of all step constants.

Shortcut

Memory clue: "two hands, firmer grip".

Spot the difference

Complexlog β
[Ag(NH₃)₂]⁺7.2
[Cu(NH₃)₄]²⁺12.7
[Cu(en)₂]²⁺20
[Fe(CN)₆]⁴⁻35

Practice questions

Easy
  1. 1What is β?
  2. 2Explain the chelate effect.
  3. 3Factors affecting stability.
Medium
  1. 4Use of EDTA.
  2. 5[Cu(en)₂]²⁺ vs [Cu(NH₃)₄]²⁺.
  3. 6What does a large β show?
Challenging
  1. 7Why is Fe³⁺ more stable?
  2. 8Which ring size is best?
Show answers (try them yourself first)
  1. 1K₁K₂…Kₙ.
  2. 2Ring ⇒ entropy ↑.
  3. 3Charge, size, ligand.
  4. 4Measuring hardness.
  5. 520 vs 12.7.
  6. 6More stable.
  7. 7Higher charge.
  8. 8Five or six.

Valence bond theory

From daily life

Guests (ligands) can sit only in empty rooms (orbitals) — their electron pairs fill them.

Simple definition

VBT: the metal hybridises empty orbitals to accept ligand pairs; hybridisation gives the shape and unpaired electrons give magnetism.

VBTValence bond theory: hybridisation and magnetismComplexHyb.Shapen (e⁻)μ (BM)[Fe(CN)₆]⁴⁻d²sp³octahedral00.00[FeF₆]³⁻sp³d²octahedral55.92[Co(NH₃)₆]³⁺d²sp³octahedral00.00[Fe(CN)₆]³⁻d²sp³octahedral11.73[Ni(CN)₄]²⁻dsp²sq. planar00.00[NiCl₄]²⁻sp³tetrahedral22.83[Ni(CO)₄]sp³tetrahedral00.00μ = √(n(n + 2)) BMInner-orbital d²sp³ (strong ligand, electrons pair); outer-orbital sp³d² (weak ligand)

In the figure : Hybridisation and magnetic moment.

Live figure

Valence bond theory

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Rule 1: Hybridisation and shape — d²sp³ / sp³d² ⇒ octahedral; sp³ ⇒ tetrahedral; dsp² ⇒ square planar.

  • Easy : [Co(NH₃)₆]³⁺ is d²sp³.
  • Medium : [Ni(CN)₄]²⁻ is dsp².
  • Exam-type : [NiCl₄]²⁻ is sp³.

Rule 2: Inner and outer orbital — Strong ligands pair electrons ⇒ d²sp³ (low spin); weak ligands ⇒ sp³d² (high spin).

  • Easy : [Fe(CN)₆]⁴⁻ is d²sp³, diamagnetic.
  • Medium : [FeF₆]³⁻ is sp³d², 5 e⁻.
  • Exam-type : [Fe(CN)₆]³⁻ has 1 e⁻.

Rule 3: Magnetic moment — μ = √(n(n+2)) BM; n = 1, 2, 3, 4, 5 ⇒ 1.73, 2.83, 3.87, 4.90, 5.92.

  • Easy : [Cu(NH₃)₄]²⁺ n = 1 ⇒ 1.73.
  • Medium : [NiCl₄]²⁻ n = 2 ⇒ 2.83.
  • Exam-type : [FeF₆]³⁻ n = 5 ⇒ 5.92.

Calculation: μ of [FeF₆]³⁻

  • Step 1: Fe³⁺ = 3d⁵; F⁻ is weak ⇒ no pairing ⇒ n = 5.
  • Step 2: μ = √(5 × 7) = √35.
  • Step 3: μ = 5.92 BM, hybridisation sp³d².

CET level (MCQ)

  • CET 1) Hybridisation of [Ni(CN)₄]²⁻? 1) dsp² 2) sp³ 3) d²sp³ 4) sp³d² — Answer: 1).
  • CET 2) μ for n = 3? 1) 3.87 2) 2.83 3) 4.90 4) 1.73 — Answer: 1).

Real-life use: Magnetic materials, MRI contrast.

Remember

d²sp³ inner, sp³d² outer, sp³, dsp²; μ = √(n(n+2)).

Common mistake

Using number of pairs in place of n.

CET trap

n = number of unpaired electrons.

Shortcut

Memory clue: "strong pairs, weak leaves".

Spot the difference

HybridisationShape
d²sp³octahedral
sp³d²octahedral
sp³tetrahedral
dsp²square planar

Practice questions

Easy
  1. 1Basics of VBT.
  2. 2Inner vs outer orbital.
  3. 3Hybridisation of [Fe(CN)₆]⁴⁻.
Medium
  1. 4μ of [FeF₆]³⁻.
  2. 5Hybridisation and magnetism of [Ni(CN)₄]²⁻.
  3. 6Shape and μ of [NiCl₄]²⁻.
Challenging
  1. 7Where is μ = √(n(n+2)) used?
  2. 8μ of [Cu(NH₃)₄]²⁺.
Show answers (try them yourself first)
  1. 1Empty orbitals + ligand pairs.
  2. 2Paired vs unpaired.
  3. 3d²sp³, diamagnetic.
  4. 45.92 BM.
  5. 5dsp², diamagnetic.
  6. 6Tetrahedral, 2.83 BM.
  7. 7From unpaired electrons.
  8. 81.73 BM.

Crystal field theory, colour and applications

From daily life

As a prism splits white light, the ligand field splits the d orbitals into two sets.

Simple definition

CFT: ligands act as point charges; d orbitals split into t₂g (lower) and e_g (upper) with gap Δ₀.

CFTCrystal field theory: octahedral splittingfree ioneg (+0.6 Δ₀)t2g (–0.4 Δ₀)Δ₀Strong fieldΔ₀ > Plow spinWeak fieldΔ₀ < Phigh spinTetrahedral: Δt = (4/9) Δ₀ (always high spin)I⁻ < Br⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < NH₃ < en < CN⁻ < COλ = 500 nm ⇒ Δ₀ = hc/λ · NA = 239 kJ/mol; seen colour = complementaryUses: EDTA (hardness), cisplatin, haemoglobin (Fe), chlorophyll (Mg), B₁₂ (Co), Mond Ni refining

In the figure : Octahedral splitting.

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Crystal field theory, colour and applications

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Rule 1: Splitting and spin — Δ₀ > P (pairing energy) ⇒ low spin; Δ₀ < P ⇒ high spin; Δt = (4/9) Δ₀ ⇒ tetrahedral always high spin.

  • Easy : CN⁻ strong ⇒ low spin.
  • Medium : F⁻ weak ⇒ high spin.
  • Exam-type : CFSE = (–0.4 n(t₂g) + 0.6 n(e_g)) Δ₀.

Rule 2: Spectrochemical series — I⁻ < Br⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO.

  • Easy : Left side weak.
  • Medium : Right side strong.
  • Exam-type : Δ₀ rises to the right.

Rule 3: Colour and uses — d–d transitions absorb a colour; the complementary colour is seen. Uses: EDTA hardness, cisplatin, haemoglobin, chlorophyll, vitamin B₁₂, Mond process for Ni, cyanide extraction of Au/Ag.

  • Easy : λ = 500 nm ⇒ Δ₀ = 239 kJ/mol.
  • Medium : Ti³⁺ is violet (d¹).
  • Exam-type : Zn²⁺ is colourless (d¹⁰).

Calculation: Δ₀ for λ = 500 nm

  • Step 1: E = hc/λ = (6.626×10⁻³⁴ × 3×10⁸)/(500×10⁻⁹) = 3.97×10⁻¹⁹ J.
  • Step 2: × 6.022×10²³ mol⁻¹ = 239 kJ/mol.
  • Step 3: Δ₀ ≈ 239 kJ/mol.

CET level (MCQ)

  • CET 1) Δt = ? 1) (4/9) Δ₀ 2) (9/4) Δ₀ 3) Δ₀ 4) 2Δ₀ — Answer: 1).
  • CET 2) Strongest ligand? 1) CO 2) H₂O 3) F⁻ 4) Cl⁻ — Answer: 1).

Real-life use: Gemstones, pigments, drugs, natural colours.

Remember

t₂g lower, e_g upper; Δ₀ > P ⇒ low spin; Δt = 4/9 Δ₀.

Common mistake

Treating absorbed and seen colours as the same.

CET trap

The seen colour is the complement.

Shortcut

Memory clue: "CO, CN⁻ strong; I⁻ weak".

Spot the difference

AspectOctahedralTetrahedral
SplittingΔ₀(4/9) Δ₀
Lowert₂ge
Spineitherhigh

Practice questions

Easy
  1. 1Basics of CFT.
  2. 2t₂g and e_g.
  3. 3Strong vs weak field.
Medium
  1. 4Relation of Δt and Δ₀.
  2. 5λ = 500 nm ⇒ Δ₀.
  3. 6Why are complexes coloured?
Challenging
  1. 7Why is Zn²⁺ colourless?
  2. 8Three uses.
Show answers (try them yourself first)
  1. 1Ligands as point charges.
  2. 2Lower and upper sets.
  3. 3Δ₀ > P vs Δ₀ < P.
  4. 44/9.
  5. 5239 kJ/mol.
  6. 6d–d transitions.
  7. 7d¹⁰ ⇒ no transition.
  8. 8EDTA, cisplatin, haemoglobin.

Quick revision of the whole lesson

ConceptKey formula / fact
EANZ – O.N. + 2 × CN
Magnetic momentμ = √(n(n+2)) BM
Tetrahedral splittingΔt = (4/9) Δ₀
CFSE(–0.4 n_t₂g + 0.6 n_eg) Δ₀
Stabilityβ = K₁K₂…Kₙ
EnergyE = hc/λ
AgCl massn × 143.5 g
Oxidation numberTotal charge – ligand charges
Live figure

🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Ligands, Werner.
  • Nomenclature.
  • EAN.
  • Isomerism.
  • Stability, chelate.
  • VBT.
  • CFT, colour, uses.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.