Werner's theory and basic terms of coordination compounds
Link to Std 11 : In Std 11 you learnt the coordinate (dative) bond and d-block metals; here a metal ion accepts electron pairs from several donors.
Haemoglobin in your blood carries oxygen using iron held by nitrogen atoms in a ring; this is a coordination compound found inside you.
In a coordination compound a central metal atom or ion is bound to a fixed number of ligands (ions or molecules that donate electron pairs). The metal and ligands together form the coordination entity, written in square brackets. The number of donor atoms bonded to the metal is the coordination number. Werner's theory: a metal has a primary valence (oxidation state, ionisable) and a secondary valence (coordination number, non-ionisable, directed in space).
Werner's theory and basic terms of coordination compounds
In the app you can move this figure with your finger.
Open in the appIn the figure : The six NH₃ ligands around Co³⁺ give an octahedral shape. Everything inside the brackets stays together when the compound dissolves; ions outside the brackets separate in water.
Rule 1: Terms to know — Central atom/ion: the acceptor (Co³⁺). Ligand: donor. Unidentate ligands use one donor atom (Cl⁻, NH₃, H₂O); didentate two (ethane-1,2-diamine, oxalate); polydentate more (EDTA, six); ambidentate have two possible donor atoms (NO₂⁻ via N or O; SCN⁻ via S or N). A chelate is a ring formed by a polydentate ligand. Homoleptic: one kind of ligand; heteroleptic: more than one.
- Easy : In [Co(NH₃)₆]³⁺ the ligand NH₃ is unidentate and the complex is homoleptic.
- Medium : [Co(NH₃)₄Cl₂]⁺ is heteroleptic with ligands NH₃ and Cl⁻.
- Exam-type : Name the denticity: oxalate C₂O₄²⁻ is didentate; EDTA⁴⁻ is hexadentate (two N and four O donors).
Rule 2: Oxidation number and coordination number — Oxidation number of the metal = (charge on the entity) − (sum of the ligand charges). The coordination number is the number of donor atoms, not the number of ligands (a didentate ligand counts twice).
- Easy : [Co(NH₃)₅Cl]²⁺: x + 0 + (−1) = +2, so Co is +3; coordination number = 6.
- Medium : [Fe(CN)₆]⁴⁻: x + 6(−1) = −4, so Fe is +2. K₃[Fe(CN)₆]: Fe is +3.
- Exam-type : [Co(en)₃]³⁺ has oxidation state +3 (en neutral) and coordination number 6, because each en gives two donor atoms (3 × 2 = 6).
Rule 3: Werner's evidence — When a complex dissolves, only the ions outside the brackets precipitate with the usual reagents. For three cobalt(III) chloride–ammonia compounds, the moles of chloride precipitated by AgNO₃ show how many chlorides are free: [Co(NH₃)₆]Cl₃ gives 3 AgCl, [Co(NH₃)₅Cl]Cl₂ gives 2, [Co(NH₃)₄Cl₂]Cl gives 1.
- Easy : Cl⁻ inside the brackets is bonded to the metal and is not precipitated by AgNO₃.
- Medium : 0.010 mol [Co(NH₃)₅Cl]Cl₂ with excess AgNO₃ gives 0.020 mol AgCl = 0.020 × 143.5 = 2.87 g.
- Exam-type : The number of ions in solution (found from conductivity) also matches: 4, 3 and 2 ions for the three compounds.
Ions inside the square bracket do not ionise; ions outside do.
Counting the number of ligands as the coordination number for didentate or polydentate ligands.
Memory clue: "Count the donors, not the donors' names" — coordination number = number of donor atoms bonded to the metal.
Spot the difference
| Point | Primary valence | Secondary valence |
|---|---|---|
| Other name | Oxidation state | Coordination number |
| Ionisable | Yes | No |
| Satisfied by | Negative ions | Ligands (ions or molecules) |
Practice questions
- 1Define a ligand.
- 2What is the coordination number of Co in [Co(NH₃)₆]³⁺?
- 3Give one example each of a unidentate and a didentate ligand.
- 4Find the oxidation state of the metal in [Fe(CN)₆]⁴⁻.
- 5How many moles of AgCl are formed when 0.010 mol [Co(NH₃)₅Cl]Cl₂ reacts with excess AgNO₃?
- 6What are ambidentate ligands? Give an example.
- 7Why does [Co(NH₃)₄Cl₂]Cl give only one mole of AgCl per mole but CoCl₃·6NH₃ gives three?
- 8What is the coordination number of the metal in [Co(en)₃]³⁺? Explain.
Show answers (try them yourself first)
- 1A species (ion or molecule) that donates a pair of electrons to the central metal atom or ion.
- 26.
- 3NH₃ (unidentate), oxalate or ethane-1,2-diamine (didentate).
- 4x + 6(−1) = −4, so x = +2.
- 50.020 mol (2 Cl⁻ are outside the bracket), i.e. 2.87 g.
- 6Ligands with two different donor atoms, either of which can bind, e.g. NO₂⁻ (N or O) or SCN⁻ (S or N).
- 7Two Cl⁻ are inside the brackets (coordinated) and cannot be precipitated; one Cl⁻ is outside. In [Co(NH₃)₆]Cl₃ all three Cl⁻ are outside.
- 86, because each of the three ethane-1,2-diamine ligands is didentate (3 × 2 = 6 donor atoms).
IUPAC naming of coordination compounds
Link to Std 11 : In Std 11 you named organic compounds with fixed rules; complexes have their own set of rules.
A well-known pigment 'Prussian blue' and the medicine cisplatin are coordination compounds with exact systematic names that tell the chemist their composition.
Names follow a fixed order: cation first, then anion; inside a complex, ligands in alphabetical order, then the metal with its oxidation state in Roman numerals in brackets. If the complex ion is an anion the metal name ends in -ate.
In the figure : The name is built left to right in the same order in which the parts are written in the formula, except that the formula writes the metal first and the name writes the metal last.
Rule 1: Rules for ligands and metal — Anionic ligands end in -o: chlorido, bromido, cyanido, hydroxido, oxalato, sulphato, nitrito-N. Neutral ligands keep their names except aqua (H₂O), ammine (NH₃), carbonyl (CO), nitrosyl (NO). Number prefixes di-, tri-, tetra-, penta-, hexa-; for complex ligand names use bis-, tris-, tetrakis-. Anionic complex: ferrate (Fe), cuprate (Cu), argentate (Ag), aurate (Au), cobaltate (Co), platinate (Pt), nickelate (Ni), zincate (Zn).
- Easy : [Co(NH₃)₆]Cl₃ = hexaamminecobalt(III) chloride.
- Medium : K₃[Fe(CN)₆] = potassium hexacyanidoferrate(III): the complex is an anion, so the metal ends in -ate.
- Exam-type : [Cr(H₂O)₄Cl₂]Cl is tetraaquadichloridochromium(III) chloride; alphabetical order ignores the prefixes (aqua before chlorido).
Rule 2: Writing a formula from the name — Write the metal first, then the ligands in alphabetical order of their names, put the complex in square brackets and add the counter ion; find the charge from the oxidation state.
- Easy : Sodium dicyanidoargentate(I): Na[Ag(CN)₂] (Ag +1, two CN⁻ = −2, charge −1).
- Medium : Tetraamminedichloridocobalt(III) chloride: [Co(NH₃)₄Cl₂]Cl (charge +3 − 2 = +1, balanced by one Cl⁻).
- Exam-type : Potassium tetrachloridopalladate(II): K₂[PdCl₄] (Pd +2, 4Cl⁻ = −4, charge −2, so two K⁺).
Rule 3: Ambidentate and bridging ligands — For ambidentate ligands the donor atom is shown: nitrito-N (NO₂⁻ bound through N) and nitrito-O (bound through O); thiocyanato-S and thiocyanato-N (or isothiocyanato). Compounds with the same formula but different donor atoms are linkage isomers.
- Easy : [Co(NH₃)₅(NO₂)]²⁺ is pentaamminenitrito-N-cobalt(III) ion.
- Medium : [Co(NH₃)₅(ONO)]²⁺ is pentaamminenitrito-O-cobalt(III) ion.
- Exam-type : Write the name of [Co(NH₃)₅(ONO)]Cl₂: pentaamminenitrito-O-cobalt(III) chloride.
Order: ligands alphabetical (prefixes ignored), metal, oxidation state; -ate for an anionic complex.
Alphabetising by the prefix (di-, tri-) instead of the ligand name itself.
Memory clue: "Ligands first, then the metal with a number" — in the name the metal comes last with its Roman numeral.
Spot the difference
| Point | Cationic complex | Anionic complex |
|---|---|---|
| Metal name | Plain name (cobalt) | Ends in -ate (cobaltate) |
| Example | [Co(NH₃)₆]³⁺ | [Fe(CN)₆]³⁻ |
| Name | hexaamminecobalt(III) | hexacyanidoferrate(III) |
Practice questions
- 1Write the IUPAC name of [Co(NH₃)₆]Cl₃.
- 2What is the name of the ligand H₂O in a complex?
- 3Write the formula of potassium hexacyanidoferrate(III).
- 4Write the name of [Cr(H₂O)₄Cl₂]Cl.
- 5Write the formula of sodium dicyanidoargentate(I).
- 6Why does the metal name end in -ate in K₄[Fe(CN)₆]?
- 7Name [Pt(NH₃)₂Cl₂] and state the oxidation state of Pt.
- 8Write the formula of tetraamminedichloridocobalt(III) chloride and the number of ions it gives in solution.
Show answers (try them yourself first)
- 1Hexaamminecobalt(III) chloride.
- 2Aqua.
- 3K₃[Fe(CN)₆].
- 4Tetraaquadichloridochromium(III) chloride.
- 5Na[Ag(CN)₂].
- 6The complex ion [Fe(CN)₆]⁴⁻ is an anion, so the metal name ends with -ate (ferrate).
- 7Diamminedichloridoplatinum(II); Pt is +2 (ligands NH₃ neutral, 2 Cl⁻ = −2).
- 8[Co(NH₃)₄Cl₂]Cl; it gives two ions in solution: [Co(NH₃)₄Cl₂]⁺ and Cl⁻.
Isomerism in coordination compounds
Link to Std 11 : In Std 11 you saw structural isomers and geometrical isomers in organic chemistry; coordination compounds show the same two families and more.
Your left and right hands have the same parts but cannot be superimposed; some complexes are like that.
Isomers have the same formula but different arrangements. Structural: ionisation, hydrate (solvate), linkage and coordination isomerism. Stereoisomers: geometrical (cis–trans) and optical (non-superimposable mirror images).
In the figure : The two Pt complexes have the same atoms but a different spatial arrangement. cis and trans have different dipole moments, solubility and reactivity; the cis isomer is the anticancer drug cisplatin.
Rule 1: Structural isomerism — Ionisation: [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄ differ in which ion is outside the bracket (AgNO₃ detects Br⁻ in one, BaCl₂ detects SO₄²⁻ in the other). Hydrate: [Cr(H₂O)₆]Cl₃ (violet) and [CrCl(H₂O)₅]Cl₂·H₂O (grey-green) differ in the number of water molecules coordinated. Linkage: [Co(NH₃)₅(NO₂)]²⁺ and [Co(NH₃)₅(ONO)]²⁺. Coordination: [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆] (ligands swap between two metal ions).
- Easy : [Co(NH₃)₅Br]SO₄ gives a white precipitate with BaCl₂; its isomer [Co(NH₃)₅SO₄]Br gives a pale yellow precipitate with AgNO₃.
- Medium : [Cr(H₂O)₆]Cl₃ and [CrCl(H₂O)₅]Cl₂·H₂O are hydrate isomers.
- Exam-type : NO₂⁻ bonded through N (nitro) or through O (nitrito) gives linkage isomers.
Rule 2: Geometrical isomerism — Occurs in square planar MA₂B₂ (cis, trans), octahedral MA₄B₂ (cis, trans) and MA₃B₃ (fac, mer). Tetrahedral complexes show no geometrical isomerism because all positions are equivalent.
- Easy : [Pt(NH₃)₂Cl₂] (square planar): cis and trans, two isomers.
- Medium : [Co(NH₃)₄Cl₂]⁺ (octahedral): cis and trans.
- Exam-type : [Co(NH₃)₃(NO₂)₃]: fac (three identical ligands on one face) and mer (three in a line).
Rule 3: Optical isomerism — A complex is optically active when its mirror image cannot be superimposed (chiral). Octahedral complexes with three didentate ligands, such as [Co(en)₃]³⁺, or cis-[CoCl₂(en)₂]⁺ are chiral; trans isomers are not, because they have a plane of symmetry. Enantiomers rotate plane-polarised light in opposite directions (dextro, laevo).
- Easy : [Co(en)₃]³⁺ has two optical isomers (d and l).
- Medium : cis-[CoCl₂(en)₂]⁺ is optically active, but trans-[CoCl₂(en)₂]⁺ is not.
- Exam-type : Tetrahedral and square planar complexes are generally not optically active because they have symmetry planes.
Check for symmetry: if a mirror plane exists, no optical activity; geometrical isomers exist only where positions are not all equivalent.
Looking for cis–trans isomers in tetrahedral complexes; there are none.
Memory clue: "Cis = Close, Trans = Across" — cis groups sit next to each other, trans groups face each other.
Spot the difference
| Point | cis isomer | trans isomer |
|---|---|---|
| Like ligands | Adjacent (90°) | Opposite (180°) |
| Polarity | Polar | Non-polar (for MA₂B₂) |
| Pt(NH₃)₂Cl₂ | Cisplatin, anticancer | Inactive |
Practice questions
- 1Name one kind of structural isomerism in complexes.
- 2Do tetrahedral complexes show geometrical isomerism?
- 3What type of isomers are [Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄?
- 4How many geometrical isomers does [Pt(NH₃)₂Cl₂] (square planar) have?
- 5Why is [Co(en)₃]³⁺ optically active?
- 6Which reagent distinguishes [Co(NH₃)₅Br]SO₄ from [Co(NH₃)₅SO₄]Br?
- 7Draw (describe) the fac and mer isomers of [Co(NH₃)₃Cl₃].
- 8Explain why trans-[CoCl₂(en)₂]⁺ is optically inactive but the cis isomer is active.
Show answers (try them yourself first)
- 1Ionisation, hydrate, linkage or coordination isomerism.
- 2No, all four positions in a tetrahedron are equivalent.
- 3Ionisation isomers.
- 4Two (cis and trans).
- 5Its mirror image cannot be superimposed on it (three didentate ligands give a chiral octahedron).
- 6AgNO₃ gives a precipitate of AgBr only with the second compound; BaCl₂ gives BaSO₄ only with the first.
- 7fac: the three Cl⁻ occupy the corners of one triangular face of the octahedron; mer: the three Cl⁻ lie in one plane through the metal, two of them trans to each other.
- 8The trans isomer has a plane of symmetry and its mirror image is superimposable; the cis isomer has no such plane and exists as two mirror-image forms.
Valence bond theory: shape and magnetic behaviour
Link to Std 11 : In Std 11 you used hybridisation (sp, sp², sp³) to predict shapes; here d orbitals also take part in hybrid orbitals.
A magnet pulls on iron(III) salts but ignores zinc salts; the number of unpaired d electrons decides, and the ligands change how many there are.
In valence bond theory the metal gives empty hybrid orbitals and each ligand donates a lone pair into one of them. Common cases: d²sp³ or sp³d² = octahedral, sp³ = tetrahedral, dsp² = square planar. Inner orbital (low spin) complexes use (n−1)d orbitals, outer orbital (high spin) complexes use nd orbitals.
In the figure : Strong-field ligands such as CN⁻ push the d electrons together so (n−1)d orbitals are freed for bonding (inner, low spin). Weak-field ligands such as F⁻ leave the d electrons unpaired and use outer nd orbitals (outer, high spin).
Rule 1: Octahedral complexes — Strong field: electrons pair (low spin, d²sp³). Weak field: electrons stay unpaired (high spin, sp³d²). Magnetic moment μ = √[n(n + 2)] BM.
- Easy : [Co(NH₃)₆]³⁺: Co³⁺ is d⁶, NH₃ pairs the electrons (n = 0), d²sp³, diamagnetic.
- Medium : [CoF₆]³⁻: Co³⁺ d⁶, F⁻ weak field (n = 4), sp³d², μ = √24 = 4.9 BM.
- Exam-type : [FeF₆]³⁻: Fe³⁺ d⁵, n = 5, sp³d², μ = √35 = 5.92 BM; [Fe(CN)₆]³⁻ pairs to n = 1, d²sp³, μ = √3 = 1.73 BM.
Rule 2: Tetrahedral and square planar — Four-coordinate Ni²⁺ (d⁸): strong field CN⁻ forces pairing and uses dsp² (square planar, diamagnetic); weak field Cl⁻ gives sp³ (tetrahedral, two unpaired electrons).
- Easy : [Ni(CN)₄]²⁻: dsp², n = 0, square planar, diamagnetic.
- Medium : [NiCl₄]²⁻: sp³, n = 2, tetrahedral, μ = √8 = 2.83 BM.
- Exam-type : [Ni(CO)₄]: Ni is 3d⁸4s² in the atom; CO forces all electrons into 3d¹⁰ and the 4s, 4p give sp³: tetrahedral, diamagnetic.
Rule 3: Limitations of the theory — It explains shape and magnetism but not colour, it does not say why the same metal can be high or low spin with different ligands (no spectrochemical ordering), and it does not give quantitative magnetic data. Crystal field theory handles these.
- Easy : VB theory cannot explain why [Ti(H₂O)₆]³⁺ is purple.
- Medium : VB theory does not predict a numerical energy for the electron's absorption of light.
- Exam-type : Why is crystal field theory needed? It explains colour and the spectrochemical effect.
Strong ligand → electrons pair → inner (low spin). Weak ligand → no pairing → outer (high spin).
Assuming every octahedral complex is d²sp³; weak-field ligands give sp³d².
Memory clue: "Strong makes them pair" — CN⁻, CO, NH₃ pair electrons; F⁻, Cl⁻ do not.
Spot the difference
| Point | Inner orbital complex | Outer orbital complex |
|---|---|---|
| Hybridisation (octahedral) | d²sp³ | sp³d² |
| Ligand field | Strong | Weak |
| Magnetism | Low spin, less paramagnetic | High spin, more paramagnetic |
Practice questions
- 1Name the hybridisation in a square planar complex.
- 2Is [Fe(CN)₆]⁴⁻ paramagnetic or diamagnetic?
- 3What is the shape of a complex with sp³ hybridisation?
- 4Calculate the magnetic moment of [FeF₆]³⁻.
- 5[Co(NH₃)₆]³⁺ is inner orbital and [CoF₆]³⁻ is outer orbital. Explain.
- 6Give the hybridisation and shape of [Ni(CN)₄]²⁻.
- 7Explain why [NiCl₄]²⁻ is paramagnetic but [Ni(CN)₄]²⁻ is diamagnetic.
- 8State two limitations of valence bond theory.
Show answers (try them yourself first)
- 1dsp².
- 2Diamagnetic (Fe²⁺ d⁶, electrons paired by strong CN⁻).
- 3Tetrahedral.
- 4n = 5, μ = √35 = 5.92 BM.
- 5NH₃ is a stronger ligand than F⁻, so in the Co³⁺ complex d⁶ electrons pair up leaving two d orbitals free for d²sp³ (inner), whereas F⁻ leaves them unpaired and uses the outer 4d orbitals (sp³d²).
- 6dsp²; square planar.
- 7Cl⁻ is weak, Ni²⁺ keeps two unpaired electrons and uses sp³ (tetrahedral); CN⁻ is strong, electrons pair, dsp² and no unpaired electrons.
- 8It cannot explain the colour of complexes, and it does not account for the spectrochemical series or give quantitative magnetic properties.
Crystal field theory, colour and spectrochemical series
Link to Std 11 : In Std 11 you learnt that d orbitals have different shapes and orientations; ligands approaching along the axes affect them differently.
A solution of copper(II) sulphate turns from pale blue to deep blue when ammonia is added because the ligands change the splitting and the colour of light absorbed.
In crystal field theory the ligands are treated as point negative charges that repel the metal's d electrons. In an octahedral field the five d orbitals split into a lower t2g set (three, lowered by 0.4Δo) and a higher eg set (two, raised by 0.6Δo), separated by Δo. In a tetrahedral field Δt = (4/9)Δo and the order is reversed.
Crystal field theory, colour and spectrochemical series
In the app you can move this figure with your finger.
Open in the appIn the figure : Whether electrons fill the upper set depends on the pairing energy P. If Δo < P (weak field), electrons spread out (high spin); if Δo > P (strong field), they pair in t2g first (low spin).
Rule 1: Spectrochemical series and spin — Ligands arranged by their ability to split d orbitals: I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO. Weak field (left side) gives high spin; strong field (right) gives low spin. Only d⁴ to d⁷ configurations in octahedral fields can be either.
- Easy : d⁶ with H₂O (weak): t2g⁴eg², n = 4, high spin.
- Medium : d⁶ with CN⁻ (strong): t2g⁶eg⁰, n = 0, low spin.
- Exam-type : d³ and d⁸ always have the same number of unpaired electrons whatever the ligand: d³ has 3 and d⁸ has 2.
Rule 2: Crystal field stabilisation energy — CFSE = [−0.4 × (electrons in t2g) + 0.6 × (electrons in eg)] Δo (pairing energy is extra). A larger CFSE means a more stable complex.
- Easy : d⁶ low spin: CFSE = −0.4 × 6 = -2.4 Δo.
- Medium : d⁶ high spin: CFSE = −0.4 × 4 + 0.6 × 2 = -0.4 Δo.
- Exam-type : d⁴ high spin (t2g³eg¹): CFSE = −1.2 + 0.6 = -0.6 Δo; Δt for Δo = 20000 cm⁻¹ is 4/9 × 20000 = 8888 cm⁻¹ (tetrahedral splitting is small).
Rule 3: Colour of complexes — A d electron absorbs a photon with energy equal to Δo and jumps from t2g to eg. The complex looks the complementary colour of the light absorbed. A stronger field shifts absorption to shorter wavelength. E = hc ÷ λ; Δo per mole = E × 6.022 × 10²³.
- Easy : [Ti(H₂O)₆]³⁺ (d¹) absorbs near 500 nm (green-blue), so it looks purple.
- Medium : For λ = 500 nm: ν̃ = 1 ÷ (500 × 10⁻⁷ cm) = 20000 cm⁻¹, E = hc ÷ λ = 239.0 kJ mol⁻¹ = Δo.
- Exam-type : Complexes with d⁰ or d¹⁰ metal ions (Sc³⁺, Zn²⁺) are colourless; a ligand that gives a larger Δo moves the absorption to shorter wavelengths.
Strong field: big Δo, electrons pair, low spin. Weak field: small Δo, electrons spread, high spin.
Forgetting that tetrahedral splitting is only 4/9 of octahedral, so tetrahedral complexes are almost always high spin.
Memory clue: "I weakest, CO strongest" — learn the ends of the series, and the middle run H₂O < NH₃ < en < CN⁻.
Spot the difference
| Point | Weak field ligand | Strong field ligand |
|---|---|---|
| Example | I⁻, Cl⁻, F⁻, H₂O | NH₃, en, CN⁻, CO |
| Δo vs P | Δo < P | Δo > P |
| Spin | High spin, more unpaired | Low spin, fewer unpaired |
Practice questions
- 1Name the two sets of d orbitals in an octahedral field.
- 2Which is the stronger field ligand, CN⁻ or Cl⁻?
- 3What is the relation between Δt and Δo?
- 4Find the CFSE for a high spin d⁶ octahedral complex.
- 5Find the CFSE for a low spin d⁶ octahedral complex.
- 6Why is [Ti(H₂O)₆]³⁺ coloured but [Sc(H₂O)₆]³⁺ colourless?
- 7[Ti(H₂O)₆]³⁺ absorbs light of 500 nm. Find Δo in kJ mol⁻¹.
- 8Explain why a tetrahedral complex of Co²⁺ is blue while its octahedral aqua complex is pink (hint: Δt = 4/9 Δo).
Show answers (try them yourself first)
- 1t2g (lower) and eg (upper).
- 2CN⁻.
- 3Δt = (4/9) Δo.
- 4t2g⁴eg²: CFSE = −0.4 × 4 + 0.6 × 2 = -0.4 Δo.
- 5t2g⁶: CFSE = −0.4 × 6 = -2.4 Δo.
- 6Ti³⁺ is d¹ and the electron can jump from t2g to eg absorbing visible light; Sc³⁺ is d⁰ and has no electron to excite.
- 7E = hc ÷ λ = (6.626 × 10⁻³⁴ × 3.0 × 10⁸) ÷ (500 × 10⁻⁹) = 3.98 × 10⁻¹⁹ J per ion; × 6.022 × 10²³ = 239.0 kJ mol⁻¹.
- 8Δt is much smaller than Δo, so the tetrahedral complex absorbs light of lower energy (longer wavelength, red) and appears blue, while the octahedral complex absorbs in the blue–green region and appears pink.
Metal carbonyls and importance of coordination compounds
Link to Std 11 : In Std 11 you met the concept of bonding using orbitals; carbonyls show a metal bond in which both donation and back donation occur.
Nickel is purified using a gas, Ni(CO)₄, that forms at low temperature and splits at high temperature, giving back pure nickel.
Metal carbonyls are homoleptic complexes in which CO is the ligand: Ni(CO)₄ (tetrahedral), Fe(CO)₅ (trigonal bipyramidal), Cr(CO)₆ (octahedral). The metal is in oxidation state zero. Bonding is synergic: the carbon lone pair is donated to the metal (σ bond) and filled metal d orbitals donate back to the empty π* orbitals of CO (π bond).
In the figure : The σ donation puts extra electron density on the metal and the π back donation takes it away again, so both bonds become stronger.
Rule 1: Carbonyls — In Ni(CO)₄ the metal is Ni(0) with 3d¹⁰ after the rearrangement and sp³ hybridisation; in Fe(CO)₅ it is dsp³; in Cr(CO)₆ it is d²sp³. Back bonding strengthens the M–C bond and lengthens (weakens) the C–O bond.
- Easy : Ni(CO)₄ is tetrahedral and diamagnetic.
- Medium : Fe(CO)₅ is trigonal bipyramidal with Fe in oxidation state 0.
- Exam-type : Why is the M–C bond in a carbonyl partly a double bond? The π back donation from the metal to the CO π* orbital adds a π component to the σ bond.
Rule 2: Importance in metallurgy and analysis — Complex formation is used to extract metals: silver and gold dissolve in NaCN as [Ag(CN)₂]⁻ and [Au(CN)₂]⁻ and are recovered with zinc. Ni is purified by the Mond process (Ni + 4CO → Ni(CO)₄ → Ni + 4CO). EDTA titrations measure hardness of water (Ca²⁺, Mg²⁺). Qualitative tests: Ni²⁺ with dimethylglyoxime gives a bright red precipitate; Fe³⁺ with SCN⁻ gives a blood-red colour; Cu²⁺ with excess NH₃ gives deep blue [Cu(NH₃)₄]²⁺.
- Easy : Gold is extracted by forming dicyanidoaurate(I).
- Medium : Photography: undeveloped AgBr dissolves in hypo as [Ag(S₂O₃)₂]³⁻.
- Exam-type : Ni + 4CO → Ni(CO)₄ at about 330–350 K and Ni(CO)₄ → Ni + 4CO at about 450–470 K: this temperature swing purifies nickel.
Rule 3: Biological and medical importance — Chlorophyll is a magnesium complex of a porphyrin ring; haemoglobin contains iron(II) in a porphyrin; vitamin B₁₂ (cyanocobalamin) contains cobalt. Cisplatin, cis-[Pt(NH₃)₂Cl₂], is used against some cancers. EDTA is used to remove lead poisoning by forming a stable chelate. Coordination compounds are also used as catalysts (Wilkinson's catalyst for hydrogenation) and in electroplating baths.
- Easy : Haemoglobin carries oxygen using an Fe²⁺ porphyrin complex.
- Medium : Chlorophyll holds Mg²⁺, vitamin B₁₂ holds Co.
- Exam-type : Why is EDTA useful in treating lead poisoning? It forms a very stable, soluble chelate with Pb²⁺ that is excreted.
In carbonyls CO is both a σ donor and a π acceptor; this synergic bonding makes the M–C bond strong.
Thinking that the oxidation state of the metal in Ni(CO)₄ is +4; CO is neutral so Ni is 0.
Memory clue: "CO gives and takes" — CO donates a lone pair to the metal and accepts electrons back into π*.
Spot the difference
| Point | σ donation | π back donation |
|---|---|---|
| Direction | Ligand → metal | Metal → ligand |
| Orbitals | Filled C lone pair → empty metal orbital | Filled metal d → empty π* of CO |
| Effect on C–O | Little | Weakens it |
Practice questions
- 1Give the oxidation state of Ni in Ni(CO)₄.
- 2Name the central metal in chlorophyll.
- 3What is the shape of Ni(CO)₄?
- 4Explain synergic bonding in metal carbonyls.
- 5How is nickel purified using Ni(CO)₄?
- 6Name the complex used in treatment of cancer.
- 7Why does the C–O bond length increase in metal carbonyls?
- 8Explain, with an example, how coordination compounds are used in the extraction of gold.
Show answers (try them yourself first)
- 10 (zero).
- 2Magnesium.
- 3Tetrahedral.
- 4CO donates a lone pair (σ bond) to the metal and the metal donates electrons from filled d orbitals into empty π* orbitals of CO; each effect strengthens the other.
- 5Impure Ni reacts with CO at a low temperature to form volatile Ni(CO)₄, which is decomposed at a higher temperature to give pure Ni.
- 6Cisplatin, cis-[Pt(NH₃)₂Cl₂].
- 7Electrons enter the π* (antibonding) orbital of CO through back donation, which weakens and lengthens the C–O bond.
- 8Gold ore is treated with NaCN in air to form [Au(CN)₂]⁻ (a soluble complex), and gold is recovered by adding zinc: 2[Au(CN)₂]⁻ + Zn → 2Au + [Zn(CN)₄]²⁻.