Chapter 14: Coordination Compounds

Coordination Compounds Introduction

Key Features

• Transition metals form complexes by coordinate bonds.

• Central metal atom binds ions or neutral molecules donating lone pairs.

• Examples: Chlorophyll (Mg), Haemoglobin (Fe), Vitamin B12 (Co).

Solved Examples

• Ex 1: Identify the central metal in haemoglobin.
  Answer: Iron (Fe).

Practice Set

• Q1: What type of bond links ligands to the central metal?
  A. Ionic
  B. Covalent
  C. Coordinate
  D. Metallic

• Q2: Which metal is central in chlorophyll?
  A. Iron
  B. Magnesium
  C. Cobalt
  D. Copper

• Q3: Coordination compounds involve:
  A. Only neutral molecules
  B. Only ions
  C. Both ions and neutral molecules
  D. Only metals

Answer Key

• Q1: C - Coordinate bonds involve donation of lone pairs from ligands to metal.

• Q2: B - Magnesium is central in chlorophyll.

• Q3: C - Both ions and neutral molecules act as ligands.

Memory Aids

• Coordination = Central metal + ligands via coordinate bonds.

• Chlorophyll = Mg; Haemoglobin = Fe; Vitamin B12 = Co.

Werner Theory

Main Postulates

• Metals show two valencies: primary (oxidation state) and secondary (coordination number).

• Primary valency: ionisable, satisfied by negative ions, non-directional.

• Secondary valency: non-ionisable, satisfied by ligands, directional, defines geometry.

• Coordination number determines spatial arrangement (coordination polyhedra).

• Common geometries: octahedral, tetrahedral, square planar.

• Examples: [Co(NH3)6]3+ (octahedral), [Ni(CO)4] (tetrahedral), [PtCl4]2− (square planar).

Solved Examples

• Ex 1: Determine coordination number of Co in [CoCl(NH3)5]2+.
  Answer: 6 (5 NH3 + 1 Cl).

Practice Set

• Q1: Primary valency corresponds to:
  A. Coordination number
  B. Oxidation state
  C. Number of ligands
  D. Geometry

• Q2: Which geometry is common for [Ni(CO)4] complex?
  A. Octahedral
  B. Tetrahedral
  C. Square planar
  D. Linear

• Q3: Secondary valency is satisfied by:
  A. Negative ions only
  B. Neutral molecules or ligands
  C. Positive ions
  D. Metals

Answer Key

• Q1: B - Primary valency equals oxidation state.

• Q2: B - [Ni(CO)4] is tetrahedral.

• Q3: B - Secondary valency satisfied by ligands (neutral or ions).

Memory Aids

• Primary valency = oxidation state, ionisable.

• Secondary valency = coordination number, defines shape.

• Octahedral = 6, Tetrahedral = 4, Square planar = 4.

Difference Between Double Salt and Complex

Key Differences

• Both formed by combination of stable compounds in fixed ratio.

• Double salts dissociate completely into ions in solution.

• Complexes do not dissociate into ions; ligands remain bonded.

• Examples of double salts: Carnallite (KCl·MgCl2·6H2O), Mohr's salt (FeSO4·(NH4)2SO4·6H2O), Potash alum (KAl(SO4)2·12H2O).

Solved Examples

• Ex 1: Identify if Mohr's salt is a double salt or complex.
  Answer: Double salt (dissociates into ions).

Practice Set

• Q1: Double salts dissociate into:
  A. Molecules
  B. Ions
  C. Complexes
  D. Atoms

• Q2: Which is a double salt?
  A. [Co(NH3)6]Cl3
  B. KAl(SO4)2·12H2O
  C. [Ni(CO)4]
  D. [PtCl4]2−

• Q3: Complex compounds differ from double salts because:
  A. They dissociate into ions
  B. Ligands remain bonded to metal
  C. They are ionic
  D. They are mixtures

Answer Key

• Q1: B - Double salts dissociate into ions.

• Q2: B - Potash alum is a double salt.

• Q3: B - Complexes have ligands bonded, do not dissociate.

Memory Aids

• Double salts = dissociate; complexes = stable coordination entities.

• Remember: Mohr's salt and alum are double salts.

Coordination Entity/Coordination Sphere

Definition

• Central metal atom/ion and directly attached ligands enclosed in square brackets.

• Collectively called coordination sphere.

• Coordination sphere is non-ionisable.

Solved Examples

• Ex 1: In K4[Fe(CN)6], coordination sphere is [Fe(CN)6]4−.

Practice Set

• Q1: What is enclosed in the coordination sphere?
  A) Only ligands
  B) Only metal ion
  C) Metal ion and ligands
  D) Counter ions

• Q2: Which of the following is non-ionisable?
  A) Coordination sphere
  B) Counter ions
  C) Both A and B
  D) None

• Q3: In K4[Fe(CN)6], what is the charge on coordination sphere?
  A) 4+
  B) 4−
  C) 2+
  D) 2−

Answer Key

• Q1: C - Coordination sphere includes metal ion and ligands.

• Q2: A - Coordination sphere is non-ionisable; counter ions are ionisable.

• Q3: B - Coordination sphere [Fe(CN)6] carries 4− charge.

Memory Aids

• Coordination sphere = [Metal + Ligands]

• Counter ions always outside brackets

Counter Ions

Definition

• Ionisable groups written outside the coordination sphere brackets.

• Balance the charge of coordination entity.

Solved Examples

• Ex 1: In K4[Fe(CN)6], K+ ions are counter ions.

Practice Set

• Q1: Where are counter ions written?
  A) Inside brackets
  B) Outside brackets
  C) Both inside and outside
  D) Not written

• Q2: Counter ions are
  A) Non-ionisable
  B) Ionisable
  C) Neutral molecules
  D) Ligands

• Q3: In [Co(NH3)6]Cl3, what are the counter ions?
  A) NH3
  B) Cl−
  C) Co3+
  D) None

Answer Key

• Q1: B - Counter ions are outside the brackets.

• Q2: B - Counter ions are ionisable.

• Q3: B - Cl− ions are counter ions.

Memory Aids

• Counter ions = charge balancers outside []

Central Metal Atom/Ion

Definition

• Metal atom/ion bonded to ligands by coordinate bonds in definite geometry.

• Acts as Lewis acid (electron pair acceptor).

Solved Examples

• Ex 1: In [NiCl2(H2O)4], central metal ion is Ni2+.

• Ex 2: In [CoCl(NH3)5]2+, central metal ion is Co3+.

Practice Set

• Q1: Central metal ion is a
  A) Lewis base
  B) Lewis acid
  C) Ligand
  D) Counter ion

• Q2: In [Fe(CN)6]3−, central metal ion is
  A) Fe2+
  B) Fe3+
  C) CN−
  D) Fe

• Q3: Which of the following is NOT a central metal ion?
  A) Ni2+
  B) Co3+
  C) NH3
  D) Fe3+

Answer Key

• Q1: B - Central metal ion is Lewis acid.

• Q2: B - Fe3+ is central metal ion.

• Q3: C - NH3 is a ligand, not metal ion.

Memory Aids

• Metal ion = Lewis acid, accepts electron pairs.

Ligands

Definition and Types

• Species donating lone pair to metal via coordinate bond.

• Types based on donor atoms:

• Unidentate: binds through 1 donor atom (e.g., Cl−, H2O, NH3).

• Didentate: binds through 2 donor atoms (e.g., ethylenediamine H2NCH2CH2NH2, oxalate C2O42−).

• Polydentate: binds through several donor atoms (e.g., EDTA4−).

• Chelate ligand: didentate or polydentate forming 5 or 6 membered ring with metal.

• Ambidentate: unidentate ligand coordinating through two different atoms (e.g., NO2−, SCN−).

Solved Examples

• Ex 1: Cl− is unidentate ligand.

• Ex 2: Ethylenediamine (en) is didentate ligand.

• Ex 3: EDTA4− is polydentate ligand.

• Ex 4: NO2− is ambidentate ligand.

Practice Set

• Q1: Which ligand binds through one donor atom?
  A) EDTA4−
  B) Cl−
  C) Ethylenediamine
  D) Oxalate

• Q2: Chelate ligands form
  A) Single bonds
  B) Five or six membered rings
  C) Ionic bonds
  D) No rings

• Q3: Ambidentate ligand example is
  A) NH3
  B) NO2−
  C) EDTA4−
  D) Cl−

Answer Key

• Q1: B - Cl− is unidentate.

• Q2: B - Chelate ligands form 5 or 6 membered rings.

• Q3: B - NO2− is ambidentate.

Memory Aids

• Unidentate = 1 donor atom

• Didentate = 2 donor atoms, forms rings

• Polydentate = many donor atoms (EDTA)

• Ambidentate = can bind via two different atoms

Coordination Number

Definition

• Total number of coordinate bonds between central metal and ligands.

• Depends on ligand denticity and number of ligands.

Solved Examples

• Ex 1: In [PtCl6]2−, coordination number = 6 (6 unidentate Cl−).

• Ex 2: In [Ni(NH3)4]2+, coordination number = 4 (4 unidentate NH3).

• Ex 3: In [Fe(C2O4)3]3−, coordination number = 6 (3 didentate oxalate ligands × 2).

• Ex 4: In [Co(en)3]3+, coordination number = 6 (3 didentate ethylenediamine ligands × 2).

Practice Set

• Q1: Coordination number is the number of
  A) Ligands
  B) Coordinate bonds
  C) Metal ions
  D) Counter ions

• Q2: Coordination number of [Ni(NH3)6]2+ is
  A) 4
  B) 6
  C) 2
  D) 3

• Q3: Coordination number of [Fe(C2O4)3]3− is
  A) 3
  B) 6
  C) 9
  D) 12

Answer Key

• Q1: B - Number of coordinate bonds.

• Q2: B - Six NH3 ligands, coordination number 6.

• Q3: B - Three didentate ligands × 2 = 6.

Memory Aids

• Coordination number = total donor atoms bonded to metal.

Oxidation Number

Definition

• Charge on central metal if ligands removed with shared electron pairs.

• Represented by Roman numeral in parentheses after metal name.

Solved Examples

• Ex 1: In [Cu(CN)4]3−, oxidation number of Cu is +1 (Cu(I)).

Practice Set

• Q1: Oxidation number is
  A) Number of ligands
  B) Charge on metal after removing ligands
  C) Coordination number
  D) Number of electrons donated by ligands

• Q2: Oxidation number of Fe in [Fe(CN)6]3− is
  A) +2
  B) +3
  C) 0
  D) −3

• Q3: How is oxidation number represented?
  A) Arabic numeral
  B) Roman numeral in parentheses
  C) Chemical symbol
  D) Charge outside brackets

Answer Key

• Q1: B - Charge on metal after removing ligands.

• Q2: B - Fe is +3 in [Fe(CN)6]3−.

• Q3: B - Roman numeral in parentheses.

Memory Aids

• Oxidation number = metal charge after ligand removal

• Represented as Metal(II), Metal(III), etc.

Homoleptic and Heteroleptic Complexes

Definitions

• Homoleptic: Metal bound to only one kind of ligand (e.g., [Co(NH3)6]3+).

• Heteroleptic: Metal bound to more than one kind of ligand (e.g., [Co(NH3)4Cl2]+).

Solved Examples

• Ex 1: [Co(NH3)6]3+ is homoleptic.

• Ex 2: [Co(NH3)4Cl2]+ is heteroleptic.

Practice Set

• Q1: Homoleptic complex has
  A) One type of ligand
  B) Two types of ligands
  C) No ligands
  D) Only counter ions

• Q2: Which is heteroleptic?
  A) [Co(NH3)6]3+
  B) [Co(NH3)4Cl2]+
  C) [Fe(CN)6]3−
  D) [Ni(CO)4]

• Q3: Homoleptic complexes are generally
  A) Less stable
  B) More stable
  C) Unstable
  D) Reactive only

Answer Key

• Q1: A - One kind of ligand.

• Q2: B - Contains NH3 and Cl− ligands.

• Q3: B - Usually more stable due to uniform ligands.

Memory Aids

• Homo = same ligands

• Hetero = different ligands

Effective Atomic Number Rule (EAN Rule)

Definition

• EAN = Atomic number of metal − electrons lost + electrons gained from ligands.

• Used to predict stability of complexes.

Solved Examples

• Ex 1: Complexes following EAN rule: [Pd(NH3)6]4+, [Pt(NH3)6]4+, [Cu(CN)4]3−.

• Ex 2: Complexes not following EAN rule but stable: [PdCl4]2− (EAN=52), [Ag(NH3)2]+ (EAN=50), [Ni(CN)4]2− (EAN=34), [Ni(en)3]2+ (EAN=38).

Practice Set

• Q1: EAN is calculated by
  A) Atomic number + electrons lost − electrons gained
  B) Atomic number − electrons lost + electrons gained
  C) Atomic number + electrons lost + electrons gained
  D) Atomic number − electrons lost − electrons gained

• Q2: Which complex follows EAN rule?
  A) [PdCl4]2−
  B) [Pd(NH3)6]4+
  C) [Ag(NH3)2]+
  D) [Ni(CN)4]2−

• Q3: EAN rule helps to predict
  A) Color
  B) Stability
  C) Solubility
  D) Magnetic properties

Answer Key

• Q1: B - EAN = Atomic number − electrons lost + electrons gained.

• Q2: B - [Pd(NH3)6]4+ follows EAN rule.

• Q3: B - EAN predicts complex stability.

Memory Aids

• EAN = Metal atomic no. − lost e− + gained e−

• Stable complexes often have EAN = noble gas configuration

Nomenclature of Coordination Compounds

Rules for Writing Formula

• Write cation formula first, then anion.

• Coordination sphere enclosed in square brackets; polyatomic ligands in parentheses.

• List ligands alphabetically, ignoring charge.

• Balance total positive and negative charges for neutrality.

Rules for Naming Coordination Compounds

• Name cation first, then anion.

• Within coordination sphere, name ligands alphabetically, then metal, then oxidation state in Roman numerals.

• Start name with lowercase; write complex part as one word.

Naming of Ligands

• Anionic ligands end with -ido (e.g., Cl-: Chlorido).

• Neutral ligands retain names (e.g., NH3: Ammine, H2O: aqua).

• Cationic ligands end with -ium (e.g., NO2+: Nitronium).

• Some ligands use abbreviations in parentheses (e.g., ethylenediamine: en).

• Ambidentate ligands named by donor atom or ligand name (e.g., SCN-: Thiocyanato-S or Thiocyanato-N).

• Use prefixes di-, tri-, tetra-, etc. for ligand numbers; if ligand name has prefix, use bis-, tris-, tetrakis-, etc. with parentheses.

• For anionic coordination spheres, metal name ends with -ate; for cationic/neutral, metal name unchanged.

Solved Examples

• Ex 1: Name [CoCl(NO2)(NH3)4]NO3
  Step 1: Ligands alphabetically: Ammine, Chlorido, Nitrito-N
  Step 2: Metal: Cobalt, Oxidation state III
  Step 3: Name: Tetraamminechloridonitrito-N-cobalt(III) nitrate

• Ex 2: Write formula for Potassium trioxalatoferrate(III)
  Step 1: Ligand: Oxalato (C2O42-)
  Step 2: Metal: Fe, oxidation state III
  Step 3: Formula: K3[Fe(C2O4)3]

Practice Set

• Q1: In naming coordination compounds, which is named first?
  A. Anion
  B. Cation
  C. Ligands
  D. Metal

• Q2: What is the correct name for [Fe(CO)5]?
  A. Pentacarbonyliron(0)
  B. Pentacarbonyliron(II)
  C. Iron pentacarbonyl
  D. Carbonyliron(V)

• Q3: What is the formula of Potassium tetrachloridomercurate(II)?
  A. K[HgCl4]
  B. K2[HgCl4]
  C. K3[HgCl4]
  D. K[HgCl3]

Answer Key

• Q1: B. Cation is named first as per nomenclature rules.

• Q2: A. Neutral complex named Pentacarbonyliron with oxidation state 0.

• Q3: B. Potassium tetrachloridomercurate(II) formula is K2[HgCl4].

Memory Aids

• Ligand naming ends: anionic -ido, neutral unchanged, cationic -ium.

• Use bis-, tris-, tetrakis- if ligand name has prefixes.

• Metal name ends with -ate if coordination sphere is anionic.

• Always write cation first in formula and name.

Examples of Nomenclature

Explanation

• [CoCl(NO2)(NH3)4]NO3: Tetraamminechloridonitrito-N-cobalt(III) nitrate

• [CoCl(ONO)(en)2]+: Chloridobis(ethane-1,2-diamine)nitrito-O-cobalt(III) ion

• K3[Al(C2O4)3]: Potassium trioxalatoaluminate(III)

• Fe4[Fe(CN)6]3: Ferric hexacyanidoferrate(II)

• [PtCl2(NH3)4][PtCl4]: Tetraamminedichloridoplatinum(IV) tetrachloridoplatinate(II)

• [Pt(NH3)4][CuCl4]: Tetraammineplatinum(II) tetrachloridocuperate(II)

• [Cr(NH3)6][Co(C2O4)3]: Hexaamminechromium(III) trioxalatocobaltate(III)

• [Ni(dmg)2]: bis(dimethylglyoximato)nickel(II)

• K2[HgCl4]: Potassium tetrachloridomercurate(II)

Solved Examples

• Ex 1: Name Fe4[Fe(CN)6]3
  Step 1: Ligand: Cyanido
  Step 2: Metal: Ferric (Fe3+) and Ferrous (Fe2+)
  Step 3: Name: Ferric hexacyanidoferrate(II)

• Ex 2: Write formula for Tetraammineplatinum(II) tetrachloridocuperate(II)
  Step 1: Cation: [Pt(NH3)4]2+
  Step 2: Anion: [CuCl4]2-
  Step 3: Formula: [Pt(NH3)4][CuCl4]

Practice Set

• Q1: What is the correct name for [PtCl2(NH3)4][PtCl4]?
  A. Tetraamminedichloridoplatinum(IV) tetrachloridoplatinate(II)
  B. Tetrachloridoplatinum(II) tetraammineplatinate(IV)
  C. Tetraamminedichloridoplatinum(II) tetrachloridoplatinate(IV)
  D. Tetraamminedichloridoplatinum(IV) tetrachloridoplatinate(IV)

• Q2: Which ligand abbreviation represents ethylenediamine?
  A. en
  B. ed
  C. eda
  D. eth

• Q3: What is the oxidation state of cobalt in [CoCl(ONO)(en)2]+?
  A. +1
  B. +2
  C. +3
  D. 0

Answer Key

• Q1: A. Correct name matches the given complex with oxidation states.

• Q2: A. 'en' is the standard abbreviation for ethylenediamine.

• Q3: C. Cobalt is in +3 oxidation state in the complex.

Memory Aids

• Abbreviations: ethylenediamine = en.

• Oxidation state found by balancing charges in complex ion.

• Complex names combine ligand names alphabetically, metal, and oxidation state.

Isomerism in Coordination Compounds

Types of Isomerism

• Structural Isomerism: Different connectivity or composition of ligands.

  • Ionisation Isomerism: Different ions released in solution.

  • Hydrate/Solvate Isomerism: Variation in water molecules inside/outside coordination sphere.

  • Linkage Isomerism: Ligands bind through different atoms (e.g., NO2- via N or O).

  • Coordination Isomerism: Ligand exchange between cationic and anionic complexes.

• Stereoisomerism: Same ligands, different spatial arrangement.

  • Geometrical Isomerism: Different ligand positions (cis/trans, fac/mer).

  • Optical Isomerism: Non-superimposable mirror images (enantiomers) that rotate plane-polarized light.

Geometrical Isomerism Details

• Square Planar Complexes:

  • Cis-isomer: Similar ligands adjacent.

  • Trans-isomer: Similar ligands opposite.

• Octahedral Complexes:

  • Facial (fac) isomer: Three identical ligands on one face.

  • Meridional (mer) isomer: Three identical ligands in a plane through the center.

Optical Isomerism

• Enantiomers are mirror images, non-superimposable.

• Rotate plane-polarized light in opposite directions: Dextro (+) and Laevo (−).

• Optically inactive forms (meso, racemic) do not rotate light due to internal or external compensation.

Solved Examples

• Ex 1: Identify cis and trans isomers in [Pt(NH3)2Cl2].

  • Cis: NH3 ligands adjacent; trans: NH3 ligands opposite.

• Ex 2: Number of optical isomers for [Pt(Py)(NH3)(NO2)(Cl)(Br)(I)].

  • 15 geometrical isomers × 2 (d and l forms) = 30 optical isomers.

Practice Set

• Q1: Which is an example of linkage isomerism?

  • A) [Co(NH3)6]Cl3

  • B) [Co(NO2)(NH3)5]Cl2

  • C) [Co(NH3)4Cl2]Cl

  • D) [Co(H2O)6]Cl3

• Q2: In octahedral complexes, the mer isomer has three identical ligands arranged:

  • A) On one face of the octahedron

  • B) In a plane passing through the center

  • C) Adjacent to each other

  • D) Opposite each other

• Q3: Which statement about optical isomers is correct?

  • A) They are superimposable mirror images.

  • B) They rotate plane-polarized light in the same direction.

  • C) They are non-superimposable mirror images.

  • D) They do not affect plane-polarized light.

Answer Key

• Q1: B - Linkage isomerism occurs when NO2- binds through N or O atoms.

• Q2: B - Mer isomer has three identical ligands in a plane through the center.

• Q3: C - Optical isomers are non-superimposable mirror images that rotate light differently.

Memory Aids

• Linkage isomerism: Think "link" changes in ligand attachment atom.

• Geometrical isomers: Cis = same side; Trans = opposite side.

• Fac and Mer: Fac = face; Mer = meridian (plane).

• Optical isomers: Like left and right hands, non-superimposable.

Classification of Isomerism

Structural Isomerism

• Ionisation Isomerism: Different ions released in solution.

• Hydrate/Solvate Isomerism: Variation in water molecules inside/outside coordination sphere.

• Linkage Isomerism: Ligands bind through different atoms.

• Coordination Isomerism: Ligand exchange between cationic and anionic complexes.

Stereoisomerism

• Geometrical Isomerism: Different spatial arrangement of ligands (cis/trans, fac/mer).

• Optical Isomerism: Non-superimposable mirror images (enantiomers) that rotate plane-polarized light.

Solved Examples

• Ex 1: Identify the type of isomerism in [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2.

  • Linkage isomerism due to NO2- binding through N or O.

• Ex 2: Differentiate fac and mer isomers in [CoCl3(NH3)3].

  • Fac: Three Cl ligands on one face; Mer: three Cl ligands in a plane.

Practice Set

• Q1: Which is an example of coordination isomerism?

  • A) [Co(NH3)6] [Cr(CN)6]

  • B) [Co(NH3)5NO2]Cl2

  • C) [Co(H2O)6]Cl3

  • D) [Co(NH3)4Cl2]Cl

• Q2: Which is true for hydrate isomerism?

  • A) Different ions in solution

  • B) Different number of water molecules inside/outside coordination sphere

  • C) Ligands attached through different atoms

  • D) Ligand exchange between complexes

• Q3: How many geometrical isomers are possible for an octahedral complex with six different monodentate ligands?

  • A) 3

  • B) 6

  • C) 15

  • D) 30

Answer Key

• Q1: A - Coordination isomerism involves ligand exchange between cationic and anionic complexes.

• Q2: B - Hydrate isomerism differs in water molecules inside or outside the coordination sphere.

• Q3: C - 15 geometrical isomers possible for six different monodentate ligands in octahedral complex.

Memory Aids

• Ionisation isomers: Different ions in solution.

• Hydrate isomers: Water molecules location changes.

• Linkage isomers: Ligand binds via different atoms.

• Coordination isomers: Ligand swap between complexes.

• Geometrical isomers: Cis/trans, fac/mer arrangements.

• Optical isomers: Mirror images, rotate light.

Bonding in Coordination Compounds

Valence Bond Theory

• Metal ion provides empty orbitals equal to coordination number.

• Orbitals involved: inner (n-1)d, ns, np or outer nd, ns, np depending on ligand strength.

• Vacant orbitals hybridize to form equivalent orbitals with definite geometry: octahedral, tetrahedral, square planar.

• Coordination number (C. No.) determines hybridisation and geometry.

• Examples: C. No. 2 → sp (linear), 4 → sp3 (tetrahedral), 6 → d2sp3 (octahedral).

• Inner orbital complexes: use (n-1)d orbitals, form low spin complexes (few unpaired electrons).

• Examples: [Co(NH3)6]3+, [Co(CN)6]4-.

• Outer orbital complexes: use nd orbitals, form high spin complexes (many unpaired electrons).

• Examples: [MnF6]3-, [CoF6]3-.

Magnetic Properties

• Low spin complexes are diamagnetic; high spin complexes are paramagnetic.

• Paramagnetism depends on number of unpaired electrons (n).

• Magnetic moment, μ = √(n(n+2)) Bohr Magneton (B.M.).

Limitations of Valence Bond Theory

• Based on many assumptions.

• Cannot distinguish between weak and strong field ligands.

• No quantitative interpretation of magnetic data.

• Does not explain thermodynamic or kinetic stability quantitatively.

• Fails to explain relative energies of different shapes.

• Cannot explain color and spectral properties.

• Fails to predict tetrahedral and square planar structures.

Solved Examples

• Ex 1: Determine hybridisation and geometry of [Ni(CN)4]2-.

• Solution: Coordination number = 4; CN- is strong field ligand → dsp2 hybridisation → square planar geometry.

Practice Set

• Q1: Which hybridisation corresponds to octahedral geometry?

• A) sp3 B) d2sp3 C) dsp2 D) sp

• Q2: Inner orbital complexes are generally:

• A) High spin B) Low spin C) Paramagnetic D) Outer orbital

• Q3: Magnetic moment of a complex with 3 unpaired electrons is:

• A) 3.87 B.M. B) 5.92 B.M. C) 4.90 B.M. D) 2.83 B.M.

Answer Key

• Q1: B) d2sp3 - Octahedral complexes have d2sp3 hybridisation.

• Q2: B) Low spin - Inner orbital complexes form low spin complexes.

• Q3: A) 3.87 B.M. - μ = √(3(3+2)) = √15 ≈ 3.87 B.M.

Memory Aids

• Strong field ligands → inner orbital → low spin.

• Weak field ligands → outer orbital → high spin.

• Octahedral → d2sp3, Tetrahedral → sp3, Square planar → dsp2.

Crystal Field Theory

Basic Concepts

• Metal-ligand bond is ionic; interaction is electrostatic.

• d-orbitals split into two groups due to ligand field: eg (higher energy) and t2g (lower energy) in octahedral complexes.

• Energy difference between these sets is ligand field splitting parameter, ∆0.

• Energy levels: t2g at -0.4∆0, eg at +0.6∆0, total energy zero.

• ∆0 depends on ligand field strength and metal ion charge.

• Spectrochemical series (increasing field strength): I- < Br- < SCN- < Cl- < S2- < F- < OH- < C2O42- < H2O < NCS- < EDTA4- < NH3 < en < NO2- < CN- < CO.

Electron Configuration and Spin States

• If ∆0 > Pairing energy (P), electrons pair in lower orbitals → low spin complex.

• If ∆0 < P, electrons occupy higher orbitals before pairing → high spin complex.

• Tetrahedral splitting ∆t = (4/9) ∆0, usually small → mostly high spin complexes.

• Tetrahedral complexes lack center of symmetry; no 'g' subscript used.

Colour in Coordination Compounds

• Colour arises from d-d electronic transitions: electron jumps from t2g to eg level.

• Absorbed light wavelength corresponds to ∆0; transmitted light is complementary colour.

• Without ligands, no splitting → colourless complexes.

• Examples: [Ti(H2O)6]3+ violet, [Cu(H2O)4]2+ blue.

• Colour absorbed and transmitted are complementary (e.g., yellow absorbed → purple seen).

Limitations of Crystal Field Theory

• Does not consider metal-ligand covalent bonding (π-bonding).

• Cannot explain relative ligand strengths fully (e.g., H2O stronger than OH-).

• No account of partial covalent character of bonds.

Calculation of Crystal Field Stabilization Energy (CFSE)

• CFSE = (-0.4 x + 0.6 y) ∆0, where x = electrons in t2g, y = electrons in eg.

• Strong field ligands → high ∆0 → low spin complexes.

• Weak field ligands → low ∆0 → high spin complexes.

• Large ∆0 absorbs higher energy (shorter wavelength) light.

• All octahedral complexes of Ni2+, Cu+, Cu2+, Zn2+, Co2+ are outer orbital complexes.

Solved Examples

• Ex 1: Calculate CFSE for d6 low spin octahedral complex.

• Solution: x=6 (t2g), y=0 (eg), CFSE = (-0.4×6 + 0.6×0)∆0 = -2.4∆0.

Practice Set

• Q1: In octahedral complexes, which orbitals form the higher energy set?

• A) t2g B) eg C) dxy, dxz, dyz D) All d orbitals equally

• Q2: Which ligand is a strong field ligand?

• A) I- B) H2O C) CN- D) F-

• Q3: For tetrahedral complexes, the crystal field splitting energy ∆t is related to octahedral ∆0 as:

• A) ∆t = ∆0 B) ∆t = 9/4 ∆0 C) ∆t = 4/9 ∆0 D) ∆t = 1/2 ∆0

Answer Key

• Q1: B) eg - eg orbitals have higher energy in octahedral field.

• Q2: C) CN- - CN- is a strong field ligand.

• Q3: C) ∆t = 4/9 ∆0 - Tetrahedral splitting is smaller than octahedral.

Memory Aids

• Spectrochemical series orders ligands by increasing ∆0.

• Low spin if ∆0 > pairing energy; high spin if ∆0 < pairing energy.

• Tetrahedral complexes usually high spin due to small ∆t.

• Colour seen is complementary to colour absorbed (use colour wheel).

Bonding in Metal Carbonyls

Overview

• Transition metal carbonyls are organometallic compounds with only CO ligands (homoleptic).

• Metal–carbon bond has both sigma (σ) and pi (π) character.

• Bonding involves:

  • σ-bond: Lone pair on carbon (CO) donates to empty metal orbital.

  • π-back bonding: Metal d-orbitals donate electrons to empty π* orbitals of CO.

• Back bonding strengthens metal–ligand interaction (synergic effect).

Examples of Metal Carbonyls

• Mononuclear: V(CO)5, Cr(CO)6, Mo(CO)6, W(CO)6, Ni(CO)4, Fe(CO)5

• Polynuclear: Co2(CO)8, Co4(CO)12, Mn2(CO)10

Structures

• Ni(CO)4: Tetrahedral
  

  

• Fe(CO)5: Trigonal bipyramidal
  

  

• Cr(CO)6: Octahedral
  

  

• Mn2(CO)10: Two square pyramidal units joined by Mn–Mn bond
  

  

• Co2(CO)8: Co–Co bond bridged by two CO ligands
  

  

Solved Examples

Ex 1

• Identify the geometry of Fe(CO)5.

• Answer: Trigonal bipyramidal due to 5 CO ligands arranged around Fe.

Practice Set

• Q1: What type of bonding is involved in metal carbonyls?

  • A. Only σ-bonding

  • B. Only π-bonding

  • C. Both σ- and π-bonding

  • D. Ionic bonding

• Q2: Which of the following metal carbonyls has octahedral geometry?

  • A. Ni(CO)4

  • B. Fe(CO)5

  • C. Cr(CO)6

  • D. Mn2(CO)10

• Q3: In back bonding, electrons are donated from

  • A. CO to metal

  • B. Metal d-orbitals to CO π* orbitals

  • C. Metal s-orbitals to CO σ orbitals

  • D. CO π orbitals to metal d-orbitals

Answer Key

• Q1: C - Metal carbonyls have both σ-donation from CO to metal and π-back bonding from metal to CO.

• Q2: C - Cr(CO)6 has octahedral geometry with 6 CO ligands.

• Q3: B - Back bonding involves metal d-orbitals donating electrons to empty π* orbitals of CO.

Memory Aids

• Remember: Metal carbonyl bonding = σ-donation + π-back bonding (synergic effect).

• Geometry clues:

  • 4 CO ligands → Tetrahedral (Ni(CO)4)

  • 5 CO ligands → Trigonal bipyramidal (Fe(CO)5)

  • 6 CO ligands → Octahedral (Cr(CO)6)

• Back bonding strengthens metal–CO bond and weakens C–O bond.

Stability of Coordination Compounds

Definition and Equilibrium

• Stability refers to the extent of complex formation in solution.

• Equilibrium: M + 4L ⇌ ML4

• Equilibrium constant (K) indicates stability magnitude.

Factors Affecting Stability

• Charge on metal ion: Higher charge → greater stability.

• Basic nature of ligand: Stronger base ligands increase stability.

• Chelate effect: Presence of chelate rings enhances stability.

• Size of metal ion: Smaller ions form more stable complexes.

• Electronegativity and polarising power: Higher values increase stability.

Irving-Williams Order

• Stability of divalent first-row transition metal complexes follows:

  • Mn(II) < Fe(II) < Co(II) < Ni(II) < Cu(II) > Zn(II)

Solved Examples

Ex 1

• Arrange the stability of complexes: Fe(II), Cu(II), Mn(II).

• Answer: Mn(II) < Fe(II) < Cu(II) (Irving-Williams order).

Practice Set

• Q1: Which factor does NOT increase complex stability?

  • A. Higher metal ion charge

  • B. Larger metal ion size

  • C. Presence of chelate rings

  • D. Strong basic ligands

• Q2: According to Irving-Williams order, which complex is most stable?

  • A. Mn(II)

  • B. Fe(II)

  • C. Cu(II)

  • D. Zn(II)

• Q3: Stability constant (K) for M + 4L ⇌ ML4 is 106. What does this imply?

  • A. Complex is unstable

  • B. Complex formation is favored

  • C. Ligand dissociation is favored

  • D. Equilibrium lies to left

Answer Key

• Q1: B - Larger metal ion size decreases stability.

• Q2: C - Cu(II) complexes are most stable in first-row divalent metals.

• Q3: B - High K means complex formation is favored.

Memory Aids

• Irving-Williams order mnemonic: "My FeCoNi Cu Zn" (Mn < Fe < Co < Ni < Cu > Zn).

• Chelate rings = more stable complexes.

• Higher metal charge and smaller size = stronger complexes.

Importance and Applications of Coordination Compounds

Qualitative and Quantitative Analysis

• Dimethyl glyoxime (DMG) forms a coordination compound with Ni(II) used to detect nickel.

• Cobalt is detected by forming a coordination compound with α-nitroso-β-naphthol.

• EDTA acts as a ligand for detection and estimation of various metal ions.

• Water hardness is estimated by titration with Na2EDTA; Ca2+ and Mg2+ form stable complexes with EDTA.

• Selective estimation of Ca2+ and Mg2+ is possible due to different stability constants.

Extraction of Metals

• Metals like Ag and Au are extracted as their cyanide complexes: [Ag(CN)2]− and [Au(CN)2]−.

• Metals can be precipitated from solution by adding Zn.

Purification of Metals

• Impure Ni forms Ni(CO)4, which decomposes to give pure nickel.

In Biological Systems

• Chlorophyll: Mg-complex with macrocyclic porphyrin; responsible for photosynthesis.

• Haemoglobin: Fe-complex with macrocyclic porphyrin; carries oxygen in blood.

• Myoglobin: Stores oxygen and regulates respiration.

• Vitamin B-12 (cyanocobalamine): Cobalt complex essential for health.

• Enzymes like carboxypeptidase-A are Zn complexes with proteins.

Catalysts

• Organometallic compounds act as catalysts.

• Ziegler-Natta catalyst [TiCl4(C2H5)3Al] polymerizes alkenes at room temperature and atmospheric pressure.

• Wilkinson's catalyst [RhCl(PPh3)3] catalyzes hydrogenation reactions.

• Metal carbonyls serve as industrial catalysts.

Other Applications

• Electroplating with silver and gold is smoother from complexes [Ag(CN)2]− and [Au(CN)2]− than from simple ions.

• In black and white photography, hypo solution fixes film by forming [Ag(S2O3)2]3− complex.

• Many medicines are coordination compounds.

Treatment of Diseases

• Excess toxic metals are removed by chelating agents like penicillamine and desferrioxime-B.

• EDTA is used to treat lead poisoning.

• cis-Platin [PtCl2(NH3)2] is used in cancer treatment.

Solved Examples

• Ex 1: Detect nickel in a sample using DMG.

  • Add DMG to the sample; formation of red precipitate confirms Ni presence.

• Ex 2: Estimate water hardness using EDTA.

  • Titrate water sample with Na2EDTA; endpoint indicates total Ca2+ and Mg2+ concentration.

• Ex 3: Purify nickel from impure sample.

  • Form Ni(CO)4 by passing CO gas; decompose to get pure Ni.

Practice Set

• Q1: Which ligand is used to detect nickel?

  • A. α-nitroso-β-naphthol

  • B. Dimethyl glyoxime (DMG)

  • C. EDTA

  • D. Penicillamine

• Q2: Which complex is used in the purification of nickel?

  • A. Ni(CO)4

  • B. [Ni(DMG)2]

  • C. [Ni(CN)4]2−

  • D. NiCl2

• Q3: Which coordination compound is used in cancer treatment?

  • A. EDTA

  • B. cis-Platin [PtCl2(NH3)2]

  • C. Penicillamine

  • D. Vitamin B-12

Answer Key

• Q1: B. DMG forms a red complex with Ni(II) for detection.

• Q2: A. Ni(CO)4 is volatile and decomposes to pure Ni.

• Q3: B. cis-Platin is a platinum coordination compound used in chemotherapy.

Memory Aids

• DMG detects Ni by red precipitate formation.

• Ni purification: Ni + CO → Ni(CO)4 (volatile) → pure Ni.

• EDTA chelates Ca2+ and Mg2+ for water hardness estimation.

• cis-Platin = cancer drug; remember Pt coordination.