Chapter 14: Coordination Compounds

Introduction to Coordination Compounds

Key Characteristics

• Transition metals form complexes by coordinate bonds.
• Central metal atom linked to ions or neutral molecules via lone pair donation.
• 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 the central metal to ligands in coordination compounds?
  A. Ionic bond
  B. Covalent bond
  C. Coordinate bond
  D. Metallic bond
• Q2: Which of the following is a coordination compound?
  A. NaCl
  B. Haemoglobin
  C. H₂O
  D. CO₂
• Q3: Which metal is central in vitamin B₁₂?
  A. Magnesium
  B. Iron
  C. Cobalt
  D. Copper

Answer Key

• Q1: C - Coordinate bonds form by donation of lone pairs from ligands to metal.
• Q2: B - Haemoglobin is a coordination compound with iron as central metal.
• Q3: C - Cobalt is the central metal in vitamin B₁₂.

Memory Aids

• Coordination compounds = Central metal + ligands via coordinate bonds.
• Remember: Chlorophyll (Mg), Haemoglobin (Fe), Vitamin B₁₂ (Co).

Werners Theory of Coordination Compounds

Main Postulates

• Metals show two valencies: primary (ionisable) and secondary (non-ionisable).
• Primary valency = oxidation state, satisfied by negative ions, non-directional.
• Secondary valency = coordination number, satisfied by ligands, directional, defines geometry.
• Coordination entities are inside square brackets; ions outside are counter ions.
• 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 the coordination number of Co in [CoCl(NH₃)₅]²⁺.
  Answer: 6 (octahedral geometry).

Practice Set

• Q1: What does the primary valency of a metal represent?
  A. Coordination number
  B. Oxidation state
  C. Number of ligands
  D. Number of counter ions
• Q2: Which geometry is common for [Ni(CO)₄] complex?
  A. Octahedral
  B. Tetrahedral
  C. Square planar
  D. Linear
• Q3: In Werner’s theory, secondary valency corresponds to:
  A. Ionisable valency
  B. Coordination number
  C. Oxidation state
  D. Number of counter ions

Answer Key

• Q1: B - Primary valency = oxidation state of metal.
• Q2: B - [Ni(CO)₄] has tetrahedral geometry.
• Q3: B - Secondary valency = coordination number, non-ionisable.

Memory Aids

• Primary valency = oxidation state (ionisable).
• Secondary valency = coordination number (defines geometry).
• Common shapes: 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 into ions in solution.
• Examples of double salts: Carnallite (KCl·MgCl₂·6H₂O), Mohr's salt (FeSO₄·(NH₄)₂SO₄·6H₂O), Potash alum (KAl(SO₄)₂·12H₂O).
• Complexes do not dissociate into ions easily; ligands remain bonded.

Solved Examples

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

Practice Set

• Q1: Which of the following is a double salt?
  A. [Co(NH₃)₆]Cl₃
  B. KAl(SO₄)₂·12H₂O
  C. [Ni(CO)₄]
  D. [PtCl₄]^2-
• Q2: Double salts in solution:
  A. Do not dissociate
  B. Dissociate into ions
  C. Form coordinate bonds
  D. Are covalent compounds
• Q3: Which property distinguishes complexes from double salts?
  A. Stoichiometric ratio
  B. Dissociation into ions
  C. Formation of coordinate bonds
  D. Presence of metal

Answer Key

• Q1: B - KAl(SO₄)₂·12H₂O is a double salt.
• Q2: B - Double salts dissociate into ions in solution.
• Q3: C - Complexes have coordinate bonds; double salts do not.

Memory Aids

• Double salts dissociate; complexes do not.
• Double salts: Mohr's salt, Carnallite, Potash alum.
• Complexes: Coordinate bonds, fixed geometry.

Coordination Entity/Coordination Sphere

Explanation

• Central metal atom/ion and ligands directly attached are enclosed in square brackets.
• Collectively called coordination sphere or coordination entity.
• Coordination sphere is non-ionisable.

Solved Examples

• 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
  • D. None
• Q3: In K₄[Fe(CN)₆], what is the charge on coordination sphere?
  • A. 4+
  • B. 4-
  • C. Neutral
  • D. 2-

Answer Key

• Q1: C - Coordination sphere includes both metal ion and ligands.
• Q2: A - Coordination sphere is non-ionisable.
• Q3: B - Coordination sphere [Fe(CN)₆]^4- has 4- charge.

Memory Aids

• Coordination sphere = [Metal + Ligands]
• Non-ionisable part of complex

Counter Ions

Explanation

• Ionisable groups written outside the coordination sphere brackets.
• Balance the charge of coordination entity.

Solved Examples

• In K₄[Fe(CN)₆], counter ion is K⁺.

Practice Set

• Q1: Where are counter ions written in a coordination compound?
  • A. Inside brackets
  • B. Outside brackets
  • C. Both inside and outside
  • D. None
• Q2: What is the role of counter ions?
  • A. Coordinate with metal
  • B. Balance charge
  • C. Donate electrons
  • D. Form coordinate bonds
• Q3: In [Co(NH₃)₅Cl]Cl₂, what are the counter ions?
  • A. NH₃
  • B. Cl^- outside brackets
  • C. Co³⁺
  • D. Cl inside brackets

Answer Key

• Q1: B - Counter ions are outside the brackets.
• Q2: B - They balance the charge of the complex.
• Q3: B - Cl^- ions outside brackets are counter ions.

Memory Aids

• Counter ions = outside brackets, ionisable
• Balance complex charge

Central Metal Atom/Ion

Explanation

• Metal atom/ion bonded to ligands by coordinate bonds in definite geometry.
• Acts as Lewis acid (electron pair acceptor).
• Examples: Ni²⁺ in [NiCl₂(H₂O)₄], Co³⁺ in [CoCl(NH₃)₅]²⁺, Fe³⁺ in [Fe(CN)₆]^3-.

Solved Examples

• Identify central metal ion in [Fe(CN)₆]^3-: Fe³⁺.

Practice Set

• Q1: What is the role of central metal ion?
  • A. Electron donor
  • B. Lewis acid
  • C. Ligand
  • D. Counter ion
• Q2: Which is central metal ion in [Co(NH₃)₆]³⁺?
  • A. NH₃
  • B. Co³⁺
  • C. Cl^-
  • D. H₂O
• Q3: Central metal ion in [NiCl₂(H₂O)₄] is?
  • A. Ni²⁺
  • B. Cl^-
  • C. H₂O
  • D. NH₃

Answer Key

• Q1: B - Central metal ion acts as Lewis acid.
• Q2: B - Co³⁺ is central metal ion.
• Q3: A - Ni²⁺ is central metal ion.

Memory Aids

• Central metal = Lewis acid
• Coordinates ligands

Ligands

Explanation

• Molecules or ions donating lone pair to central metal via coordinate bond.
• Types based on donor atoms:
  • Unidentate: single donor atom
  • Didentate: two donor atoms
  • Polydentate: several donor atoms
  • Ambidentate: can coordinate through two different atoms

Solved Examples

• Unidentate ligands: Cl^-, H₂O, NH₃
• Didentate ligands: Ethylenediamine (H₂NCH₂CH₂NH₂), Oxalate (C₂O₄^2-)

Practice Set

• Q1: What type of ligand is NH₃?
  • A. Unidentate
  • B. Didentate
  • C. Polydentate
  • D. Ambidentate
• Q2: Which ligand binds through two donor atoms?
  • A. Cl^-
  • B. Ethylenediamine
  • C. H₂O
  • D. NO₂^-
• Q3: Ambidentate ligand example is?
  • A. NH₃
  • B. NO₂^-
  • C. EDTA^4-
  • D. C₂O₄^2-

Answer Key

• Q1: A - NH₃ is unidentate ligand.
• Q2: B - Ethylenediamine is didentate ligand.
• Q3: B - NO₂^- is ambidentate ligand.

Memory Aids

• Unidentate = 1 donor atom
• Didentate = 2 donor atoms
• Ambidentate = 2 possible donor atoms, binds through one

Polydentate Ligand

Explanation

• Ligand binds through several donor atoms to metal ion.
• Example: EDTA^4-.

Solved Examples

• EDTA^4- binds through multiple donor atoms forming stable complexes.

Practice Set

• Q1: Polydentate ligand binds through how many donor atoms?
  • A. One
  • B. Two
  • C. Several
  • D. None
• Q2: Which is an example of polydentate ligand?
  • A. NH₃
  • B. EDTA^4-
  • C. Cl^-
  • D. NO₂^-
• Q3: Polydentate ligands form complexes that are?
  • A. Unstable
  • B. Stable
  • C. Neutral
  • D. Ionised

Answer Key

• Q1: C - Polydentate ligands bind through several donor atoms.
• Q2: B - EDTA^4- is polydentate ligand.
• Q3: B - Polydentate ligands form stable complexes.

Memory Aids

• Polydentate = many donor atoms
• EDTA = common polydentate ligand

Chelate Ligand

Explanation

• Didentate or polydentate ligand forming two or more coordinate bonds with metal.
• Forms 5 or 6 membered ring with metal ion.

Solved Examples

• Ethylene diamine forms chelate ring with metal ion.

Practice Set

• Q1: Chelate ligand forms how many coordinate bonds?
  • A. One
  • B. Two or more
  • C. None
  • D. Only one donor atom
• Q2: Chelate ring size is usually?
  • A. 3 or 4 membered
  • B. 5 or 6 membered
  • C. 7 or 8 membered
  • D. 2 membered
• Q3: Which ligand forms chelate complex?
  • A. NH₃
  • B. EDTA^4-
  • C. Cl^-
  • D. NO₂^-

Answer Key

• Q1: B - Chelate ligands form two or more coordinate bonds.
• Q2: B - Chelate rings are 5 or 6 membered.
• Q3: B - EDTA^4- forms chelate complexes.

Memory Aids

• Chelate = ring formation with metal
• 5 or 6 membered rings are most stable

Ambidentate Ligand

Explanation

• Unidentate ligand that can coordinate through two different atoms.
• Examples: NO₂^-, SCN^-.

Solved Examples

• NO₂^- can bind through N or O atom.

Practice Set

• Q1: Ambidentate ligand binds through?
  • A. One fixed atom
  • B. Two different atoms alternatively
  • C. Several atoms simultaneously
  • D. None
• Q2: Which is an ambidentate ligand?
  • A. NH₃
  • B. NO₂^-
  • C. EDTA^4-
  • D. Cl^-
• Q3: SCN^- binds through?
  • A. S atom
  • B. N atom
  • C. Both S or N
  • D. O atom

Answer Key

• Q1: B - Ambidentate ligands can bind through two different atoms alternatively.
• Q2: B - NO₂^- is ambidentate ligand.
• Q3: C - SCN^- can bind through S or N atom.

Memory Aids

• Ambidentate = two possible donor atoms
• Binding site depends on conditions

Coordination Number

Explanation

• Total number of coordinate bonds between central metal and ligands.
• Examples:
  • [PtCl₆]^2- coordination number = 6
  • [Ni(NH₃)₄]²⁺ coordination number = 4
  • [Fe(C₂O₄)₃]^3- coordination number = 6 (oxalate is didentate)
  • [Co(en)₃]³⁺ coordination number = 6 (en = ethane-1,2-diamine, didentate)

Solved Examples

• Find coordination number of Ni in [Ni(NH₃)₄]²⁺: 4

Practice Set

• Q1: Coordination number is number of?
  • A. Ligands
  • B. Coordinate bonds
  • C. Metal atoms
  • D. Counter ions
• Q2: Coordination number of Fe in [Fe(C₂O₄)₃]^3- is?
  • A. 3
  • B. 6
  • C. 4
  • D. 2
• Q3: Coordination number of Pt in [PtCl₆]^2- is?
  • A. 4
  • B. 6
  • C. 2
  • D. 1

Answer Key

• Q1: B - Coordination number counts coordinate bonds.
• Q2: B - Oxalate is didentate, total 6 bonds.
• Q3: B - Pt has 6 coordinate bonds with Cl ligands.

Memory Aids

• Coordination number = total coordinate bonds
• Didentate ligands count twice

Coordination Polyhedron

Explanation

• Spatial arrangement of ligands around central metal ion.
• Determines geometry of complex.

Solved Examples

• Octahedral, tetrahedral, square planar are common coordination polyhedra.

Practice Set

• Q1: Coordination polyhedron describes?
  • A. Ligand types
  • B. Spatial arrangement of ligands
  • C. Metal ion charge
  • D. Counter ions
• Q2: Which geometry is common for coordination number 6?
  • A. Tetrahedral
  • B. Octahedral
  • C. Square planar
  • D. Linear
• Q3: Coordination polyhedron affects?
  • A. Stability
  • B. Color
  • C. Magnetic properties
  • D. All of the above

Answer Key

• Q1: B - It describes spatial arrangement of ligands.
• Q2: B - Octahedral geometry is common for coordination number 6.
• Q3: D - Geometry affects stability, color, magnetism.

Memory Aids

• Octahedral = 6 ligands
• Tetrahedral = 4 ligands
• Square planar = 4 ligands, different geometry

Oxidation Number

Explanation

• Charge on central metal if ligands removed with shared electron pairs.
• Represented by Roman numeral in parentheses after name.
• Example: Cu in [Cu(CN)₄]^3- is +1, written as Cu(I).

Solved Examples

• Oxidation number of Fe in [Fe(CN)₆]^3- is +3.

Practice Set

• Q1: Oxidation number represents?
  • A. Total electrons in complex
  • B. Charge on metal after ligand removal
  • C. Number of ligands
  • D. Number of counter ions
• Q2: Oxidation number of Cu in [Cu(CN)₄]^3- is?
  • A. +1
  • B. +2
  • C. 0
  • D. -1
• 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: A - Cu is +1 in given complex.
• Q3: B - Roman numeral in parentheses.

Memory Aids

• Oxidation number = metal charge after ligand removal
• Use Roman numerals in names

Homoleptic Complex

Explanation

• Metal ion bound to only one kind of ligand.
• Example: [Co(NH₃)₆]³⁺.

Solved Examples

• [Co(NH₃)₆]³⁺ is homoleptic complex.

Practice Set

• Q1: Homoleptic complex has how many ligand types?
  • A. One
  • B. Two
  • C. More than two
  • D. None
• Q2: Which is homoleptic?
  • A. [Co(NH₃)₆]³⁺
  • B. [Co(NH₃)₄Cl₂]⁺
  • C. [Fe(CN)₆]^3-
  • D. [Ni(H₂O)₆]²⁺
• Q3: Homoleptic complexes are generally?
  • A. Mixed ligands
  • B. Single ligand type
  • C. Unstable
  • D. Ionised

Answer Key

• Q1: A - Only one ligand type.
• Q2: A - [Co(NH₃)₆]³⁺ has only NH₃ ligands.
• Q3: B - Homoleptic means single ligand type.

Memory Aids

• Homo = same ligands
• Single ligand type only

Heteroleptic Complex

Explanation

• Metal ion bound to more than one kind of ligand.
• Example: [Co(NH₃)₄Cl₂]⁺.

Solved Examples

• [Co(NH₃)₄Cl₂]⁺ is heteroleptic complex.

Practice Set

• Q1: Heteroleptic complex has how many ligand types?
  • A. One
  • B. More than one
  • C. None
  • D. Only metal ion
• Q2: Which is heteroleptic?
  • A. [Co(NH₃)₆]³⁺
  • B. [Co(NH₃)₄Cl₂]⁺
  • C. [Fe(CN)₆]^3-
  • D. [Ni(H₂O)₆]²⁺
• Q3: Heteroleptic complexes are generally?
  • A. Mixed ligands
  • B. Single ligand type
  • C. Unstable
  • D. Ionised

Answer Key

• Q1: B - More than one ligand type.
• Q2: B - [Co(NH₃)₄Cl₂]⁺ has NH₃ and Cl ligands.
• Q3: A - Heteroleptic means mixed ligands.

Memory Aids

• Hetero = different ligands
• More than one ligand type

Effective Atomic Number Rule (EAN Rule)

Explanation

• EAN = Atomic number of metal - electrons lost + electrons gained from ligands.
• Used to predict stability of complexes.
• Examples following EAN: [Pd(NH₃)₆]⁴⁺, [Pt(NH₃)₆]⁴⁺, [Cu(CN)₄]^3-.
• Examples not following EAN but stable: [PdCl₄]^2-, [Ag(NH₃)₂]⁺, [Ni(CN)₄]^2-, [Ni(en)₃]²⁺.

Solved Examples

• Calculate EAN for [Cu(CN)₄]^3-:
  • Cu atomic no. = 29
  • Electrons lost = +1 (Cu(I))
  • Electrons from 4 CN ligands = 8
  • EAN = 29 - 1 + 8 = 36

Practice Set

• Q1: EAN formula is?
  • A. Atomic no. + electrons lost - electrons from ligands
  • B. Atomic no. - electrons lost + electrons from ligands
  • C. Atomic no. + electrons lost + electrons from ligands
  • D. Atomic no. - electrons lost - electrons from ligands
• Q2: Which complex follows EAN rule?
  • A. [PdCl₄]^2-
  • B. [Pd(NH₃)₆]⁴⁺
  • C. [Ag(NH₃)₂]⁺
  • D. [Ni(en)₃]²⁺
• Q3: EAN of [Cu(CN)₄]^3- is?
  • A. 36
  • B. 34
  • C. 50
  • D. 52

Answer Key

• Q1: B - EAN = Atomic no. - electrons lost + electrons from ligands.
• Q2: B - [Pd(NH₃)₆]⁴⁺ follows EAN.
• Q3: A - EAN calculated as 36.

Memory Aids

• EAN = Atomic no. - lost + gained electrons
• Stable complexes often follow EAN

Isomerism in Coordination Compounds

Types of Isomerism

• Geometrical Isomerism: Different spatial arrangements of ligands around the metal center.
• 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: Non-superimposable mirror images (enantiomers) that rotate plane-polarized light differently.

Solved Examples

• Ex 1: Identify cis and trans isomers in [Pt(NH₃)₂Cl₂] (square planar).
  Step 1: Locate similar ligands.
  Step 2: If adjacent, it is cis; if opposite, trans.
• Ex 2: Distinguish fac and mer isomers in [Co(NH₃)₃Cl₃] (octahedral).
  Step 1: Check positions of three Cl ligands.
  Step 2: If on one face, fac; if in a plane, mer.

Practice Set

• Q1: In a square planar complex, which arrangement corresponds to trans isomer?
  A. Similar ligands adjacent
  B. Similar ligands opposite
  C. All ligands different
  D. None of these
• Q2: Which is true for facial (fac) isomer in octahedral complexes?
  A. Three identical ligands in a plane
  B. Three identical ligands on one face
  C. Two ligands opposite
  D. Mirror images
• Q3: Optical isomers are:
  A. Superimposable mirror images
  B. Non-superimposable mirror images
  C. Geometrical isomers
  D. Ionisation isomers

Answer Key

• Q1: B - Trans isomer has similar ligands opposite each other.
• Q2: B - Fac isomer has three identical ligands on one face of octahedron.
• Q3: B - Optical isomers are non-superimposable mirror images.

Memory Aids

• "Cis" = "Side by side"; "Trans" = "Across"
• Facial = Face of octahedron; Meridional = Meridian plane
• Optical isomers = Like left and right hands

Classification of Isomerism

Structural Isomerism

• Different ligands or ions within coordination spheres.
• 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: Cis/trans arrangements in octahedral and square planar complexes.
• Optical Isomerism: Non-superimposable mirror images (enantiomers) rotating plane-polarized light differently.

Solved Examples

• Ex 1: Identify ionisation isomers in [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br.
  Step 1: Check ions released in solution.
  Step 2: Different ions indicate ionisation isomers.
• Ex 2: Determine linkage isomerism in [Co(NH₃)₅NO₂]ClO₄.
  Step 1: Identify ligand binding atom (N or O).
  Step 2: Different binding sites indicate linkage isomers.

Practice Set

• Q1: Which is an example of coordination isomerism?
  A. [Co(NH₃)₆][Cr(CN)₆]
  B. [Co(NH₃)₅Br]SO₄
  C. Cis-[Pt(NH₃)₂Cl₂]
  D. Optical isomers
• Q2: Hydrate isomerism differs in:
  A. Number of ligands
  B. Number of water molecules
  C. Metal center
  D. Charge on complex
• Q3: Linkage isomerism occurs due to:
  A. Different ligand atoms binding
  B. Different ions released
  C. Different spatial arrangement
  D. Ligand exchange between ions

Answer Key

• Q1: A - Coordination isomerism involves ligand exchange between cation and anion complexes.
• Q2: B - Hydrate isomers differ in water molecules inside/outside coordination sphere.
• Q3: A - Linkage isomerism arises from ligands binding through different atoms.

Memory Aids

• Ionisation = Different ions in solution
• Hydrate = Water molecule variation
• Linkage = Ligand binds via different atoms
• Coordination = Ligand swap between ions

Optical and Geometrical Isomerism Details

Optical Isomerism

• Non-superimposable mirror images called enantiomers.
• Rotate plane-polarized light in opposite directions: Dextro (+) and Laevo (-).
• Optically inactive forms (meso, racemic, dl) do not rotate light due to internal or external compensation.
• Example: Octahedral complex [Pt(Py)(NH₃)(NO₂)(Cl)(Br)(I)] has 15 possible geometrical isomers; each has d and l optical forms, total 30 optically active isomers.

Geometrical Isomerism

• Different spatial arrangements of ligands in octahedral and square planar complexes.
• 15 theoretical geometrical isomers possible for octahedral complex with six different monodentate ligands.
• Only 3 geometrical isomers isolated experimentally for [Pt(Py)(NH₃)(NO₂)(Cl)(Br)(I)].

Solved Examples

• Ex 1: Calculate total optically active isomers for [Pt(Py)(NH₃)(NO₂)(Cl)(Br)(I)].
  Step 1: 15 geometrical isomers.
  Step 2: Each has d and l forms.
  Step 3: Total = 15 × 2 = 30.

Practice Set

• Q1: Optical isomers are:
  A. Superimposable mirror images
  B. Non-superimposable mirror images
  C. Geometrical isomers
  D. Ionisation isomers
• Q2: An optically inactive isomer is called:
  A. Enantiomer
  B. Meso form
  C. D form
  D. L form
• Q3: How many geometrical isomers are theoretically possible for [Pt(Py)(NH₃)(NO₂)(Cl)(Br)(I)]?
  A. 3
  B. 6
  C. 15
  D. 30

Answer Key

• Q1: B - Optical isomers are non-superimposable mirror images.
• Q2: B - Meso form is optically inactive due to internal compensation.
• Q3: C - 15 geometrical isomers are theoretically possible.

Memory Aids

• Optical isomers = Left and right hands
• Meso = Optically inactive
• Geometrical = Cis/trans arrangements

Bonding in Metal Carbonyls

Overview

• Transition metal carbonyls are organometallic compounds with only CO ligands (homoleptic carbonyls).
• Metal-carbon bond has both sigma (σ) and pi (π) character.
• Bond formation involves sigma donation from CO to metal and pi back-donation from metal to CO (back bonding).
• Back bonding stabilizes the metal-ligand interaction (synergic effect).

Solved Examples

• Ex 1: Describe bonding in Fe(CO)₅.
  • CO donates lone pair to Fe forming σ bond.
  • Fe d-orbitals back-donate electrons to CO π* orbitals forming π bond.
  • Result: strong Fe–CO bond with synergic effect.

Practice Set

• Q1: What type of bonding is present in metal carbonyls?
  • A) Only σ bonding
  • B) Only π bonding
  • C) Both σ and π bonding
  • D) Ionic bonding
• Q2: Back bonding in metal carbonyls involves overlap of
  • A) Metal d-orbitals with CO lone pair
  • B) Metal d-orbitals with CO π* orbitals
  • C) CO lone pair with metal s-orbitals
  • D) Metal s-orbitals with CO π orbitals
• Q3: Synergic effect in metal carbonyls refers to
  • A) Strengthening of metal-ligand bond by combined σ and π bonding
  • B) Weakening of metal-ligand bond
  • C) Only σ bonding effect
  • D) Only π bonding effect

Answer Key

• Q1: C - Metal carbonyls have both σ and π bonding.
• Q2: B - Back bonding is metal d-orbitals overlapping with CO π* orbitals.
• Q3: A - Synergic effect is the combined strengthening by σ and π bonding.

Memory Aids

• Remember: CO donates electrons via σ bond; metal returns electrons via π back bonding.
• Synergic effect = mutual strengthening of bonds.

Examples of Metal Carbonyls

Mononuclear Carbonyls

• V(CO)₅ - Pentacarbonylvanadium (0)
• Cr(CO)₆ - Hexacarbonylchromium (0)
• Mo(CO)₆ - Hexacarbonylmolybdenum (0)
• W(CO)₆ - Hexacarbonyltungsten (0)
• Ni(CO)₄ - Tetracarbonylnickel (0)
• Fe(CO)₅ - Pentacarbonyliron (0)

Polynuclear Carbonyls

• Co₂(CO)₈ - Octacarbonyldicobalt (0)
• Co₄(CO)₁₂ - Dodecacarbonyltetracobalt (0)
• Mn₂(CO)₁₀ - Decacarbonyldimanganese (0)

Solved Examples

• Ex 1: Identify the formula of hexacarbonylchromium.
  • Answer: Cr(CO)₆
• Ex 2: Name Ni(CO)₄.
  • Answer: Tetracarbonylnickel (0)

Practice Set

• Q1: Which is a mononuclear carbonyl?
  • A) Co₂(CO)₈
  • B) Ni(CO)₄
  • C) Co₄(CO)₁₂
  • D) Mn₂(CO)₁₀
• Q2: The formula of pentacarbonyliron is
  • A) Fe(CO)₄
  • B) Fe(CO)₅
  • C) Fe(CO)₆
  • D) Fe₂(CO)₁₀
• Q3: Which polynuclear carbonyl has 12 CO ligands?
  • A) Co₂(CO)₈
  • B) Co₄(CO)₁₂
  • C) Mn₂(CO)₁₀
  • D) Ni(CO)₄

Answer Key

• Q1: B - Ni(CO)₄ is mononuclear.
• Q2: B - Pentacarbonyliron is Fe(CO)₅.
• Q3: B - Co₄(CO)₁₂ has 12 CO ligands.

Memory Aids

• Mononuclear: Ni(CO)₄, Fe(CO)₅, Cr(CO)₆
• Polynuclear: Co₂(CO)₈, Co₄(CO)₁₂, Mn₂(CO)₁₀

Structures of Metal Carbonyls

Common Structures

• Ni(CO)₄: Tetrahedral geometry.
• Fe(CO)₅: Trigonal bipyramidal geometry.
• Cr(CO)₆: Octahedral geometry.
• 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 geometry of Ni(CO)₄.
  • Tetrahedral
• Ex 2: Describe bonding in Mn₂(CO)₁₀.
  • Two Mn atoms bonded; each Mn bonded to 5 CO groups.
  • Structure: two square pyramids joined by Mn–Mn bond.

Practice Set

• Q1: Geometry of Fe(CO)₅ is
  • A) Tetrahedral
  • B) Square planar
  • C) Trigonal bipyramidal
  • D) Octahedral
• Q2: Mn₂(CO)₁₀ consists of
  • A) Two octahedrons
  • B) Two square pyramids joined by Mn–Mn bond
  • C) One tetrahedron and one octahedron
  • D) One square planar and one tetrahedron
• Q3: Co₂(CO)₈ has
  • A) Co–Co bond bridged by two CO ligands
  • B) No metal-metal bond
  • C) Single Co atom
  • D) Octahedral geometry

Answer Key

• Q1: C - Fe(CO)₅ is trigonal bipyramidal.
• Q2: B - Mn₂(CO)₁₀ has two square pyramids joined by Mn–Mn bond.
• Q3: A - Co₂(CO)₈ has Co–Co bond bridged by two CO ligands.

Memory Aids

• Ni(CO)₄ - Tetrahedral
• Fe(CO)₅ - Trigonal bipyramidal
• Cr(CO)₆ - Octahedral
• Mn₂(CO)₁₀ - Two square pyramids joined
• Co₂(CO)₈ - Metal-metal bond bridged by CO

Stability of Coordination Compounds

Factors Affecting Stability

• Stability refers to the degree of association in equilibrium: M + 4L ⇔ ML₄
• Magnitude of equilibrium constant (K) expresses stability.
• Factors affecting stability:
  • Charge on metal ion: Higher charge increases stability.
  • Basic nature of ligand: Stronger base ligands increase stability.
  • Chelate rings: Presence increases stability.
  • Size of metal ion: Smaller size increases stability.
  • Electronegativity and polarising power: Higher values increase stability.

Irving-Williams Order

• Stability order of divalent first-row transition metal ions:
  • Mn(II) < Fe(II) < Co(II) < Ni(II) < Cu(II) > Zn(II)

Solved Examples

• Ex 1: Which metal ion forms the most stable complex?
  • Answer: Cu(II) as per Irving-Williams order.

Practice Set

• Q1: Stability of coordination complex increases with
  • A) Decreasing metal ion charge
  • B) Increasing ligand basicity
  • C) Larger metal ion size
  • D) Decreasing polarising power
• Q2: Which factor does NOT increase complex stability?
  • A) Chelate ring formation
  • B) Smaller metal ion size
  • C) Lower metal ion charge
  • D) Higher electronegativity of metal ion
• Q3: According to Irving-Williams order, which ion forms the least stable complex?
  • A) Mn(II)
  • B) Cu(II)
  • C) Ni(II)
  • D) Zn(II)

Answer Key

• Q1: B - Increasing ligand basicity increases stability.
• Q2: C - Lower metal ion charge decreases stability.
• Q3: A - Mn(II) forms the least stable complex.

Memory Aids

• Irving-Williams order mnemonic: "My Fe Co Ni Cu Zn" (increasing stability)
• Chelate rings always increase stability.

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 using coordination compound with α-nitroso-β-naphthol.
• EDTA acts as a ligand for detection and estimation of many 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

• Many metals are extracted as coordination compounds from ores.
• Silver and gold are extracted as cyanide complexes: [Ag(CN)₂]^- and [Au(CN)₂]^-.
• Metals can be precipitated from solution by adding zinc.

Purification of metals

• Purification involves formation and decomposition of coordination compounds.
• Example: 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 pressure.
• Wilkinson's catalyst [RhCl(PPh₃)₃] catalyzes hydrogenation.
• Metal carbonyls serve as industrial catalysts.
• Electroplating with silver and gold is smoother from complexes [Ag(CN)₂]^- and [Au(CN)₂]^-.
• In black and white photography, hypo solution fixes film by forming [Ag(S₂O₃)₂]^3- complex.

Treatment of diseases

• Medicines often used as coordination compounds.
• Excess toxic metals removed by chelating agents like penicillamine and desferrioxime-B.
• EDTA treats lead poisoning.
• Cis-platin [PtCl₂(NH₃)₂] used in cancer treatment.

Solved Examples

• Ex 1: Detection of Ni using DMG
  • Mix sample with DMG solution.
  • Formation of red precipitate confirms Ni presence.
• Ex 2: Estimation of water hardness
  • Titrate water sample with Na₂EDTA.
  • Calculate Ca²⁺ and Mg²⁺ concentration from volume used.
• Ex 3: Purification of Ni
  • Impure Ni reacts with CO to form Ni(CO)₄ gas.
  • Decompose Ni(CO)₄ to get pure Ni metal.

Practice Set

• Q1: Which ligand is used to detect Ni(II)?
  A. α-nitroso-β-naphthol
  B. Dimethyl glyoxime
  C. EDTA
  D. Penicillamine
• Q2: Which complex is used in the extraction of gold?
  A. [Au(CN)₂]^-
  B. Ni(CO)₄
  C. [RhCl(PPh₃)₃]
  D. [Ag(S₂O₃)₂]^3-
• Q3: Which coordination compound is used in cancer treatment?
  A. EDTA
  B. Cis-platin
  C. Vitamin B-12
  D. Ziegler-Natta catalyst

Answer Key

• Q1: B - Dimethyl glyoxime forms a red precipitate with Ni(II) for detection.
• Q2: A - Gold is extracted as the cyanide complex [Au(CN)₂]^-.
• Q3: B - Cis-platin [PtCl₂(NH₃)₂] is used in cancer treatment.

Memory Aids

• DMG detects Ni by red precipitate formation.
• EDTA titration differentiates Ca²⁺ and Mg²⁺ by stability constants.
• Ni(CO)₄ formation purifies nickel.
• Ziegler-Natta catalyst polymerizes alkenes at room temperature.
• Cis-platin is a platinum coordination compound used in chemotherapy.