Coordination Compounds : Class 12 Chemistry Notes For Board Exams, NEET, JEE, CUET and others

Coordination Compounds

Coordination Compounds are an important part of Class 12 Chemistry because they connect several concepts at once—coordination number, ligands, nomenclature, oxidation state, isomerism, bonding theories, magnetic properties, colour, stability and applications.

The easiest way to understand this chapter is not to memorise complex formulas separately. First understand who is connected to whom, how many donor atoms are attached to the central metal, and how electrons are involved in bonding. Once these ideas are clear, nomenclature, isomerism and bonding become much easier.

1. What is a Coordination Compound?

A coordination compound contains a central metal atom or ion surrounded by ions or molecules called ligands, which are attached to the metal through coordinate bonds.

A general representation is:

[M(L)n]

where:

M = central metal atom or ion
L = ligand
n = number of ligand units directly attached to the central metal

Example

[Cu(NH₃)₄]SO₄

In this compound:

Cu²⁺ is the central metal ion.
NH₃ molecules are ligands.
Four NH₃ molecules are directly attached to Cu²⁺.
SO₄²⁻ is outside the coordination sphere and acts as the counter ion.

The part inside square brackets:

[Cu(NH₃)₄]²⁺

is called the coordination entity or coordination sphere.

2. Important Terms in Coordination Compounds

Understanding the terminology is the first step in this chapter.

2.1 Central Atom or Central Ion

The metal atom or ion to which ligands are directly attached is called the central atom or central ion.

Examples:

[Co(NH₃)₆]³⁺ → Co³⁺ is the central ion.
[Fe(CN)₆]⁴⁻ → Fe²⁺ is the central ion.
[Cu(NH₃)₄]²⁺ → Cu²⁺ is the central ion.

Transition-metal ions are especially suitable for forming coordination compounds because they have suitable vacant orbitals and can accept electron pairs from ligands.

2.2 Ligands

A ligand is an ion or molecule that donates at least one lone pair of electrons to the central metal atom or ion to form a coordinate bond.

In simple words:

Ligand = electron-pair donor

Examples

Ligand	Name
H₂O	aqua
NH₃	ammine
CO	carbonyl
Cl⁻	chlorido
Br⁻	bromido
I⁻	iodido
OH⁻	hydroxido
CN⁻	cyanido
NO₂⁻	nitro / nitrito depending on linkage
C₂O₄²⁻	oxalato

Example

In:

[Co(NH₃)₆]³⁺

NH₃ donates a lone pair from nitrogen to Co³⁺.

The bond formed is a coordinate bond.

3. Types of Ligands

Ligands can be classified according to the number of donor atoms through which they attach to the central metal.

3.1 Monodentate Ligands

Ligands that donate through one donor atom are called monodentate ligands.

Examples:

NH₃
H₂O
Cl⁻
CN⁻
CO

Example

[Ag(NH₃)₂]⁺

Each NH₃ ligand attaches through one nitrogen atom.

3.2 Bidentate Ligands

Ligands that have two donor atoms and can attach to the same metal ion through both donor atoms are called bidentate ligands.

Examples:

Ethane-1,2-diamine (en):

H₂N–CH₂–CH₂–NH₂

It has two nitrogen donor atoms.

Another important example:

Oxalate ion, C₂O₄²⁻

It can coordinate through two oxygen atoms.

Example

[Co(en)₃]³⁺

Three bidentate en ligands are attached to Co³⁺.

Although there are only three ligand molecules, there are six donor atoms attached to the metal.

3.3 Polydentate Ligands

Ligands having several donor atoms are called polydentate ligands.

Example:

EDTA⁴⁻

EDTA can coordinate through six donor atoms and is therefore called a hexadentate ligand.

4. Chelating Ligands

A ligand that attaches to the same metal ion through two or more donor atoms and forms one or more rings is called a chelating ligand.

The resulting ring is called a chelate ring.

Examples

en
oxalate
EDTA

Example

[Ni(en)₃]²⁺

The en ligands bind through their two nitrogen atoms.

Why are chelate complexes often more stable?

When a multidentate ligand binds through several donor atoms, it can hold the metal ion strongly and form stable rings.

This increased stability is called the chelate effect.

5. Ambidentate Ligands

An ambidentate ligand contains two different possible donor atoms, but it generally coordinates through only one of them at a time.

Important examples:

NO₂⁻

It can coordinate through:

N → nitro
O → nitrito

SCN⁻

It can coordinate through:

This property leads to linkage isomerism.

6. Coordination Number

The coordination number (C.N.) is the number of donor atoms of ligands directly bonded to the central metal atom or ion.

Do not confuse coordination number with the number of ligand molecules.

Example 1

[Co(NH₃)₆]³⁺

Six NH₃ molecules each donate one donor atom.

Therefore:

Coordination number = 6

Example 2

[Co(en)₃]³⁺

Each en ligand is bidentate.

Therefore:

3 × 2 = 6 donor atoms

So:

Coordination number = 6

Example 3

[PtCl₂(NH₃)₂]

There are four donor atoms directly attached to Pt.

Therefore:

Coordination number = 4

7. Coordination Sphere

The central metal and the ligands directly attached to it are collectively called the coordination sphere.

It is represented inside square brackets.

Example:

[Co(NH₃)₆]Cl₃

Coordination sphere:

[Co(NH₃)₆]³⁺

Counter ions:

3Cl⁻

Important distinction

In:

[Co(NH₃)₆]Cl₃

the three Cl⁻ ions are outside the coordination sphere.

In:

[Co(NH₃)₅Cl]Cl₂

one Cl⁻ is directly coordinated to Co³⁺, while two Cl⁻ ions remain outside.

This difference becomes important in ionisation isomerism and precipitation reactions.

8. Oxidation State of the Central Metal

The oxidation state of the central metal can be calculated by considering the charge of the entire coordination entity and the charges of the ligands.

Example

Find the oxidation state of Co in:

[Co(NH₃)₆]Cl₃

NH₃ is neutral.

There are three Cl⁻ ions outside the coordination sphere.

Therefore, the coordination entity has charge +3.

Let oxidation state of Co = x.

x + 6(0) = +3

Therefore:

x = +3

Co is in the +3 oxidation state.

Another Example

Find the oxidation state of Fe in:

K₄[Fe(CN)₆]

The complex ion has charge:

−4

CN⁻ has charge −1.

Let oxidation state of Fe = x.

x + 6(−1) = −4

x − 6 = −4

x = +2

Therefore:

Fe = +2

9. Werner’s Theory of Coordination Compounds

The theory proposed by Alfred Werner was one of the earliest successful explanations of coordination compounds.

Werner proposed two types of valencies:

Primary valency
Secondary valency

9.1 Primary Valency

Primary valency corresponds to the oxidation state of the metal.

It is generally satisfied by negative ions.

Example:

In [Co(NH₃)₆]Cl₃, Co has oxidation state +3.

Therefore:

Primary valency = 3

9.2 Secondary Valency

Secondary valency corresponds to the coordination number of the metal.

It is satisfied by ligands directly attached to the central metal.

For:

[Co(NH₃)₆]Cl₃

Co³⁺ has:

Primary valency = 3
Secondary valency = 6

Important difference

Primary valency	Secondary valency
Related to oxidation state	Related to coordination number
Usually ionisable	Usually non-ionisable
Satisfied by negative ions	Satisfied by ligands
Directional nature not emphasised	Has definite spatial arrangement

10. Werner’s Explanation of CoCl₃·6NH₃

Werner used compounds of cobalt chloride and ammonia to establish his theory.

Different compounds were found:

CoCl₃·6NH₃

CoCl₃·5NH₃

CoCl₃·4NH₃

These can be represented as:

[Co(NH₃)₆]Cl₃

[Co(NH₃)₅Cl]Cl₂

[Co(NH₃)₄Cl₂]Cl

The number of chloride ions outside the coordination sphere differs.

Therefore, they produce different amounts of AgCl when treated with AgNO₃.

Example

[Co(NH₃)₆]Cl₃

All three Cl⁻ ions are outside the coordination sphere.

Therefore:

[Co(NH₃)₆]Cl₃ + 3AgNO₃ → Co(NH₃)₆₃ + 3AgCl↓

11. Nomenclature of Coordination Compounds

Naming coordination compounds is a very important board-exam and entrance-exam topic.

Follow the steps in order instead of trying to memorise complete names.

Basic Rules

Rule 1: Name the cation first

Then name the anion.

This is similar to ordinary ionic compounds.

Rule 2: Name ligands before the central metal

Within the coordination sphere, ligands are named first and the metal is named afterwards.

Rule 3: Ligands are arranged alphabetically

The alphabetical order is based on the ligand names.

Prefixes such as di-, tri-, tetra- generally do not decide the alphabetical order.

12. Names of Common Ligands

Formula	Ligand name
NH₃	ammine
H₂O	aqua
CO	carbonyl
NO	nitrosyl
F⁻	fluorido
Cl⁻	chlorido
Br⁻	bromido
I⁻	iodido
OH⁻	hydroxido
CN⁻	cyanido
NO₂⁻	nitro / nitrito
C₂O₄²⁻	oxalato

Important spelling

For NH₃ ligand, the name is:

ammine

with double m.

It should not be written as “amine” in coordination nomenclature.

13. Numerical Prefixes for Ligands

For simple ligands:

2 → di
3 → tri
4 → tetra
5 → penta
6 → hexa

Example

[Co(NH₃)₆]³⁺

Name:

hexaamminecobalt(III) ion

14. Naming Complex Anions

If the coordination entity is an anion, the name of the metal generally ends in -ate.

Examples:

Fe → ferrate
Cu → cuprate
Ag → argentate
Au → aurate
Co → cobaltate
Ni → nickelate
Cr → chromate
Mn → manganate

Example

K₄[Fe(CN)₆]

Name:

potassium hexacyanidoferrate(II)

Because the complex ion is negatively charged, Fe is named ferrate.

15. Examples of Coordination Nomenclature

Example 1

[Co(NH₃)₆]Cl₃

Name:

hexaamminecobalt(III) chloride

Example 2

[Co(NH₃)₅Cl]Cl₂

Name:

pentaamminechloridocobalt(III) chloride

Example 3

K₄[Fe(CN)₆]

Name:

potassium hexacyanidoferrate(II)

Example 4

K₃[Fe(CN)₆]

Name:

potassium hexacyanidoferrate(III)

Example 5

[Pt(NH₃)₂Cl₂]

Name:

diamminedichloridoplatinum(II)

16. Isomerism in Coordination Compounds

Coordination compounds can show different types of isomerism.

Two broad categories are:

Structural isomerism
Stereoisomerism

17. Structural Isomerism

Structural isomers have the same molecular formula but different arrangements of bonds.

Important types include:

ionisation isomerism
hydrate/solvate isomerism
linkage isomerism
coordination isomerism

17.1 Ionisation Isomerism

Ionisation isomerism occurs when an ion inside the coordination sphere exchanges position with an ion outside the coordination sphere.

Example

[Co(NH₃)₅Br]SO₄

and

[Co(NH₃)₅SO₄]Br

These produce different ions in solution.

First compound gives:

SO₄²⁻

as an ion outside the coordination sphere.

Second compound gives:

Br⁻

as an ion outside the coordination sphere.

Therefore, they are ionisation isomers.

18. Solvate or Hydrate Isomerism

This occurs when solvent molecules, especially water molecules, occupy different positions inside or outside the coordination sphere.

For example, chromium(III) chloride hydrates can have different numbers of water molecules inside and outside the coordination sphere.

A commonly used representation is:

[Cr(H₂O)₆]Cl₃

[Cr(H₂O)₅Cl]Cl₂·H₂O

[Cr(H₂O)₄Cl₂]Cl·2H₂O

These compounds can show different numbers of ionisable chloride ions.

19. Linkage Isomerism

Linkage isomerism occurs when an ambidentate ligand coordinates through different donor atoms.

Example: NO₂⁻

NO₂⁻ can coordinate through:

N → nitro
O → nitrito

Therefore:

[Co(NH₃)₅(NO₂)]Cl₂

and

[Co(NH₃)₅(ONO)]Cl₂

are linkage isomers.

Important memory point

NO₂⁻ has two possible donor atoms: N and O.

20. Coordination Isomerism

Coordination isomerism occurs in compounds containing both complex cations and complex anions, when ligands exchange between the two metal coordination spheres.

Example:

[Co(NH₃)₆][Cr(CN)₆]

and

[Cr(NH₃)₆][Co(CN)₆]

These have the same overall composition but different ligand arrangements around the metals.

21. Stereoisomerism

Stereoisomers have the same connectivity but differ in the spatial arrangement of ligands.

Two important types are:

geometrical isomerism
optical isomerism

22. Geometrical Isomerism

Geometrical isomerism occurs when the same ligands occupy different relative positions in space.

The most common terms are:

cis
trans

cis

Similar ligands are adjacent to each other.

trans

Similar ligands are opposite to each other.

Example: [Pt(NH₃)₂Cl₂]

This compound has two geometrical isomers:

cis-[Pt(NH₃)₂Cl₂]

The two Cl ligands are adjacent.

trans-[Pt(NH₃)₂Cl₂]

The two Cl ligands are opposite.

This is an important Class 12 example.

23. Geometrical Isomerism in Octahedral Complexes

Consider:

[Co(NH₃)₄Cl₂]⁺

It can exist as:

cis
trans

cis form

The two Cl ligands are next to each other.

trans form

The two Cl ligands are opposite each other.

24. fac-mer Isomerism

Some octahedral complexes of the type:

[MA₃B₃]

can show fac-mer isomerism.

fac

The three identical ligands occupy one face of the octahedron.

mer

The three identical ligands lie along a meridian.

Example:

[Co(NH₃)₃Cl₃]

can exist as:

fac-[Co(NH₃)₃Cl₃]
mer-[Co(NH₃)₃Cl₃]

25. Optical Isomerism

Some coordination compounds exist as non-superimposable mirror images.

Such compounds are called optically active.

The two mirror-image forms are called enantiomers.

Important example

[Co(en)₃]³⁺

This complex exists as two non-superimposable mirror images.

These forms rotate plane-polarised light in opposite directions.

Important concept

If two structures are mirror images but cannot be superimposed, they are optical isomers.

26. Coordination Compounds and Bonding

To understand the properties of coordination compounds, we need to understand how the metal and ligand are bonded.

Important theories are:

Valence Bond Theory (VBT)
Crystal Field Theory (CFT)

At Class 12 level, both are important, but they explain different aspects.

27. Valence Bond Theory

According to VBT, the central metal ion provides suitable vacant orbitals that undergo hybridisation.

Ligands donate lone pairs into these vacant hybrid orbitals.

Thus coordinate bonds are formed.

Important hybridisations

Coordination number	Common hybridisation	Geometry
2	sp	linear
4	sp³	tetrahedral
4	dsp²	square planar
6	d²sp³	octahedral
6	sp³d²	octahedral

28. Inner Orbital and Outer Orbital Complexes

For octahedral complexes, VBT distinguishes between:

Inner orbital complex

Uses inner d-orbitals:

d²sp³

Outer orbital complex

Uses outer d-orbitals:

sp³d²

Example

[Co(NH₃)₆]³⁺

Co³⁺:

[Ar] 3d⁶

Depending on ligand strength and electron pairing, the complex may involve pairing of d-electrons and use inner d-orbitals.

29. Strong and Weak Field Ligands

Ligands differ in their ability to cause pairing of d-electrons.

Weak-field ligands

They generally produce smaller crystal-field splitting and often do not cause pairing in suitable cases.

Examples:

F⁻
Cl⁻
Br⁻
I⁻

Strong-field ligands

They produce larger splitting and can promote pairing.

Examples:

CN⁻
CO
NH₃
en

Important point

The exact behaviour depends on the metal ion, oxidation state and ligand environment. Do not treat the ligand list as an absolute rule for every complex.

30. Crystal Field Theory

Crystal Field Theory (CFT) explains the bonding between metal ions and ligands by considering the electrostatic interaction between them.

According to CFT:

Ligands are treated as point charges or point dipoles that create an electric field around the central metal ion.

In a free metal ion, the five d-orbitals have the same energy.

This is called degeneracy.

When ligands approach the metal ion, this degeneracy is removed.

The d-orbitals split into groups of different energies.

This is called crystal-field splitting.

31. Splitting of d-Orbitals in an Octahedral Complex

In an octahedral complex, six ligands approach along the x, y and z axes.

The five d-orbitals split into two groups:

Higher-energy set

d(x²−y²)
d(z²)

These point directly toward the approaching ligands.

Therefore, they experience greater repulsion and have higher energy.

Lower-energy set

t₂g

These lie between the axes and experience less repulsion.

Therefore:

t₂g < eg

32. Crystal Field Splitting Energy

The energy difference between the two sets is called crystal field splitting energy.

For an octahedral complex: Δ₀ or Δoct is used. The lower t₂g set is stabilised by: −0.4Δ₀ per electron

The higher eg set is destabilised by: +0.6Δ₀ per electron

For a complete set:

3(−0.4) + 2(+0.6) = 0

This shows that the average energy remains unchanged.

33. High-Spin and Low-Spin Complexes

Whether electrons pair or occupy higher orbitals depends on the competition between:

crystal field splitting energy, Δ₀
pairing energy, P

If Δ₀ < P

Electrons prefer to occupy higher-energy orbitals rather than pair.

This gives a high-spin complex.

If Δ₀ > P

Electrons prefer to pair in lower-energy orbitals.

This gives a low-spin complex.

Simple Hindi explanation

अगर orbital splitting की energy कम है और electron pairing ज्यादा costly है, तो electron ऊपर वाले orbital में चला जाएगा।

अगर splitting energy ज्यादा है, तो electron lower orbital में pair कर सकता है।

34. Tetrahedral Complexes

In tetrahedral complexes, four ligands approach the metal between the axes.

The splitting pattern is opposite to the octahedral case.

The two orbitals:

d(z²), d(x²−y²)

form the higher-energy e set in tetrahedral geometry, while:

dxy, dyz, dzx

form the lower-energy t₂ set.

The splitting energy is:

Δt

and is smaller than octahedral splitting.

Approximately:

Δt ≈ 4/9 Δ₀

Therefore, tetrahedral complexes are usually high spin because Δt is relatively small.

35. Square Planar Complexes

Square-planar complexes are common for certain d⁸ metal ions, particularly Pt²⁺ and some Ni²⁺/Pd²⁺ complexes.

Example:

[Pt(NH₃)₂Cl₂]

The arrangement of ligands is in one plane around the central metal.

Square-planar complexes can show cis-trans isomerism.

36. Magnetic Properties of Coordination Compounds

Magnetic behaviour depends on the number of unpaired electrons.

Paramagnetic

A complex with one or more unpaired electrons is paramagnetic.

Diamagnetic

A complex with all electrons paired is diamagnetic.

The spin-only magnetic moment is:

μ = √[n(n + 2)] BM

where:

n = number of unpaired electrons

Example

If n = 4:

μ = √[4(4 + 2)]

μ = √24 BM

37. Colour of Coordination Compounds

Many coordination compounds are coloured because of electronic transitions.

When light is absorbed, an electron can move from a lower-energy d-orbital to a higher-energy d-orbital.

The absorbed wavelength corresponds to the energy difference between the orbitals.

The colour we see is related to the light that is not absorbed.

Important point

The colour depends on:

metal ion
oxidation state
ligand
geometry
crystal-field splitting

Therefore, changing the ligand can change the colour.

38. Why are Some Coordination Compounds Colourless?

Complexes with:

d⁰ or d¹⁰

configurations generally do not show d–d transitions.

Therefore they are often colourless.

Examples:

Sc³⁺ → d⁰

Zn²⁺ → d¹⁰

However, colour can still arise through other mechanisms such as charge-transfer transitions.

39. Stability of Coordination Compounds

The stability of a coordination compound depends on several factors:

charge on the metal ion
size of the metal ion
nature of ligand
metal–ligand interaction
chelate effect
crystal-field stabilisation
solvent effects

Chelate effect

Multidentate ligands often form more stable complexes than comparable monodentate ligands.

For example, en can bind through two nitrogen atoms and form a chelate ring.

40. Formation of Complexes and Applications

Coordination compounds are not merely theoretical. They are important in:

metallurgy
qualitative analysis
biological systems
medicine
photography
extraction of metals
analytical chemistry

41. Coordination Compounds in Biological Systems

Haemoglobin

Haemoglobin contains an iron-containing coordination system.

The Fe centre in haem is responsible for binding oxygen.

Chlorophyll

Chlorophyll contains Mg²⁺ at its centre.

Vitamin B₁₂

Vitamin B₁₂ contains a cobalt coordination system.

These examples show how coordination chemistry is closely connected with biological processes.

42. Coordination Compounds in Medicine

Cisplatin

Cisplatin: [Pt(NH₃)₂Cl₂]

is an important platinum coordination compound used in cancer treatment.

Its biological action is related to its ability to interact with DNA.

Important chemistry point

Cisplatin and transplatin have the same composition but different spatial arrangements.

Therefore, they demonstrate geometrical isomerism.

43. Coordination Compounds in Metallurgy

Coordination chemistry helps in the extraction and purification of metals.

Gold extraction

Gold forms soluble complexes with cyanide under suitable conditions.

A simplified representation is:

4Au + 8CN⁻ + O₂ + 2H₂O → 4[Au(CN)₂]⁻ + 4OH⁻

The complex formed is dicyanidoaurate(I).

This property is used in gold extraction processes.

44. Coordination Compounds in Qualitative Analysis

Complex formation can help identify metal ions.

Example: Cu²⁺ and NH₃

When NH₃ is added to a Cu²⁺ solution, a deep blue complex can form:

[Cu(NH₃)₄]²⁺

This characteristic colour is useful in qualitative analysis.

45. Coordination Compounds in Photography

Silver halides are important in traditional photographic processes.

Unreacted AgBr can be dissolved using sodium thiosulfate because it forms a soluble silver-thiosulfate complex.

A simplified representation is:

AgBr + 2S₂O₃²⁻ → [Ag(S₂O₃)₂]³⁻ + Br⁻

This complex formation helps remove unreacted silver halide.

46. Important Differences Students Commonly Confuse

Coordination Number vs Oxidation State

Coordination number = number of donor atoms directly attached to the metal.

Oxidation state = formal charge assigned to the metal after considering ligand charges.

Example:

[Co(en)₃]³⁺

Coordination number = 6
Oxidation state of Co = +3

Ligand vs Counter Ion

In:

[Co(NH₃)₆]Cl₃

NH₃ = ligand
Cl⁻ = counter ion

In:

[Co(NH₃)₅Cl]Cl₂

five NH₃ + one Cl⁻ = ligands
two Cl⁻ = counter ions

Coordination Entity vs Coordination Compound

[Co(NH₃)₆]³⁺ is a coordination entity.

[Co(NH₃)₆]Cl₃ is a coordination compound.

47. Common Student Mistakes

Mistake 1: Counting ligand molecules instead of donor atoms

In:

[Co(en)₃]³⁺

There are 3 ligand molecules but coordination number is 6, because en is bidentate.

Mistake 2: Calling NH₃ “amine” in nomenclature

For coordination compounds:

NH₃ → ammine

with double m.

Mistake 3: Forgetting the charge of the complex ion

For:

K₄[Fe(CN)₆]

the complex ion is:

[Fe(CN)₆]⁴⁻

not neutral.

Mistake 4: Confusing coordination number with oxidation state

A metal can have:

Oxidation state = +3

and:

Coordination number = 6

at the same time.

Mistake 5: Assuming every complex is octahedral

Coordination number alone does not always determine geometry.

For example, coordination number 4 can give:

tetrahedral
square planar

depending on the metal and electronic situation.

Mistake 6: Treating all ligands as monodentate

en is bidentate.

Oxalate is bidentate.

EDTA⁴⁻ is hexadentate.

Mistake 7: Ignoring the charge of ligands while finding oxidation state

Always write the charge of every ligand before calculating the metal’s oxidation state.

48. Board Exam Important Points

For board examinations, prepare these topics particularly well:

Definitions

Coordination compound
Coordination entity
Ligand
Coordination number
Chelating ligand
Ambidentate ligand
Coordination sphere
Primary valency
Secondary valency

Important theories

Werner’s theory
Valence Bond Theory
Crystal Field Theory

Nomenclature

Practise:

cationic complexes
anionic complexes
neutral complexes
different ligand types
oxidation state calculation

Isomerism

Know examples of:

ionisation isomerism
hydrate/solvate isomerism
linkage isomerism
coordination isomerism
geometrical isomerism
optical isomerism
fac-mer isomerism

Bonding

Understand:

hybridisation
geometry
inner-orbital complex
outer-orbital complex
strong-field ligand
weak-field ligand
high-spin complex
low-spin complex
d-orbital splitting

49. NEET/JEE Important Concept Points

For competitive examinations, focus on:

calculating oxidation state quickly
determining coordination number
identifying ligand denticity
recognising ambidentate ligands
predicting geometrical isomerism
identifying optical isomerism
determining unpaired electrons
calculating magnetic moment
distinguishing high-spin and low-spin complexes
identifying inner- and outer-orbital complexes
understanding octahedral and tetrahedral splitting
recognising square-planar complexes
comparing ligand strengths
understanding colour and d–d transitions

Particularly important examples

[Co(NH₃)₆]³⁺

[Co(NH₃)₅Cl]Cl₂

[Co(NH₃)₄Cl₂]Cl

[Pt(NH₃)₂Cl₂]

[Co(en)₃]³⁺

K₄[Fe(CN)₆]

K₃[Fe(CN)₆]

50. Solved Examples

Example 1: Oxidation State

Find the oxidation state of Co in:

[Co(NH₃)₅Cl]Cl₂

The complex ion has charge:

NH₃ is neutral.

Cl inside the coordination sphere has charge −1.

Let oxidation state of Co = x.

x + 5(0) − 1 = +2

Therefore:

x = +3

So Co is in the +3 oxidation state.

Example 2: Coordination Number

Find the coordination number of Co in:

[Co(en)₂Cl₂]⁺

Each en ligand is bidentate:

2 en × 2 donor atoms = 4

Two Cl⁻ ions contribute:

2 × 1 = 2

Therefore:

Coordination number = 4 + 2 = 6

Example 3: Magnetic Moment

An ion has three unpaired electrons. Calculate its spin-only magnetic moment.

Formula:

μ = √[n(n + 2)] BM

n = 3

Therefore:

μ = √[3(3 + 2)]

μ = √15 BM

Example 4: Oxidation State in an Anionic Complex

Find the oxidation state of Fe in:

[Fe(CN)₆]³⁻

Let Fe oxidation state = x.

CN⁻ = −1

Therefore:

x + 6(−1) = −3

x − 6 = −3

x = +3

Hence:

Fe = +3

The complex ion is therefore:

hexacyanidoferrate(III).

51. Practice Questions

Try these yourself before checking your notes.

Conceptual Questions

Define a coordination compound with an example.
What is a ligand? Give four examples.
What is coordination number?
Distinguish between primary and secondary valencies.
What is a chelating ligand?
What is an ambidentate ligand?
Why is coordination number different from the number of ligand molecules in some complexes?
Explain the chelate effect.
State the main postulates of Werner’s theory.
What is crystal-field splitting?

Nomenclature Questions

Write the name of [Co(NH₃)₆]Cl₃.
Write the name of K₄[Fe(CN)₆].
Write the name of K₃[Fe(CN)₆].
Write the name of [Pt(NH₃)₂Cl₂].
Write the formula of tetraamminecopper(II) sulfate.

Isomerism Questions

Explain ionisation isomerism with an example.
Explain linkage isomerism.
What is geometrical isomerism?
Which type of isomerism is shown by [Pt(NH₃)₂Cl₂]?
Explain optical isomerism using [Co(en)₃]³⁺.

Numerical Questions

Calculate the oxidation state of Fe in K₄[Fe(CN)₆].
Calculate the coordination number of Co in [Co(en)₃]³⁺.
Calculate the spin-only magnetic moment of an ion with 5 unpaired electrons.
Calculate the oxidation state of Cr in [Cr(H₂O)₄Cl₂]Cl.
Determine the number of donor atoms in three molecules of EDTA.

52. 20 Important MCQs

1. In a coordination compound, a ligand generally acts as:

(A) Electron-pair acceptor
(B) Proton donor
(C) Electron-pair donor
(D) Oxidising agent only

Correct Answer: (C) Electron-pair donor

A ligand donates a lone pair to the central metal ion to form a coordinate bond.

2. The coordination number of Co in [Co(en)₃]³⁺ is:

(A) 3
(B) 4
(C) 6
(D) 9

Correct Answer: (C) 6

en is bidentate, so three en ligands provide six donor atoms.

3. Which of the following is an ambidentate ligand?

(A) NH₃
(B) H₂O
(C) NO₂⁻
(D) en

Correct Answer: (C) NO₂⁻

NO₂⁻ can coordinate through N or O.

4. The oxidation state of Fe in K₄[Fe(CN)₆] is:

(A) +1
(B) +2
(C) +3
(D) +4

Correct Answer: (B) +2

x + 6(−1) = −4, so x = +2.

5. The correct name of [Co(NH₃)₆]Cl₃ is:

(A) Hexaamminecobalt(III) chloride
(B) Hexaaminecobalt(III) chloride
(C) Hexachloridocobalt(III) ammine
(D) Cobalt hexammine chloride

Correct Answer: (A) Hexaamminecobalt(III) chloride

NH₃ as a ligand is named ammine.

6. Which ligand is bidentate?

(A) NH₃
(B) Cl⁻
(C) H₂O
(D) en

Correct Answer: (D) en

Ethane-1,2-diamine has two nitrogen donor atoms.

7. The type of isomerism shown by [Pt(NH₃)₂Cl₂] is:

(A) Ionisation isomerism only
(B) Geometrical isomerism
(C) Coordination isomerism only
(D) Solvate isomerism only

Correct Answer: (B) Geometrical isomerism

It exists in cis and trans forms.

8. In an octahedral complex, the lower-energy d-orbitals are:

(A) eg
(B) t₂g
(C) sp³
(D) d²sp³

Correct Answer: (B) t₂g

The t₂g orbitals lie between the axes and experience less repulsion from ligands.

9. The crystal-field splitting energy in an octahedral complex is represented by:

(A) Δt
(B) Δ₀
(C) P
(D) E₀

Correct Answer: (B) Δ₀

Δ₀ or Δoct represents octahedral crystal-field splitting.

10. Which of the following generally acts as a strong-field ligand?

(A) I⁻
(B) Br⁻
(C) Cl⁻
(D) CN⁻

Correct Answer: (D) CN⁻

CN⁻ produces relatively large crystal-field splitting.

11. A complex containing unpaired electrons is:

(A) Diamagnetic
(B) Paramagnetic
(C) Non-magnetic
(D) Always ferromagnetic

Correct Answer: (B) Paramagnetic

Unpaired electrons produce paramagnetic behaviour.

12. The spin-only magnetic moment for two unpaired electrons is:

(A) √3 BM
(B) √8 BM
(C) √15 BM
(D) √24 BM

Correct Answer: (B) √8 BM

μ = √[2(2+2)] = √8 BM.

13. Which of the following is a hexadentate ligand?

(A) NH₃
(B) Cl⁻
(C) en
(D) EDTA⁴⁻

Correct Answer: (D) EDTA⁴⁻

EDTA can coordinate through six donor atoms.

14. In [Co(NH₃)₅Cl]Cl₂, the number of chloride ions outside the coordination sphere is:

(A) 1
(B) 2
(C) 3
(D) 5

Correct Answer: (B) 2

Two Cl⁻ ions are outside the coordination sphere; one Cl⁻ is coordinated to Co.

15. Linkage isomerism is possible because of:

(A) Chelating ligands
(B) Ambidentate ligands
(C) Monodentate ligands only
(D) Counter ions only

Correct Answer: (B) Ambidentate ligands

An ambidentate ligand can coordinate through different donor atoms.

16. Which complex can show optical isomerism?

(A) [Co(en)₃]³⁺
(B) [Co(NH₃)₆]³⁺
(C) [Pt(NH₃)₂Cl₂]
(D) [Co(NH₃)₄Cl₂]⁺

Correct Answer: (A) [Co(en)₃]³⁺

The arrangement of three bidentate en ligands can produce non-superimposable mirror images.

17. In an octahedral complex, the eg orbitals have higher energy because they:

(A) Lie between the axes
(B) Point directly toward the ligands
(C) Have no electrons
(D) Are completely filled

Correct Answer: (B) Point directly toward the ligands

They experience greater electrostatic repulsion.

18. Which compound contains an anionic coordination entity?

(A) [Co(NH₃)₆]Cl₃
(B) [Cu(NH₃)₄]SO₄
(C) K₄[Fe(CN)₆]
(D) [Pt(NH₃)₂Cl₂]

Correct Answer: (C) K₄[Fe(CN)₆]

The coordination entity is [Fe(CN)₆]⁴⁻.

19. The geometry commonly associated with [Pt(NH₃)₂Cl₂] is:

(A) Linear
(B) Tetrahedral
(C) Square planar
(D) Trigonal planar

Correct Answer: (C) Square planar

Pt(II) commonly forms square-planar complexes.

20. Which statement about a chelating ligand is correct?

(A) It coordinates through only one donor atom
(B) It cannot form rings
(C) It coordinates through two or more donor atoms to the same metal ion
(D) It is always negatively charged

Correct Answer: (C) It coordinates through two or more donor atoms to the same metal ion

Chelating ligands form one or more rings with the central metal.

53. Frequently Asked Questions

1. What is the easiest way to identify a ligand?

Look for an ion or molecule capable of donating a lone pair of electrons to the central metal ion.

2. Is NH₃ a neutral ligand?

Yes. NH₃ is a neutral ligand and its ligand charge is taken as zero while calculating oxidation state.

3. Is Cl⁻ always a counter ion?

No. It can be either a ligand or a counter ion.

In:

[Co(NH₃)₅Cl]Cl₂

one Cl⁻ is a ligand and two are counter ions.

4. What is the difference between coordination number and denticity?

Denticity tells how many donor atoms one ligand uses to bind to the metal.

Coordination number counts all donor atoms directly attached to the central metal.

5. Why is [Co(en)₃]³⁺ coordination number 6?

Each en ligand is bidentate. Three en ligands therefore provide:

3 × 2 = 6 donor atoms.

6. What is the difference between cis and trans?

In a cis isomer, identical ligands are adjacent.

In a trans isomer, identical ligands are opposite.

7. Why does [Pt(NH₃)₂Cl₂] show cis-trans isomerism?

Its square-planar structure allows the two Cl ligands to occupy either adjacent or opposite positions.

8. Why are many coordination compounds coloured?

Many have partially filled d-orbitals. Absorption of visible light can promote d–d transitions between split d-orbitals.

9. What is the difference between high-spin and low-spin complexes?

High-spin complexes have a greater number of unpaired electrons, whereas low-spin complexes have more paired electrons because of larger crystal-field splitting.

10. Why are coordination compounds important in daily life and science?

They are involved in biological systems such as haemoglobin and chlorophyll and are also important in medicine, metallurgy, qualitative analysis and industrial processes.

54. Quick Revision / Key Takeaways

Basic Terms

Central metal ion → accepts electron pairs.

Ligand → donates electron pair.

Coordination number → number of donor atoms directly attached to the metal.

Coordination sphere → species inside square brackets.

Counter ion → ion outside the coordination sphere.

Ligand Types

Monodentate → one donor atom

Examples: NH₃, H₂O, Cl⁻

Bidentate → two donor atoms

Examples: en, oxalate

Hexadentate → six donor atoms

Example: EDTA

Ambidentate → can coordinate through different donor atoms

Examples: NO₂⁻, SCN⁻

Werner Theory

Primary valency → oxidation state

Secondary valency → coordination number

Isomerism

Structural

Ionisation
Solvate/hydrate
Linkage
Coordination

Stereoisomerism

Geometrical
Optical
fac-mer in suitable octahedral complexes

Bonding

Octahedral

t₂g lower

eg higher

Δ₀ = octahedral splitting

Tetrahedral

e higher

t₂ lower

Δt ≈ 4/9 Δ₀

Magnetic moment

μ = √[n(n + 2)] BM

Important Examples

[Co(NH₃)₆]Cl₃

→ hexaamminecobalt(III) chloride

K₄[Fe(CN)₆]

→ potassium hexacyanidoferrate(II)

[Pt(NH₃)₂Cl₂]

→ square-planar; cis-trans isomerism

[Co(en)₃]³⁺

→ optical isomerism

[Cu(NH₃)₄]²⁺

→ characteristic deep-blue complex

Final Exam Checklist

Before your Class 12 Chemistry examination, make sure you can confidently:

Define a coordination compound.
Identify the central metal ion and ligands.
Calculate the oxidation state of the central metal.
Calculate coordination number.
Distinguish monodentate, bidentate, polydentate and ambidentate ligands.
Explain Werner’s theory.
Write correct IUPAC names of coordination compounds.
Write formulae from coordination-compound names.
Identify ionisation, hydrate, linkage and coordination isomerism.
Identify cis-trans and fac-mer isomerism.
Understand optical isomerism.
Explain VBT and hybridisation.
Distinguish inner-orbital and outer-orbital complexes.
Understand strong-field and weak-field ligands.
Explain crystal-field splitting.
Distinguish high-spin and low-spin complexes.
Calculate the number of unpaired electrons.
Calculate magnetic moment using μ = √[n(n+2)] BM.
Explain the origin of colour in coordination compounds.
Remember important applications such as cisplatin, haemoglobin, chlorophyll, gold extraction and photographic fixing.

The central idea of this entire chapter is simple:

A coordination compound is built around a central metal ion, and its properties depend strongly on which ligands are attached, how they are attached, how many donor atoms are involved, and how the resulting d-orbitals are arranged in energy.

Once these relationships are understood, nomenclature, isomerism, bonding, magnetic behaviour and colour stop looking like separate topics and become different consequences of the same coordination chemistry.