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Class 12 Chemistry · NCERT Chapter 5 (Part 2)

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Coordination Compounds Class 12 Chemistry Notes

Coordination compounds run from haemoglobin and chlorophyll to cis-platin and vitamin B12. This chapter covers Werner's theory, nomenclature, isomerism, valence bond and crystal field theories, magnetic and spectral properties and the biological and industrial importance of complexes.

Class:12Subject:ChemistryCovers:Nomenclature · Isomerism · VBT · CFTExams:CBSE · CUET · NEET
6 Key Formulas2 Solved Examples40 Practice MCQs
DWritten byDeep Narayan
Updated
Key Concept Summary

Why is [Co(NH₃)₆]³⁺ diamagnetic while [CoF₆]³⁻ is paramagnetic?

NH₃ is a strong-field ligand: in [Co(NH₃)₆]³⁺ the 3d electrons pair up into an inner-orbital d²sp³ complex (low spin, all paired, diamagnetic). F⁻ is weak-field so [CoF₆]³⁻ uses outer 3d orbitals (sp³d², high spin), leaving four unpaired electrons — paramagnetic.

01

Coordination Compounds and Basic Terms

Definition

Coordination compound

A compound containing a central metal atom or ion bonded to a fixed array of surrounding neutral molecules or anions (ligands) — e.g. K₄[Fe(CN)₆] or [Co(NH₃)₆]Cl₃. The metal–ligand bonds are coordinate (dative) covalent bonds.

  • Coordination entity: the whole metal-plus-ligand unit, written in square brackets [Fe(CN)₆]⁴⁻.
  • Ligand: any molecule/ion that donates a lone pair — NH₃, Cl⁻, H₂O, CN⁻, en, EDTA.
  • Monodentate vs polydentate: en (bidentate) and EDTA (hexadentate); a ring formed with the metal is a chelate.
  • Coordination number: number of donor atoms, not ligands, attached to the metal.
  • Counter ions: ions outside the brackets that neutralise the complex; they are ionisable in solution.
Definition

Oxidation state of the metal

The charge left on the metal when all ligands are removed with their electrons — in [Co(NH₃)₆]Cl₃ it is +3, making the coordination entity [Co(NH₃)₆]³⁺.

First step

Before any naming or magnetic problem, fix the metal oxidation state from the overall charge on the complex ion. Write x + (sum of ligand charges) = net charge.

MCQ Questions

Q1
The coordination number of the metal in [Fe(CN)₆]⁴⁻ is:
Q2
EDTA is a ligand with denticity:
02

Werner's Theory of Coordination Compounds

Werner solved the puzzle of constant-valence coordination compounds by proposing two valencies for the metal: a primary (ionisable) valency satisfied by anions and a secondary valency (the fixed coordination number) satisfied by ligands, arranged in space around the metal.

  • Primary valency = oxidation state; it is non-directional and satisfied by counter ions.
  • Secondary valency = coordination number; it is directional and fixes the geometry — 4 gives tetrahedral/square planar, 6 gives octahedral.
  • Werner's isomers: CoCl₃·5NH₃ (1 Cl⁻ free), CoCl₃·4NH₃ (2 Cl⁻ free) etc. — the number of free Cl⁻ ions corresponds to the primary valency.
  • Coordination compounds preserve their secondary valency in solution; only the primary valency ionises.

Experiment he used

Adding AgNO₃ to CoCl₃·xNH₃ precipitates only the ionisable chlorides: 1 for 5NH₃, 2 for 4NH₃, 3 for 3NH₃ — while the coordination number of cobalt stays 6 throughout.

MCQ Questions

Q3
According to Werner, the primary valency equals the:
Q4
The secondary valency of cobalt in CoCl₃·4NH₃ is:
03

Nomenclature of Coordination Compounds

IUPAC naming follows a fixed order: write ligands alphabetically (cation before anion in a complex), then the metal. Prefixes number identical ligands: di/tri for simple (chloro, ammine) but bis/tris/tetrakis for ligands with built-in numbers (en, EDTA).

  • Order: alphabetical by ligand name (ignoring di/tri prefixes), then metal, then oxidation state in Roman numerals.
  • Cationic/neutral complexes: metal name unchanged. Anionic complexes: metal ends in −ate (ferrate, cobaltate, cuperate).
  • Examples: [Co(NH₃)₆]Cl₃ = hexaamminecobalt(III) chloride; K₃[Fe(CN)₆] = potassium hexacyanidoferrate(III).
  • Counter ion named as usual (chloride, sulphate).

Frequent error

'Ammine' (with two m's) is the ligand name in a complex; 'amine' is the organic amine. And for anionic complexes use ferrate/cobaltate, not iron/cobalt.

MCQ Questions

Q5
The correct IUPAC name of [Co(NH₃)₆]Cl₃ is:
Q6
K₄[Fe(CN)₆] is named:
04

Isomerism in Coordination Compounds

Definition

Structural isomerism

Different connectivity of ligands: ionisation ([Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br), linkage (M–NO₂ vs M–ONO), coordination ([Co(NH₃)₆][Cr(CN)₆] vs [Co(CN)₆][Cr(NH₃)₆]), and hydrate/solvate ([Cr(H₂O)₆]Cl₃ vs [Cr(H₂O)₅Cl]Cl₂·H₂O) isomerism.

Definition

Stereoisomerism

Same connectivity, different spatial arrangement: geometrical (cis/trans in square planar and octahedral; fac/mer in octahedral [MA₃B₃]) and optical (non-superimposable mirror images, e.g. [Co(en)₃]³⁺).

  • cis-[Pt(NH₃)₂Cl₂] (cis-platin) is an anti-cancer drug; trans-platin is not.
  • Square planar MA₂B₂: two geometric isomers. Octahedral MA₃B₃: fac and mer.
  • Optical isomers rotate plane-polarised light in opposite senses; a 1:1 mixture is racemic.

Quick test

To tell ionisation isomers, add AgNO₃: the isomer with the extra Cl⁻ outside the bracket precipitates AgCl first.

MCQ Questions

Q7
Geometrical isomerism in square planar complexes arises when the complex is:
Q8
[Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are examples of:
05

Bonding: Valence Bond Theory

Valence bond theory explains complex geometry by hybridisation of the metal's d, s and p orbitals. The metal accepts ligand lone pairs into vacant hybrid orbitals, and the number of such empty hybrid orbitals fixes the coordination number and shape.

  • CN 4 → sp³ (tetrahedral, e.g. [Ni(CO)₄]) or dsp² (square planar, e.g. [Ni(CN)₄]²⁻).
  • CN 6 → d²sp³ inner orbital (low spin) or sp³d² outer orbital (high spin).
  • Inner-orbital complexes pair electrons using the inner (n−1)d orbitals; outer-orbital use nd orbitals.
  • [Co(NH₃)₆]³⁺: Co³⁺ d⁶ + strong NH₃ → d²sp³, all electrons paired, diamagnetic.
  • [CoF₆]³⁻: Co³⁺ d⁶ + weak F⁻ → sp³d², four unpaired electrons, paramagnetic.

Key driver

Whether electrons pair depends on the ligand field strength: strong-field ligands (CN⁻, CO, NH₃) force pairing (low spin); weak-field ligands (F⁻, H₂O) keep unpaired electrons (high spin).

MCQ Questions

Q9
The hybridisation in [Co(NH₃)₆]³⁺ is:
Q10
[Ni(CO)₄] is tetrahedral with hybridisation:
06

Crystal Field Theory and Applications

Crystal field theory views metal–ligand bonding as purely electrostatic: the ligand field splits the d orbitals into lower-energy t₂g (dxy, dyz, dzx) and higher-energy eg (dx²−y², dz²) sets in octahedral complexes. The splitting energy Δₒ decides electron arrangement.

Crystal field stabilisation energy (octahedral, ignoring pairing energy)
  • Spectrochemical series (weak→strong): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < py < NH₃ < en < NO₂⁻ < CN⁻ < CO.
  • Weak field → high spin (e.g. [CoF₆]³⁻); strong field → low spin (e.g. [Co(CN)₆]³⁻).
  • Colour: d–d transition between t₂g and eg absorbs visible light.
  • Magnetic moment from unpaired electrons: μ = √(n(n+2)) BM.
  • cis-platin — anti-cancer drug.
  • EDTA — chelation therapy for lead/mercury poisoning, water softening.
  • Chlorophyll — Mg porphyrin complex; haemoglobin — Fe porphyrin complex; vitamin B12 — cobalt complex.
  • Metal carbonyls, dyes, photographic fixing, dyeing and tanning industries.

Colour logic

Small Δₒ absorbs low-energy (red) light → complex looks green/blue; large Δₒ absorbs violet light → complex looks yellow/red. Both are d–d transitions.

MCQ Questions

Q11
In CFT, a strong-field ligand gives a complex that is:
Q12
Which natural molecule has an Fe²⁺ complex at its centre?
07

NCERT Exercise Questions

From the textbook

NCERT questions for this chapter, in textbook order. Answers are given at a glance below each set.

Exercises (NCERT Q1–Q3)

  1. 1
    Explain Werner's theory of coordination compounds with the help of a suitable example.
  2. 2
    What is the coordination number and oxidation state of cobalt in [Co(NH3)5Cl]Cl2?
  3. 3
    Draw the geometrical isomers for [Pt(NH3)2Cl2].

Answers at a glance

  1. 1
    Werner's theory distinguishes primary (ionisable) and secondary (non‑ionisable, coordination) valencies; e.g., [Co(NH3)6]Cl3 shows 6 NH3 in coordination sphere and 3 ionisable Cl.
  2. 2
    Coordination number = 6; oxidation state of Co = +3.
  3. 3
    Cis‑ and trans‑isomers exist for [Pt(NH3)2Cl2].
08

CBSE Pattern Questions

How CBSE asks this chapter

CBSE generally asks 2-3 questions from Coordination Compounds, focusing on Werner’s theory, nomenclature, VBT, CFT splitting, isomerism and stability constants.

MCQ Questions

Q13
The IUPAC name of [Pt(NH3)2Cl(NO2)] is:
Q14
The coordination number of Co in [Co(NH3)6]3+ is:
Q15
Which of the following complexes is expected to be diamagnetic?
Q16
The complex [NiCl4]2‑ is paramagnetic with two unpaired electrons. Its geometry is:

1 Mark Questions

  1. 1
    Define ligand and give an example.
  2. 2
    What is coordination number?
  3. 3
    Write the formula of tetraamminezinc(II) sulphate.

2 Mark Questions

  1. 1
    Using VBT, explain why [Ni(CN)4]2‑ is diamagnetic.
  2. 2
    What is chelate effect? Give an example.

3 Mark Questions

  1. 1
    Explain crystal field splitting in an octahedral field. How does the magnitude of Δo depend on the ligand?
  2. 2
    Discuss the geometrical isomerism in [Co(NH3)4Cl2]+.

5 Mark Questions

  1. 1
    Compare VBT and CFT with respect to their ability to explain magnetic properties, colour and isomerism in coordination compounds.
  2. 2
    A coordination compound CrCl3·6H2O is mixed with AgNO3 solution. It gives a precipitate of AgCl. On adding excess NH3, the precipitate dissolves. Identify the complex ion present and write its formula.
09

PDF: Assignments & Tests

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Assignment 1 — Topic-wise Practice Set A

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Assignment 2 — Topic-wise Practice Set B

Topic-wise practice covering the second half of the chapter, with answer key.

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CBSE Pattern Test 1

Full CBSE-pattern test (90 minutes, 40 marks) with answer key.

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Second full CBSE-pattern test (90 minutes, 40 marks) with answer key.

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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Coordination number

Count donor atoms, not ligands.

Coordination number charge balance

Find the metal oxidation state from the charge on the complex ion.

Magnetic moment

n = unpaired electrons from VBT/CFT.

CFSE (octahedral)

Plus pairing energy when electrons pair.

Splitting stabilisation (weak d⁴)

Weak ligands: Δₒ < pairing energy.

Splitting stabilisation (strong d⁶)

Strong ligands: Δₒ > pairing energy.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • Naming rules: ligand alphabetical order, di/tri vs bis/tris, Roman-numeral oxidation state, anionic complexes ending in −ate (e.g. ferrate, cobaltate).
  • Determining oxidation state and coordination number from a formula — the starting point of almost every question.
  • IUPAC naming and formula writing for typical complexes (hexaamminecobalt(III) chloride, potassium hexacyanidoferrate(III)).
  • Structural and geometrical (cis/trans, fac/mer) isomerism; optical isomers of [Co(en)₃]³⁺.
  • Werner's primary and secondary valency — 1-mark and MCQ staple.
  • VBT hybridisation with inner/outer complexes; CFT high-spin vs low-spin with Δₒ vs pairing energy.
  • Colour of complexes from d–d transitions and spectrochemical series ordering of ligands.
  • Applications: cis-platin, EDTA, chlorophyll, haemoglobin and vitamin B12 as natural complexes.

Practice Numericals

Step-by-step worked examples

Standard exam-level problems solved with clear conceptual breakdowns.

1

Nomenclature of a complex ion

Problem StatementName [Co(NH₃)₄Cl₂]⁺ and give the oxidation state of cobalt.

Step-by-Step Solution

  1. 1Charge: x + 4(0) + 2(−1) = +1, so x = +3.
  2. 2Ligands in order: ammine, chloro.
  3. 3Name: tetraamminedichloridocobalt(III) ion.

Final Answer

tetraamminedichloridocobalt(III) ion; Co = +3

2

CFT high-spin vs low-spin for d⁴

Problem StatementFor [Mn(CN)₆]³⁻ the d⁴ electron arrangement is t₂g⁴ eg⁰ (low spin). Why does CN⁻ force pairing?

Step-by-Step Solution

  1. 1CN⁻ is a strong-field ligand with a large Δₒ.
  2. 2Since Δₒ > pairing energy, the 4th electron enters the t₂g level instead of going to eg.
  3. 3Configuration t₂g⁴ gives a single pairing and a diamagnetic/less paramagnetic complex.

Final Answer

Strong-field CN⁻ gives a low-spin d⁴ complex.

FAQ

Frequently asked questions

What is the difference between a ligand's denticity and its coordination number?

Denticity is the number of donor atoms a single ligand uses to bond to the metal (bidentate EDTA uses six). The coordination number of the complex is the total number of donor atoms attached to the metal, which equals the sum of denticities of all ligands.

Why does EDTA form very stable complexes?

EDTA is a hexadentate ligand furnishing six donor atoms, so it wraps around the metal in many chelate rings. Chelate complexes are extra stable (chelate effect) because the entropy rises when many tightly bound water ligands are released into free solution. This is why EDTA is used to treat lead and mercury poisoning and in hard-water softening.

How do you decide between inner- and outer-orbital complexes in VBT?

Look at the metal's electron count and the ligand field. Inner-orbital (d²sp³) complexes use inner 3d orbitals, form when a strong-field ligand pairs electrons (low spin), and the complex tends to be diamagnetic/less paramagnetic. Outer-orbital (sp³d²) complexes use 4d orbitals with weak-field ligands (high spin) and more unpaired electrons.

Why are most transition metal complexes coloured?

Visible light promotes an electron between the t₂g and eg levels of the split d orbitals (d–d transition). The transmitted light shows the complementary colour. If the complex has no d electrons (d⁰) or a filled subshell (d¹⁰), no such transition is possible and the complex is colourless.

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