Coordination Compounds

Covers NCERT Class 12 Chemistry, Chapter "Coordination Compounds" — Werner's theory, nomenclature, isomerism, VBT and Crystal Field Theory. A consistently very-high-yield chapter for NEET.

Foundation Advanced High-Yield Revision
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⭐ NTA-favourite subtopics in this chapter

IUPAC nomenclature of complexes, calculating EAN and oxidation state of the central metal, CFT splitting (Δ₀ vs Δₜ) and high-spin/low-spin configuration, magnetic behaviour from unpaired electrons, isomerism (geometrical & optical in octahedral/square planar complexes), and the difference between double salts and coordination compounds (ionisation in water).

Exam-oriented notes

NCERT-tagged theory, simplified

NCERT §9.1

Werner's theory: metals show two types of valency — primary (ionisable, satisfied by anions) and secondary (non-ionisable, satisfied by ligands, fixed = coordination number, directional in space).

NCERT §9.2

A ligand donates at least one electron pair to the central atom/ion. Denticity = number of donor atoms from one ligand binding the same metal — monodentate (NH₃), bidentate (en, ox²⁻), polydentate (EDTA⁴⁻, hexadentate).

NCERT §9.2

Coordination number = number of ligand donor atoms directly bonded to the central metal — 4 usually gives tetrahedral or square planar, 6 usually gives octahedral geometry.

NCERT §9.3

Nomenclature order inside the coordination sphere: ligands named alphabetically (ignoring multiplying prefixes), anionic ligands end in "-o", then the metal name, then its oxidation state in Roman numerals in brackets. Complex anions end in "-ate".

NCERT §9.4

Two structurally distinct classes of isomerism dominate NEET questions: (a) geometrical (cis/trans, fac/mer) — needs at least two different ligand types on a square planar or octahedral centre; (b) optical — non-superimposable mirror images, common in octahedral complexes with bidentate ligands.

NCERT §9.5

Valence Bond Theory: hybridisation of the metal decides geometry — sp³ (tetrahedral), dsp² (square planar), sp³d² (outer orbital octahedral, high spin), d²sp³ (inner orbital octahedral, low spin).

NCERT §9.6

Crystal Field Theory: in an octahedral field, the d-orbital set splits into a lower t₂g (3 orbitals) and higher eg (2 orbitals) set, separated by Δ₀. Strong-field ligands (CN⁻, CO) give large Δ₀ → pairing favoured → low spin; weak-field ligands (F⁻, H₂O, halides) give small Δ₀ → high spin.

Spectrochemical series (memorise order)

I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ ≈ CO (weak → strong field, increasing Δ₀).

Double salt vs complex

A double salt (e.g. Mohr's salt, potash alum) ionises completely into all its simple ions in water and loses its identity; a coordination compound retains its complex ion largely intact in solution (e.g. [Fe(CN)₆]⁴⁻ does not test positive for free Fe²⁺ or free CN⁻).

Printable

Formula & reference sheet

Coordination Compounds — key relations

EAN
EAN = Z − oxidation state + 2×C.N.
Magnetic moment
μ = √n(n+2) Bohr Magneton
Octahedral splitting
Δ₀ separates t₂g (lower) & eg (upper)
Tetrahedral splitting
Δₜ ≈ (4/9)Δ₀ — always weaker
Common geometries
C.N.4 → tetrahedral / square planar; C.N.6 → octahedral
HybridisationGeometrySpin typeExample
sp³Tetrahedral[NiCl₄]²⁻
dsp²Square planarLow spin[Ni(CN)₄]²⁻
sp³d²Octahedral (outer orbital)High spin[FeF₆]³⁻
d²sp³Octahedral (inner orbital)Low spin[Fe(CN)₆]³⁻
PYQ-style practice

Practice set, NEET pattern

Original questions modelled on recurring NEET question types — not verbatim reproductions of any official paper.

Q1. The oxidation state of Fe in [Fe(CN)₆]⁴⁻ is:
  • +2
  • +3
  • +4
  • 0
Show solution
CN⁻ contributes −6 total; overall charge −4, so Fe + (−6) = −4 ⇒ Fe = +2. Answer: A
Q2. Which pair correctly represents a strong-field and a weak-field ligand respectively?
  • CN⁻, CO
  • F⁻, I⁻
  • CN⁻, F⁻
  • H₂O, NH₃
Show solution
By the spectrochemical series, CN⁻ is strong field, F⁻ is weak field. Answer: C
Q3. [Ni(CO)₄] is diamagnetic while [NiCl₄]²⁻ is paramagnetic because:
  • CO is a strong-field ligand forcing pairing (dsp², no unpaired e⁻); Cl⁻ is weak field, sp³, unpaired e⁻ retained
  • Ni has different oxidation states in the two complexes
  • CO is negatively charged while Cl⁻ is neutral
  • [NiCl₄]²⁻ is octahedral
Show solution
Both have Ni in the same (0/+2 respectively, but the key driver is) ligand field strength: CO forces electron pairing (square planar/dsp², diamagnetic); Cl⁻ is weak field, retains unpaired electrons in the sp³ tetrahedral complex, hence paramagnetic. Answer: A
Q4. Which of the following ligands is bidentate?
  • NH₃
  • Cl⁻
  • Ethylenediamine (en)
  • H₂O
Show solution
Ethylenediamine has two donor N atoms and can bind the same metal at two sites. Answer: C
Q5. The IUPAC name of [Co(NH₃)₅Cl]Cl₂ is:
  • Pentaamminechloridocobalt(III) chloride
  • Pentaamminechloridocobalt(II) chloride
  • Chloridopentaamminecobalt(III) chloride
  • Pentaamminecobalt(III) chloride
Show solution
Ligands alphabetically: ammine before chlorido. Charge balance: complex ion charge +2 (2 Cl⁻ outside), Co + 0(NH₃) − 1(Cl) = +2 ⇒ Co = +3. Answer: A
Q6. Potash alum, K₂SO₄·Al₂(SO₄)₃·24H₂O, is best classified as:
  • A coordination compound
  • A double salt — ionises completely into simple ions in water
  • A chelate
  • An organometallic compound
Show solution
Potash alum fully dissociates into K⁺, Al³⁺ and SO₄²⁻ in solution and each ion can be tested for individually — the hallmark of a double salt, not a complex. Answer: B
Daily Practice Problems

DPP — Coordination Compounds (15 questions)

Scaled from direct NCERT application (Q1–Q6) to mixed NEET-level difficulty (Q7–Q15).

1.Define coordination number with one example.
2.Find the oxidation state of Cr in K₂Cr₂O₇ and in [Cr(NH₃)₆]³⁺.
3.Name the type of isomerism shown by [Co(NH₃)₄Cl₂]⁺ (cis and trans forms).
4.Write the formula for tetraamminecopper(II) sulphate.
5.State whether EDTA is monodentate or polydentate; give its denticity.
6.State Werner's postulate distinguishing primary and secondary valency.
7.Predict the geometry and magnetic nature of [CoF₆]³⁻ (weak field F⁻, Co³⁺ is d⁶).
8.Calculate the EAN of Co in [Co(NH₃)₆]³⁺ (Z of Co = 27).
9.Explain why [Ni(H₂O)₆]²⁺ is coloured but [Zn(H₂O)₆]²⁺ is colourless.
10.Which shows optical isomerism: [Pt(NH₃)₂Cl₂] (square planar) or [Co(en)₃]³⁺ (octahedral)? Justify.
11.Arrange in increasing Δ₀: [Co(H₂O)₆]³⁺, [Co(NH₃)₆]³⁺, [Co(CN)₆]³⁻.
12.Give an everyday/biological example of a naturally occurring coordination compound.
13.Distinguish ionisation isomerism from linkage isomerism with one example each.
14.Why is [NiCl₄]²⁻ tetrahedral (sp³) while [Ni(CN)₄]²⁻ is square planar (dsp²)?
15.Assertion: [Fe(CN)₆]³⁻ is low spin. Reason: CN⁻ is a strong-field ligand causing large Δ₀ and pairing of electrons in t₂g. Judge the pair.
Show answer key
1. Number of donor atoms bonded to metal, e.g. 6 in [Fe(CN)₆]⁴⁻  |  2. +6 in dichromate, +3 in the ammine complex  |  3. Geometrical (cis-trans) isomerism  |  4. [Cu(NH₃)₄]SO₄  |  5. Polydentate, hexadentate  |  6. Primary = ionisable, satisfied by anions; secondary = fixed, non-ionisable, satisfied by ligands  |  7. Octahedral, high spin, paramagnetic (4 unpaired e⁻)  |  8. EAN = 27 − 3 + 12 = 36  |  9. Ni²⁺ (d⁸) has d-d transitions possible; Zn²⁺ (d¹⁰) has a completely filled d-subshell, no d-d transition possible  |  10. [Co(en)₃]³⁺ — octahedral with bidentate ligands lacks a plane of symmetry; the square planar complex generally does not show optical isomerism  |  11. [Co(H₂O)₆]³⁺ < [Co(NH₃)₆]³⁺ < [Co(CN)₆]³⁻  |  12. Haemoglobin (Fe) or chlorophyll (Mg)  |  13. Ionisation: different ions in solution, e.g. [Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br; linkage: ambidentate ligand binds via different atom, e.g. -NO₂ vs -ONO  |  14. Cl⁻ is weak field (high spin, sp³); CN⁻ is strong field, forces pairing, leaving a d-orbital free for dsp²  |  15. Both true, reason correctly explains assertion
Story mode

A memory-only mythology narrative for coordination chemistry

Purely a recall device — not a reinterpretation of the source stories.

📖 Indra's court — the coordination sphere

Picture the central metal ion as a king holding court, like Indra. Only a fixed number of courtiers are allowed in the inner circle at once — that fixed number is the coordination number (commonly 4 or 6). Each courtier (ligand) offers a pair of electrons as tribute — a coordinate bond, donor to acceptor, one-directional in origin even though the resulting bond is as strong as any covalent bond. A courtier skilled enough to serve the king from two posts at once (two donor atoms, like ethylenediamine) is the bidentate ligand; one who can serve from six posts (EDTA) is prized above all — a hexadentate advisor.

📖 The court's seating order — Crystal Field splitting

In an octahedral court, six courtiers stand along the x, y and z axes. Two of the king's own attendants (the eg orbitals) happen to be seated exactly along those axes and feel the crowding most; three others (t₂g) sit in between the axes and are comparatively undisturbed. That difference in "how crowded each seat feels" is Δ₀. Persuasive courtiers (strong-field ligands like CN⁻) crowd the king's attendants enough to force them to pair up in the calmer seats rather than spread out — giving a low-spin arrangement.

🧠 Spectrochemical series, one line

"I Bring Cats Fully Home, Watching Nightly, Eating Cheese, Constantly" → I⁻, Br⁻, Cl⁻, F⁻, H₂O, NH₃, en, CN⁻/CO — weak field to strong field, left to right.