Why is CO a strong field ligand but H2O is weak?

CO is a strong field ligand because it can accept pi-back-donation from metal d-orbitals into its empty pi* antibonding orbital (synergic bonding), greatly stabilising the complex and increasing delta-o. H2O has no low-lying empty orbital for back-bonding, so it acts as a sigma-donor only and gives a weaker field.

Why is [Ni(CN)4]2- square planar while [NiCl4]2- is tetrahedral?

Ni2+ is 3d8. In [Ni(CN)4]2-, CN- is a strong field ligand that pairs the 3d electrons, freeing one 3d orbital (dsp2 hybridisation, square planar, diamagnetic). In [NiCl4]2-, Cl- is weak field and cannot pair the electrons, so outer orbitals (4s, 4p) are used (sp3, tetrahedral, paramagnetic with 2 unpaired electrons).

How many questions come from coordination compounds in JEE Mains?

On average 1-2 questions (4-8 marks) appear from Coordination Compounds in every JEE Main session. Key topics: IUPAC nomenclature, hybridisation & magnetic behaviour (VBT), isomerism, CFSE calculations, and the spectrochemical series.

Coordination Compounds JEE Notes 2026 | IUPAC, VBT, CFT, CFSE, PYQs & Trends

1

What are Coordination Compounds?

A coordination compound is a molecule/ion in which a central metal atom/ion is surrounded by a fixed number of ions or neutral molecules (called ligands) held by coordinate (dative) bonds. The branch of chemistry dealing with these compounds is called Coordination Chemistry, founded by Alfred Werner.

Example: In [Co(NH₃)₆]Cl₃, Co³⁺ is the central metal ion, NH₃ molecules are ligands, the bracketed portion [Co(NH₃)₆]³⁺ is the coordination sphere, and the 3 Cl⁻ ions outside the sphere form the ionisation sphere.

Coordination compounds are everywhere — chlorophyll (Mg complex), haemoglobin (Fe complex), vitamin B₁₂ (Co complex), cisplatin (anticancer Pt drug), and the Wilkinson's catalyst for hydrogenation. In analytical chemistry, the detection of Cu²⁺, Ni²⁺, Fe³⁺, etc., all rely on complex formation.

2

Werner's Coordination Theory

In 1893, Alfred Werner (Nobel Prize 1913) proposed the first successful explanation of complexes like CoCl₃·6NH₃. His postulates:

In coordination compounds, metals show two types of valencies:
- Primary (ionisable) valency — corresponds to oxidation state; satisfied by negative ions; non-directional; lost in solution as ions.
- Secondary (non-ionisable) valency — corresponds to coordination number; satisfied by ligands (neutral or negative); directional → decides geometry.
Every metal has a fixed number of secondary valencies (coordination number, usually 6 or 4).
Ligands satisfying secondary valencies are inside the coordination sphere and do NOT ionise.
Ions outside the sphere (primary valencies) ionise and give tests in solution (e.g., AgNO₃ test for Cl⁻).

★ IMPORTANT (Most Repeated PYQ Topic)

Werner's classification of CoCl₃·nNH₃ adducts based on number of Cl⁻ ions precipitated by AgNO₃ is among the most repeated questions in NEET and JEE. Memorise this table.

Werner's Experimental Series of Co(III)–NH₃–Cl Complexes

Complex (old form)	Modern Formula	Cl⁻ precipitated by AgNO₃	Ions in solution	Colour
CoCl₃·6NH₃	[Co(NH₃)₆]Cl₃	3	4 ([Co(NH₃)₆]³⁺ + 3Cl⁻)	Yellow
CoCl₃·5NH₃	[Co(NH₃)₅Cl]Cl₂	2	3 ([Co(NH₃)₅Cl]²⁺ + 2Cl⁻)	Purple
CoCl₃·4NH₃	[Co(NH₃)₄Cl₂]Cl	1	2 ([Co(NH₃)₄Cl₂]⁺ + Cl⁻)	Green / Violet
CoCl₃·3NH₃	[Co(NH₃)₃Cl₃]	0	0 (non-electrolyte)	—

Tip: Number of Cl⁻ ions precipitated = number of Cl⁻ outside the coordination sphere. The total Cl⁻ in formula = primary valency ions + secondary valency Cl⁻ inside sphere.

▶ PYQ 2024 (NEET)

Q. The number of chloride ions precipitated by AgNO₃ when 1 mol of CoCl₃·4NH₃ is treated with excess silver nitrate is:

A. CoCl₃·4NH₃ → [Co(NH₃)₄Cl₂]Cl. Only the Cl⁻ outside the sphere ionises.

Answer: 1 mol of Cl⁻ is precipitated (per mole of complex).

3

Important Terms & Definitions

2.1 Ligand

A ligand is an ion or molecule that can donate a lone pair of electrons to the central metal atom/ion. Classifications by denticity (number of donor sites):

Type	# Donor atoms	Examples
Unidentate	1	H₂O, NH₃, Cl⁻, CN⁻, OH⁻, CO
Bidentate	2	en (ethane-1,2-diamine), ox (oxalate C₂O₄²⁻), acac
Polydentate	≥3	EDTA (hexadentate, 6 donor atoms), dien (tridentate)
Ambidentate	1 (but 2 different sites possible)	NO₂⁻/ONO⁻, CN⁻/NC⁻, SCN⁻/NCS⁻
Chelating ligand	≥2 (forms ring)	en, ox, EDTA

★ IMPORTANT (Most Repeated PYQ Topic)

Identifying ambidentate ligands and distinguishing chelate complexes is asked very frequently. The classic examples — NO₂⁻ (nitrito-N vs nitrito-O), SCN⁻ (thiocyanato-S vs isothiocyanato-N) — must be memorised.

2.2 Coordination Number (CN)

The number of σ-bonds between the central metal and ligands. Note: only sigma bonds count, not π-bonds.

CN of [Co(NH₃)₆]³⁺ = 6 (six σ-bonds to N of NH₃)
CN of [PtCl₄]²⁻ = 4
CN of [EDTA]⁴⁻ complexes = 6 (hexadentate)

2.3 Coordination Sphere & Ionisation Sphere

In K₄[Fe(CN)₆], the part [Fe(CN)₆]⁴⁻ (in square brackets) is the coordination sphere; the 4 K⁺ counter ions form the ionisation sphere. Double salts (e.g., Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O) dissociate completely into constituent ions in solution, whereas complex salts retain their complex ion in solution.

2.4 Oxidation Number of Central Metal

The charge the metal would carry if all ligand donor atoms were removed along with their electron pairs. Found by charge balance.

For [Co(NH₃)₆]Cl₃: Total charge = 0 → [Co(NH₃)₆] + 3Cl⁻ = 0
→ [Co(NH₃)₆] = +3, NH₃ is neutral → Co = +3

For K₄[Fe(CN)₆]: 4K⁺ + [Fe(CN)₆] = 0 → [Fe(CN)₆] = –4; 6 CN⁻ give –6 → Fe = +2

2.5 Homoleptic vs Heteroleptic

Homoleptic — only one type of ligand (e.g., [Co(NH₃)₆]³⁺). Heteroleptic — more than one type of ligand (e.g., [Co(NH₃)₄Cl₂]⁺).

▶ PYQ 2023 (JEE Main)

Q. Which of the following is an ambidentate ligand?

(a) H₂O (b) NH₃ (c) NO₂⁻ (d) en

Answer: (c) NO₂⁻

Reasoning: NO₂⁻ can donate through N (nitrito-N, –NO₂) or O (nitrito-O, –ONO). It has two different donor atoms.

4

IUPAC Nomenclature of Coordination Compounds

★ IMPORTANT — #1 Most Repeated Topic

IUPAC nomenclature is the single most frequently tested concept in NEET & JEE for Coordination Compounds (asked almost every year). Mastering the rules below guarantees marks.

3.1 Rules for Naming Complexes

Cation first, then anion (as in simple salts). For [Co(NH₃)₆]Cl₃, name the cation [Co(NH₃)₆]³⁺ first, then "chloride".
Within the complex sphere: ligands are named first (alphabetically), then the metal.
Ligand names:
- Anionic ligand → end in -o (Cl⁻ = chlorido, CN⁻ = cyanido, OH⁻ = hydroxido, NO₂⁻ = nitrito-N, SO₄²⁻ = sulfato)
- Neutral ligand → keep name (NH₃ = ammine, H₂O = aqua, CO = carbonyl, NO = nitrosyl)
- Organic molecules: ethane-1,2-diamine = en (still name in full)
Number of ligands indicated by prefixes: di, tri, tetra, penta, hexa for simple; bis, tris, tetrakis for complex ligand names (those already containing a numerical prefix).
Metal name:
- If complex is cationic/neutral → metal name unchanged.
- If complex is anionic → metal name ends in -ate (Co → cobaltate, Cr → chromate, Pt → platinate, Fe → ferrate, Cu → cuprate, Ag → argentate, Au → aurate, Sn → stannate, Pb → plumbate).
Oxidation state of metal written in Roman numerals in parentheses after metal name (Stock notation).
No space between ligand name and metal name; one word for the whole complex ion.

3.2 Worked Examples

Formula	IUPAC Name
[Co(NH₃)₆]Cl₃	Hexaamminecobalt(III) chloride
[Pt(NH₃)₂Cl₂]	Diamminedichloridoplatinum(II)
K₄[Fe(CN)₆]	Potassium hexacyanidoferrate(II)
K₃[Fe(CN)₆]	Potassium hexacyanidoferrate(III)
[Cr(en)₃]Cl₃	Tris(ethane-1,2-diamine)chromium(III) chloride
[Co(NH₃)₄Cl₂]Cl	Tetraamminedichloridocobalt(III) chloride
Na₃[Co(NO₂)₆]	Sodium hexanitrito-N-cobaltate(III)
[Ni(CO)₄]	Tetracarbonylnickel(0)
Fe(CO)₅	Pentacarbonyliron(0)
[PtCl₄]²⁻	Tetrachloridoplatinate(II) ion

Tricks to avoid mistakes:

"Ammine" (NH₃) vs "amine" (organic). Always "ammine" with double m.
For ligands already containing prefix (e.g., ethane-1,2-diamine), use bis, tris, tetrakis.
Alphabetical order: use the first letter of the ligand's English name, NOT the prefix. So "diammine" alphabetises under "a", not "d".
For ambidentate ligands, specify donor atom: nitrito-N (-NO₂) or nitrito-O (-ONO); thiocyanato-S (-SCN) or isothiocyanato (-NCS).

▶ PYQ 2024 (NEET)

Q. The IUPAC name of [Pt(NH₃)₂Cl(NO₂)] is:

Answer: Diamminechloridonitrito-N-platinum(II)

Explanation: Two NH₃ → diammine, Cl⁻ → chlorido, NO₂⁻ bound via N → nitrito-N. Pt is in +2 oxidation state (NH₃ = 0, Cl = –1, NO₂ = –1, complex neutral). Alphabetical order: ammine < chlorido < nitrito-N.

5

Isomerism in Coordination Compounds

★ IMPORTANT — Top-3 Repeated PYQ Topic

Isomerism (especially geometrical (cis-trans) and optical isomerism in octahedral complexes) is heavily tested. NEET 2022, 2020, 2018, 2016 + JEE Main all featured this. You MUST know which combinations give rise to optical isomers.

4.1 Structural Isomerism

Ionisation isomerism: exchange of ions between spheres. e.g., [Co(NH₃)₅SO₄]Br (gives Br⁻ test with AgNO₃) vs [Co(NH₃)₅Br]SO₄ (gives SO₄²⁻ test with BaCl₂).
Hydrate/Solvate isomerism: exchange of water. [Cr(H₂O)₆]Cl₃ (violet, 3 Cl⁻ precipitated) vs [Cr(H₂O)₅Cl]Cl₂·H₂O (blue-green, 2 Cl⁻) vs [Cr(H₂O)₄Cl₂]Cl·2H₂O (dark green, 1 Cl⁻).
Linkage isomerism: ambidentate ligands bind through different atoms. [Co(NH₃)₅NO₂]²⁺ (nitrito-N, yellow) vs [Co(NH₃)₅ONO]²⁺ (nitrito-O, red).
Coordination isomerism: both cation and anion are complexes; ligand exchange between them. [Co(NH₃)₆][Cr(CN)₆] vs [Cr(NH₃)₆][Co(CN)₆].
Coordination position isomerism: in polynuclear complexes, ligands swap positions.

4.2 Stereoisomerism

(A) Geometrical (cis-trans) Isomerism: arises when ligands occupy different relative positions.

Square planar [Ma₂b₂]: e.g., [Pt(NH₃)₂Cl₂] — cis (Peyrone's salt, anti-tumour) and trans.
Octahedral [Ma₄b₂]: e.g., [Co(NH₃)₄Cl₂]⁺ — cis & trans.
Octahedral [Ma₃b₃]: e.g., [Rh(py)₃Cl₃] — facial (fac) & meridional (mer).
Tetrahedral complexes do NOT show geometrical isomerism.

(B) Optical Isomerism: complexes that are non-superimposable on their mirror image (chiral) show optical activity. They rotate plane-polarised light.

Octahedral complexes showing optical isomerism:

[M(en)₃]ⁿ⁺ type — e.g., [Co(en)₃]³⁺, [Cr(en)₃]³⁺ (most famous)
[M(AA)₂a₂] — with bidentate symmetrical ligands
[M(AA)₂bc] — with two bidentate + two unidentate
[M(AA)b₂c₂]
[Mabcdef] — all six ligands different (15 pairs!)

Octahedral complexes NOT showing optical isomerism: [Ma₆], [Ma₅b], [Ma₄b₂], [Ma₃b₃], [M(AA)₃] only with symmetric AA — wait, [M(AA)₃] DOES show optical activity.

Caution: [M(AA)₃] type (like [Co(en)₃]³⁺) is optically active — exists as d and l enantiomers. Cis-[M(AA)₂a₂] is optically active but trans is NOT. This distinction is the heart of many PYQs.

▶ PYQ 2022 (NEET)

Q. Which of the following will show optical isomerism?

(a) [Co(NH₃)₃Cl₃] (b) [Pt(NH₃)₂Cl₂] (c) [Co(en)₃]³⁺ (d) [Co(NH₃)₅Cl]²⁺

Answer: (c) [Co(en)₃]³⁺

Explanation: Only complexes with two or more bidentate chelating ligands (without plane of symmetry) exhibit optical isomerism. [Co(en)₃]³⁺ has three bidentate ethane-1,2-diamine ligands, has no plane of symmetry, and exists as d- and l-forms.

▶ PYQ 2021 (JEE Main)

Q. The complex that can show fac-mer isomerism is:

(a) [Ma₂b₂] (b) [Ma₃b₃] (c) [Ma₄b₂] (d) [Mabcdef]

Answer: (b) [Ma₃b₃]

Explanation: fac (facial — three identical ligands on one face of octahedron) and mer (meridional — three identical ligands in a plane containing the metal) are possible only when 3 of each ligand type are present, i.e., [Ma₃b₃] octahedral.

6

Valence Bond Theory (VBT)

Proposed by Linus Pauling. VBT explains bonding, geometry, and magnetic properties of complexes by considering hybridisation of metal orbitals.

5.1 Main Postulates

The central metal ion makes available a number of empty s, p and d orbitals (of appropriate energy) to accept lone pairs from ligands.
These orbitals hybridise to give a set of equivalent orbitals with definite geometry.
Each ligand donates a pair of electrons to the empty hybridised orbital → coordinate bond (dative σ-bond).
If inner (n–1)d orbitals are used → inner orbital / low spin complex; if outer nd orbitals are used → outer orbital / high spin complex.
Magnetic moment μ (in BM) is related to number of unpaired electrons n by: μ = √[n(n+2)] BM.

Hybridisation ↔ Geometry table

sp³ → Tetrahedral (CN = 4)
dsp² → Square planar (CN = 4)
sp³d² → Octahedral, outer (CN = 6, high spin)
d²sp³ → Octahedral, inner (CN = 6, low spin)

5.2 VBT Analysis — Famous Examples

Case 1: [Ni(CO)₄] — Tetrahedral, diamagnetic

Ni(0): 3d⁸4s². CO is a strong field ligand → pairs up 4s electrons, vacates 4s. Hybridisation: sp³ (one 4s + three 4p). Geometry = tetrahedral. Magnetic moment = 0 (diamagnetic).

Case 2: [Ni(CN)₄]²⁻ — Square planar, diamagnetic

Ni(II): 3d⁸. CN⁻ is strong field → pairs up 3d electrons, vacates one 3d orbital. Hybridisation: dsp² (one 3d + 4s + two 4p). Geometry = square planar. μ = 0.

Case 3: [NiCl₄]²⁻ — Tetrahedral, paramagnetic

Ni(II): 3d⁸. Cl⁻ is weak field → no pairing. Hybridisation: sp³ (using 4s, 4p). Geometry = tetrahedral. μ = √[2(2+2)] = √8 ≈ 2.83 BM (2 unpaired e⁻).

Case 4: [Co(NH₃)₆]³⁺ — Inner orbital octahedral, diamagnetic

Co(III): 3d⁶. NH₃ strong field → pairs up all six 3d electrons into 3 orbitals, leaving two 3d vacant. Hybridisation: d²sp³ (inner orbital). μ = 0.

Case 5: [CoF₆]³⁻ — Outer orbital octahedral, paramagnetic

Co(III): 3d⁶. F⁻ weak field → no pairing. Hybridisation: sp³d² (outer orbital). 4 unpaired e⁻. μ = √[4(6)] = √24 ≈ 4.90 BM.

Case 6: [Fe(CN)₆]⁴⁻ — Inner orbital octahedral, diamagnetic

Fe(II): 3d⁶. CN⁻ strong field → pairs up. d²sp³. μ = 0.

Case 7: [Fe(H₂O)₆]²⁺ — Outer orbital octahedral, paramagnetic

Fe(II): 3d⁶. H₂O weak field → no pairing. sp³d². 4 unpaired e⁻. μ ≈ 4.90 BM.

★ IMPORTANT — Most Repeated VBT PYQ

"Why is [Ni(CN)₄]²⁻ square planar but [NiCl₄]²⁻ tetrahedral?" or "Why does [Ni(CO)₄] have tetrahedral geometry?" — asked in NEET 2018, JEE 2019, NEET 2020. Memorise Cases 1, 2, 3 above.

5.3 Magnetic Moment Formula

μ = √[n(n+2)] BM (spin-only)

where n = number of unpaired electrons.

n	1	2	3	4	5
μ (BM)	1.73	2.83	3.87	4.90	5.92

▶ PYQ 2019 (NEET)

Q. The hybridisation and magnetic behaviour of [Co(CN)₆]³⁻ is:

(a) d²sp³, paramagnetic (b) sp³d², diamagnetic (c) d²sp³, diamagnetic (d) sp³d², paramagnetic

Answer: (c) d²sp³, diamagnetic

Explanation: Co(III) = 3d⁶. CN⁻ is a strong field ligand → pairs up all six electrons into 3 orbitals of 3d, leaving two 3d orbitals vacant. These combine with 4s and two 4p → d²sp³ hybridisation. Zero unpaired electrons → diamagnetic.

7

Crystal Field Theory (CFT)

Proposed by Hans Bethe (1929) and refined by van Vleck. CFT treats ligands as point negative charges that interact electrostatically with metal d-orbitals. No covalent bonding is assumed.

6.1 Key Assumptions

Ligands are point charges (anions) or point dipoles (neutral molecules like NH₃, H₂O).
Metal–ligand interaction is purely electrostatic (ionic).
The five d-orbitals of the free metal ion are degenerate (same energy).
When ligands approach, the d-orbitals experience repulsion and split according to the geometry of the complex.

6.2 Crystal Field Splitting in Octahedral Field

In an octahedral complex, six ligands approach along ±x, ±y, ±z axes. The two orbitals pointing along axes — d(x²–y²) and d(z²) (called e_g set) — experience greater repulsion and are raised in energy. The three orbitals lying between axes — d(xy), d(yz), d(xz) (called t_2g set) — experience less repulsion and are lowered.

Splitting:

t_2g (3 orbitals) — lowered by 0.4Δ_o
e_g (2 orbitals) — raised by 0.6Δ_o

Where Δ_o (or 10Dq) = octahedral crystal field splitting energy.

Energy conservation: 3(–0.4Δ_o) + 2(+0.6Δ_o) = 0 (barycentre rule).

6.3 Crystal Field Splitting in Tetrahedral Field

In a tetrahedral complex, four ligands approach between the axes. Here, the e set (d(z²), d(x²–y²)) is lowered and t₂ set (d(xy), d(yz), d(xz)) is raised.

Δ_t = (4/9) Δ_o

Tetrahedral splitting is smaller (about 4/9 of octahedral) because:

Fewer ligands (4 vs 6)
Ligands do not approach along axes

Order of orbitals is reversed compared to octahedral: e (low) & t₂ (high).

6.4 Strong Field vs Weak Field — High Spin vs Low Spin

Property	Strong Field (SFL)	Weak Field (WFL)
Δ_o vs pairing energy P	Δ_o > P	Δ_o < P
Electron filling	Pairing in t_2g first	Fill singly across all 5
Spin state	Low spin (LS)	High spin (HS)
Unpaired electrons	Fewer	More
Magnetic moment	Lower / 0	Higher
Examples	CN⁻, CO, NO₂⁻, NH₃	F⁻, Cl⁻, H₂O, OH⁻

★ IMPORTANT — Most Repeated CFT PYQ

Spectrochemical series and high-spin vs low-spin d-electron configurations for d⁴–d⁷ octahedral complexes are tested almost every year.

6.5 The Spectrochemical Series

I⁻ < Br⁻ < S²⁻ < SCN⁻(S) < Cl⁻ < NO₃⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻(N) < CH₃CN < py < NH₃ < en < bpy < phen < NO₂⁻ < PPh₃ < CN⁻ < CO

Weak field ← — — — — — — — — — — — — — → Strong field

Mnemonic: "I Broke Sister's Car, Now Father Only Helps, Came Home Nervous, Now Everyone's Pretty, N-C-O Becomes Stronger"

OR simply: halides < oxygen donors < nitrogen donors < carbon donors.

▶ PYQ 2023 (NEET)

Q. Which of the following ligands will produce a low-spin octahedral complex with Fe³⁺ (d⁵)?

(a) F⁻ (b) H₂O (c) CN⁻ (d) Cl⁻

Answer: (c) CN⁻

Explanation: CN⁻ is one of the strongest field ligands. With CN⁻, Δ_o > P → electrons pair up in t_2g → low-spin d⁵ gives t_2g⁵ e_g⁰ → 1 unpaired electron. F⁻, H₂O, Cl⁻ are weak-field → high-spin d⁵ = t_2g³ e_g² → 5 unpaired e⁻.

8

CFSE (Crystal Field Stabilisation Energy) Calculations

CFSE is the net energy gained by the d-electrons due to splitting of d-orbitals in a crystal field, relative to the barycentre (unsplit state).

For octahedral complex:

CFSE = (–0.4 × n_t2g + 0.6 × n_eg) Δ_o + P × n_pairs

For tetrahedral complex:

CFSE = (–0.6 × n_e + 0.4 × n_t2) Δ_t + P × n_pairs

where n = number of electrons in that set, P = pairing energy, n_pairs = number of pairs formed due to splitting (not the original pairs).

★ IMPORTANT

CFSE calculations for d¹–d¹⁰ systems (both octahedral HS and LS) are routinely asked in JEE Main. The "two-peak" stability curve across the 3d series (max at d³ and d⁸ LS, d⁵ HS and d¹⁰ HS being minima) is a frequent conceptual question.

7.1 CFSE Table for Octahedral Complexes

dⁿ	High-spin (HS) config	CFSE (HS)	Low-spin (LS) config	CFSE (LS)
d¹	t_2g¹	–0.4Δ_o	—	—
d²	t_2g²	–0.8Δ_o	—	—
d³	t_2g³	–1.2Δ_o	—	—
d⁴	t_2g³ e_g¹	–0.6Δ_o	t_2g⁴	–1.6Δ_o + P
d⁵	t_2g³ e_g²	0	t_2g⁵	–2.0Δ_o + 2P
d⁶	t_2g⁴ e_g²	–0.4Δ_o + P	t_2g⁶	–2.4Δ_o + 3P
d⁷	t_2g⁵ e_g²	–0.8Δ_o + 2P	t_2g⁶ e_g¹	–1.8Δ_o + 3P
d⁸	t_2g⁶ e_g²	–1.2Δ_o + 3P	—	—
d⁹	t_2g⁶ e_g³	–0.6Δ_o + 3P	—	—
d¹⁰	t_2g⁶ e_g⁴	0 + 4P	—	—

▶ PYQ 2022 (JEE Main)

Q. Calculate the CFSE (in terms of Δ_o) for the complex [Fe(CN)₆]⁴⁻.

A. Fe(II) = d⁶. CN⁻ strong field → low spin. Configuration: t_2g⁶ e_g⁰.

CFSE = (–0.4 × 6 + 0.6 × 0) Δ_o + 3P (three pairs formed in t_2g as compared to free ion which had 1 pair)

CFSE = –2.4Δ_o + 3P

Answer: CFSE = –2.4Δ_o + 3P (energy stabilisation, hence the negative sign).

9

Colour of Coordination Compounds

Most transition metal complexes are coloured because of d–d electronic transitions. When white light falls on a complex, photons with energy equal to Δ_o are absorbed → electrons jump from t_2g to e_g. The remaining (transmitted/reflected) light is what we perceive as colour.

Energy–wavelength relation:

E = hν = hc/λ ⇒ Δ_o = hc/λ

Therefore: larger Δ_o → smaller λ absorbed → higher energy colour absorbed.

★ IMPORTANT — Frequently Asked

Explaining why a particular complex is coloured, predicting colour from complementary colour wheel, and the trend in colour across a series (e.g., [Ti(H₂O)₆]³⁺ is purple) is asked regularly.

8.1 Complementary Colours

Colour absorbed	λ (nm)	Colour observed
Violet	400–450	Yellow-green
Blue	450–500	Yellow
Blue-green	500–530	Red
Green	530–560	Magenta
Yellow-green	560–580	Violet
Yellow	580–600	Blue
Orange	600–640	Cyan
Red	640–700	Blue-green

8.2 Why Some Complexes are Colourless

d⁰ or d¹⁰ configurations: no d–d transition possible → colourless. e.g., [Sc(H₂O)₆]³⁺ (d⁰), [Zn(NH₃)₄]²⁺ (d¹⁰), [Cu(I)] complexes, [TiCl₄] (Ti⁴⁺, d⁰).
Some complexes show colour due to charge transfer (e.g., MnO₄⁻ is intensely purple despite Mn being d⁰; this is ligand-to-metal charge transfer, LMCT).

▶ PYQ 2020 (NEET)

Q. Which of the following complexes is colourless?

(a) [Ti(H₂O)₆]³⁺ (b) [Cu(H₂O)₆]²⁺ (c) [Sc(H₂O)₆]³⁺ (d) [Ni(H₂O)₆]²⁺

Answer: (c) [Sc(H₂O)₆]³⁺

Explanation: Sc³⁺ has d⁰ configuration. With no d-electrons, no d–d transition can occur → complex is colourless. [Ti(H₂O)₆]³⁺ (d¹) is purple; [Cu(H₂O)₆]²⁺ (d⁹) is blue; [Ni(H₂O)₆]²⁺ (d⁸) is green.

▶ PYQ 2018 (JEE Main)

Q. The colour of [Ti(H₂O)₆]³⁺ is due to:

(a) d–d transition (b) charge transfer (c) ligand excitation (d) metal oxidation

Answer: (a) d–d transition

Explanation: Ti³⁺ is d¹. The single electron in t_2g is excited to e_g by absorbing yellow-green light (≈ 500 nm, Δ_o ≈ 20,300 cm⁻¹) → complex appears purple (complementary colour). This is the textbook example of d–d transition.

10

Limitations of Crystal Field Theory

Although CFT successfully explains structures, colours, and magnetic properties, it has notable shortcomings:

Treats metal–ligand bond as purely ionic — but covalent character clearly exists, especially with π-acceptor ligands like CO, CN⁻.
Cannot explain π-bonding — e.g., the famous synergic bonding in metal carbonyls (M–C σ-bond + M→CO π back-bond) is outside CFT's scope.
Does not account for the order of spectrochemical series — why CO and CN⁻ are strong field while halides are weak cannot be explained by pure electrostatics. In fact, point-charge theory predicts F⁻ (small, high charge density) should be stronger than I⁻, which is correct, but predicts H₂O > NH₃ in field strength, which is incorrect.
Cannot explain relative stabilities of complexes with similar d-electron counts quantitatively.
Spectra of complexes show fine structure (bands splitting) which CFT cannot predict — requires ligand field theory (LFT) or molecular orbital theory (MOT).

Resolution: Modern Ligand Field Theory (LFT), an extension that incorporates covalent character via MO theory, overcomes most of these limitations. However, CFT remains the cornerstone for board-level and JEE/NEET preparation.

▶ PYQ 2017 (NEET)

Q. One of the limitations of CFT is that it does not account for:

(a) geometry (b) magnetic moment (c) colour (d) covalent character of M–L bond

Answer: (d) covalent character of M–L bond

Explanation: CFT considers ligands as point charges/dipoles and the metal–ligand bond as purely electrostatic. Hence it cannot explain the partial covalent character (especially with π-acceptor ligands like CO).

11

Bonding in Metal Carbonyls

Metal carbonyls are complexes of transition metals with CO. They exhibit a characteristic synergic bonding — simultaneous σ-donation and π-back-donation — which strengthens the M–C bond.

10.1 Mechanism of Synergic Bonding

σ-bond (M ← CO): The carbon atom of CO donates its lone pair (on the sp-hybridised C) into an empty metal orbital → forms a σ-coordinate bond.
π-back-bond (M → CO): The metal donates electron density from its filled d-orbital into the empty π* antibonding orbital of CO → forms a π-bond.
Synergic effect: The σ-donation increases electron density on metal, which favours more π-back-donation, which in turn further strengthens σ-donation — a self-reinforcing loop.

Consequence on CO bond: Electron density in π* (antibonding) of CO weakens the C–O bond → bond order decreases from 3 (in free CO) to ~2.5 in carbonyls.

Evidence: C–O stretching frequency in free CO = 2143 cm⁻¹; in metal carbonyls it shifts to 1850–2125 cm⁻¹.

10.2 EAN Rule (Effective Atomic Number Rule)

Proposed by Sidgwick. For many metal carbonyls (and other stable complexes), the total electrons on the metal atom after accepting electron pairs from ligands equals the atomic number of the next noble gas.

EAN = (Atomic number of metal) – (Oxidation state) + 2 × (number of ligands)

Examples:

[Cr(CO)₆]: Cr (Z=24) → EAN = 24 – 0 + 2(6) = 36 (Kr) ✓
[Fe(CO)₅]: Fe (Z=26) → EAN = 26 – 0 + 2(5) = 36 (Kr) ✓
[Ni(CO)₄]: Ni (Z=28) → EAN = 28 – 0 + 2(4) = 36 (Kr) ✓
[Mn₂(CO)₁₀]: Mn (Z=25) → with Mn–Mn bond, EAN = 25 + 2(5) + 1 = 36 (Kr) ✓

Exceptions: Many stable complexes do NOT satisfy the EAN rule — e.g., [V(CO)₆] (EAN = 35), [Ni(en)₂Cl₂], [Co(NH₃)₆]Cl₃. So EAN rule is a useful guideline, not a strict rule.

▶ PYQ 2021 (NEET)

Q. The EAN of metal 'M' in [M(CO)₆] is 36. The metal M is:

(a) Cr (b) Mn (c) Fe (d) Co

Answer: (a) Cr

Explanation: EAN = Z – 0 + 2(6) = 36 → Z = 24 → Chromium (Z = 24). Carbonyls of Cr (Cr(CO)₆), Fe (Fe(CO)₅), Ni (Ni(CO)₄) all satisfy EAN = 36 (Kr).

12

Stability & Importance of Coordination Compounds

11.1 Stability of Complexes

The stability of a complex in solution is measured by its stability constant (formation constant), K_f or β:

For M + nL ⇌ [ML_n]:

K_f = [[ML_n]] / ([M][L]ⁿ)

Higher K_f → more stable complex. Often expressed as log K_f.

Factors affecting stability:

Higher charge density on metal ion → more stable (smaller size + higher charge).
Chelate effect: chelating ligands (like en, EDTA) form more stable complexes than analogous monodentate ligands. e.g., [Ni(en)₃]²⁺ (log K ≈ 18.6) is more stable than [Ni(NH₃)₆]²⁺ (log K ≈ 8.6).
Ligand basicity: more basic ligands form stronger bonds.
Larger aromatic chelating ligands (phen, bpy) show extra stability due to macrocyclic effect.
HSAB principle: hard acids (Fe³⁺, Cr³⁺, Al³⁺) prefer hard bases (F⁻, O-donors); soft acids (Ag⁺, Hg²⁺, Pt²⁺) prefer soft bases (I⁻, S-donors, CN⁻, CO).

11.2 Importance and Applications

Biological systems: Haemoglobin (Fe-porphyrin, O₂ transport), chlorophyll (Mg-porphyrin, photosynthesis), vitamin B₁₂ (Co–corrin), carbonic anhydrase (Zn-enzyme).
Medicine: Cisplatin cis-[Pt(NH₃)₂Cl₂] — anticancer drug; DTPA & EDTA — heavy metal (Pb, Hg) poisoning antidotes; Au(I) complexes for arthritis.
Extraction of metals: Cyanide leaching of gold (formation of [Au(CN)₂]⁻); purification of Ni via Mond process (Ni(CO)₄).
Analytical chemistry: EDTA titrations (hardness of water); detection of Cu²⁺ (deep blue [Cu(NH₃)₄]²⁺), Ni²⁺ (red Ni-DMG), Fe³⁺ (blood red [Fe(SCN)]²⁺), etc.
Electroplating: Au, Ag, Cu complexes used in plating baths for smooth deposition.
Catalysis: Wilkinson's catalyst [RhCl(PPh₃)₃] — hydrogenation of alkenes; Ziegler-Natta catalyst for polymerisation.
Photography: [Ag(S₂O₃)₂]³⁻ formation (hypo) dissolves unexposed AgBr in fixing.

▶ PYQ 2016 (NEET)

Q. Which complex is used in the treatment of cancer?

(a) [Co(NH₃)₆]Cl₃ (b) cis-[Pt(NH₃)₂Cl₂] (c) [Ni(CO)₄] (d) [Cu(NH₃)₄]SO₄

Answer: (b) cis-[Pt(NH₃)₂Cl₂] — Cisplatin

Explanation: Cisplatin binds to DNA at N-7 of guanine bases, causing cross-linking that inhibits DNA replication → triggers apoptosis in rapidly dividing cancer cells. Note: only the cis isomer is active; the trans isomer is biologically inactive.

Previous Year Questions Analysis

The most repeated concepts from NEET & JEE Main (2015–2024). Mastering these ensures ~90% coverage of likely questions.

The following questions represent the most repeated concepts in NEET & JEE Main from 2015–2024. Mastering them ensures 90% coverage of likely questions in the upcoming exams.

PYQ #1 · NEET 2024 · IUPAC Nomenclature

Q. The IUPAC name of K₃[Co(NO₂)₆] is:

A. Potassium hexanitrito-N-cobaltate(III)

K⁺ cations named first. NO₂⁻ via N → "nitrito-N", metal Co is in anionic complex → "cobaltate". Co is +3 (3 K⁺ balance 3– on complex; 6 NO₂⁻ give 6–; Co = +3).

PYQ #2 · JEE Main 2023 · VBT Hybridisation

Q. The correct statement about [Ni(CN)₄]²⁻ is:

(a) tetrahedral, paramagnetic (b) square planar, diamagnetic (c) tetrahedral, diamagnetic (d) square planar, paramagnetic

A. (b) square planar, diamagnetic

Ni²⁺ = 3d⁸. CN⁻ is strong field → pairs the 3d electrons, freeing one 3d orbital. Hybridisation = dsp² → square planar, no unpaired e⁻ → diamagnetic.

PYQ #3 · NEET 2022 · Isomerism

Q. The number of geometrical isomers of [Pt(NH₃)₂Cl₂] is:

A. 2 (cis and trans)

Pt(II) forms square planar complexes. For [Ma₂b₂] square planar, two geometrical isomers exist: cis (Cl⁻ adjacent) and trans (Cl⁻ opposite). cis is the active anticancer drug "cisplatin".

PYQ #4 · JEE Main 2021 · CFT Splitting

Q. The crystal field splitting energy (Δ_o) for [Cr(H₂O)₆]³⁺ is 17,400 cm⁻¹. Calculate the wavelength of light absorbed.

A. λ = 1/Δ̄ν = 1 / 17400 cm⁻¹ = 5.747 × 10⁻⁵ cm = 574.7 nm

The complex absorbs in the yellow-green region (~575 nm) → appears purple/violet. This is why [Cr(H₂O)₆]³⁺ is violet.

PYQ #5 · NEET 2020 · Magnetic Moment

Q. The spin-only magnetic moment of [Fe(H₂O)₆]²⁺ is:

A. 4.90 BM

Fe²⁺ = d⁶; H₂O weak field → high spin → 4 unpaired e⁻. μ = √[4(4+2)] = √24 = 4.90 BM.

PYQ #6 · NEET 2019 · Spectrochemical Series

Q. Arrange the following in increasing order of field strength: F⁻, CN⁻, H₂O, NH₃, I⁻

A. I⁻ < F⁻ < H₂O < NH₃ < CN⁻

This follows the spectrochemical series directly. I⁻ is the weakest, CN⁻ is among the strongest field ligands.

PYQ #7 · JEE Main 2018 · Linkage Isomerism

Q. Which of the following pairs are linkage isomers?

(a) [Co(NH₃)₅Br]SO₄ & [Co(NH₃)₅SO₄]Br (b) [Co(NH₃)₅NO₂]²⁺ & [Co(NH₃)₅ONO]²⁺

A. (b)

NO₂⁻ is ambidentate — binds through N (-NO₂, nitrito-N) or O (-ONO, nitrito-O). Option (a) is an ionisation isomer pair.

PYQ #8 · NEET 2018 · Geometries Comparison

Q. [NiCl₄]²⁻ is tetrahedral while [Ni(CN)₄]²⁻ is square planar. Explain.

A. Both have Ni²⁺ (3d⁸). Cl⁻ is weak field ligand → cannot pair 3d electrons → uses 4s and 4p (outer) → sp³ → tetrahedral, paramagnetic (2 unpaired e⁻, μ = 2.83 BM). CN⁻ is strong field → pairs 3d electrons, vacates one 3d → uses one 3d + 4s + two 4p (inner) → dsp² → square planar, diamagnetic.

PYQ #9 · NEET 2017 · EAN Rule

Q. The metal carbonyl that obeys EAN rule is:

(a) V(CO)₆ (b) Cr(CO)₆ (c) Mn(CO)₅ (d) Fe(CO)₄

A. (b) Cr(CO)₆

EAN of Cr(CO)₆ = 24 + 2(6) = 36 (Kr) — satisfies EAN. V(CO)₆: 23+12=35, doesn't satisfy. Cr(CO)₆ is the textbook example of EAN rule.

PYQ #10 · NEET 2016 · CFT Colour

Q. Why do transition metal complexes show colour?

A. Due to d–d electronic transitions. The crystal field splits the d-orbitals into t_2g and e_g sets. Visible light photons with energy = Δ_o are absorbed, exciting electrons from t_2g to e_g. The transmitted/reflected light (complementary colour) is observed. d⁰ and d¹⁰ complexes are colourless (no d–d transition).

PYQ #11 · JEE Main 2015 · CFSE

Q. The CFSE for d⁶ low-spin octahedral complex is:

(a) –0.4Δ_o + P (b) –1.2Δ_o + 3P (c) –2.4Δ_o + 3P (d) 0

A. (c) –2.4Δ_o + 3P

d⁶ low-spin: t_2g⁶ e_g⁰. CFSE = (–0.4 × 6 + 0.6 × 0) Δ_o + (3 pairs formed – 0 originally) P = –2.4Δ_o + 3P.

Quick Revision — Formula Cheat Sheet

Geometry ↔ Hybridisation

Tetrahedral (CN=4) → sp³
Square planar (CN=4) → dsp²
Octahedral inner (CN=6) → d²sp³
Octahedral outer (CN=6) → sp³d²

Magnetic Moment

μ = √[n(n+2)] BM

n: 1→1.73, 2→2.83, 3→3.87, 4→4.90, 5→5.92

CFSE (Octahedral)

CFSE = (–0.4 n_t2g + 0.6 n_eg)Δ_o + P·n_pairs

Δ_t vs Δ_o

Δ_t = (4/9) Δ_o

EAN Rule

EAN = Z – OS + 2 × (no. of ligands)

Target = noble gas Z

Stability Constant

K_f = [ML_n] / ([M][L]ⁿ)

★ TOP 5 MOST REPEATED PYQ TOPICS

IUPAC Nomenclature (every year!)
Hybridisation & magnetic behaviour from VBT (e.g., [Ni(CN)₄]²⁻ vs [NiCl₄]²⁻)
Isomerism — optical in [M(AA)₃], geometrical cis-trans, fac-mer
CFSE calculation for dⁿ configurations
Colour explanation — d–d transition, complementary colours, why d⁰/d¹⁰ are colourless

> FAQ

Frequently Asked Questions

What is the difference between a double salt and a coordination compound?

A double salt (e.g., Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O, carnallite KCl·MgCl₂·6H₂O) dissociates completely into its constituent ions in solution and loses its identity. A coordination compound (e.g., K₄[Fe(CN)₆]) retains the complex ion [Fe(CN)₆]⁴⁻ in solution; only the counter-ions dissociate.

Why is CO a strong field ligand but H₂O is weak, even though both have lone pairs on O?

CO is a strong field ligand due to its ability to accept π-back-donation from metal d-orbitals into its empty π* antibonding orbital. This synergic bonding greatly stabilises the complex and increases Δ_o. H₂O has no low-lying empty orbital for back-bonding, so it acts as a σ-donor only → weaker field.

Why do tetrahedral complexes rarely show optical isomerism?

Tetrahedral complexes (with 4 different ligands, Mabcd) can theoretically show optical isomerism, but such complexes are extremely rare and difficult to synthesise. Most tetrahedral complexes have repeated ligands, leading to a plane of symmetry → not optically active.

What is the chelate effect? Give an example.

Chelate effect refers to the enhanced stability of complexes containing chelating (multidentate) ligands compared to those with analogous monodentate ligands. It is primarily an entropy-driven effect. Example: [Ni(en)₃]²⁺ (log K_f = 18.6) is far more stable than [Ni(NH₃)₆]²⁺ (log K_f = 8.6), even though both have 6 N-donor atoms around Ni²⁺.

Why is [Ni(CN)₄]²⁻ square planar while [NiCl₄]²⁻ is tetrahedral?

Ni²⁺ = 3d⁸. In [Ni(CN)₄]²⁻, CN⁻ is a strong field ligand that forces the 3d electrons to pair, freeing one 3d orbital. Hybridisation is dsp² → square planar, diamagnetic. In [NiCl₄]²⁻, Cl⁻ is weak field and cannot pair the electrons, so 3d remains fully occupied → outer orbitals (4s, 4p) are used → sp³ → tetrahedral, paramagnetic (2 unpaired e⁻).

What is the spectrochemical series?

The spectrochemical series arranges ligands in order of increasing crystal field splitting (Δ). It is determined experimentally from absorption spectra. Order: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO. Weak-field ligands (left) give high-spin complexes; strong-field ligands (right) give low-spin complexes.

Coordination Compounds notes, VBT, CFT & & PYQs for JEE 2026

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