Coordination Compounds — Concepts, Reactions & Solved Examples

What Are Coordination Compounds — and Why Do They Matter?

Take a copper sulphate solution. Now add excess ammonia. That vivid deep blue colour? That’s not CuSO₄ anymore — it’s $[\text{Cu(NH}_3)_4]^{2+}$ , a coordination compound. The copper ion has grabbed four ammonia molecules and formed a completely new species with different colour, different behaviour, different chemistry.

This is the central idea of coordination chemistry: a central metal atom or ion surrounded by ligands (molecules or ions that donate electron pairs), forming a complex. Understanding this lets us explain why blood is red (haemoglobin), why plants are green (chlorophyll), why platinum compounds cure cancer (cisplatin), and why so many JEE questions target this chapter.

In boards, expect 5–8 marks. In JEE Main, coordination compounds consistently contribute 1–2 questions. The chapter rewards conceptual clarity — memorising formulae without understanding the underlying logic is how students drop marks here.

Key Terms and Definitions

Central Metal Atom/Ion: The metal at the heart of the complex. Almost always a transition metal (Fe, Cu, Co, Ni, Pt, Cr). It accepts electron pairs from ligands.

Ligand: Any molecule or ion that has a lone pair and donates it to the central metal. The bond formed is a coordinate (dative) bond. Ligands are classified by how many donor atoms they have:

Monodentate: One donor atom — $\text{NH}_3$ , $\text{H}_2\text{O}$ , $\text{Cl}^-$ , $\text{CN}^-$
Bidentate: Two donor atoms — ethylenediamine (en), oxalate ( $\text{C}_2\text{O}_4^{2-}$ )
Polydentate: Multiple donor atoms — EDTA (hexadentate, 6 donor atoms)

Coordination Number (CN): Total number of ligand donor atoms bonded to the central metal. Not the number of ligands — a bidentate ligand contributes 2 to the CN.

Coordination Sphere: The central metal + its ligands, enclosed in square brackets in the formula. Everything outside the brackets is the counter ion (or ionisation sphere).

Oxidation State: Determined by treating all ligands as neutral molecules (except ionic ligands which carry their charge). Then use the overall charge of the complex.

To find oxidation state: let the metal’s oxidation state be $x$ . Sum of (metal OS + ligand charges) = overall charge of the coordination sphere. For $[\text{Fe(CN)}_6]^{3-}$ : $x + 6(-1) = -3$ , so $x = +3$ .

Chelate: A complex formed when a polydentate ligand bonds to the metal through multiple donor atoms, forming a ring. Chelates are more stable than similar complexes with monodentate ligands — this is the chelate effect.

Ambidentate Ligand: A monodentate ligand that can donate through two different atoms. Classic examples: $\text{NO}_2^-$ (bonds through N as nitro, through O as nitrito), $\text{SCN}^-$ (bonds through S or N).

IUPAC Nomenclature — The System Behind the Names

Naming coordination compounds follows a strict order. Learn the logic, not just rules.

Naming Order Within the Complex

Ligands first, metal second
Ligands listed alphabetically (ignore multiplying prefixes like di-, tri-)
Anionic ligands end in -o ( $\text{Cl}^-$ → chlorido, $\text{CN}^-$ → cyanido, $\text{OH}^-$ → hydroxido)
Neutral ligands keep their name (exceptions: $\text{H}_2\text{O}$ → aqua, $\text{NH}_3$ → ammine, CO → carbonyl, NO → nitrosyl)
Metal name, then oxidation state in Roman numerals in parentheses
If the complex is an anion, metal gets -ate suffix (cobalt → cobaltate, iron → ferrate, copper → cuprate)

Overall Compound Naming

Cation named first, then anion — same as any ionic compound.

Example: $\text{K}_4[\text{Fe(CN)}_6]$

$\text{K}^+$ is outside — potassium (cation, named first)
Inside: 6 CN⁻ ligands (cyanido), Fe²⁺
Complex is an anion → ferrate(II)
Name: Potassium hexacyanidoferrate(II)

Students write “hexacyanoferrate” — this is the older name. IUPAC now uses “cyanido” not “cyano”. Both appear in textbooks; CBSE accepts both, but JEE prefers the IUPAC form.

Bonding Theories

Werner’s Theory (Historical)

Alfred Werner (1893) proposed two types of valency:

Primary valency = oxidation state (satisfied by counter ions)
Secondary valency = coordination number (satisfied by ligands, directed in space)

This explained why $\text{CoCl}_3 \cdot 6\text{NH}_3$ conducts like a 1:3 electrolyte (gives 4 ions) while $\text{CoCl}_3 \cdot 3\text{NH}_3$ gives no ions.

Valence Bond Theory (VBT)

The metal uses hybridised orbitals to accept electron pairs from ligands. The hybridisation determines the geometry.

Coordination Number	Hybridisation	Geometry
2	sp	Linear
4	sp³	Tetrahedral
4	dsp²	Square planar
6	sp³d²	Octahedral
6	d²sp³	Octahedral (inner orbital)

Inner orbital complex: Uses inner (n-1)d orbitals → low spin, diamagnetic (usually)

Outer orbital complex: Uses nd orbitals → high spin, paramagnetic (usually)

Strong field ligands (CN⁻, CO, NO⁺, en) force inner orbital complex formation. Weak field ligands (F⁻, Cl⁻, H₂O, OH⁻) give outer orbital complexes. Remember the spectrochemical series: $\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{NO}_2^- < \text{CN}^- \approx \text{CO}$

Limitation of VBT: Cannot explain colour of complexes, and has no quantitative basis for “strong” vs “weak” field — this is why Crystal Field Theory exists.

Crystal Field Theory (CFT)

CFT treats ligands as point charges. It explains how ligands split the d-orbitals of the metal.

In an octahedral field: Ligands approach along x, y, z axes. The $d_{x^2-y^2}$ and $d_{z^2}$ orbitals (called $e_g$ set) point directly at ligands → higher energy. The $d_{xy}$ , $d_{xz}$ , $d_{yz}$ orbitals (called $t_{2g}$ set) point between ligands → lower energy.

\Delta_o = \text{Crystal Field Splitting Energy (CFSE)}

e_g \text{ orbitals: } +\frac{3}{5}\Delta_o \text{ (or } +0.6\Delta_o)

t_{2g} \text{ orbitals: } -\frac{2}{5}\Delta_o \text{ (or } -0.6\Delta_o \text{ per electron... wait)}

t_{2g} \text{ energy: } -0.4\Delta_o \text{ per electron}

e_g \text{ energy: } +0.6\Delta_o \text{ per electron}

Colour explanation: When white light hits a complex, it absorbs certain wavelengths (causing $t_{2g} \to e_g$ electron transitions). The complementary colour is what we see.

$[\text{Ti(H}_2\text{O)}_6]^{3+}$ absorbs green light (~500 nm), so we see purple.

Magnetic properties from CFT: If $\Delta_o >$ pairing energy → electrons pair up first (low spin). If $\Delta_o <$ pairing energy → electrons occupy all orbitals first (high spin, Hund’s rule).

Isomerism in Coordination Compounds

Isomers are compounds with the same molecular formula but different arrangement. This is a major source of JEE questions.

Structural Isomers

Ionisation isomers: Counter ion and ligand exchange positions.

$[\text{Co(NH}_3)_5\text{Br}]\text{SO}_4$ vs $[\text{Co(NH}_3)_5\text{SO}_4]\text{Br}$

Linkage isomers: Ambidentate ligand bonds through different atoms.

$[\text{Co(NH}_3)_5\text{(NO}_2)]\text{Cl}_2$ (nitro, N-bonded) vs $[\text{Co(NH}_3)_5\text{(ONO)}]\text{Cl}_2$ (nitrito, O-bonded)

Solvate/Hydrate isomers: Water inside vs outside the coordination sphere.

$[\text{Cr(H}_2\text{O)}_6]\text{Cl}_3$ vs $[\text{Cr(H}_2\text{O)}_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O}$

Stereoisomers

Geometrical (cis-trans) isomerism:

In square planar $\text{MA}_2\text{B}_2$ : cis (like groups adjacent) and trans (like groups opposite).

Cisplatin $[\text{Pt(NH}_3)_2\text{Cl}_2]$ (cis) is an anticancer drug. Transplatin (trans) is not.

In octahedral $\text{MA}_4\text{B}_2$ : cis (B’s adjacent, 90° apart) and trans (B’s opposite, 180°).

Optical isomerism:

A complex is optically active if it is non-superimposable on its mirror image (chiral). The two forms are called enantiomers — one rotates plane-polarised light left (levo), the other right (dextro).

$[\text{Co(en)}_3]^{3+}$ is the classic example — three bidentate ligands create a propeller-like structure with no plane of symmetry.

For JEE Main: Expect one question on naming OR isomerism OR magnetic moment calculation. For CBSE boards: Chapter contributes 5 marks typically — one long answer covering VBT/CFT and one nomenclature/isomerism question.

Solved Examples

Example 1 — Easy (CBSE Level)

Q: Write the IUPAC name of $[\text{Co(NH}_3)_4\text{Cl}_2]\text{Cl}$ .

Step 1: Identify what’s inside the coordination sphere: 4 ammine ligands (NH₃), 2 chlorido ligands (Cl⁻), central metal Co.

Step 2: Find oxidation state. Charge of sphere = +1 (since Cl⁻ outside). So: $x + 4(0) + 2(-1) = +1$ , giving $x = +3$ .

Step 3: Name ligands alphabetically: ammine before chlorido. Multiply: tetraammine, dichloro.

Step 4: Metal with OS: cobalt(III). Complex is cation, no -ate suffix.

Name: Tetraamminedichloridocobalt(III) chloride

Example 2 — Medium (JEE Main Level)

Q: Find the magnetic moment of $[\text{Fe(CN)}_6]^{4-}$ .

Step 1: Oxidation state of Fe. $x + 6(-1) = -4$ , so $x = +2$ . Fe²⁺ has configuration $[\text{Ar}]3d^6$ .

Step 2: CN⁻ is a strong field ligand → inner orbital complex → $d^2sp^3$ hybridisation → low spin.

Step 3: In low spin $d^6$ : all 6 electrons pair up in $t_{2g}$ → 0 unpaired electrons.

\mu = \sqrt{n(n+2)} \text{ BM}

where $n$ = number of unpaired electrons

Step 4: $\mu = \sqrt{0(0+2)} = 0$ BM. The complex is diamagnetic.

Example 3 — Hard (JEE Advanced Level)

Q: Among $[\text{CoCl}_2(\text{en})_2]^+$ , how many geometrical and optical isomers exist?

Analysis: Co³⁺, CN = 6, octahedral. Two bidentate en ligands + 2 Cl⁻.

Geometrical isomers:

trans: both Cl at opposite positions (180°)
cis: both Cl at adjacent positions (90°)

So 2 geometrical isomers.

Optical isomers:

trans isomer has a plane of symmetry → optically inactive (no optical isomers)
cis isomer has no plane of symmetry → exists as d and l forms → 2 optical isomers

Total: 3 stereoisomers (1 trans + 2 cis enantiomers)

General rule: trans isomers of $[\text{MA}_4\text{B}_2]$ type always have a plane of symmetry and are optically inactive. Only cis forms can be chiral.

Exam-Specific Tips

For CBSE Class 12 Boards

The marking scheme rewards showing work clearly. For 3-mark questions on naming: write the formula identification (1 mark), name the ligands with multiplying prefixes (1 mark), complete IUPAC name (1 mark). Don’t skip steps.

CFSE calculation is asked as: “Calculate the CFSE for $d^4$ in strong field octahedral complex.” Write electronic configuration in $t_{2g}$ and $e_g$ , then multiply by $-0.4\Delta_o$ and $+0.6\Delta_o$ .

For JEE Main

Weightage is 1-2 questions, mostly from:

Magnetic moment calculation (calculate $n$ , then $\mu$ )
Number of isomers
Identify the complex type (inner/outer orbital)
Werner’s theory and primary/secondary valency

The 2023 JEE Main January session had a question on linkage isomerism — specifically identifying which pairs are linkage isomers vs ionisation isomers.

EFfective Atomic Number (EAN) = atomic number of metal − oxidation state + 2 × coordination number. Stable complexes often have EAN = 36 (Kr), 54 (Xe), or 86 (Rn). This is occasionally asked in JEE.

Common Mistakes to Avoid

Mistake 1 — Confusing coordination number with number of ligands. If a complex has 3 bidentate ligands, the coordination number is 6, not 3. Always count donor atoms.

Mistake 2 — Wrong alphabetical order for naming. Alphabetise by the ligand name, not the formula. “ammine” comes before “bromo” — but students often list them by formula order in the complex.

Mistake 3 — Applying $\mu = \sqrt{n(n+2)}$ to all complexes. This formula is valid only for spin-only magnetic moments. It doesn’t account for orbital contribution (relevant for some heavy metals, but safe for Class 12).

Mistake 4 — Forgetting that trans isomers are usually optically inactive. Students correctly identify 2 geometrical isomers but then add optical isomers for the trans form too. Check for plane of symmetry first.

Mistake 5 — Getting hybridisation wrong for square planar complexes. Square planar is $dsp^2$ (uses one inner d orbital), not $sp^3$ . Both have CN = 4 but completely different geometry and magnetic properties.

Practice Questions

Q1. Write the IUPAC name of $\text{K}_2[\text{PtCl}_4]$ .

Pt oxidation state: $x + 4(-1) = -2$ , so $x = +2$ . Complex is anion → platinate(II). Ligands: tetrachloriodo. Cation K⁺ = potassium.

Answer: Potassium tetrachloridoplatinate(II)

Q2. How many ions does $[\text{Co(NH}_3)_6]\text{Cl}_3$ produce in water?

The complex cation $[\text{Co(NH}_3)_6]^{3+}$ stays intact. Three Cl⁻ ionise. Total ions = 1 + 3 = 4 ions. It conducts like a 1:3 electrolyte.

Q3. Calculate the magnetic moment of $[\text{MnCl}_6]^{3-}$ .

Mn³⁺ has $3d^4$ configuration. Cl⁻ is weak field → high spin → 4 unpaired electrons.

$\mu = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90$ BM

Q4. Which of these will show geometrical isomerism: $[\text{Pt(NH}_3)_2\text{Cl}_2]$ or $[\text{Zn(NH}_3)_2\text{Cl}_2]$ ?

Pt²⁺ forms square planar ( $dsp^2$ ) complexes → cis and trans isomers exist → shows geometrical isomerism.

Zn²⁺ forms tetrahedral ( $sp^3$ ) complexes → $\text{MA}_2\text{B}_2$ tetrahedral has no geometrical isomers.

Q5. Give the formula of a complex that shows ionisation isomerism with $[\text{Pt(NH}_3)_4\text{Cl}_2]\text{Br}_2$ .

Ionisation isomers swap a ligand inside the sphere with a counter ion outside.

Answer: $[\text{Pt(NH}_3)_4\text{BrCl}]\text{BrCl}$ or $[\text{Pt(NH}_3)_4\text{Br}_2]\text{Cl}_2$

Q6. An octahedral complex of $\text{Cr}^{3+}$ with ethylenediamine — write its formula and predict if it is optically active.

Cr³⁺ has CN = 6, en is bidentate. Three en ligands fill all positions: $[\text{Cr(en)}_3]^{3+}$ .

This is a tris-bidentate complex with propeller geometry — no plane of symmetry. It is optically active (d and l forms exist).

Q7. Calculate CFSE for $[\text{Fe(H}_2\text{O)}_6]^{2+}$ given $\Delta_o$ .

Fe²⁺ is $d^6$ . H₂O is weak field → high spin. Electronic configuration: $t_{2g}^4 e_g^2$ .

CFSE $= 4 \times (-0.4\Delta_o) + 2 \times (+0.6\Delta_o) = -1.6\Delta_o + 1.2\Delta_o = -0.4\Delta_o$

Q8. How many stereoisomers does $[\text{CoCl}_2(\text{NH}_3)_4]^+$ have?

Octahedral $\text{MA}_4\text{B}_2$ type. Two geometrical isomers: cis (Cl adjacent) and trans (Cl opposite).

Trans has a plane of symmetry → optically inactive. Cis — check for symmetry: it has a mirror plane through the two Cl atoms → also optically inactive.

Total: 2 stereoisomers (2 geometrical isomers, no optical isomers).

FAQs

Why do coordination compounds show colour?

Colour comes from d-d transitions. Ligands split the d-orbitals into two energy levels ( $t_{2g}$ and $e_g$ ). Electrons absorb specific wavelengths of visible light to jump between these levels. The colour we see is the complement of what’s absorbed. Complexes with no unpaired electrons or no d-electrons are often colourless — $[\text{Sc(H}_2\text{O)}_6]^{3+}$ ( $d^0$ ) and $[\text{Zn(NH}_3)_4]^{2+}$ ( $d^{10}$ ) are colourless.

What is the difference between a double salt and a complex salt?

A double salt (like alum, $\text{KAl(SO}_4)_2\cdot 12\text{H}_2\text{O}$ ) dissociates completely in water to give all its constituent ions. A complex salt contains a coordination entity that remains intact in solution — $[\text{Cu(NH}_3)_4]\text{SO}_4$ gives $[\text{Cu(NH}_3)_4]^{2+}$ and $\text{SO}_4^{2-}$ , not free Cu²⁺.

Why is EDTA such an effective ligand?

EDTA is hexadentate — it has 2 nitrogen and 4 oxygen donor atoms. It wraps around a metal ion and forms 5 chelate rings simultaneously. The chelate effect makes this complex extraordinarily stable. This is why EDTA is used in water softening, food preservation, and as an antidote for heavy metal poisoning.

How do we decide if a complex is inner orbital or outer orbital?

Look at the ligand field strength. Strong field ligands (CN⁻, CO, en, NH₃) force the metal to use inner (n-1)d orbitals → $d^2sp^3$ hybridisation → inner orbital complex (low spin, usually diamagnetic or low $n$ ). Weak field ligands (Cl⁻, F⁻, H₂O) → outer orbital, $sp^3d^2$ hybridisation (high spin, more unpaired electrons).

Why is cisplatin an anticancer drug but transplatin is not?

Cisplatin has both Cl⁻ ligands on the same side. Inside a cell, water replaces the Cl⁻ ligands, and the resulting complex binds to two adjacent guanine bases on DNA (cross-links the same DNA strand). This blocks DNA replication, killing the cancer cell. Transplatin’s geometry (Cl⁻ trans to each other) doesn’t allow it to form this intrastrand crosslink — it binds to guanines on different strands in a way that the cell can repair.

What is effective atomic number (EAN)?

EAN = (atomic number of metal) − (oxidation state) + 2 × (coordination number). This gives the total electron count around the metal after ligands donate electrons. Stable complexes tend to have EAN equal to the nearest noble gas (36 for Kr, 54 for Xe). It’s the electron-counting version of the “18-electron rule” used in organometallic chemistry.

Can coordination number exceed 6?

Yes. Coordination numbers of 7, 8, and even 9 are known for lanthanides and actinides, which have larger ionic radii and available f orbitals. For transition metals in Class 12, we focus on CN = 2, 4, and 6.