Coordination Compounds

1. Werner's Theory & Key Terms

A coordination compound has a central metal atom/ion bonded to a fixed number of ions or molecules (ligands) by coordinate bonds.

Werner's postulates (1893)

1. A metal shows two kinds of valency: primary (ionisable, satisfied by anions, shown as dashed line/outside the bracket) and secondary (non-ionisable, equals the coordination number, satisfied by ligands).
2. Secondary valencies have fixed directions in space, so each complex has a definite geometry (CN 4: tetrahedral or square planar; CN 6: octahedral).

Terms you must define with an example

Chelate effect: a chelating ligand forms a more stable complex than comparable unidentate ligands (more rings, entropy gain, ΔS > 0). [Co(en)₃]³⁺ is far more stable than [Co(NH₃)₆]³⁺. EDTA uses this to estimate hardness of water (Ca²⁺, Mg²⁺).

Counting ions. Add the ions produced on dissolving. [Co(NH₃)₆]Cl₃ gives 1 + 3 = 4 ions; [Cr(NH₃)₅Cl]Cl₂ gives 3; [Cr(NH₃)₃Cl₃] gives 0. More ions means higher molar conductivity.

2. IUPAC Nomenclature

Rules in order

1. Cation first, then anion, as in simple salts. Complex ions are named as one word without spaces.
2. Ligands before metal, in alphabetical order of ligand name. Prefixes di, tri, tetra are ignored while alphabetising.
3. Use bis, tris, tetrakis for ligand names that already contain a prefix or are complex: bis(ethane-1,2-diamine).
4. Metal name: unchanged in a cation or neutral complex; ends in -ate in an anion (ferrate, cuprate, argentate, aurate, platinate, zincate, cobaltate, nickelate).
5. Oxidation state of the metal in Roman numerals inside brackets, directly after the metal name.
6. In a formula: metal symbol first, then anionic ligands, then neutral ligands, each alphabetically by the first symbol; the whole entity goes in [ ].

Finding the oxidation state: sum of charges = charge of the entity. In [Pt(NH₃)₂Cl₂]²⁺: x + 0 + 2(−1) = +2, so x = +4.

Worked names (all from board papers)

• [Co(NH₃)₅Cl]Cl₂ : Pentaamminechloridocobalt(III) chloride
• K₃[Fe(C₂O₄)₃] : Potassium trioxalatoferrate(III)
• [Ag(NH₃)₂][Ag(CN)₂] : Diamminesilver(I) dicyanidoargentate(I)
• K₂[Zn(OH)₄] : Potassium tetrahydroxidozincate(II)
• Fe₄[Fe(CN)₆]₃ : Iron(III) hexacyanidoferrate(II)
• [Co(NH₃)₄(H₂O)Cl]Cl₂ : Tetraammineaquachloridocobalt(III) chloride

3. Isomerism

Structural isomers (different connectivity)

Stereoisomers (same connectivity, different spatial arrangement)

Geometrical (cis/trans, fac/mer)

• Square planar MA₂B₂: cis and trans (both exist). MA₄ or MA₃B: no geometrical isomers. [Pt(NH₃)(Br)(Cl)(py)] (Mabcd): 3 isomers.
• Octahedral MA₄B₂: cis and trans. MA₃B₃: fac and mer.
• Tetrahedral: none, all positions are equivalent.

Optical isomerism needs a non-superimposable mirror image (no plane of symmetry).

• Octahedral [M(AA)₃]: [Co(en)₃]³⁺, [Cr(C₂O₄)₃]³⁻ → 2 optical isomers, no geometrical.
• cis-[M(AA)₂B₂]: cis is optically active (d and l), trans is not. [CoCl₂(en)₂]⁺ has 3 stereoisomers in total: cis-d, cis-l, trans.
• [Co(NH₃)₃Cl₃]: fac and mer, both optically inactive, so 2 isomers.

4. Valence Bond Theory (VBT)

The metal provides empty orbitals, hybridises them, and each accepts a lone pair from a ligand. The hybridisation decides shape, and the number of unpaired d electrons decides magnetism.

• Strong field ligand (CN⁻, CO, en, NH₃) pairs up the 3d electrons, freeing 3d orbitals: inner orbital (d²sp³), usually low spin.
• Weak field ligand (F⁻, Cl⁻, H₂O) cannot pair them: outer orbital (sp³d²), high spin, more unpaired electrons.

Spin-only magnetic moment: $\mu=\sqrt{n(n+2)}$ BM.

Standard complexes to memorise

Why do Ni(CO)₄ and [NiCl₄]²⁻ differ although both are tetrahedral? In Ni(CO)₄, nickel is Ni(0); CO (strong field) pushes the 4s electrons into 3d, giving 3d¹⁰ with no unpaired electrons. In [NiCl₄]²⁻, Ni²⁺ is 3d⁸ and Cl⁻ (weak field) cannot pair them, so 2 unpaired electrons remain.

Limitations of VBT: it cannot explain colour, gives no quantitative magnetic data, does not distinguish strong and weak ligands, and cannot predict tetrahedral vs square planar exactly. CFT covers these.

5. Crystal Field Theory (CFT) & Colour

CFT treats the M-L bond as purely electrostatic: ligands are point charges/dipoles. In the free ion the five d orbitals are degenerate; ligands remove this degeneracy.

Octahedral field: the ligands sit on the axes, so d_x²−y² and d_z² (pointing at ligands) rise to the e_g level (+0.6 Δₒ), and d_xy, d_yz, d_xz (between ligands) fall to the t_2g level (−0.4 Δₒ). The gap is Δₒ (crystal field splitting energy).

Tetrahedral field: the pattern is inverted (e below t₂) and Δₜ = 4/9 Δₒ. Because Δₜ is small and never exceeds the pairing energy, tetrahedral complexes are almost always high spin; low-spin tetrahedral complexes are rare (2023 board AR, both A and R true, R explains A).

Spectrochemical series (increasing ligand field strength):

I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO

High spin or low spin (d⁴ to d⁷ only) depends on Δₒ versus the pairing energy P:

• Δₒ < P (weak field): electrons occupy e_g before pairing, giving high spin (d⁴ = t₂g³e_g¹).
• Δₒ > P (strong field): electrons pair in t_2g first, giving low spin (d⁴ = t₂g⁴e_g⁰).
• d¹ to d³ and d⁸ to d¹⁰ have only one possible arrangement.

Colour of complexes. A complex absorbs one colour of white light for a d-d transition (t₂g electron jumps to e_g) and we see the complementary colour.

• [Ti(H₂O)₆]³⁺ (d¹, t₂g¹e_g⁰): absorbs blue-green (~500 nm), promoting t₂g¹ → e_g¹, so it looks violet (2024 board, NCERT 9.25).
• [Cu(H₂O)₄]²⁺ absorbs red, looks blue. [Co(NH₃)₆]³⁺ absorbs blue, looks yellow-orange.
• No ligands, no splitting, no colour: anhydrous CuSO₄ is white, CuSO₄·5H₂O is blue.
• d⁰ and d¹⁰ ions (Sc³⁺, Ti⁴⁺, Zn²⁺, Cu⁺) are colourless: no d-d transition is possible.
• A different ligand changes Δₒ, so the colour changes: [Ni(H₂O)₆]²⁺ is green, [Ni(en)₃]²⁺ is violet; [Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺ differ in colour for the same reason.

Explaining common magnetism questions (CFT)

• [Fe(H₂O)₆]³⁺ (d⁵, weak field): t₂g³e_g², 5 unpaired, strongly paramagnetic; [Fe(CN)₆]³⁻ (strong field): t₂g⁵, 1 unpaired.
• Mn²⁺ with H₂O: 5 unpaired; with CN⁻ (as [Mn(CN)₆]⁴⁻): only 1 unpaired.

Limitations of CFT: it treats ligands as point charges, so it wrongly predicts anions as strongest ligands; it ignores M-L covalent character.

6. Metal Carbonyls & Applications

Homoleptic carbonyls (only CO ligands; metal is in zero oxidation state).

Synergic bonding: CO donates a lone pair from carbon into an empty metal orbital (M ← C σ bond), and the metal donates electrons from a filled d orbital into the empty antibonding π* orbital of CO (M → C π bond, back-bonding). Each bond strengthens the other, so the M-C bond is strong and the C≡O bond is weakened.

Applications of coordination compounds

7. Quick Sheet & Last-Minute Checklist

Before the exam, check:

1. Can I name and write formulae for five complexes, including one anionic and one with en?
2. Can I count the ions from a formula and match it with AgNO₃ or conductivity data?
3. Can I list every isomer type for a given complex and say which are optically active?
4. Can I give hybridisation, shape and magnetism of [Co(NH₃)₆]³⁺, [CoF₆]³⁻, [Ni(CN)₄]²⁻, [NiCl₄]²⁻ and Ni(CO)₄?
5. Can I draw the octahedral splitting diagram and fill electrons for weak and strong fields?
6. Can I explain the violet colour of [Ti(H₂O)₆]³⁺ in three lines?