Chemistry · Ch 5 — Coordination Chemistry
Valence Bond Theory
Valence Bond Theory
VBT explains the metal-ligand bond as arising from the overlap of a filled ligand orbital (holding a lone pair) with a vacant hybrid orbital on the central metal atom. Its main assumptions: (1) the ligand-to-metal bond is covalent, formed by the ligand sharing its electron pair with the metal; (2) each ligand must have at least one filled orbital holding a lone pair; (3) the central metal ion supplies a number of vacant orbitals equal to its coordination number, to receive those donated pairs; (4) these vacant orbitals undergo hybridisation -- mixing of comparable-energy atomic orbitals into an equal number of new, equal-energy hybrid orbitals; (5) the hybrid orbitals overlap linearly with the filled ligand orbitals to form coordinate covalent sigma bonds; (6) because the hybrid orbitals are directional, their spatial orientation fixes the geometry of the complex ion; (7) if the hybridisation uses (n-1)d orbitals the complex is an inner-orbital / low-spin / spin-paired complex, whereas if it uses nd orbitals it is an outer-orbital / high-spin / spin-free complex (n being the principal quantum number of the outermost shell); (8) a complex with any unpaired electron on the metal is paramagnetic, and one with all electrons paired is diamagnetic; (9) strong-field ligands such as CO, CN⁻, en and NH₃ force pairing of the metal's electrons; and (10) greater overlap between the ligand and hybridised metal orbitals means greater bond strength. VBT's central shortcoming, despite explaining hybridisation-linked geometry and a spin-only magnetic picture, is that it cannot explain colour, only accounts for the spin contribution to magnetic moment (ignoring orbital con …
Coordination number | Hybridisation | Geometry | Examples
2 | sp | Linear | [CuCl₂]⁻, [Ag(CN)₂]⁻
3 | sp² | Trigonal planar | [HgI₃]⁻
4 | sp³ | Tetrahedral | [Ni(CO)₄], [NiCl₄]²⁻
4 | dsp² | Square planar | [Ni(CN)₄]²⁻, [Pt(NH₃)₄]²⁺
5 | dsp³ (dx²-y² involved) | Trigonal bipyramidal | Fe(CO)₅
6 | d²sp³ (inner dz², dx²-y² involved) | Octahedral (inner orbital) | [Ti(H₂O)₆]³⁺, [Fe(CN)₆]²⁻, [Fe(CN)₆]³⁻, [Co(NH₃)₆]³⁺ …
Complex | Metal config | Ligand field effect | Hybridisation/Geometry | Unpaired e⁻ / Magnetic property / μs
[Ni(CO)₄] | Ni: 3d⁸4s² | CO strong field, pairs the 4s electrons into 3d (3d¹⁰4s⁰4p⁰) | sp³, Tetrahedral | 0, diamagnetic, μs = 0
[Ni(CN)₄]²⁻ | Ni²⁺: 3d⁸4s⁰ | CN⁻ strong field, pairs the 3d electrons (3d⁸4s⁰4p⁰) | dsp², Square planar | 0, diamagnetic, μs = 0
[Fe(CN)₆]³⁻ | Fe³⁺: 3d⁵4s⁰ | CN⁻ strong field, pairs 3d electrons (inner orbital) | d²sp³, Octahedral (inner orbital) | 1, paramagnetic, μs = 1.732 BM
[CoF₆]³⁻ | Co³⁺: 3d⁶4s⁰ | F⁻ weak field, no pairing (outer orbital, 4d used) | sp³d², Octahedral (outer orbital) | 4, paramagnetic, μs = 4.899 BM …
Worked out. Three self-check problems. Q7: the spin-only magnetic moment of tetrachloridomanganate(II) ion, [MnCl₄]²⁻, is 5.9 BM (5 unpaired electrons, consistent with high-spin Mn²⁺, d⁵, sp³ hybridisation, tetrahedral geometry, since a coordination number of 4 with a weak-field ligand like Cl⁻ favours the tetrahedral sp³ outer arrangement over square planar). Q8: asks for the number of unpaired electrons in [CoCl₄]²⁻ (Co²⁺, d⁷, weak-field Cl⁻, tetrahedral sp³, giving 3 unpaired electrons). Q9: a composition Co(en)₂Cl₂Br isolated as two forms A and B, where B reacts with AgNO₃ to give a white precipitate readily soluble in ammonium hydroxide (characteristic of AgCl, so B has ionisable Cl⁻ and is [Co(en)₂ClBr]Cl... more precisely the isomer with Cl⁻ as the free counter ion) while A gives a pale-yellow precipitate (characteristic of AgBr, so A has ionisable Br⁻); the student must write the formulas of A and B, state cobalt's hybridisation in each, and calculate their spin-only magnetic …