Lesson 7 of 13 · 6 min
Valence bond theory
NCERT §5.5.1
Why is [Co(NH₃)₆]³⁺ an octahedron, and why does it not stick to a magnet while [CoF₆]³⁻ does? The first bonding model answers with hybrid orbitals.
The lesson in notes
In short
Werner's theory could not explain why only some elements form complexes, why the bonds are directional, or where the magnetic and optical properties come from. VBT and CFT are the two models treated at this level.
In VBT the metal mixes (n−1)d, ns, np orbitals, or ns, np, nd orbitals, to make a set of identical hybrid orbitals pointing in fixed directions. Each hybrid orbital accepts an electron pair from a ligand.
Hybridisation and shape: CN 4, sp³ tetrahedral; CN 4, dsp² square planar; CN 5, sp³d trigonal bipyramidal; CN 6, sp³d² or d²sp³ octahedral.
[Co(NH₃)₆]³⁺: Co³⁺ is 3d⁶. The six electrons pair into three 3d orbitals, leaving two inner 3d orbitals for d²sp³ hybridisation. No electron is unpaired, so the octahedral complex is diamagnetic: an inner orbital, low spin or spin paired complex.
[CoF₆]³⁻ uses the outer 4d orbitals (sp³d²), keeps four unpaired electrons and is paramagnetic: an outer orbital, high spin or spin free complex.
[NiCl₄]²⁻: Ni²⁺ is 3d⁸, the four Cl⁻ pairs go into sp³ hybrids, the complex is tetrahedral and two unpaired electrons make it paramagnetic. [Ni(CO)₄] is also tetrahedral but diamagnetic, because nickel is at zero oxidation state with no unpaired electron.
[Ni(CN)₄]²⁻: the 3d⁸ electrons pair up to free one 3d orbital, giving dsp² hybrids. The complex is square planar and diamagnetic.
Hybrid orbitals are not real objects; hybridisation is a mathematical treatment of the atomic wave functions.
Magnetic data often reveal the geometry in VBT: a paramagnetic four-coordinate Ni²⁺ complex is tetrahedral, while a diamagnetic one is square planar.