Valence Bond Theory

Valence Bond Theory pictures the formation of a coordination entity as a hybridisation event on the metal atom or ion. Under the influence of the approaching ligands, the metal makes available a set of empty orbitals drawn from its $(n-1)d$, $ns$, $np$ orbitals, or alternatively from its $ns$, $np$, $nd$ orbitals. These orbitals mix ("hybridise") to give a new set of orbitals that are:

• equal in number to the coordination number of the metal, and
• equivalent in energy and shape, arranged in a definite geometry — tetrahedral, square planar, trigonal bipyramidal, octahedral, and so on.

Each of these empty hybrid orbitals can then accept one electron pair donated by a ligand, forming a coordinate (dative) bond. Because the hybrid orbitals point in fixed directions, VBT automatically explains why coordination entities have definite, predictable shapes.

Hybridisation and geometry

The type of hybridisation used by the metal fixes both its coordination number and the resulting shape of the entity — for instance, four orbitals mixed as $sp^3$ give a tetrahedral arrangement, while the same four orbitals mixed instead as $dsp^2$ give a square planar one; five orbitals as $sp^3d$ give a trigonal bipyramid; and six orbitals give an octahedron, reached either as $sp^3d^2$ (using outer, higher-energy $d$ orbitals) or as $d^2sp^3$ (using inner, lower-energy $d$ orbitals already present on the metal).

The choice between an inner-orbital route ($d^2sp^3$, drawing on the $(n-1)d$ set) and an outer-orbital route ($sp^3d^2$, drawing on the $nd$ set) is not arbitrary — it depends on how many $d$ electrons the metal ion has and how the ligands influence their pairing. This choice is exactly what decides whether a complex turns out to be low spin or high spin (see the discussion of magnetic behaviour next).

Octahedral complexes — inner vs. outer orbital

Two classic octahedral cobalt(III) complexes illustrate the two routes:

• In $[\text{Co(NH}_3)_6]^{3+}$, the $\text{Co}^{3+}$ ion ($3d^6$) rearranges its six $3d$ electrons so that three $3d$ orbitals are freed up and combine with the $4s$ and the three $4p$ orbitals to give six $d^2sp^3$ hybrid orbitals. All six electron pairs supplied by the ammonia ligands are accommodated without leaving any unpaired electron, so the complex is diamagnetic. Because an inner ($3d$) orbital is used for hybridisation, this is called an inner orbital, low spin, or spin-paired complex.

• In $[\text{CoF}_6]^{3-}$, the fluoride ligands instead leave the $3d^6$ configuration of $\text{Co}^{3+}$ largely undisturbed (only paired as far as Hund's rule allows), and hybridisation draws on the empty $4d$ orbitals together with $4s$ and $4p$ to give $sp^3d^2$ hybrids. This leaves unpaired electrons in the $3d$ set, so the complex is paramagnetic. Because an outer ($4d$) orbital is used, this is called an outer orbital, high spin, or spin-free complex.

Tetrahedral and square planar complexes

For a tetrahedral entity such as $[\text{NiCl}_4]^{2-}$, the $\text{Ni}^{2+}$ ion ($3d^8$) hybridises one $4s$ and three $4p$ orbitals into four equivalent $sp^3$ hybrid orbitals oriented tetrahedrally. Each chloride ion donates one electron pair into a hybrid orbital; since the $3d^8$ configuration retains two unpaired electrons, this complex is paramagnetic.

By contrast, $[\text{Ni(CO)}_4]$ is also tetrahedral but is diamagnetic, because here nickel is in the zero oxidation state and its $3d$ configuration ($3d^{10}$ as the neutral atom's relevant count) contains no unpaired electron to begin with — the geometry is the same, but the electron count on the metal is different, so the magnetic outcome differs. …