Question

How to determine coordination number in transition metal complexes?

Answer 1

In undergraduate courses, it ends to be either octahedral, tetrahedral, or square planar geometries, which can only have coordination numbers of `6`, `4`, or `4` respectively...

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In general we classify the transition metal complexes by the number of `bbd` electrons that enter the crystal field orbitals.

Since they have significant metal character, we just approximate them as metal `d` orbitals in the splitting diagram, and say we have a "`bb(d^n)` complex".

COORDINATION NUMBER 4: TETRAHEDRAL

This is common with `d^8` complexes that contain small central atoms like `"Ni"^(2+)` (or sometimes `d^10` complexes, like with `"Cu"^+`).

Also, boron can form tetrahedral fluorides, and manganese also form tetrahedral oxides. The above examples are tetrafluoroborate, permanganate, nickel tetracarbonyl, and tetrakis(pyridine)copper(I), respectively.

COORDINATION NUMBER 4: SQUARE PLANAR

This is common with `d^8` complexes that contain large central metal atoms like `"Pd"` and `"Pt"`.

The above examples are cis-diamminedichloroplatinum(II) and tetrachloroplatinate(II), respectively.

COORDINATION NUMBER 5: TRIGONAL BIPYRAMIDAL/SQUARE PYRAMIDAL

This is not very common to see, but trigonal bipyramidal and distorted square pyramidal structures can form with `"Cu"^(2+)` (`d^10`) and `"Ni"^(2+)` (`d^8`). Again, these are rare.

The above examples are pentachlorocuprate(II) and pentacyanonickelate(II), respectively.

COORDINATION NUMBER 6: OCTAHEDRAL

This is the most common geometry, and occurs typically with `d^6` complexes (like `"Co"^(3+)`-based and `"Cr"^0`-based complexes).

Often, these complexes have `18` electrons in the ligand field (2 from each of 6 ligands, and 6 from the metal).

The above examples are tris(ethylenediamine)cobalt(III) and hexanitritocobaltate(III), respectively (the nitro groups are bonded via the nitrogen, not an oxygen).