What would a six coordinate metal-ligand complex structure look like?

2 Answers
Sep 27, 2017

The bonds are in an octahedral shape.

Explanation:

Let's take Sulphur hexaflouride (SF#""_6#) for example. Electron repulsion theory would say that for 6 bonds and 0 lone pairs of electrons, the four floruine atoms would be spread out evenly into a octahedron shape.
http://pd.chem.ucl.ac.uk/pdnn/symm2/sf6.htm

Since a complex is just molecules forming dative covalent bonds with a metal, the idea works the same way. Molecules will spread out evenly, for example in hexaaquairon(II) there are 6 water molecules dative covalently bonded to the Fe#""^(2+)# ion. The water molecules will evenly spread out.
http://www.lookchem.com/cas-153/15365-81-8.html

Another example is EDTA, its a polydentate ligand that forms 6 dative covalent bonds. When it bonds to a metal ion, all 6 bonds are created, and are also spread out too.
http://commons.wikimedia.org/wiki/File:Metal-EDTA(.)png

Sep 27, 2017

Most likely, octahedral.


There are exceptions, but for the most part, a simple metal-ligand complex is usually octahedral.

https://upload.wikimedia.org/

This just forms the most symmetric geometry, with equal bond angles for all adjacent ligands. It therefore minimizes bond-pair-bond-pair electron repulsions in most cases.

The main exception is trigonal prismatic:

https://upload.wikimedia.org/

This particular compound is air-sensitive, subliming at #-30^@ "C"#.

It is thermodynamically stabilized due to tungsten's #d^0# configuration, which is stabilized by the six strong #sigma# donors (#"H"_3"C":^(-)# is a strong base and nucleophile, and thus an excellent electron-donating group).

Apparently, the trigonal prismatic shape is a result of a so-called second-order Jahn-Teller distortion.