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With reference to crystal field splitting in octahedral complexes, my textbook shows the following table of distribution of $d$-orbital electrons in octahedral complexes, based on the energy difference of $t_{2g}$ and $e_g$ orbitals:

Table

Do ALL the $d$ orbital configurations (i.e. pertaining to all examples given) form octahedral complexes? I doubt it because not all of them have two vacant $d$ orbitals, especially in the case of weak field ligands where Hund's rule is never violated.

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A metal ion doesn't actually need vacant d orbitals to form a coordination complex according to Crystal field theory. All the ions mentioned in the table form complexes with water (which is a weak ligand) - see this page.

Crystal field theory assumes that all bonding between ligand and metal ion in a complex is electrostatic (ionic). That means there is no exchange of electrons between the metal ion and ligand. Instead, the negatively charged (or polar in case of neutral ligands like $NH_3$) ligand is attracted to the positively charged metal ion by purely electrostatic forces. This is why vacant d orbitals are not required for bonding.

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  • $\begingroup$ Do they all form relatively stable octahedral complexes, though? For most of the intermediate $d^n$ configurations, octahedral compounds are abundant, but what about the extremes? $\endgroup$ – Nicolau Saker Neto Apr 21 '13 at 15:09
  • $\begingroup$ So,if I want to check the structure of a coordination compound,both the configuration of electrons in $ t_2g $ and $ e_g $ and the field strength of the ligand are required isn't it?Else is all about ionic bonds,no orbital overlap and hybridisation $\endgroup$ – scienceauror Apr 23 '13 at 10:34

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