# Is the Co-ordination number of Cobalt always 6 in every co-ordination compound?

For example in the following compounds

1. CoCl3.6H2O
2. CoCl3.6NH3
3. CoCl3.5NH3
4. CoCl3.4NH3

in all of these complexes, we took the

a)oxidation state of Cobalt as +3 ( since there are 3 Cl- ions and NH3 and H2O are neutral ligands )

b) Co-ordination number of Cobalt as 6

and the answer to writing the compounds was:

1.[Co(H2O)6]Cl3 or [Co(H2O) 5Cl]Cl2.H2O

1. [Co(NH3)6]Cl3

2. [Co(NH3)5Cl]Cl2

3. [Co(NH3)4Cl2]Cl

My question is this: Why do we take the co-ordination number of Cobalt always six?

Can't Co take other co-ordination numbers like 4? or 5?

So for the following compound

CoCl3.4NH3

Can't we write it as [Co(NH3)4]Cl3 instead of

[Co(NH3)4Cl2]Cl

Also how to predetermine the C.N of the central metal atom just by looking at anions and ligands? (cases when no coordination sphere is marked like in my above examples)

• This link includes a tetrahedral coordiation complex of cobalt, where the coordination number is 4. Mar 13, 2021 at 6:12
• So how do we pre-determine the C.N of the central metal atom ? Mar 13, 2021 at 6:21
• @TanyaGupta We need to know the structure (for example, crystal structure). Even then there might be issues with the determination of C.N. — that's why a concept of fractional coordination numbers exists. Mar 13, 2021 at 8:53
• Could you give an example that uses the concept of fractional co-ordination number for a compound? Mar 13, 2021 at 12:05

With these compounds, you have coordination number $$6$$, but like other transition metals cobalt can have different coordination numbers with the right ligands. In this answer the $$\ce{CoO4^{3-}}$$ ion is discussed as an example of a tetrahedral complex with the rare case of high spin despite the central ion having an oxidation state of $$+5$$. (The tetrahedral geometry and strongly pi-donating ligands favor the high-spin electron confuguration.) Five-coordination is seen in Co(II) complexes with pi-accepting organic ligands such as $$\ce{o-(C6H4)(SCH3)(P(C6H5)2)}$$ [1].