# What will be the coordination number in Fe(0)?

Fe(0) has its electronic configuration $\ce{3d^6 4s^2}$ in the ground state. In this case multiplicity will be either 1 or 3 during excitation. If multiplicity is 1, the vacant excited orbitals are $\ce{dsp^3}$. So, in this case it should be trigonal bipyramidal structure with the attachment of 5 ligands.

But in the case of multiplicity 3, the vacant excited orbitals will be $\ce{sp^3}$, excluding 2 singly occupied $\ce{3d}$ orbitals. If 4 ligands are there in attachment, what will be the structure -- tetrahedral or square planner?

Another additional information is also needed, is there any possibility to increase the number of ligand in this structure where multiplicity is 3? If yes, can you explain please?

• You'll have hard time finding hi-spin iron (0) complexes. I'm unaware of any hi-spin metal (0) complexes. I suspect this is because hi-spin mean little covalent interaction with ligands, resulting in formation of pure metal. – permeakra Jul 11 '16 at 12:41
• Fe(0) can have at least six ligands. See diiron nonacarbonyl. It does not answer your question(s), but then the answer to "What will be the coordination number of Fe(0)?" is It depends. – Ben Norris Jul 11 '16 at 12:43