# Do Ligands electrons fill the d-orbitals of the metal ion? [duplicate]

We know that ligands make coordination bonds with d-orbitals of transition metals forming complexes. My question is, if that happened, all d-orbitals will be filled and no electron transition could happen between eg and t2g orbitals to be responsible for magnetism or colour formation.

I need clarification.

You are mistaken. The electron pairs donated in the coordinate bonds by the ligands do not fill the d orbitals of the metal ion.

The basis of crystal field theory is the electrostatic repulsion between the d orbitals of the central metal ion and the electron cloud of the ligands. This repulsion then leads to the breaking of the degeneracy of the orbitals, giving rise to disparity in energy between different sets of d orbitals. This arises due to the different spatial orientations of the 5 d orbitals (i.e. $xy, yz, xz, x^2-y^2, z^2$). Thus, they experience different amounts of repulsion from the ligands when the ligands are oriented in different geometric configurations (i.e. octahedral, square planar, tetrahedral etc.) around the central metal ion.

For more details on crystal field theory and an alternative explanation known as ligand field theory (based on molecular orbital theory), you can refer to the following discussion:

In an octahedral complex, what happens to the electrons donated by the ligand?

Since the d orbitals are not filled up by any ligand electrons, the electronic transitions between the orbitals of different energy levels (such as in the case of $t2g$ and $eg*$) are certainly possible.

Now, you may wonder: Where do the donated electrons go?

Well... based on valence bond theory, they are used in the formation of covalent bonds between the central metal ion and the ligands. These covalent bonding interactions are described as the overlap of the ligands' atomic orbitals with the electron pairs with the hybrid atomic orbitals of the central metal ion.

For example, in the case of a hexaaquairon (III) complex, the iron (III) ion has 5 electrons, each singly occupying one d orbital. As the d orbitals are occupied, the vacant 4s, 4p and 4d orbitals are hybridised to give 6 $sp^3d^2$ orbitals to be used to overlap with the atomic orbitals of the ligands.

Thus, the donated electrons actually are "shared" between the valence atomic orbitals of the ligand and the valence hybrid orbitals of the metal ion.

Hope I have clarified your doubts.

• No, hybridisation can and should not be used to describe coordinate bonds. – Jan Jun 15 '17 at 18:08
• This is useful. But in case of [Cr(H2O)6]+3. there will be 3 single electrons in 3 orbitals t2g. and all the other orbitals will be hybridized d2sp3. And according to what I understand, there will be coordinate covalent bonds between the lone pair of electrons of the ligands and these hybridized orbitals, meaning that these orbitals will be filled. So, how the electrons will transfer from the t2g orbitals to any higher filled orbital? thus, How could this compound have a color? – Ibram Mikhail Jun 15 '17 at 22:58
• @Jan Why not? Although valence bond theory is limited, it suffices in this case right? Is there anything that it fails to provide an explanation for? – Tan Yong Boon Jun 16 '17 at 0:24
• @IbramMikhail Sorry if I didn't make it clear but the orbitals used to make the hybrid orbitals are the 4 d orbitals, not the 3 d orbitals. The 3 d orbitals remain as they are, in terms of electron population. – Tan Yong Boon Jun 16 '17 at 0:31
• Valence bond theory is never a good idea for complexes, full stop. If anything, use crystal field als a low-level theory to describe them. Many of valence bond theory’s assumptions are drastically incorrect for complexes. – Jan Jun 16 '17 at 21:34