# How can ionic bonding in lanthanides occur without valence orbitals available for overlap?

I've recently been taught that the 4f orbitals in lanthanides are "core-like", supposedly meaning they have radius smaller than the 4d orbitals, therefore they are not available on the outside of the atom to form covalent bonds with ligands. The lecturer then went on to say that, therefore, the bonding of lanthanides to ligands is entirely ionic.

It is my understanding that a bond is deemed ionic when there is a large electronegativity difference between the two atoms, i.e. the bonding molecular orbital is near-identical in energy to the atomic orbital of the more electronegative atom, as shown below for sodium fluoride:

However, if this is the case, how can lanthanides form any bonds if their valence orbitals are not available? Surely orbital overlap with the ligand is still required to form a molecular orbital, even if the resulting orbital is ionic in nature?

• There's no such thing a entirely ionic bond in chemistry, as that wouldn't be chemical bond, but a two separate charged particles. – Mithoron Mar 7 '20 at 18:54

Lanthanide atoms also have $$5d$$ and $$6s$$ valence orbitals available for bonding, and these can overlap effectively with ligands. Whether this overlap leads to a bonding orbital encompassing both metal and ligand (covalent bond) or transfer if the electrons to a much low-energy ligand orbital (ionic bond), a bond is formed either way. The bond may not be entirely ionic but is entirely (or nearly so) derived from the $$5d$$ and $$6s$$ orbitals of the lanthanide atom instead of the $$4f$$ orbitals.

Lanthanides do tend to form complexes with high coordination numbers, exceeding the coordination number of six commonly seen with $$d$$-block transition metal complexes. From here:

Lanthanides share many similar characteristics. These characteristics include the following:

...

Adoption of coordination numbers greater than 6 (usually 8-9) in compounds

...

A preference for more electronegative elements (such as O or F) binding

...

The coordination number thereby exceeds the number of separate covalent bonds that can be formed from $$5d$$ and $$6s$$ valence orbitals alone, so a limited number of covalent bonds must be de-localized over the many linkages. This would lead to the bonds defaulting towards a more predominantly ionic mode, requiring the ligand orbitals to be low-energy and associated with highly electronegative ligand atoms (also noted in the above excerpt).

• Thank you for your response, and my apologies for the late reply - would you be able to clarify what you mean by delocalised covalent bonds over multiple linkages? I can't think about how that works from a molecular orbital standpoint. – Sensei_Stig Mar 16 '20 at 10:53
• For a simple example, look at 3-center 4-electron bonds. Note carefully that while the overall three-center 4-electron bond is stronger than an ordinary covalent bond, it's distributed over two linkages and thus each individual linkage is weaker. – Oscar Lanzi Mar 16 '20 at 11:19
• Ah yes of course, thank you for your answer! – Sensei_Stig Mar 17 '20 at 20:08