9
$\begingroup$

In the solid state, LiCp adopts a polymeric multidecker structure where Li atoms are sandwiched between two Cp rings (shown in this diagram from Organometallics 1997, 16 (17), 3855–3858):

Solid-state structure of LiCp

If we consider the η5-binding mode of the Cp rings, then it seems that the number of electrons around Li is more than eight. However, Li can't extend its octet. How can the structure and bonding be explained?

|improve this question
$\endgroup$
  • 2
    $\begingroup$ It is ionic bonding not covalent bonding. A lithium cation is between two cyclopentadienyl anions in the 3D structure of the solid, but the chemical formula is 1:1. $\endgroup$ – MaxW Sep 12 '18 at 15:17
  • 2
    $\begingroup$ I think the question may well be more interesting than "it's ionic bonding". MeLi and other alkyllithiums, for example, can hardly be described as purely ionic - the structure and bonding is quite complicated. I wouldn't be surprised if there was significant covalency in CpLi, too, although I might of course be wrong, and a cursory flip through my usual bunch of references doesn't turn up any more useful info. $\endgroup$ – orthocresol♦ Sep 12 '18 at 15:36
  • $\begingroup$ Well, as you can easily count it's six electrons per 1 Li as all rings are bound to two Li atoms. This type of boding is present in multidecker sandwich complexes. $\endgroup$ – Mithoron Sep 12 '18 at 16:01
  • $\begingroup$ @Mithoron...I understand your point but in that case, its octet is not fulfilled, then how this structure is so stable? why the molecule does not adapt any more complicated structures? $\endgroup$ – Deepta Chattapadhyay Sep 12 '18 at 16:27
  • 1
    $\begingroup$ It didn't ping me (because of dots presumably). Li makes electron deficient molecules almost all the time. It's middle ground between, say, Rb with close to zero valence electrons and carbon with octet most of time. $\endgroup$ – Mithoron Sep 12 '18 at 19:08
4
$\begingroup$

This summary identifies several calculations indicating that covalent interactions between lithium and cyclopentadienyl moieties do occur; the interaction is mostly but not purely ionic.

Draw the occupied $\pi$ orbitals of the cyclopentadienyl groups and, for comparison, draw the $s, p_x, p_y, p_z$ orbitals on lithium with the $z$ axis along the length of the chain. Observe:

• The lithium $s$ and $p_z$ orbitals overlap the lowest energy cyclopentadienyl $\pi$ orbitals.

• The lithium $p_x$ and $p_y$ orbitals overlap the bonding degenerate pair on the cyclopentadienyl ring.

With these overlaps, we can construct three lithium-cyclopentadienyl bonding combinations per cyclopendienyl group, one corresponding to each cyclopentadienyl $\pi$ bonding orbital. We then occupy these orbitals with six electrons per ring. In effect we have extended the aromatic coupling of the cyclopentadienyl ion into the third dimension with the lithium orbitals.

Heavier alkali metals do not work as well because their orbitals are too diffuse, but they could overlap larger rings.

|improve this answer
$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.