# Why is CO a Lewis base?

##### 2 Answers
Jun 4, 2018

Because it can be a $\sigma$ donor Lewis base... but it is not just a Lewis base.

It's also a $\pi$ acceptor Lewis acid. And it can readily be demonstrated by examining the MO diagram:

Although $: \text{C"-="O} :$ has a lone pair of electrons to donate energetically owned primarily by oxygen (which could indicate Lewis basicity), it also has an empty LUMO that is energetically owned by carbon that will accept electron pairs (which could indicate Lewis acidity).

• If you examine the HOMO, which is the $3 {a}_{1}$, even though it is energetically owned by oxygen, the carbon lobe is bigger (the wave function coefficient is larger), lessening its Lewis basic behavior.

Even so, it donates electrons from its $\sigma$ MO (on the carbon side) as a $\sigma$ donor in Ligand Field Theory.

• If you examine the LUMO, which is the $2 {b}_{1}$ or $2 {b}_{2}$, it is energetically owned by carbon, AND the lobes on carbon are bigger (the wave function coefficient is larger).

Thus, it accepts electrons into its ${\pi}^{\text{*}}$ MOs (on the carbon side) as a $\pi$ acceptor in Ligand Field Theory.

As a result, CO is both a Lewis acid and a Lewis base, and it demonstrates that in Ligand Field Theory as a sigma donor and pi acceptor.

Jun 4, 2018

#### Answer:

Well, what is a $\text{Lewis base}$?

#### Explanation:

It is an electron pair DONOR....i.e. carbon monoxide contains a lone pair of electrons that can BIND to transition metals for example. $C O$ is an exceptionally good ligand for low-valent transition metals give that it can ALSO act a Lewis acid, a so-called $\pi - \text{acid}$...because the VACANT $\pi \cdot$ orbitals on the carbon monoxide are the right SYMMETRY, and right ENERGY to accept electron density from the metal centre by back-donation.

The back donation is ANTIBONDING with respect to the $C \equiv O$ molecule ... it is BONDING with respect to the complex...