3
$\begingroup$

According to molecular orbital theory, $\ce{Li2}$ fills up two electrons in the $\sigma_\mathrm{g2s}$ molecular orbital; $\ce{Li2+}$ fills up one electron in the $\sigma_\mathrm{g2s}$ MO, and $\ce{Li2-}$ fills up two electrons in the $\sigma_\mathrm{g2s}$ MO and one electron in $\sigma^*_\mathrm{u2s}$, resulting in the same bond order for the ions.

I first thought that $\ce{Li2+}$ is stable, because they don't have paired electrons in $\sigma_\mathrm{g2s}$, therefore there is no pairing energy. I soon found that $\ce{Li2+}$ is stable, but someone said it is due to the unfilled antibonding orbital. Is this a right explanation?

Because $\ce{Li2}$ is homonuclear, the weight that imposed on each $\mathrm{2s}$ $\ce{Li}$ atomic orbital would be almost the same. Even though $\ce{Li2-}$ is unstable by filling an antibonding orbital, it will be offset by filling two electrons into the bonding orbital and it has the same bond order as $\ce{Li2+}$ does.

What am I missing now?

$\endgroup$
8
$\begingroup$

First of all some visual aid: an MO Diagram for $\ce{Li2}$.

MO Diagram for dilithium

The concept of bond order allows a very rough estimate of stability. Here we get $\mathrm{BO}(\ce{Li2})=1$, $\mathrm{BO}(\ce{Li2+})=0.5$, and $\mathrm{BO}(\ce{Li2-})=0.5$. So in first approximation, $\ce{Li2-}$ and $\ce{Li2+}$ are equally stable, while $\ce{Li2}$ more stable. But we can consider some more details:

I first thought that $\ce{Li2+}$ is stable, because they don't have paired electrons in $\sigma_{g2s}$, therefore there is no pairing energy.

Adding a second electron to some orbital, would at first double the energy of that bond. But it also introduces some electron-electron repulsion between the two. This however is usually smaller than the gained bonding energy. So the contribution of the $\sigma_\mathrm{g2s}$ contribution in $\ce{Li2}$ is a little less than double its contribution in $\ce{Li2+}$. Please take this last statement with care, because things are more complex than that.

I soon found that $\ce{Li2+}$ is stable, but someone said it is due to the unfilled antibonding orbital. Is this a right explanation?

Basically yes. The thing is, that the energy split of bonding and anti-bonding orbitals is not symmetric with respect to the originating atomic orbitals. The anti-bonding orbital is a little bit higher, than what the bonding orbital is lowered:

$$ E(\sigma^*_\mathrm{u2s}) - E(\mathrm{2s}) > E(\mathrm{2s}) - E(\sigma^*_\mathrm{g2s}) $$

This is because if you add more electrons, you add more electron-electron repulsion. In case of $\ce{Li2-}$ this means the electron in the anti-bonding orbital is more destabilizing (with respect to $\ce{Li2}$), than what the additional electron in the bonding orbital is stabilizing (when going from $\ce{Li2+}$ to $\ce{Li2}$).

Because $\ce{Li2}$ is homonuclear, the weight that imposed on each $\mathrm{2s}$ $\ce{Li}$ atomic orbital would be almost the same.

Not just almost. The electronic wave function must follow the symmetry of the molecular structure, therefore the weights of the orbitals have to be exactly the same. But this argument is not really helpful for this question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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