# How are there pi bonds in B_2 molecule without sigma bonds?

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Nov 27, 2017

There shouldn't be, and that's why ${\text{B}}_{2}$ is very unstable; it has two orthogonally-localized half-$\pi$ bonds, which is quite a weak bond.

Imagine that each $\text{B}$ takes the place of a $\text{CH}$ on $\text{HC"-="CH}$.

Then, take away the $\sigma$ bond, the two $\text{H}$ atoms, and one electron from each of the two full $\pi$ bonds, and you have a structure similar to ${\text{B}}_{2}$.

What about ${\text{Li}}_{2}$ "molecule"? Is that stable? (Its bond length is $\text{267.3 pm}$, over twice the length of an average bond.)

The first chemical bond made in a molecule is preferentially a $\boldsymbol{\sigma}$ bond.

$\sigma$ bonds are formed from a direct atomic orbital overlap. In comparison, $\pi$ bonds are sidelong overlaps and thus, $\sigma$ overlaps are made preferentially because they form the stronger bond.

${\text{B}}_{2}$ contains two boron atoms, which each use a basis of a $1 s$, a $2 s$, and three $2 p$ atomic orbitals.

For ${\text{Li}}_{2}$ through ${\text{N}}_{2}$, there exists an orbital mixing effect that makes the $\sigma$ molecular orbital for a $2 {p}_{z} - 2 {p}_{z}$ overlap (${\sigma}_{g \left(2 p\right)}$) higher in energy than the $\pi$ molecular orbitals for a $2 {p}_{x / y} - 2 {p}_{x / y}$ overlaps (${\pi}_{u \left(2 p\right)}$).

So, this particular orbital energy ordering takes place in the molecular orbital diagram, similar to the ${\text{C}}_{2}$ molecule:

where $| {E}_{{\sigma}_{2 {p}_{z}}} | > \left\{| {E}_{{\pi}_{2 {p}_{x}}} | = | {E}_{{\pi}_{2 {p}_{y}}} |\right\}$.

Since $\text{C}$ has one more electron, ${\text{B}}_{2}$ would have a similar MO diagram, EXCEPT for two fewer electrons. This, at first glance, seems to suggest that ${\text{B}}_{2}$ has two half-$\pi$ bonds, with a molecular electron configuration of:

$\textcolor{g r e e n}{{\left({\sigma}_{1 s}\right)}^{2} {\left({\sigma}_{1 {s}^{\text{*"))^2(sigma_(2s))^2(sigma_(2s^"*}}}\right)}^{2} {\left({\pi}_{2 {p}_{x}}\right)}^{1} {\left({\pi}_{2 {p}_{y}}\right)}^{1}}$

Since ${\text{B}}_{2}$ is paramagnetic with two electrons, in order to make that bond, there are indeed two half-$\boldsymbol{\pi}$ bonds, which form what we represent improperly in line notation as a $\sigma$ bond... and it isn't actually a $\sigma$ bond.

$: \text{B"-"B} :$

That isn't a $\sigma$ bond, but two half-$\pi$ bonds. We would then expect ${\text{B}}_{2}$ to be very unstable, which it is. (The lone pairs are from the ${\sigma}_{2 s}$ and ${\sigma}_{2 s}^{\text{*}}$, since bonding + antibonding filled = nonbonding.)

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