# Increase in bond length of BF₄⁻ compared to BF₃

Why does $\ce{BF4^-}$ have a larger bond length than $\ce{BF3}?$ Usually in molecules the bond angle increases to reduce repulsion but, here the bond length increases. Why is it so? Does the bond length increase with the bond angle in other cases of such kinds of molecules?

• Isn't this then a simple case of the higher bond length of $Sp^3$ hybrid orbitals in $\ce {BF_4^-}$ than the $Sp^2$ hybrid orbital in boron trifluoride? – Satwik Pasani Sep 23 '13 at 14:21
• @SatwikPasani hybrid orbitals don't make for good answers in general since they don't actually exist. The fact that they remain in the chemistry vernacular has been discussed recently – bobthechemist Sep 24 '13 at 1:38
• @bobthechemist as much as I personally dislike the overuse of hybrid orbitals in chemical education, they have their use… (and lots of limitations, yes) – F'x Sep 24 '13 at 12:33

There seems to be a bit of confusion between bond length and bond angle here. It is true that VSEPR theory predicts that bonding pairs of electrons (and lone pairs, for that matter) will repel one another as much as possible, thereby increasing bond angles, but this has little to do with bond length.

One way to explain the different bond lengths in $\ce{BF_3}$ and $\ce{BF_4^-}$ is by looking at the molecular orbital diagram of $\ce{BF_3}$:

The linked diagram isn't great since it does not show the filled orbitals: the $1a_2'$ $1e''$ and $4e'$ orbitals are all filled, making the lowest unoccupied orbital (LUMO) the $2a_2''$. If you haven't wrapped your head around molecular orbital diagrams of this complexity, don't sweat it. The bottom line is, the $2a_2''$ orbital has antibodning interactions with the $2p_z$ orbitals of boron and the $2p_z$ orbitals of the surrounding fluorines. Thus, adding electrons to this orbital makes the $\ce{B-F}$ bonds weaker, and therefore longer.

• Sorry if I didn't understand well. but I imagine all orbital symmetries and energies are going to shift around once a fluoride ion is added to $\ce{BF3}$, so to me it seems a bit strange to argue about the bonding properties of $\ce{BF4^{-}}$ by considering a simple electron addition to $\ce{BF3}$ MOs. Is this really safe? – Nicolau Saker Neto May 26 '15 at 11:42
• @NicolauSakerNeto You are correct that the orbital symmetries will change and the energies will shift. We'd need better computational methods to get at quantitative explanations. I'm making an assumption that the HOMO of $BF_{4}^{-}$ will still be higher in energy (and somewhat antibonding) even in a tetrahedral conformation; however all I've seen is MO diagrams for methane or tetrahedral d-metal complexes, so I can't be certain. – bobthechemist May 27 '15 at 1:09

$\ce{B}$ in $\ce{BF3}$ is $sp^2$ hybridized (since the molecular geometry is trigonal planar), and therefore the unhybridized $p$ orbital in $\ce{B}$ can overlap with the $p$ orbitals in $\ce{F}$. You can draw resonance structures, and from the valence bond theory perspective, the bond order of $\ce{B-F}$ in $\ce{BF3}$ will be approximately 1.33 due to the partial $\pi$ bond character in the $\ce{B-F}$ bond in $\ce{BF3}$.

For $\ce{BF4-}$, the $\ce{B}$ is $sp^3$ hybridized and hence p orbital overlap is not possible. Hence, the $\ce{B-F}$ bond is simply a conventional 2-centred, 2 electron bond with bond order 1. Comparing bond orders, it becomes clear that $\ce{BF4}$ has a longer bond.

If you wish to, you can draw a simplified MO diagram showing stabilization due to overlap of the $\ce{B}$ $2p$ and $\ce{F}$ $2p$ orbitals, but the valence bond theory argument suffices.

Can we explain this on the basis of charge density?

Like, in $\ce{BF3}$, the number of $\ce{e-}$ around $\ce{B}$ are 6 while in $\ce{BF4-}$ there are more number of $\ce{e-}$ around $\ce{B}$.

Since we know that $\ce{B}$ has small atomic size, so more number of $\ce{e-}$ means reduced effective nuclear charge on the valence $\ce{e-}$, due to which the bond length increases.

Thus, $\ce{BF4-}$ has longer bond length than that in $\ce{BF3}$.