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Why is $\ce{BF3}$ considered a Lewis acid despite backbonding with fluorine and still accepting an electron pair?

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    $\begingroup$ It's mesomeric stabilization not back-bonding and bonds are still polarized to fluorine. $\endgroup$
    – Mithoron
    Oct 7, 2015 at 20:09
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    $\begingroup$ I have a feeling that fluorine in $\ce{BF3}$ does that thing almost reluctantly, like "OK, you may use my spare electron pair for a short while, but then give it back to me a.s.a.p." Probably this has something to do with the fact that we rarely hear of fluorine in $sp^2$. $\endgroup$ Oct 7, 2015 at 20:17

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A Lewis acid is defined as an electron-pair acceptor. So for something to act as a Lewis acid, it needs to want electrons. Prime examples are $\ce{H+}$, the hardest Lewis acid around (zero polarisability, very high charge per volume ratio) and practically every metal cation out there: $\ce{Al^3+, Zn^2+, Fe^3+, Ag+}$ just to name a few.

Consider boron, a rather electropositive element — it counts as a metalloid so it is somewhere between non-metals and metals. We are binding it to fluorine, the most electronegative element, and we’re doing that three times. It should be evident that there is hardly any electron density left on boron. How happy would it be, if some other atom gladly donated their electron pair to share?

Now what are we going to do if there is no Lewis base around? Well, initially boron will still be there, depleted of all its valence electrons by fluorine (or nearly at least). This is where fluorine discovers its charity side: All three fluorines donate just a tad of electron density so that the baby boron in the middle will stop crying. This is what you referred to as ‘back bonding’ and Ivan calls ‘mesomeric stabilisation’. But the point is: That doesn’t help against the electron deficiency in any way, it’s more like boron’s final counter-measure against loosing electrons.

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    $\begingroup$ Chemists please forgive me for turning atoms into humans. At least I didn’t explain the whole electron-withdrawing thing with socialism … $\endgroup$
    – Jan
    Oct 7, 2015 at 21:15
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A Lewis acid can accept a pair of electrons from a Lewis base. The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid.

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    $\begingroup$ IF you have a strong chemistry background I could go further and say that Fluorine is a more electronegative atom than Boron so the electron density in the BF3 would shift more towards the F atom. This creates the greatest electron deficiency on the B atom allowing it to accept another electron pair with ease. $\endgroup$ Oct 7, 2015 at 20:41
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    $\begingroup$ Well, this is not a full answer since it doesn't talk about the $\text{p}\pi$-$\text{p}\pi$ bonding explicitly mentioned in the question. OP is asking why $\ce{BF3}$ is still a Lewis acid in spite of this bonding. $\endgroup$ Oct 8, 2015 at 8:56
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Even though there is 2pπ-2pπ backbonding in $\ce{BF3}$ molecule, it still acts as a Lewis acid. This is because the lone pairs that are donated through this way are better and permanently donated to the boron atom than backbonding. Backbonding is just a concept for temporary internal stability of a molecule.

When any Lewis base approaches the $\ce{BF3}$ molecule, then the π-electron density of 2pπ-2pπ backbond start to shift towards the fluorine atom because there is now no need for backbonding to stabilise the molecule (as electron are now donated to boron permanently through a better way).

This situation happens many time when the molecule in which the central atom is bonded to atoms of different electronegativities through one or more multiple bonds behaves as a Lewis acid, such as $\ce{CO2},$ $\ce{CS2},$ $\ce{SO3}.$ $\ce{CO2}$ molecule acts as a Lewis acid since as the Lewis base approaches to carbon atom, the π-electron density shifts towards the oxygen atom, and carbon accepts lone pairs of a Lewis base.

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A lewis acid is defined as an electron pair acceptor. The boron in BF3 is electron deficient & has an empty d orbital,so it can accept a pair of electrons,making it a lewis acid.Also it contains only 6 electrons in outermost shell making it able to accept an electron pair to complete its octet.Thus it's lewis acid

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    $\begingroup$ This has nothing to do with d orbitals. I think you mean an empty p orbital, not a d orbital. $\endgroup$
    – bon
    Mar 16, 2016 at 16:00

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