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$\ce{F-}$ can displace $\ce{OH}$ in $\ce{B(OH)3}$ and form $\ce{BF3}$. $\ce{BF3}$ is stable due to resonance, but donation of oxygen should be stronger, as -I power of fluorine dominates its lone pair giving power (+M).

Why is $\ce{BF3}$ more stable than B(OH)3?

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  • $\begingroup$ Can you define exactly what your measure of stability is in this case? $\endgroup$
    – Andrew
    Mar 16 at 18:12

2 Answers 2

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The question might he better posed as "Why does fluorine have a greater affinity for just about everything (including boron) than oxygen?

The answer is first off, technically fluorine does not have a greater affinity for everything than oxygen. This answer explores counterexamples involving (early) transition metal complexes. However, in most settings the exceptionally low valence orbitals of fluorine and the availability of a vacancy in the fluorine atom to accept another electron facilitates the formation of bonding molecular orbitals with low energy. Meaning stronger, more stable bonds.

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There are competing effects of resonance and steric hindrance in $\ce{B(OH)3}$ and $\ce{BF3}$. It is the reduced steric hindrance, not resonance, that dominates stability in this case. Boron has small enough ionic radius for this to happen.

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    $\begingroup$ Color me sceptical - F and O have very similar covalent radii, I think. Is the question's premise even true, when BF3 hydrolyses both to $\ce{B(OH)3}$ and fluoroboric acid? $\endgroup$
    – Mithoron
    Mar 15 at 21:38
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    $\begingroup$ Boron forms covalent bonds, so why will we look at the ionic radii? Also steric hinderence will be close to negligible as bond angles are big(120 deg) and as said by Mithoron, O and F have nearly same radii $\endgroup$ Mar 15 at 21:49
  • $\begingroup$ Both O and F are more electronegative than B. That is why ionic radius is relevant. Covalent radii of F and O vary a lot in reality, values given in tables are some kind of average of different cases. Hydrolysis involves additional hydrogen. 120 deg is not big enough for ions as small as Boron's. $\endgroup$
    – Paul Kolk
    Mar 16 at 12:48
  • $\begingroup$ How will we use ionic radii though? Boron doesn't form ionic compounds and if we take ionic radii of O and F, we are doing the opposite as they are radii of F- and O-. But due to backbonding, the partial charge on O and F will be positive(which reduces their size). Also Boron is bigger than O and F. So I don't think steric hinderance will be significant. My main doubt was that, many times in organic chemistry, we take halogens as electron withdrawing group, but O is electron giving group(with lone pair). So oxygen should make boron more stable than fluorine does. $\endgroup$ Mar 17 at 6:13
  • $\begingroup$ @EagerToLearn Why not to model B3+ as an ion? It is in fact much smaller than OH- and F- . Backbonding may change their shape a bit, but doesn't make them smaller than B3+. $\endgroup$
    – Paul Kolk
    Mar 17 at 18:31

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