Since BF3 has the shape of tetrahedral, and in the provided solution, Boron is sp2, why is Fluorine sp3 hybridized?

enter image description here

  • $\begingroup$ Consider the lone pairs and bonds of fluorine simulatensously for first ding the hybridization $\endgroup$ – Yusuf Hasan Dec 6 '18 at 7:34
  • 2
    $\begingroup$ Tetrahedral is one thing BF3 is not. $\endgroup$ – Ivan Neretin Dec 6 '18 at 9:28

First things first: $\ce{BF3}$ is not tetrahedral. It cannot be, as a tetrahedral structure requires a total of five atoms (one in the centre and four in the corners).[1] Instead, it is trigonal planar. However, that is of minor importance.

Secondly, fluorine is most certainly not sp3 hybridised. In general, hybridisation is determined by geometry and not vice-versa, thus we only need to establish the trigonal planar geometry to arrive at boron’s sp2 hybridisation. However, this doesn’t help us for terminal atoms as there is no ‘geometry’—only a single binding partner. In some cases, constraints may exist due to multiple bonds which require π-type bonding orbitals. Again, this is not the case for single bonds which rely solely on their σ orbital.

Instead, the basic assumption should be to assume the hybridisation as low as possible. In that way, fluorine’s s type orbital can be as low as possible in energy—which means a higher energy gain. It is easy and possible to assume fluorine bonds solely with a correctly aligned p orbital and so it does.


[1]: This sentence depends on the system of classification you were taught. I was taught only to label those molecules as tetrahedral which have five atoms that correspond to the four corners plus the centre. If there are four atoms as in the case of ammonia ($\ce{NH3}$), it is possible for the overall structure to be a tetrahedron with one corner removed. We were taught not to call this tetrahedral but a trigonal pyramid. Your teacher’s mileage may vary.

  • $\begingroup$ I object. For example, NH3 has but four atoms, yet is tetrahedral all right. $\endgroup$ – Ivan Neretin Dec 6 '18 at 17:23
  • $\begingroup$ @IvanNeretin I object! We used the term trigonal pyramidal for compounds where the electron lone pair is the ‘fifth atom’, since electrons are not visible in structures and thus the atom positions give a flatter trigonal pyramid, not a tetrahedron. $\endgroup$ – Jan Dec 6 '18 at 17:34
  • $\begingroup$ All right, I stand corrected. $\endgroup$ – Ivan Neretin Dec 6 '18 at 19:14
  • $\begingroup$ This dispute about tetrahedral and trigonal pyramidal can be easily resolved. The former describes the molecular geometry (i.e. with respect to ligands around given central atom) while the latter describes the electronic geometry (i.e. with respect to regions of electron density around a given centre). $\endgroup$ – Tan Yong Boon Dec 7 '18 at 2:46

The answer is - they are not really sp3 hybridized. Usually (for low-level textbooks) reasoning goes like "since we have 1 bond and 3 lone pairs, fluorine has 4 electron-ish things => fluorine is sp3". But actually, the closer element is to 18 group (and F is in 17), the less preferable becomes hybridization of s and p orbitals. For fluorine, this is (almost completely) not preferable process, so you shouldn't consider it sp, sp2 or sp3 hybridized.


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.