What is the hybridisation of $\ce{F}$ in $\ce{BF3}$ or $\ce{CH3F}$ or $\ce{PF5}$ or $\ce{SF6}$? My book states that the p orbitals of fluorine overlaps, but I feel that sp3 hybridisation occurs in $\ce{F}$, so its sp3 hybrid orbital overlaps with the empty p orbital of boron.

According to Ron's answer, all atoms with p or d orbitals can hybridise.

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    $\begingroup$ It's unhybridized. $\endgroup$
    – xasthor
    Aug 7, 2017 at 3:55
  • $\begingroup$ @xasthor please see Ron's answer in the link! $\endgroup$
    – jonsno
    Aug 7, 2017 at 4:39
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    $\begingroup$ Just because it’s Ron doesn’t mean it must be correct ;) $\endgroup$
    – Jan
    Aug 7, 2017 at 9:42
  • $\begingroup$ @Jan Ok I understand that there are no general rules in chemistry! I mostly refer to his answer because he explains things in very elementary manner :) $\endgroup$
    – jonsno
    Aug 7, 2017 at 13:34
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    $\begingroup$ I was once very puzzled with this as well. When I consulted a Chemistry expert the other day, what I learnt was that hybridisation is a concept that is used to explain features of a molecule, such as geometry. Thus, determining hybridisation states for terminal atoms would not be useful and would be unnecessary. What implications would there be if it were sp or sp3? There would likely be not a lot of difference in the consequences. Thus, I would like to argue that it is unnecessary to determine the hybridisation states of terminal atoms. Please feel free to correct me if I'm wrong. $\endgroup$ Aug 12, 2017 at 10:56

1 Answer 1


TL;DR: As a rule of thumb, terminal (heavy) atoms are almost always best described as having sp hybrid orbitals (at the most).

For more on this, I refer you to my answer on What is the hybridization of chlorine in vinyl chloride?

Here are the relevant visualisations for $\ce{BF3}$ at the DF=BP86/def2-SVP level of theory. First I'll start with the canonical orbitals obtained from a molecular orbital point of view. The molecule has $D_\mathrm{3h}$ symmetry, therefore some orbitals are degenerated.

canonical orbitals of BF3

With appropriate transformations, we can interpret these orbitals in a localised fashion, which will give us hybrid orbitals. In this view the σ bond orbitals, and the lone pairs are triply degenerated. Note that there is a significant amount of delocalisation of the fluorine lone pairs into the empty boron p orbital.

nbo of BF3

Here is the representative section from the natural bond orbital analysis:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
(core and symmetry equivalent orbitals skipped)
   5. (1.99345) LP ( 1) F  2            s( 63.79%)p 0.57( 36.19%)d 0.00(  0.01%)
   6. (1.94124) LP ( 2) F  2            s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
   7. (1.86896) LP ( 3) F  2            s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
  14. (1.99778) BD ( 1) B  1- F  2
               ( 17.29%)   0.4158* B  1 s( 33.25%)p 1.98( 65.98%)d 0.02(  0.76%)
               ( 82.71%)   0.9094* F  2 s( 36.24%)p 1.76( 63.62%)d 0.00(  0.14%)
 ---------------- non-Lewis ----------------------------------------------------
  17. (0.38551) LV ( 1) B  1            s(  0.00%)p 1.00(100.00%)d 0.00(  0.00%)
  18. (0.05374) BD*( 1) B  1- F  2
               ( 82.71%)   0.9094* B  1 s( 33.25%)p 1.98( 65.98%)d 0.02(  0.76%)
               ( 17.29%)  -0.4158* F  2 s( 36.24%)p 1.76( 63.62%)d 0.00(  0.14%)

Here you can clearly see the approximate two sp orbitals and the remaining p lone pair orbitals.

The above remains true for (almost) all terminal atoms, because of the local linear symmetry (only one bonding partner, negligible external field), although with lesser extend for heavier atoms (like bromine). This is because hybridisation in general becomes a less reliable description due to the increase in the s-p gap.

  • $\begingroup$ Thanks! Is it true for all Halogen atoms like $F,Cl$ etc in such compounds like $PCl_3$ or $CH_3Cl$ ? $\endgroup$
    – jonsno
    Aug 7, 2017 at 6:27
  • $\begingroup$ @samjoe it is true for all terminal atoms. To a lesser extent for very heavy elements because of the general lack of hybridisation due to a large s p gap. $\endgroup$ Aug 7, 2017 at 6:30
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    $\begingroup$ Hehe, I started writing up an answer but then I had to go back to the wet lab. You obviously beat me to it ;) $\endgroup$
    – Jan
    Aug 7, 2017 at 9:41
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    $\begingroup$ @Jan I hope you would have been saying the same, and I also hope I didn't ruin too much of your work. Knowing that, I'm glad I hit send before I went to the seminar then, it would most likely have been the other way around ;) $\endgroup$ Aug 7, 2017 at 12:04
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    $\begingroup$ @samjoe It is a computational method to calculate approximations for the wave function. In this case it is a density functional with a specific basis set. This level usually give reasonable results for not much computational cost. The details of the methodology are only interesting for those who are regularly working with it. For the intents of this answer, you can ignore it. It's just a more accurate way to generate orbitals than guessing. $\endgroup$ Aug 7, 2017 at 13:29

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