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In my chemistry class we learned the so-called "8N rule" (very similar to the Octet rule) which helps one determine the number of bonding electron pairs and thus the number of bonds in certain molecules:

$$\text{No: of bonding electrons} = 2\cdot\text{No: of hydrogen atoms} + 8\cdot \text{No: of other atoms} - \text{Total no: of valence electrons}$$

This tends to work for most molecules I've tried. The lecture notes don't really mention any exceptions or limitations. Therefore, I assumed that it would work for a wide variety of molecules, especially ones that involve p-block atoms only.

However, the rule doesn't work with $\ce{BF3}$. Here I get $2\cdot0 + 8\cdot 4 - (3+3\cdot7) = 8$ bonding electrons which should translate to $8/2 = 4$ bonds (or one of the $\ce{B-F}$ bonds is a double bond). But this is apparently not the case - there are three single bonds or 6 bonding electrons. So where is my mistake?

I should say that this is not the first time I've encountered a rather simple molecule where the rule doesn't apply which is what gives confidence that it's not because of a mistake I made but because of a limitation of this rule that I don't know.

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  • $\begingroup$ Octect rule is more than 100 years old. It does not apply to many compounds. People have come up with better explanations and bonding theories. $\endgroup$
    – AChem
    Oct 4, 2022 at 16:50
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    $\begingroup$ Then again, the bonds in BF3 are at least somewhat double. $\endgroup$ Oct 4, 2022 at 17:05
  • $\begingroup$ In that case, is there a better and yet as handy alternative for calculating the number of bonds or the bond order? $\endgroup$
    – Hrach
    Oct 4, 2022 at 18:31
  • $\begingroup$ The octet rule is a somewhat surprising model. It holds quite well for molecules as long as you are reading the 8 as an upper limit. Since it had never been formalised, this role of thumb is very open for interpretation. $\endgroup$ Dec 28, 2022 at 13:48

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Boron is one of the exceptions to the octet rule in which the octet is incomplete, and thus it only has 6 valence electrons instead of 8 to become stable. To put it simply, because Boron has 3 valence electrons, they react to only form 3 bonds which stabilises the atom with 6 electrons. I assume there is a more complex way to explain each facet of this incomplete octet that I don't know of, though.

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  • $\begingroup$ The octet rule has a lot of exceptions. Examples : $\ce{B2H6}$, $\ce{NO}$, $\ce{NO2}$, $\ce{SF6}$, $\ce{H2SiF6}$, plus the majority of compounds containing transition metals. $\endgroup$
    – Maurice
    Oct 4, 2022 at 19:21

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