I'm learning how to draw Lewis diagrams. Everything I've read emphasizes the octet rule. However, to the best of my knowledge, that rule only applies to elements in the first three periods.

On our exam, I expect our professor will ask us to draw some Lewis diagrams of molecules whose atoms do not comport with the octet rule. However, I haven't seen any diagrams of these; it'd be useful to get a sense of what they look like so that my gut doesn't disapprove of what will otherwise be strange looking diagrams.

What are some relatively common molecules that comprise atoms that comprise a great number of valence electrons?

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    $\begingroup$ I don't think that it's really that any molecules have over 8 valence electrons, but some electrons do get shoved into the d orbitals. One example is $\ce{XeF4}$, a square planar with single bonds on all fluorines and two electron pairs on xenon. $\endgroup$ – Sparkery May 31 '15 at 17:23
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    $\begingroup$ Haha...Whenever anyone talks about violating the octet, one of the first molecules they shove into our face is $\ce{BF3}$. Or if they fail in that mission, we'll get hit with $\ce{BCl3}$. $\endgroup$ – M.A.R. May 31 '15 at 17:23
  • $\begingroup$ @M.A.Ramezani I'm afraid that I don't get your point here. $\endgroup$ – MarcoB May 31 '15 at 17:43
  • $\begingroup$ @Macro The title is asking for a common and easily understandable molecule that violates octet. I gave him two. :) $\endgroup$ – M.A.R. May 31 '15 at 17:45
  • $\begingroup$ chemistry.stackexchange.com/questions/444/… $\endgroup$ – Mithoron Jun 1 '15 at 0:41

There are 3 types of octet rule "violations" or exceptions

  • molecules with an odd number of electrons, such as nitric oxide

NO molecule

(image source)

  • molecules with less than 8 electrons around an atom, $\ce{BeCl2}$ and $\ce{BH3}$ serve as examples


(image source)

  • molecules with more than 8 electrons around an atom, such as $\ce{PCl5}$ or $\ce{SF6}$


Take a look at the image source links for other examples as well as some practice problems.

The bonding around electron deficient atoms such as the boron in $\ce{BH3}$ is best explained in terms of 3-center 2-electron bonds (reference; SE Chem example_1, SE Chem example_2)).

Sometimes, arguments based on d-orbital involvement are advanced to explain the bonding in atoms surrounded by more than 8 electrons. However d-orbital involvement in non-transition metals is now generally considered to be unlikely. Instead hypercoordinated (or hypervalent) bonding in which 3-center 4-electron bonds are formed has become a more generally accepted explanation. A number of examples have previously been discussed here on SE Chem (for example, see here).

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    $\begingroup$ Nitrous oxide, N2O also violates the octet rule by most counts. $\endgroup$ – J. LS May 31 '15 at 19:27
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    $\begingroup$ Nice answer. All I have to add is that molecules like $\ce{PCl5}$ can be drawn in accordance with the octet rule. $\endgroup$ – Jan Jun 1 '15 at 17:32
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    $\begingroup$ @Jan Do you mean using a resonance structure where one of the $\ce{P-Cl}$ bonds is shown as ionic? $\endgroup$ – ron Jun 1 '15 at 17:46

In a rather different vein are aromatic carbocationd such as the cyclopropenyl cation

enter image description here

Source: https://I.stack.imgur.com/XdJKq.png

As can be seen, each of the contributing structures has a positively charged carbon atom that violates the octet rule.

What really happens is, the molecular orbitals are so constructed that the $p$ orbitals lying outside the plane of the ring are shared among all three carbon atoms in the ring. Then one orbital is strongly stabilized and the other two orbitals are destabilized. You think you need two pairs of p-orbital electrons to "meet the octet rule" with ordinary valence bond structures, but only one pair is actually stable so the octet rule has to give way.

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