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In my experience, most texts that address hypervalency say that it only occurs from elements in the 3rd period and onwards. This explains the occurrence of $\ce{Cl2O7}$ or chlorine heptoxide. However, some 2nd period nonmetals like $\ce{C}$ and $\ce{O}$ show hypervalency.

Examples:

  • $\ce{CH5}$ - This is unlikely to occur but it does sometimes happen that carbon bonds to 5 atoms instead of 4.
  • $\ce{H3O+}$ - Here oxygen is hypervalent.

How is it possible for carbon and oxygen to each have 9 electrons if each orbital only holds 2 electrons? Do they switch between electrons or something?

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  • $\begingroup$ I have improved the formatting of your question using $\LaTeX$. For more information on how to do this yourself please see here. $\endgroup$
    – bon
    Commented Jun 1, 2015 at 18:41
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    $\begingroup$ Do you mean $\ce{CH_5^{+}}$? Also, why do you call $\ce{H3O^{+}}$ hypervalent? There are just 8 electrons around the central oxygen; 6 (3x2) in the 3 $\ce{O-H}$ bonds and 2 residing in a lone pair. $\endgroup$
    – ron
    Commented Jun 1, 2015 at 18:54
  • $\begingroup$ But it has 3 bonds which oxygen doesn't normally have since it hates being positive and would rather be negative if charged at all. Wouldn't that extra bond cause oxygen to have 9 electrons and thus be hypervalent? $\endgroup$
    – Caters
    Commented Jun 1, 2015 at 20:14
  • $\begingroup$ And yes I do mean the methyl cation. $\endgroup$
    – Caters
    Commented Jun 1, 2015 at 20:18

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Normally when we talk about a single covalent bond, we are referring to a 2-centre 2-electron bond, which means that there are two electrons holding two atoms together.

Carbon never forms 5 bonds. The only exception that I know of is the $\ce{CH5+}$ methanium cation, the bonding in which can be explained by a 3-centre-2-electron bond. The same kind of bond appears in diborane ($\ce{B2H6}$). In both cases, the octet rule (or duplet rule in the case of the bridging hydrogens in diborane) is not violated. It is just that those 2 electrons are shared amongst 3 different atoms, so each "bond" is effectively half a bond (in MO theory parlance we say that the bond order is 0.5). You could think of it as three of the C-H bonds being normal 2-electron bonds, and two of the C-H bonds being half-bonds (having one electron each). The total number of electrons around carbon is therefore $3 \times 2 + 1 + 1 = 8$.

The neutral species $\ce{CH5}$ does not exist, because it has one more electron than the $\ce{CH5+}$ cation. That would mean that you either have to put 9 electrons around carbon, or put 3 electrons around hydrogen, both of which are of course not allowed.

The hydronium ion $\ce{H3O+}$ is not actually hypervalent. It is similar to the ammonium ion $\ce{NH4+}$ in that a dative bond is formed from the lone pair on O to a $\ce{H+}$ ion.

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    $\begingroup$ Nice answer! BTW, $\ce{CH5^{+}}$ is not the only example of "5 coordinate" carbon. All non-classical carbocations fall into this category of "5 bonds to carbon" through the use of the 3-center 2-electron bond. $\endgroup$
    – ron
    Commented Jun 1, 2015 at 21:54
  • $\begingroup$ Whether the carbon in $\ce{CH5^+}$ has five bonding linkages is ambiguous. There could be four bonds to carbon with one having side-on H2 at the other end. See the description of the ions in this answer. $\endgroup$
    – Oscar Lanzi
    Commented Feb 3, 2022 at 16:05

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