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In a covalent bond between two atoms, an electron from one of the either atom is shared by overlapping of their orbitals.

So, Why can't three atoms share an electron and overlap their orbitals?

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The normal case for a covalent bond is indeed a 2-centre 2-electron bond.

There are however cases, where 3 centres (= atoms) either share 2 or 4 electrons.

  • In organic chemistry, the most famous case for an electron-deficient 3-centre 2-electron bond is the 2-norbornyl cation.
  • In inorganic chemistry, boranes with their banana bonds are the role models.

Hypervalent molecules with 3-centre 4-electron bonds, like $\ce{SF6}$ or the bifluoride anion $\ce{HF2-}$ have been discussed here too.

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    $\begingroup$ @AwalGarg Maybe I misunderstood your question then. In a "normal" covalent bond, the two atoms (centres) share two electrons, one valence electron from each partner. They are not sharing one, but two electrons. It appeared to me that you were asking for the "less normal" cases. $\endgroup$
    – Klaus-Dieter Warzecha
    Commented Mar 25, 2014 at 8:20
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    $\begingroup$ @AwalGarg In organic chemistry, I'm not aware of anything more electron-deficient than the non-classical cations (3-centre 2-electrons) and I doubt that something like 5-centre 2-electrons exists even as an extremely short-lived intermediate. On the other hand, what about metallic bonds in a crystal lattice: Is this the case for an $\infty$-centre $\infty$-electrons system? $\endgroup$
    – Klaus-Dieter Warzecha
    Commented Mar 25, 2014 at 8:48
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    $\begingroup$ Ok, then let's take a step back and look at the organic electron deficient cases. In the case of the norbornyl cation, we could imagine that to be like a normal $\ce{C-C}$ bond plus a cationic centre near by that wants to mix in. For a larger number of centres, sharing the same number (2) of electrons, I see two problems (i) lack of binding force: what will hold them together and (ii) geometrical restrictions: how will all these atoms manage sufficient orbital overlap. So, unless proven otherwise, I'd say 3-centres 2-electrons is the end of the road here. $\endgroup$
    – Klaus-Dieter Warzecha
    Commented Mar 25, 2014 at 9:02
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    $\begingroup$ But I'm sure some guys here, well-versed in theoretical chemistry, might want to give some (less pictorial) answers too :) $\endgroup$
    – Klaus-Dieter Warzecha
    Commented Mar 25, 2014 at 9:06

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