As far as I know, when writing Lewis structures for molecules, we need to consider the octet rule and the valence electrons each atom has.

But, if we take for example, the molecule $ Cl O_2^- $ , we have the following Lewis structure: (as it appears here ) O - Cl - O with 6 valence electrons surrounding each oxygen atom, and 4 valence electrons surrounding the chlorine atom.

My problem is: The oxygen atom has 6 valence electrons. By forming a chemical bond with the chlorine atom, it shares one electron, which leaves it with a total of 5 non-bonded electrons. So, how can it be that in the Lewis structure of this molecule, it has 6 non-bonded electrons and one shared electron? (or a total of 7 electrons! )


1 Answer 1


If you consider the $\ce{ClO2-}$ as coming together from neutral atoms (plus the extra electron), all the bonding electrons "come from" the chlorine. This is commonly called a 'dative covalent bond' (or just 'dative bond' for short), and is sometimes drawn like this, instead of just a line:

Dative Bond

As this image suggests, you can think of it as an atom with a lone pair bonding with an atom with no spare valence electron.


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