Does "C"_2 exist, and if so, describe its electron distribution?

1 Answer
Jun 7, 2015

Actually, diatomic carbon, or C_2, does exist, but only as a gas and at extremely high temperatures.

This article from the literature discusses it:
http://pubs.acs.org/doi/abs/10.1021/ja00194a042

(You can view that in full if you have an ACS subscription.)

You can also prove this by constructing carbon's molecular orbital diagram.

![http://www.reddit.com/r/askscience/comments/2o2emd/is_there_such_a_thing_as_a_quadruple_bond/](useruploads.socratic.org)

The C_2 molecule has a total of 8 valence electrons, 4 from each carbon atom. As you can see, 6 valence electrons occupy bonding orbitals - sigma_(2s)^(2), pi_(2px)^(2), and pi_(2py)^(2), and only 2 occupy an antibonding orbital - sigma_(2s)^(star 2).

The diatomic carbon molecule has a bond order equal to

"B.O." = 1/2("bonding electrons"-"antibonding electrons")

"B.O." = 1/2(6-2) = 2

This suggests that the two carbon atoms are bonded via a double bond. However, as you can see, it's not a "classical double bond", meaning that it is not comprised of a sigma and a pi bond

If you go by this explanation, the Lewis structure of diatomic carbon looks like:

:"C"="C":

Those lone pairs of electrons are highly reactive, which is why diatomic carbon can only exist at temperatures close to 4000^@"C".

The literature article above describes how computational chemistry was able to predict the structure of C_2, which would be a mixture of the ""^1 Sigma_g and ""^3 Pi_g states. These energy states are similar in energy (E_(""^3 Pi_g) > E_(""^1 Sigma_g)), so thermal excitation would allow both to exist more-or-less at the same time.

The ground-state singlet would look like a diradical C_2 with a triple bond:

cdot"C"-="C"cdot

And the ground-state triplet would have two electrons in carbon 1's bonding p-orbital lobe and one electron each in carbon 2's bonding and antibonding p-orbital lobes.

(the Lewis structure of the standard prediction of diatomic carbon above IS this structure, but doesn't indicate which lobes are being occupied)