Natural Science Forum / Chemistry / Organic Synthesis / September 2003

I believe C2 is a known molecule.

John Smith wrote:

Sure!  It's emission is the blue color in comet tails.  It isn't as
single (closed shell) molecule.

John Smith <jsmith@application.com> wrote in message news:<bjiblg$rpa@panther.Gsu.EDU>...

Is it stable though?

In article <bjiblg$rpa@panther.Gsu.EDU>,
  John Smith <jsmith@application.com> wrote:

It is. But according to Herzberg's book, the ground state is a triplet - so you
can't regard that as a quadruple bond.

*** To reply by e-mail, make double u single in address ***

The only way you're going to be able to form four bonds in C2 (normally
a diradical or biradical, depending on which terminology you prefer) is
to promote an electron on each C to a higher energy level (n=3+).  There
aren't enough bonding orbitals available in the n=2 energy level to
allow for four bonds (actually, four bonds and one anti-bond with n=2,
which only gives a triple bond).  Unfortunately, I doubt very strongly
that the bonding energy, even with the 3s-sigma bond, will be high
enough to counteract the energy needed to promote the 2s-sigma*
electrons into the 3s-sigma bonding orbital (and keep it there).  It
might be just possible to have something like this last a nanosecond or
two, but it would be even less stable than neutral triplet C2 (which is
saying something).

By the way, the heavier elements aren't the exception to the bonding
rules.  Carbon is the exception (along with other 2nd row elements)
because of its small size.

As far as free software goes, are you familiar with the Linux OS?
You're much more likely to find free software for Linux than for any
other OS.

Dick Brown

Ian Gay wrote:

Dick Brown <dikbrown@qwest.net> wrote in message news:<bks5d5$7gi@panther.Gsu.EDU>...

Thanks for clearing this out for me, it does make perfect sense. I
have another question though. Suppose you have an sp3 carbon-carbon
bond. According to VSEPR theory the e- clouds repel each other, thus
forming angles of 109.5 degress. Trying to disrupt these angles
requires much energy, and is therefore not stable. But what is the
case, for example, of cyclopropane? (three carbons forming a triangle,
with 2 hydrogens attached to each). This is a stable molecule, even
though the angle strain is great (in this case both carbons are sp3
hybridized). This molecule allows the strain in its angles because it
lacks the one extra hydrogen at the end of each carbon (thus forming
two methyl groups). My question is the following: why are we not able
to disrupt the strain in the angles of an sp3 hybrized-carbon to form
a quadruple bond with another carbon? the energy required to do so,
theoretically, wouldnt be much greater than that found in
cyclopropane. I'll appreciate any guidance...

I meant to say that heavier elements have a "less structured", "more
flexible", electron clouds. Such is the case of Sulfur (e.g. SF6),
which can form an expanded valence shell becuase of its electron cloud
density. Also, this is the reason why Silicon, being a 4th group
element, and having similar properties as Carbon, does not behave
exactly as carbon--the "elektron" cloud is larger and therefore a bit
more stable (more electrons to satify the oppostie charge)

Yes, i do have Linux OS. Though, i found a good software for WIN OS
called ACD/ChemSketch.

The problem with making a quadruple bond with four sp3 orbitals is that
if you point three of them at another atom, the fourth is pointed
directly away from that atom.  Let's look at your example of
cyclopropane.  Firstly, it's not that stable.  This molecule has high
angle strain and torsional strain.  Secondly, you've got bonds heading
in two separate directions (albeit only 60 degrees apart).  In the case
of a quadruple bond, all bonds have to head in the same direction,
causing all manner of problems.  Using hybridization (which is only an
approximation, and not real), there is one way that you might envision a
C-C quadruple bond, using two sp-hybridized atoms, with pi-type overlap
of the sp-orbitals and one of the p-orbitals, and sigma-overlap of the
last p-orbital.  Unfortunately, this would be a very high energy
structure, and, as noted previously, hybridization is not real.  Look at
the MO diagram sometime (a better approximation) and you'll see that the
only way to form a quadruple bond is to promote electrons into an n=3
orbital from a lower-level anti-bonding orbital.

Regarding heavier elements, d-orbital-based expanded octets are another
of those imaginary concepts that, unfortunately, persists even at high
levels of education.  The "expanded octet" is actually due more to ionic
bonding forces and closest packing.  The electron cloud isn't really
more stable (the opposite is true, actually).  The main reason that Si
does not behave exactly like C lies in the size of the respective atoms
and the poor overlap between the bonding orbitals.  With the lower
elements, it's even hard to find double bonds, and those that do exist
bear little similarity to C=C bonds.  (The Pb=Pb bond is not planar, for
example.)

Dick Brown

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