Re: Does carbon form quadruple bond?

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

Yuri wrote:

>Dick Brown <[EMAIL PROTECTED]> wrote in message news:<[EMAIL PROTECTED]>...

>>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).

>
>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...

>
>>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.

>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)

>Dick Brown

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