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Name:  Nadya
Status: other
Grade: 9-12
Location:N/A
Country: N/A
Date: 5/20/2005


Question:
How does hybridization account for the bonding of carbon tetrachloride?


Replies:
It is not necessary to invoke "hybridization" especially for CCl4. It must be invoked for all tetrahedral bonds of carbon and other atoms. The electron configuration of an isolated carbon atom in its ground state is: (1s2) (2s2)(2p2). Based solely on the simplest Aufbau principle one would predict that carbon in the ground state should behave a lot like oxygen with two unpaired electrons available for bonding with a bond angle somewhat greater than 90 degrees. The problem is that this prediction is wrong, very wrong, because the neglect of filled valence shells (or sub-shells) oversimplifies the problem to such an extent that it makes a prediction that does not match the experimental data. That is, how we "say" electrons are behaving is inconsistent with "how" electrons are actually behaving. One might suggest that the electronic state of carbon could be: (1s1)(2s2)(2px2)(2py)(2pz). This would lead one to the conclusion that carbon would have 2 lone pairs (2s2), (2px2) --similar to nitrogen, but non-equivalent, and 3 bonding electrons (2 equivalent (2py)(2pz), and the other different (2s2)). However, that reasoning too does not match the data that carbon has 4 bonding electrons and that they are identical.

The correct reasoning, arguing from the experimental data, is that electronic states form hybrid bonds represented by wave functions that are linear combinations of atomic wave functions. This is not "black magic" as it might appear. If a differential equation (and this is important, ANY DIFFERENTIAL EQUATION) has several solutions, then any linear combination of those solutions meeting the boundary conditions is also a valid solution to the differential equation. Whether the other linear combinations are useful is a different matter, but one cannot just throw away possible solutions "without cause".

In the case of carbon the linear combination of one 2s and three 2p orbitals that result in four equivalent bonds is T1 = 1/2(2s+2px+2py+2pz); T2 = 1/2(2s-20x-2py+2pz); T3 = 1/2(2s+2px-2py-2pz); T4 = 1/2(2s-2px+2py-2pz). The shorthand symbols: s2, 2px, 2py, and 2pz stand for actual mathematical formulas of the position of the electrons. Writing those out explicitly would make formulas for T1, T2, T3, and T4 very messy looking so the letter shorthand is used.

It is important to understand that other possible solutions cannot be excluded a priori for mathematical reasons. The one that best fits the observations is the one that is retained. That imposes a certain humility on our understanding of chemical bonding -- or it should.

Vince Calder



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