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Hybridization and Carbon Tetrachloride
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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Update: June 2012
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