Electron delocalization frequently reveals itself
through a distorted molecular geometry. The most frequently cited
situation is a symmetric set of bond distances in a molecule that
does not have a symmetric Lewis structure, but delocalization can
also create other kinds of geometrical distortions.
The first two sections in this essay examine distortions
in bond distances and bond angles. The last section looks at recent
research on the "real" factors that make benzene (and
possibly many other molecules) symmetric.
| Bond Angles | Benzene
Ozone, O3, is an
excellent example of a symmetrical resonance hybrid with distorted
bond distances. If ozone was a normal molecule with localized bonds,
that is, if it was adequately described by Lewis structure I, we
might expect to find one short OO bond of ~122 pm (this distance
is observed in O=O )
and one long bond of ~147 pm (this distance is observed in
As it happens, the OO distances in ozone are identical,
and at 128 pm, they tends toward "short", but we could
call them "in between." To explain these observations,
chemists say ozone's electron pattern is a "hybrid" of
resonance forms I and II as shown in the dashed-line formula on
the left, and ozone's geometry reflects its partial double bonds.
This kind of analysis is applied to all resonance
hybrids. We always expect a hybrid's bond distances to reflect the
hybrid's bond pattern. If the hybrid contains a partial double
bond, the bond distance should be somewhat longer than a complete
double bond, and somewhat shorter than a single bond. Analogous
predictions would be made for partial single bonds and partial
VSEPR rules link bond angles to the number of electron
pairs surrounding an atom. 4 pairs imply a tetrahedral atom with
bond angles of ~109o, 3 pairs
imply a trigonal atom with bond angles of 120o,
and 2 pairs imply a linear atom.
The counting of electron pairs, and the prediction
of bond angles, becomes uncertain when a lone pair is delocalized.
(VSEPR rules treat "lone pair + sigma bond pair" as two
electron pairs, and "sigma bond pair + pi bond pair" as
only one; a delocalized lone pair with partial pi bonding properties
falls somewhere in between.) Therefore, a resonance hybrid with
a delocalized lone pair may display distorted bond angles.
Consider the N atoms in compounds III-V. These atoms
have 4 electron pairs, one of which is a lone pair. We expect to
see tetrahedral bond angles of ~109.5o
if the lone pair electrons are localized on N.
Molecular models show that only one N (compound III)
is tetrahedral (angles = 109.6, 109.6, 105.8o).
The N in compound V is trigonal (angles = 121.8, 119.2, 119.0o)
and the N in compound IV is somewhere "in between" (angles
= 114.6, 114.6, 111.0o). Distorted
bond angles like these indicate that IV and V may be resonance hybrids
with delocalized lone pairs.
More information can be obtained by looking at CN
bond distances. Compound III, with its normal bond angles, also
contains a normal CN single bond: 146 pm. The CN bond in compound
V (trigonal N) is much shorter: 136 pm. And the CN bond in compound
IV ("in between" N) is somewhere "in between":
140 pm. These data suggest that the CN bonds in V and IV have some
partial double bond character, consistent with delocalization of
the N lone pair.
These ideas can be described by the resonance forms
shown below. Compound V's bond angles and distances are highly distorted,
so its minor form must make an appreciable contribution. Compound
IV's geometry is closer to normal, and its minor forms must make
a smaller contribution. (You might have anticipated that "minor"
V would be more important than "very minor" IV because
"minor" V places negative charge on electronegative O,
while "very minor" IV places it on C.)
Compounds IV and V show how misleading a single Lewis
structure can be. If we didn't know the geometries of these compounds,
we might never have suspected these compounds of being resonance
hybrids. However, now that we know these compounds are resonance
hybrids, we can be on the "look out" for other similar
compounds. Any time a N (equipped with a lone pair) lies next to
a pi system, it may be possible to delocalize the N lone pair by
converting it into a pi bond pair.
Revised VSEPR: partial electron domains. If you try to apply VSEPR to the superposition shown above, you will probably get confused. The major form with an N lone pair shows 4 electron domains around N, while the minor form with a YN double bond shows just 3. The hybrid is a single molecule with a single geometry; how can this be consistent with two different electron domain values?
To solve this problem, we might pay attention to the fact that the YN bond domain exists in both forms, but the N lone pair domain does not. We might think of the latter as a partial domain, making the total number of N domains fall somewhere between 4 and 3. (Note: for reasons that I won't go into here, the N lone pair domain effectively vanishes even when the "Y=N" form is only a minor form. One can do well by ignoring delocalized lone pairs when counting electron domains.)
(Every now and then, someone announces that the "theory
of chemistry" is complete. Whoever thinks this is true might
consider this next story...)
Several lines of evidence identify benzene as a resonance
hybrid. One of these is benzene's symmetric geometry. All 6 CC bond
distances are identical, and at 140 pm they lie in between the distances
observed for normal CC single bonds (153 pm) and double bonds (134
Chemists expect a hybrid's bond distances to reflect
its bond pattern. In the case of benzene, a partial double bond
should produce an "in between" bond distance.
Chemists have noted for a long time that benzene's
symmetric geometry is ideal from another point-of-view: the symmetric
geometry makes the energies of the resonance forms identical, the
precise condition needed to create maximum energy stabilization
(see Resonance Effects on Energy). Based
on this (correct!) analysis, chemists then jumped to a logical,
but incorrect conclusion: benzene is symmetric because
of electron delocalization and the energy gain it provides.
Beginning in the 1980's, two chemists, Sason
Shaik of Israel and Philippe
Hiberty of France, published a series of research papers that
showed that the conventional wisdom regarding benzene's geometry
cannot be true. Their counter-theory, which is now supported by
several kinds of experimental and computational evidence, shows
that it would be better (energetically) for benzene if its pi
electrons remained localized in 3 short double bonds instead
of delocalizing to form 6 long partial double bonds. Their theory
also identifies benzene's localized sigma bonds as the real
driving force behind benzene's symmetric geometry; the sigma bonds
benefit greatly from identical CC bond distances.
How can 3 short double bonds be preferred when a symmetric
geometry provides maximum pi electron delocalization and stabilization?
The answer to this paradox lies in the two factors that contribute
to electron energy (see Resonance Effects
= Ewrong + interactions
When benzene moves from a "3 short double bond"
geometry to a symmetric geometry, two things happen simultaneously:
stabilizing interactions are maximized because electron delocalization
is maximized, but Ewrong
is destabilized because the symmetric geometry is far from ideal
for the double bond pattern seen in each resonance form. Ewrong
turns out to be the deciding factor. Hence, Shaik and Hiberty's
conclusion: it would be better for benzene if its pi electrons
remained localized in 3 short double bonds.
This theory may seem strange (it has yet to find its
way into chemistry textbooks), but one can easily find analogous
systems that prefer localized bonds over delocalized ones.
The old argument might have suggested that 6 H atoms
would form an H6 resonance hybrid
with 6 identical delocalized bonds (see above), but it is well known
that they prefer to form 3 H2
molecules with 3 short localized bonds.
Shaik and Hiberty's analysis of benzene almost certainly
applies to other resonance hybrids. One can only wonder what chemistry
textbooks will look like 50 years from now.
 OO does not contain
a double bond in its ground state. The cited distance is observed
in the first excited singlet state, which does contain a double