Even (perhaps especially) if you believe in quantum mechanics right down to your toes, you should be skeptical of a crude, approximate application like MO theory. "Especially" because it tries to treat strongly interdependent electrons as if they were independent.

When we assert that you can get some surprisingly realistic insights from the perspective of MO theory, you should demand examples, specific evidence that it agrees with reality. A great preliminary example is

Hybridization and Structure of XH3 Molecules

Experimental Evidence


Question
Theoretical Prediction
Facts
Evidence from Infrared
Evidence from ESR - CH3/CD3
Why CF3 is Pyramidal

Question:

You have often been told that bond angles at a carbon atom are related to its hybridization (sp linear ; sp2 trigonal ; sp3 tetrahedral). The basis of this relationship is that it is widely thought (not so easily proven) that the hybrids on a single atom must be orthogonal to one another, that is, they must have no net overlap.

[It is easy to see that the overlap of two pure p orbitals on the same atom depends on the angle between their axes. If they are parallel (that is, if they are one and the same p orbital) the overlap integral is 1; if they are antiparallel, it is -1 ; if they are perpendicular, it is 0. So you will not be surprised to learn that the integral is cos(θ), where θ is the angle between the axes of the two p orbitals. It is a small step from here to derive the angle between orbitals of arbitrary hybridization (e.g. the angle between sp1.2 and sp7.3 hybrids on the same atom), but all we need to know is that there is such a relationship. Click here if you just must know what it is]

If we assume that bonds are straight (which is often true), this gives us a way of using bond angles to test our prediction of hybridization.

For molecules of the type XH3, there are two obvious symmetrical structures:

Planar with X in the center of an equilateral triangle of H atoms.
This would give 3 sp2 hybrids at 120 and a 4th, purely p AO.

Pyramidal with the X a certain distance above (or below) the center of the triangle formed by the three H atoms.
This would give angles as small as 90 (for pure p orbitals) and leave s-character for the 4th hybrid AO.

The Question is whether XH3 should be planar or pyramidal for a particular X.


Theoretical Prediction:

The X atom must prepare itself for bonding to three H atoms by hybridizing to give three orbitals with a single electron each. To prepare the best such hybrid, from the isolated atom's point of view, the principle is do not waste any of the low-energy s-orbital.

When the central atom is B, with only 3 valence electrons, one atomic hybrid will be vacant. This lone orbital should be p, not an sp hybrid, so as not to waste s-character. So in preparing to bond, B should use sp2 hybridization for the three bonding orbitals. This uses up 3 * 1/3 = all of the s-orbital.

When the central atom is C, with 4 valence electrons, it makes no difference how the s-character is distributed over the 4 hybrid orbitals. Since all four hybrids are equally occupied and together consume all of the s orbital, none of the s-character will go to waste.

When the central atoms is N, with 5 valence electrons, there are 2 electrons for the nonbonding orbital. The atom would prefer to put its s-character in this heavily occupied orbital. Nitrogen's idea of an ideal hybridization would be pure s for the pair and a bonding electron in each of 3 p orbitals.

But atomic energy isn't all there is to it. Bonding helps lower the energy of the shared electrons, especially when overlap is good, and the best overlap for 3 orbitals comes with sp2 hybridization.

In two cases the prediction is clear. By both atomic and bonding considerations BH3 should be planar and hard to deform out of the plane. CH3, which doesn't care from an atomic point of view should be planar to give the strongest bonds, but it should be easier to deform than BH3.

The excitement comes with NH3. Where the atom wants to be pyramidal while the bonds want to be planar. Which consideration is more important? Here experiment will teach us something, rather than just confirming our qualitiative theory.

A very simple alternative theory for structure (but not reactivity), Valence State Electron Pair Repulsion VSEPR, considers only what MO theory "ignores", namely electron repulsion. It correctly predicts planar for BH3 and pyramidal for NH3.  It also predicts pyramidal CH3.


Facts:

XH3
# of Electrons in
nonbonding AO
Atomic Hybrid
Preference
Best Overlap
for Bonding
Observed
Spectroscopic
Evidence
BH3
0
3 x sp2; vacant p
120
3 x sp2
120
"strongly"
planar
120
IR
1141/cm
CH3
1
no preference
"
"weakly"
planar
120
IR, 606/cm

ESR CH3 38G
CD3 36G

NH3
2
3 x p ; pair in s
90
"
pyramidal
107
IR
932, 968/cm
CF3
1
3 x sp>2 ; s>0p
<120
"
pyramidal
111
ESR
271G


Evidence from Infrared Spectroscopy

The table says that BH3 and CH3 are predicted to be planar, and that they are planar. How do we know this?

It is not practical to measure the angle by x-ray diffraction for several reasons:
(1) It is hard to find H locations by x-ray (why?).
(2) BH3 is a gas, not a crystal.
(3) As we'll see very shortly, BH3 molecules bond with one another to form a dimer, so BH3 doesn't even exist, except as a dilute gas. CH3 is worse. Whenever two of them get near one another their SOMOs mix to form C2H6 irreversibly.

It is possible to observe absorption of infrared light by a dilute gaseous sample of BH3, or by CH3 trapped in an inert solid at low temperature. The IR spectrum of BH3 shows absorption of light whose wavelength gives 1141 waves per centimeter (the more waves per cm, the higher the frequency and energy). CH3 absorbs lower-energy light at 606/cm.

Long experience with many molecules (and quantum mechanical calculation) has shown that this kind of absorption is due to the "umbrella" vibration of XH3, the vibration that distorts a pyramid through planarity to the mirror-image pyramid (think of an umbrella in a high wind, or see the diagrams below the graphs).

We can use one-dimensional quantum mechanics to deal with this problem, but the case is a little different from tye type to which we are accustomed. We know the mass, but what IR give us is

not potential energy as a function of bond distance
but rather energy separation of the first two quantum levels for umbrella vibration

What we need to find is the potential energy as a function of bending distortion. We assume it will be an harmonic oscillator and adjust the steepness of the parabola until the energy separation of the levels matches the IR frequency.

The figures below show for each molecule:

potential energy as a function of deformation (dashed parabola)
lowest-energy vibrational wave function (red nodeless wave)
energy for the lowest wavefunction (horizontal dashed
red line)
second vibrational wave function (blue wave with one node)
energy for this second wavefunction (horizontal dashed
blue line)

The greater energy difference for BH3, measured by IR absorbance, shows that it meets four times more resistance to deforming from planarity than CH3 does. This confirms our theory that BH3 cares more about being planar (to leave p vacant and fully use s) than CH3 does.

But how do we know that the potential energy functions for BH3 and CH3 should be parabolas, which we assumed in choosing to fit an harmonic potential? After all, choosing a parabola assumes that the planar form is lowest in energy!

We don't know that they are precisely parabolas, but we can be certain that they are not the alternative double-minimum displayed by NH3. The curious thing about the IR spectrum of NH3 is that there are two closely spaced lines at 932 and 968/cm rather than a single line as in BH3 and CH3. This results from a double minimum, where the planar structure is a local maximum in energy between two pyramidal minima, as approximated below [the simple form of the double minimum potential in Erwin won't do the job quite right]:

The lowest allowed total energy level (black dashed horizontal line) represents two degenerate wavefunctions (not shown) which are the sum and difference of nodeless functions in the two wells.

The functions shown are the favorable (red) and unfavorable (blue) combination of single-node, single-well functions. These are not very different in energy, because overlap of the single-well functions is small. The red/blue energy difference gives rise to the double absorption of IR light, which is a certain sign that planar NH3 is an energy maximum between the two pyramidal minima. Together with other data, this result shows that NH3 is almost tetrahedral (the HNH angles are 107 vs. 109.5 for tetrahedral), and that the planar barrier between the two wells is about 3 kcal/mole above the pyramidal energy minima.

Incidentally, with microwave spectroscopy it is possible to measure the energy separation of the two lowest-energy "out-of-plane" vibrational states for the ammonia molecule (the ones whose energies are indicated by the single lowest dashed line in the figure above). The difference is about 2 cal/mole (not kcal/mole) or about 1/cm. This is obviously much less that the the difference of about 36/cm between the third and fourth levels (where the overlap is greater). In the lowest levels it takes ~5 x 10-14 / 0.002 = 2 x 10-11 seconds to "tunnel" from one well to the other. Tunneling between the excited levels is 36 times faster.

In this case the quantum theory of molecular vibration has been used to help interpret experimental data. It provides a quantitative measure of molecular energetics and confirms the geometric prediction of a very approximate (MO) form of the quantum theory of electron distribution.


Evidence from Electron Spin Resonance Spectroscopy - CH3/CD3

Other types of spectroscopy confirm the favored planarity of CH3. For example in ESR spectra there is a separation or splitting of peaks due to interaction between a magnetic nucleus and an "odd" electron (one not existing in an electron pair, bonding or unshared. Normally this interaction can be observed only when the electron is actually on the nucleus, not just nearby. Thus splitting would be observed for an electron in an s or sp-hybrid orbital, which has density on the nucleus, but not for one in a pure p orbital, which has a node at the nucleus.

There is one "odd" electron in CH3, the one not involved in forming one of the three bonds. In samples with a magnetic 13C nucleus (common 12C in not magnetic), one can use the splitting to measure what fraction of its time the odd electron spends in the 2sC orbital.

The two figures below show the odd-electron MO of CH3 [calculated by Macspartan, 6-31G(*)] for a structure with tetrahedral bond angles. This orbital is often called the SOMO (Singly Occupied MO). In the figure on the right i the chosen density contour has been made transparent to reveal the nuclei. Obviously the Carbon atom lies near the node of this orbital. Because of nonplanarity, the H 1s orbitals overlap with the mostly-2p orbital of Carbon and create small bulges in the red lobe. This keeps the red lobe from appearing smaller than the blue one, as one might have anticipated for a Carbon sp hybrid that points up.

 

 

The contour nets below provide a SOMO comparison of planar and pyramidal CH3
Hybridization of the pyramidal carbon has caused the red lobe poke up and engulf the carbon nucleus (the radius of which has been exaggerated 10,000-fold).
(It is interesting to think why the bottom lobe bulges up to the nucleus, rather than the top lobe's bulging down.)
Slices through the center of the carbon nucleus show how it lies on the node for the planar structure
but in the lower lobe for the pyramidal one.
(It makes no difference that red and blue were accidentally interchanged from the figures above)
closeup flat CH3

So in planar CH3 the electron should spend no time on the 13C nucleus, and in bent CH3 it should spend only a modest amount of time.

As we expect the magnetic interaction measured by EPR is small; 38 Gauss is not much compared with 400 G calculated for an electron in a pure 2s orbital of carbon. So CH3 must be nearly planar with an almost pure 2p SOMO.

That CH3 gives even 38 G of interaction is because it is not static. Rather it vibrates back and forth from planarity in the umbrella mode, spending most of its time slightly bent, as you already know from the vibrational wave function calculated above for CH3.

The reason CD3 gives a smaller splitting of 36 G, is that its heavier mass keeps the vibrational wave function closer to planarity.


Why CF3 is Pyramidal

Since CH3 is planar, one might suppose that CF3 should be planar as well.

The F atoms have more electrons than H, and are also partially charged because of uneven sharing of the bonding electrons with C. On both counts they should repel one another more strongly than H atoms do. Keeping the F atoms as far apart as possible should favor planarity.

In fact CF3 is observed by ESR to be pyramidal. The splitting by 13C is 272 G, 8 times as large as in CH3!

Here is the SOMO of the minimum energy structure [calculated by MacSpartan]. The orbital contours mostly obscure the green Fs, the black C, and the cream-colored bonds. The net contour and the central detail show the same kind of bulge over the nucleus as in pyramidal CH3.

 

 

CF3 is observed to be pyramidal, and calculated to be so by complicated programs like MacSpartan, but how can we understand the fact, when we predicted planarity? Surprising cases like this offer us the chance to learn something new.

The explanation derives from the poor energy-match in the C-F bond. Whereas C and H have orbitals of about the same energy and share bonding electrons about 50:50, the bonding orbitals in CF3 are largely on F giving a partially ionic bond.

Consider the Carbon atom's point of view. In CH3 it didn't care about hybridization, since each of its 4 hybrid orbitals would have one electron. Each bonding orbital would at any typical time house one member of the electron pair shared with H. In CF3, however, the shared pairs spend most of their time on F, so the Carbon is not so responsible for them. The Carbon figures, "Why not put my s orbital where it will do the most good, stabilizing the SOMO electron, which I am mostly responsible for (as shown above)?"

The situation is analogous to NH3, where nitrogen is responsible for 2 electrons in its unshared pair orbital, but only (a little more than) one electron in each bond. In CF3 carbon is responsible for 1 electron in the SOMO and less than one electron in each bond.


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copyright 2001-2003 J. M. McBride