Organic Chemistry CHE230-001
Thursday, August 30, 2001
Lecture 5
The equal exploration of space.
- Two equal orbitals in principle quantum number 2 give rise to linear molecules upon
mixing.
* The px points in two directions, -x and +x.
* The two sp orbitals point in opposite directions because the new orbitals inherit
their directionality from the p orbital.
- Three equal orbitals in principle quantum number 2 give rise to trigonal molecules upon
mixing.
* The px and py orbitals lie in the xy plane.
* Since the new orbitals get their directionality from these they can not explore
space outside the xy plane and they must explore directionality in the xy plane equally.
- Likewise the sp3 orbitals need to explore all space equally.
* directionality to do this is inherited from the Cartesian 2px, 2py and 2pz orbitals.
Let's break out the balloons.
- Instructor connected four balloons sequentially to demonstrate this process.
- The balloons fight for space equally.
The structure of molecules can be explained by two major facts:
1. Electrostatic attraction between electron and nucleus.
2. The shape and nature of atomic orbitals
- The s orbital gets the electron closer to the nucleus.
- Orbitals rich in s character get the electron closer to the nucleus than orbital rich in p
character.
- You need to understand this issue because it will recur in the subjects we encounter later
on in the text. The above explains more than just structure, it will also explain reactivity in later
chapters.
- Why does BF3 have a flat, trigonal structure instead of a bent structure? Why not form
the bonds with p-orbitals and reserve the s orbital for the empty orbital?
* We might be inclined to say that the F atoms repel one another . . . while
plausible this hypothesis does not explain as many phenomena as orbital mixing.
Hybridization of atomic orbitals:
 We tackled this issue when we discussed the structure of methane. Let's pursue it further.
 When we mix orbitals the number of orbitals going in equals the number of orbitals coming
out.
- Thus mixing one 2s orbital and the 2px orbital gives two sp orbitals.
- Likewise one 2s and the 2px and 2py orbital gives rise to three sp2 orbitals
- If you take the whole lot (all orbital space in principle quantum number 2) you end up
with four sp3 orbitals from the 2s, 2px, 2py and 2pz
- There are four orbitals available in principle quantum number 2. Therefore from orbital
conservation it follows that if you use two in hybridization you have two left that remain the
way they started out. Likewise if you hybridize three orbitals (as making three sp2 hybrid
orbitals) there will be one left.

Further the orbitals trade characteristics.
- The s orbital is not directional, like a sphere it points in all directions at once.
* The p orbital is very directional and is labeled based in its directionality x, y and z.
* The new hybrid orbitals are directional. They explore space anisotropically.
- The s orbital penetrates space better to the nucleus. Electrons occupying this orbital can get
closer to the nuclei than electrons occupying p orbitals.
* The new hybrid orbitals penetrate better than plain vanilla p orbitals.
The last point above is the reason why molecules take on the shapes that they do.
 Let's go back to BF3 and think this over.
- We have three choices . . . (at least) for a molecular bonding scheme for BF3.
1.
construct the molecule from 2s, 2px, 2py and 2pz
2.
construct the molecule from two p and one s (non-hybridized).
3.
construct the molecule from three sp2 hybrid orbitals.
The choice that explains what is seen in spectral studies of BF3 is choice number 3.
The reason choice number 3 is better is because the electrons can get closer to the nucleus
in orbitals with increased s character.
The same reasons drive methyl cation and methyl radical to planarity.
Hybridization is not necessarily integral
 Let's consider methyl anion.
CH3(-) bond angles 107.5˚ between the H-C-H
How does this compare to 109.47?
- if electron seek orbitals rich in s character, these two electrons might well exist in an
orbital that is richer in s character than the bonded orbitals.
- in the bonded orbitals the electrons are shared between two nuclei, namely the C and H
atoms.
- The lone pair only has the C nucleus to stabilize its charge.
 We do not have to construct pure sp3 or sp2 orbitals.
- These new mixed orbitals simply need to be some linear combination of the s and p shells.
- We can make . . . three orbitals sp4 or some other such fractional contribution to the
orbitals from s and p.
* in this case we would have to have one other orbital that we could label s0.4p0.6
= sp1.5.
- the math looks something like the following
- When all is said and done the numbers when taken together must sum to 1s and 3p.
- in the case in which we have four sp3 orbitals.
0.25s 0.75p
0.25s 0.75p
0.25s 0.75p
0.25s 0.75p
--------------total: one s and three p
In the example above we have
0.2
0.8
0.2
0.8
0.2
0.8
------------total 0.6s
2.4p with one more orbital to consider!
1s-0.6s = 0.4s
3p-2.4p=0.6p
thus s0.4p0.6 or more briefly sp1.5.
* the three orbitals would possess 20% s and 80% p character.
- contrast this to four sp3 possessing 25% s and 75% p character. Our
three orbitals have more p character and less s character.
* the remaining orbital would possess 40% s character and 60% p character.
- this is a situation in which we have more s character and less p character than
the perfectly sp3 hybridized four orbitals.
- the affinity between nucleus and electron is what drives this unequal hybridization.
- simply put, electrons want to be in orbitals greatest s character.

Compare methyl cation to methyl radical. Which should be more planar which one might
exist as a shallow pyramidal species or dynamic species?
- which one should be perfectly sp2 and why?
- which one should be more p hybridized and why?
Answer! The methyl cation would be expected to be perfectly sp2 hybridized, because it reserves
one of the orbitals to be empty and equalizes the rest. If one is left empty and that one is
purely p, Why? then there are only three orbitals to play with, one s and two p orbitals.
Answering the other part of the question, the radical should be expected to include some the
lone, pure p orbital in its hybridization scheme because the unpaired electron wants to get
close to the nucleus, and to do this it needs some s character.
- However this orbital does not have two electrons like the orbitals involved in the bonds
have.
- Thus it should tend to grab up precious s character less than the filled, bonding orbitals.