VALENCE BOND THEORY

It is easier than it sounds. The typical atom encountered in organic chemistry has four valence orbitals (hydrogen has just one).  They are straight atomic orbitals, or some kind of hybrids, but there are four of them.  Some of them may not be available for bonding (they will hold lone pairs); the rest will overlap with orbitals of the neighboring atom(s)–one orbital with another, pairwise. They may do it head on (σ bonds), or sideways (π bonds), but from each pair a new bond will form that will hold paired electrons of low energy.  These bonding combinations are shown below. Here, you can see several ways in which σ bonds can be made, and one way to form a π bond from p orbitals. The antibonding combinations (not shown here) of the modified VB theory will become important only when reactivity of molecules is analyzed.

Typical organic molecules have all their electrons paired and all their valence orbitals utilized for bonding or holding lone pairs. There are, however, numerous examples of usually unstable and reactive molecules that may have unpaired electrons (radicals) or empty orbitals..

The atoms are like politicians: they COMPROMISE all the time. Well, they really settle into an energy minimum. OK, here is the game they may play: they want to be noble. So they share. Electrons, that is. To share well they want to overlap their orbitals as much as possible. They want to get close, but not too close. The big problem– if they get too close–is the nuclear repulsion. These positively multi-charged balls just cannot stand each other.  But, electrons too want get as far from each other as possible: so the bonding or non-bonding (lone) electron pairs will try to occupy spaces as far from each other as possible.  Look at  methane: the bonds point toward corners of a tetrahedron (109.5^.o apart). That is why the carbon formed four equivalent sp³ hybrids in the first place!

And that is not all the "compromising"!  Look for example at ammonia. Now not all electron pairs are equal: one is not shared (the lone pair). These electrons would like to have a little more s fraction (we call it a character!) in their hybrid, because s orbital is lower in energy.  But if they "steal" some of the s from other hybrids, these other hybrids now will have more p character, and the angle between them will get smaller. Yeah!  You guessed it: a compromise is reached. The angle (H-N-H) is only 107^o, but the lone pair orbital has more s character (it is sp^2.1 instead of sp³; check it under hybridization). You may think about it as if the lone pair needed more space and pushed bonding pairs away.  Such instantaneous compromises are made, bond after bond, keeping the molecule at the bottom of an energy well.

σ bonds have cylindrical symmetry, i.e. the rotation around such a bond does not change the overlap of the contributing atomic orbitals. They are of lower energy than π bonds.  Here, you can see a cross-section (contour and "dot" representations) of a σ bond made of sp hybrids. The red area in the middle is where there is increased electron density between the nuclei.

π bonds are not cylindrically symmetric. A rotation around the axis of the bond would lead to diminished overlap reaching zero at 90^o and breaking that bond.  Because of the sideways approach of the orbital lobes, the overlap of the contributing orbitals is weaker in π bonds than in σ bonds.  Here, you can see a cross-section (contour and "dot" representations) of a π bond made from p orbitals. There is no electron density on the line connecting the nuclei (black dots), but there is an  increase in electron density above and below this line.

Here is a simple demonstration. Look at the electron density in σ bonds of HF and LiH. In both cases the atoms involved differ in electronegativity (H: 2.1; Li: 1.0, F: 4.0). In HF the fluorine (green nucleus) pulls most of the bonding electrons toward itself. In LiH the hydrogen (white nucleus) is "stealing" the electron density leaving lithium on the "edge" of the electron cloud. The bond in this case is almost ionic (Li⁺H⁻).  (By now you know, that the nuclei are shown much too big to be on the correct scale with orbital sizes.)

We replace each bonding pair with a line. The lone electrons can be shown as two dots or a line. So instead of drawing all these orbitals and sketching electron density we just draw lines. It is quick and informative. Do not be mislead, however, in reality there is nothing (no stick, no rope, no spring, etc.) connecting the nuclei.  It's sharing of electrons that keeps them together.

Troubles with simple Lewis notation start to appear when we have electrons that are delocalized over more than two atoms. For example when an electron pair is shared between three carbons as in the allyl cation. In such cases we draw two (or more) Lewis structure to account properly for the electron distribution. We call these drawings resonance structures, and the phenomenon of delocalization is known as resonance or conjugation. Now, there is nothing wrong with molecules that require two or more structures to be correctly represented in simple Lewis notation. It is a problem of the notation itself!  These resonance structures are "hypothetical models" that do not represent two (or more) structures of one molecule. There is only one molecular structure that we (for lack of a better method) describe as a superposition of these hypothetical "models". The resonance structures will be especially important in molecules with π systems. Here is an example of the carbonate ion (CO₃⁻²):

Each Lewis structure shows that one oxygen is different that the other two (check the electron count and charges). In reality, all oxygen atoms are equivalent, all bond lengths are the same (corresponding to 1 and ⅓ bond), and the negative charge is distributed equally (–⅔ per oxygen). Thus none of the individual formulas is correct, but a superposition of the three (with equal weights in this case) gives a good description of the carbonate ion.  Note that the double-headed arrow (↔) does not represent the equilibrium between the forms; it is the resonance sign. [The equilibrium sign has two arrows ( or )]. If we were to probe the electron density distribution in this ion by measuring the electrostatic interactions between a positive (+1) charge used as a probe and the electron density on the "surface" of  the molecule, the observed picture (in color!) would be indeed very symmetrical (note the positive charge accumulated in the center of the molecule; here blue color implies repulsion of the positive probe).

So, maybe a better Lewis structure would be one shown on the right? The positive carbon in the center does not have an octet, however. We stress again, it is a notation problem - and not a problem with the electronic structure.