Bonds (σ or π) are made by overlap (electron sharing) of atomic (s, p)
or hybridized (spx) orbitals. In the modified VB theory, formation of a bonding orbital is accompanied by the formation of the
corresponding localized antibonding orbital (σ* or π*) which remain
unoccupied and does not contribute to the structure of the molecule.
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.
2. Paired electrons are localized in specific
internuclear spaces between bonded atoms or remain unshared (lone pairs).
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..
3. The electronic structure and
geometry is arrived at (qualitatively) by finding the best compromise between the maximum
overlap (electron-nucleus attraction) and repulsion (electron-electron and
The atoms are like politicians: they
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.
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
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 107o, but the lone pair orbital has more s character (it
is sp2.1 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 are formed between a pair of atoms within the
molecule and are characterized by the increased electron density on the
internuclear axis. In the modified VB, the corresponding σ* also
form, but are unoccupied by any electrons.
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
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.
5. π bonds may cover more than two nuclei (see resonance) and are characterized by the increased
electron density above and below the internuclear axis (but not on the
axis itself). In the modified VB, the corresponding π* orbitals also form, but
usually remain unoccupied.
π bonds are not cylindrically
symmetric. A rotation around the axis of the bond would lead to diminished overlap
reaching zero at 90o 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.
The electrons are not always shared equally. More electronegative
atoms attract electrons more, and the electron density increases around the
more electronegative atom at the expense of the less electronegative atom.
In the extreme, ionic bonds form.
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
7. Skeletal structures used in chemistry (Lewis structures) correspond
to the valence bond model.
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.
are used to represent molecules that are not adequately described by a single
structure, because the electrons are shared by more than two nuclei (resonance =
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 (
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.