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10.a. Molecular bonds distribution
10.b. Excited states of molecules
10.c. Emission spectra of molecules
10.d. The covalent bond
10.e. Sigma bond, pi bond, and hybridization
There are several different mechanisms of chemical bonding. Most of matter on earth is not in elementary form (atoms) but molecules. The types of bonds we will investigate are the ionic, covalent, Van der Waals and hydrogen bonding. While the ionic and Van der Waals bonding can be explained in terms of simple elecrostatic terms, the hydrogen and mostly the covalent bonding, which is the most prevalent in molecules, owes its existence solely based on quantum effects.
Excitation energies of molecules are much lower than those of atoms. This is so because molecules have more degrees of freedom that can be excited than atoms for which there are electronic excitations only. For the sake of simplicity we will only consider two-atomic molecules. Let us view the molecules as a dumbell, first with a rigid separation, R, then we will allow the distance of the two atoms to change.
If the distance of the two atoms, R, is fixed, then the possible motions of the system, beyond the trivial translational motion are rotations. We know that angular momentum operator can have a sharp value in addition to energy. We also know that the angular momentum is quantized. Then the rotational motion will give discrete rotationl energy levels. The rotational energy of the two atomic molecule is
Erot = L2 / 2I,
where I is the moment of inertia around the center of mass. It is easy to calculate I. Suppose that the distance of atom 1 of mass m1 from the center of mass is r1. Then the distance of the other atom of mass m2 is r2 = R- r1. Then r1 can be found from the center of mass condition m1r1 = m2(R- r1). Solving for r1 we obtain
r1 = m2R /( m1 + m2 ), r2 = m1R /( m1 + m2 ).
The moment of inertia is obtained as
I = S miri2 = [R2 /( m1 + m2 )2] ( m1 m22 + m2 m12) = m R2 .
where m is the reduced mass, m = m1 m2 /( m1 + m2 ).
Then the rotational enrgy levels are
Erot = h2 l(l+1) / (2 m R2 ),
because the rotational angular momentum is quantized the same way
as any other angular momentum, L2 =
h2 l(l+1), where l is called the
rotational quantum number. Let us compare now these excitation
energies to those of resulting from electronic excitations. In the
ground state of the hydrogen atom the contribution of kinetic energy
is just 1/2 of the contribution of the potential energy and equals in
magnitude to the binding energy. Its value is |E| = p2/2m,
where m is the electron mass. Using the uncertainty relation, we
obtain
E = h2/ 2mr2.
Notice that then the electronic energy is in the same form as the rotational energy, except for the mass and the distance. The distance r should be in the same order of magnitude as R, because the binding of atoms and molecules is caused by same types of quantum mechanical effects. The mass however is the electron mass that is about 1000-10000 times smaller than the reduced mass of of two-atomic systems. Consequently, it is expected that the electronic energies are 1000-10000 times larger than rotational energies. Therefore instead of being in the visible range, rotational spectra of molecules lie in the microwave range.
If we allow the distance of the atoms to change then, in addition to the rotational excitations, we will have vibrational excitations as well. Let us study now vibrational energy levels. As we will see when we study the covalent bond in more details, but for other types of bonds as well, the interaction of two atoms leads to an effective potential between them that has the shape of the function like the one for the hydrogen atom at nonzero angular momantum. This potential has a minimum at some distance R0. If one assumes that the oscillations are small then one can expand the potential around R= R0. One obtains V = V( R0) + 1/2 K (R- R0)2 + ...
where K = V" ( R0). Until now R was regarded as a classical coordinate. We know , however, that we live in a quantum world and every system must be quantized. The vibrations of the two atom system in their CMS are desribed by the Schrodinger equation
[ p2/ 2 m1 + p2/ 2 m2 + 1/2 K (R- R0)2 ] y(R) = Evibr y(R).
The momenta are equal because of the CMS system. Using 1/ m1 + 1 /m2 = 1/ m, and introducing the oscillation distance x = R- R0, we obtain the harmonic oscillator equation
- h2/2 m
y"(x) + 1/2 m w2 x2 y(x) = Evibr y(x).
Notice that the moment are due to the oscillatory motion of x, so
indeed p=( h/i) d/dx. Then we know the spectrum of
the oscillation energy is
Evibr = (n+1/2) h w = (n+1/2) h [K / m]1/2.
Now K = V" ( R0). it is reasonable to assume that K
µ V( R0) /
R02 µ
h2 / 2m R04
(here we use that R0 is essentially the scale of the
change of the potential). This is so because the the minimum value of
the potential is an electronic energy, V(
R0) µ
h2 / 2m R02,
and it also changes on the scale of R0 itself, so V" (
R0) µ V( R0) /
R02. Then substituting into the vibrational
energy formula we obtain
Evibr µ (n+1/2)
h2 / [ (2m m
)1/2R02].
This is again in the same form as the electronic energy or the rotational energy except the mass parameter is not m or m, but (m m )1/2, which is somewhere in between m and m (geometric mean). Consequently the vibrational spectra are uniformly spaced and the spacing is about a 100 times denser than that of the electronic levels, but at the same time rotational levels are a hundred times denser then vibratonal levels. The transition between vibrational levels will then be in the infrared range of the spectrum.
Return to toper of
In the previous section we derived the form for the vibrational and rotational spectra of molecules. These levels have the form
E(l,n) = E0 + (h2/ 2 I ) l
(l+1) + (n+1/2) hw.
These levels lead to an emission spectrum. If we excite a molecule
by heat or other forms of radiation into excited states then it will
spontaneosly emit photons while going back to lower level excited
states or the ground state. The following selection rules apply: As
for other excitations, due to the nature of the photon (spin one
particle) The angular momentum can only change by one unit, |Dl |= 1. Due to the nature of harmonic forces
since the operator causing the transition is a dipole operator, x, it
can only cause transitions between neighboring vibrational
excitations, so |Dn|=1 as well. This
implies the peculiar fact that normally pure rotational or pure
vibrational transitions are forbidden and the low lying rotational
spectrum alone cannot be observed. Since the vibrational energy scale
is much larger than the rotational, a an optical absorbtin must
always be accompanied by the increase of the quantum number n
by one unit, corresponding to the energy hw. Then the rotational quantum number can either
increase or decrease by one unit giving the following two energies
for spectral absorbtion lines:
DE = hw + (h2/ 2 I )(2l
+2), when l also increases (l->l+1), original l=0,1,2,...
DE = hw - (h2/ 2 I )2l,
when l decreases (l->l-1), original l=1,2,...
The covalent bond will be represented on the simplest example or the H2 molecule. The aim of the calculation is to calculate the energy of the system as a function of R, the sepatation between the two nuclei. Then the minimum of that energy as function of R will serve as the binding energy and the distance R0 where the minimum occurs is the equilibrium distance. The exact wavefunction cannot be determined by analytic methods so one is forced to use approximate wavefunctions to model the system. This is of course not a novelty, we were forced to do the same for the He atom or heavier atoms. In fact there is a physical boundary condition on the wavefunction and on the energy of the system: at large distances of the two hydrogen atoms it should go over into the product of wavefunctions of two H atoms, while at short distances it will become a He atom. To simplify the problem we will first consider the ionized H2 molecule, H2+. The advantage of considering such a molecule is that one needs to deal with a single electron wave function, or as it is usually called molecular orbital, only. For the case of the H2+ ion one has the conditions that for short distances the ground state wavefunction and the ground state energy should go over into the well known He+ wavefunction (H-like) and energy (-4Ry). The potential energy is
V(r) = -ke2/|r| - ke2/|R-r|.
At large distances on has a good wavefunction, that is a superposition of the two wavefunctions corresponding to the electron around one or the other proton:
y+(r) µ y0(r) + y0(R-r),
where y0(r) is the ground state wavefunction of the H atom centered around the origin. This wavefunction is called a bonding orbital. It clearly satisfies the constraint that at large distance the electron is around either one or the other proton, with equal probability. Unfortunately at short distances it goes over to the wavefunction of the H atom rather then the He+ ion.
There is an alternative wavefunction, that could also serve as an approximation to the wavefunction of the H2+ ion. It is antisymmetrized between the two H wavefunctions
y-(r) µ y0(r) - y0(R-r),
and it is called an antibonding orbital. The reason for this nomenclature becomes clear if we calculate the expectation values of energies associated with these wavefunctions:
E = Ú
y* H y dV = Ú y* [
-h2/2m D -ke2/|r| -
ke2/|R-r| ] y
dV + k e2/|R|.
These integrals can easily be calculated and get
E+ = -1 -(2/1+ D)[ 1/R - e-2R(1+R)/R + e-R (1+R) ] and
E- = -1 -(2/1- D)[ 1/R - e-2R(1+R)/R - e-R (1+R) ]
in units of Ry. Here D is the overlap integral (overlap of y0(r) and y0(R-r) ),
D = (1+R + R2) e-R.
The energy E+ goes to -1Ry when R goes to infinity, as it should. For short distances, however it goes to -3Ry instead of -4Ry, the energy of the He ion. Adding the repulsion of the protons, ke2/R, we obtain a function that has a broad minimum at R0=2.5 a0 between the two extreme limits. the minimum value is E+ ( R0) = -2.13Ry. 2/ R0 Ry, we obtain the the net molecular energy of -15.37 eV. At infinite R the energy is just -1Ry, so the electron creates a -.13 Ry binding energy. The true binding energy is more like -.2 Ry . The antibonding, E- , energy as a function of distance is also goes to the limit of -1Ry. as R goees to infinity. After adding the repulsion energy of the protons, at shorter distances, instead of heaving a minimum, the energy is keep rising. Consequently it cannot create a bound state. That is why the two orbitals are called bonding and antibonding.
The simplest model for the H2 molecule would be obtained if we neglect the Coulomb repulsion between the two electrons is two put both into a bonding orbital with opposite spins, i.e. to use the wavefunction
y(r1, r2) µ [y0(r1) + y0(R-r1)] [y0(r2) + y0(R-r2)]
This wavefunction has now two problems. In addition to not tending to the wavefunction of the He atom when R goes to zero (this is a problem we already had with the orbitals themselves), but in the large R limit it does not tend to the product of wavefunction of two H atoms as it should. When we multiply the wavefunction out, there are extra terms, y0(r1) y0(r2) and y0(R-r1) y0(R-r2) that correspon to the situation when both of the electrons stay around one or the other proton. This is of course very unphysical. This should never happen. Instead of this wavefunction one can use and get much better energies as well the valence bond or Heitler-London wavefunctions. These are obtained if we omit the above mention bad terms
y(r1, r2) µ y0(r1) y0(R-r2) + y0(r2) y0(R-r1)
This is called a valence bond wavefunction because the two electrons create a single valence bond between the two H atoms. The valence bond method uses similar wavefunctions for pairs of electrons for every covalent bond. The spins of the two elctrons must be antiparallel. they are in the total spin s=0 state. Again, using the negative sign one can obtain alternative orbitals. They do not bond the two H atoms. They would require symmetric spin states and an s=1 state.
The covalent bond we discussed above was the so-called sigma bond. A sigma bond is axially symmetric, i.e. the system is invariant for the rotation around the connecting axis between the two bonded atoms. Valence electrons are either s-electrons or p electrons. The H molecule of course has s valance electrons that always create sigma bonds. If the valence electrons are in the p state (orbital angular momentum 1), then the bond can either be axially symmetric or axially non-symmetric, in which case it is called a pi bond. If there is only one valence electron in the p state then it always creates a sigma bond because a sigma bond is stronger (has a larger overlap integral) than the pi bond. There cannot be two sigma bonds in the p state, becuse recall that each valance bond involves two electrons in identical states and a third in the same state not allowed because of the Pauli exclusion principle. If we choose the quantization axis (z-axis) as the connecting axis between the two atoms than subsequent p electrons must be in the px or py state not the pz state. These states are "perpendicular" to the connecting axis of the two atoms and as such have less overlap with the elctron wave functions of the other atom, consequently generate a weaker coupling. They create a bond that is called pi bond. A pi bond is not symmetric for a rotation around the connecting axis.
An example for the two kind of bonds is the N2 molecule. N has three p valence electrons. The two N atoms together have 6 valence electrons. They form 3 filled valence bonds. One of these is a sigma bond and two perpendicular bonds are pi bonds.
Heteronuclear bonds, such as HF or H2O are bonds between p and s electrons. Theya are also sigma bonds. The lost reflection symmetry between the two atoms implies that the electron distribution is not going to be symmetric either. If the electron distribution is not symmetric then the the probability that the electron is around the F atom is not the same (it happens to be higher) and that creates an electric dipole momentum for the molecule. The water molecule also has a nonvanishing electric dipole moment.
Another important phenomenon, mostly in organic compounds is hybridization. We have discussed hybridization earlier. Carbon has two 2p valence electrons. In the shell n=2 the energy difference of 2s and 2p electrons is not very large (n=2 electrons do not have a lot of screening). Thus, most of the time the s electrons also participate in creating molecular bonds making Carbon atoms of valence 4. With valance 4 the molecular binding of C atoms gets stronger, the molecular states are stronger bound. The energetically most favorable orientation of the four valence electrons of C is in fact obtained by promoting one of the s electrons into a p electrons to a configuration of 1s22s2p3. Then, appropriate combination of the four n=2 two wavefunction form four valence bonds that are directed towards the vertices of a regular tetrahedron the center of which is the C nucleus. This geometry insures the least overlap of different bonds, thus minimizing the Coulomb repulsion energy.