Molecular Structure and Spectra

How are molecules and solids held together?

BondsIonic: Coulomb attraction of a positive and a negative ionCovalent: sharing of electrons between two atomsDipole-Dipole: electric forces between - two polar atoms/molecules

- one polar and one non-polar atoms/molecules- two non-polar atoms/molecules

Metallic: electrons shared by all the atoms in a solid (not discussed until chap. 10)

The type of bonding also plays a big role in determining the structure of solids.

To be stable, a molecule or solid must have lower energy than the constituents.

Ionic BondingAn electron is transferred from one atom to the other to create two ions,

one positive and one negative, which requires energy. The Coulomb force provides the attraction and lowers the energy of the system.

The repulsion of the electron clouds increases the energy if the atoms get too close.So there is a minimum in the energy at a particular separation.

Ionic Bond Energy

One important measure of the stability of the bond is U(r), the total potential energy of thesystem compared to separate atoms, which is a function of the separation of the ions.

This energy comes from three contributions:Eion is the energy to transfer an electron from one atom to the other, which is positive.

It is the difference between the ionization energy of one atom (the energy required to remove the highest energy electron from the atom)

and the electron affinity of the other atom(the binding energy of the electron added to the other atom).

ECoulomb is the energy due to the attractive electric force, which is negative.Eexclusion is the energy due to the repulsion of the electron clouds, which is positive.

This repulsion is not due to the Coulomb force between the electrons, but is instead due to the Pauli exclusion principle. We can’t have overlap between two electrons in the same state, that is, with the same quantum numbers. This results in an energy cost which increases steeply as the overlap of the electron orbitals increases.

U r E r E r Eker

Ar

ECoulomb ex ion n ion( ) ( ) ( )= + + =−

+ +2

Since alkali metals (Group I) have the lowest ionization energies and halogens (GroupVII) have the highest electron affinities, the strongest ionic bonds are formed fromone alkali and one halogen, NaCl for example.

U(r), the total potential energy of the system

Covalent BondingElectrons are shared by two atoms.

The energy decrease is not from Coulomb interaction but a purely quantum-mechanical effect.

As an example, let’s look at twosquare wells.

If one electron is sharedbetween the wells, it will beequally likely to be found ineither one.

Either a symmetric or anti-symmetric wave function willgive the same probabilities ifthe wells are separated.

However, if the wells are close together thesymmetric and anti-symmetric wave function

will give different probabilities and theenergy of the states are now different.

Notice that the symmetric wave function isgetting to look like the ground state for onewell and the anti-symmetric state is getting

to look like the first excited state.

The symmetric state is lower in energy thanthe anti-symmetric state if the wells are

close together.

The symmetric state has a high probability of finding the electron between the wellswhile the anti-symmetric state has a low probability.

The idea of two degenerate states being split into two different energy states by aninteraction will be a general theme.

We can add a second electron to the ground state of the two well system as long asthey have opposite spins (so S=0) so the total wave function is anti-symmetric. Two

electrons in the anti-symmetric state would need the same spin (S=1).

If we replace the square wells with protons, our Hamiltonian will be

.Hpm

ker r rop

op= + − − +

22

1 2 021 1 1

Symmetric and anti-symmetric combinations of the ground state solutions to Schrodinger’s equation for two separations of the protons:

At large separation of the protons, theenergy of both the symmetric and anti-symmetric solutions is the same as theenergy of the ground state of one hydrogenatom, -13.6 eV.

As the protons separation is decreased, thesymmetric and anti-symmetric solutionssplit in energy. At small separations, theanti-symmetric solution looks like the firstexcited state of helium (Z=2, n=2, E=-13.6eV) while the symmetric solution looks likethe ground state (Z=2, n=1, E=-54 eV).

To both of these we need to add the energydue to the repulsion of the protons, Up.

The result gives a minimum in energy forthe symmetric state, but no minimum for theanti-symmetric state.

Symmetric solution - bonding stateAnti-symmetric solution - non-bonding state

We can add a second electron to thesymmetric state as long as it has theopposite spin so that the total wavefunction is anti-symmetric, so S=0 in theground state, a singlet state.

To add a second electron to the anti-symmetric state, the spins must bealigned, so S=1, a triplet state.

Hydrogen is s-bondedsince both electrons are ins states.

Flourine is p-bonded since both electrons are in p states.

Hybridization - mixing of atomic orbitals

ψ

ψ

ψ

ψ

112

212

312

412

= + + +

= + − −

= − − +

= − + −

( )

( )

( )

( )

s p p p

s p p p

s p p p

s p p p

x y z

x y z

x y z

x y z

Dipole-Dipole Bonding (two polar molecules)The electric field from a dipole is

�

� � � �

E kp

rp r rrd = −•

3 5

3( )

If the magnitude of the dipole moment isexpressed as p=qa and for r>>a, then

Ekqar

kprd = =3 3

The potential energy of a second dipole in the electric field of the first is

U p Ed= − •�

�

The potential energy is proportional to 1/r3 and the force is proportional to 1/r4.The force between polar molecules is generally a few tenths of an electron volt.If hydrogen is involved its called hydrogen bonding. This force is responsible forpolar molecules condensing to liquids and solids at relatively high temperatures, e.g.water. Directionality of bonding results in ordered crystals.

Induced Dipole (one polar, one non-polar molecule)

The degree to which a non-polar molecule will become polarized in an electric fielddepends on its polarizability, , such that . α

�

�

p E= α

So if a non-polar molecule is near a polar molecule, it will have an induced dipolemoment and there will be a potential energy,

U p E E E E k p rd d d d= − • = − • = − = −�

� � �

22 2

12 6α α α /

and a force which is proportional to 1/r7, so it is weak and very short-range.

Fluctuating Dipole (two non-polar molecules)

Molecules which have no permanent dipole moment still have an instantaneous dipolemoment that fluctuates quickly. If the fluctuations of two neighboring non-polarmolecules (e.g. He) becomes correlated, there is a net attractive force and anegative energy. Just as above, the potential energy is proportional to 1/r6 and theforce is proportional to 1/r7. This force is called the van der Waals force or theLondon dispersion force.

Rotational Energy LevelsWe will consider only diatomic molecules for simplicity,

but the ideas extend to polyatomic molecules.

ELI

l lI

l l E r= =+

= +2 2

021

21

( )( )

�

E=12E0r

E=6E0r

E=2E0r

E=0

I m r m r rm m

m m

= + =

=+

1 12

2 22

02

1 2

1 2

µ

µ

Vibrational Energy LevelsSince the curve of potential energy is close to a parabola near the minimum,

a diatomic molecule has quantized energy levels similar to a harmonic oscillator.

E v hf v v Kv = + = + = +( ) ( ) ( ) /12

12

12� �ω µ

More accurately, we have an anharmonic oscillator and the spacings of the levels are not quite equal.

Since the separation of the atoms is larger in the excited state than the ground state, the moment of inertia is different and the rotational energies are different.

Since the shape of the potential energy curve is different, the force constant is different and the vibrational energies are different.

Emission SpectraWhen a molecule makes a transition from one vibrational state to another, the change in thevibrational quantum number must be +1 or -1, so the energy of the photon is hf. When themolecule makes a transition from an excited electronic state to a lower energy state, it mayalso change its vibrational state, but in this case there is no selection rule for changes in v.

If a molecule does not have a permanent dipole moment(e.g. any homonuclear diatomic), it can’t change

rotational states by emitting a photon . If it does have apermanent dipole moment, it can make a purely

rotational transition and emit a photon as long as thechange in the rotational quantum number is +1 or -1.

All molecules can change rotational states while makingtransitions between electronic energy levels as long as the

change in the rotational quantum number is +1 or -1.

Absorption SpectraAt room temperature, most molecules are in the electronic and vibrational ground states butin a rotationally excited state. These molecules can absorb an infrared photon and transitionto the first vibrationally excited state. However, they must also change rotational states such

that or .∆ l = +1 ∆ l = −1

E hf l l Ev l r= = + +012 01, ( )

E hf l l Ev l r= + = + + +1 132 01 2, ( )( )

E hf l lEv l r= − = + −1 112 01, ( )

∆ E hf l El l r→ + = + +1 02 1( )∆ E hf lEl l r→ − = −1 02

[ ]n E g E e

n E l e

n E l e

lIkT

l lE kT

lhf l l E kT

ll l IkT

l

r

( ) ( )

( ) ( )

( ) ( )

/

/ ( ) /

( ) /

max

=

= +

∝ +

= −

−

− + +

− +

2 1

2 1

12

41

2 1

1 2

2

0

2�

�

Interaction of radiation with molecules

FluorescenceAbsorption of a photon (orsome other process) putsmolecule into electronicallyexcited state. Photon(s) aregiven off as moleculetransitions to lower electronicstate.

Resonance AbsorptionExcited state is first excited state, so emitted photon issame energy as absorbed photon.

Rayleigh ScatteringScattering of a photon from a molecule inwhich there is no change in frequency. Theintensity of the scattered radiation isproportional to f 4 (its why the sky is blue).

Compton Scattering Energy of photon is large enough to ionize themolecule, so electron ends up as free particle.Conservation of energy and momentum gives theCompton shift in frequency.

Raman Scattering The scattered photon leaves themolecule in a higher (lower)vibrational or rotational state andtherefore the scattered photon has alower (higher) frequency. The energyof the incident photon does not need tomatch a characteristic energy of themolecule.

Photoelectric effect Absorption of a photon and ionization of themolecule. For a molecule this can’t conserveenergy and momentum without another particlebeing involved.

Stimulated Emission - The presence ofa photon with an energy matching thetransition energy, increases theprobability of emission. The emittedphoton goes in the same direction as theexisting photon and with the samephase.

Calling the lower state 1 and the upper state 2, the energy difference between the two states is. The probability per unit time of absorption if the system is in state 1 isE E hf2 1 12− =

, where B12 is called the Einstein coefficient of absorption and u(f) is the energy densityB u f12 ( )at a frequency f.

The probability per unit time of spontaneous emission if the system is in state 2 is . TheA21

lifetime of the state is .t As = 1 21/

The probability per unit time of stimulated emission at a frequency f is , where B21 isB u f21 ( )called the Einstein coefficient of stimulated emission.

If the two states have the same degeneracy and are in equilibrium, then

.N N e eE E kT hf kT2 1

2 1/ ( )/ /= =− − −

In equilibrium the number of absorptions per unit time must be equal to the number of emissions,both spontaneous and stimulated.

N B u f N A B u f1 12 2 21 21( ) ( ( ))= +

Rearranging to solve for u(f) gives

.u f

AB

NN

BB

AB

BB

ehf kT( )

/=

−=

−

21

21

1

2

12

21

21

21

12

211 1

Recall Planck’s law that says .u fhf c

ehf kT( )/

/=−

81

3 3π

This implies that and .B B21 12=AB

hfc

21

21

3

3

8=

π

Notice that the ratio of spontaneous emission to stimulated emission is proportional to f 3.When hf>>kT, spontaneous emission dominates over stimulated emission. When hf<<kT, stimulated emission is the dominant de-excitation mechanism.

LasersLight Amplification by Stimulated Emission of Radiation

To get functioning laser in a medium (gas or crystal), you need more photons emitted thanabsorbed. So you need more stimulated emission than absorption. But the relative probabilitycoefficients of the two are equal, so you need to get a much higher population in the excited state,

, so that . N N2 1> N B u f N B u f2 21 1 12( ) ( )>

Since in thermal equilibrium the lower state is more populated than the upper state, this is called apopulation inversion.