Transcript Document

Rotational Spectroscopy
To a stationary
observer, a rotating
polar molecule
looks like an
oscillating dipole
which can stir the
electromagnetic
field into oscillation.
This picture is the
classical origin of
the gross selection
rule for rotational
transitions.
When a photon is
absorbed by a
molecule, the
angular momentum
of the combined
system is
conserved. If the
molecule is rotating
in the same sense
as the spin of the
incoming photon,
which has angular
momentum of 1,
then J increases by
1.
From the diagram of the allowed
rotational energy levels of a rigid
diatomic molecule:
For J = 0, we have EJ = 0 and
therefore the molecule is not
rotating at all.
For J = 1, the rotational energy is EJ
= 2B from J(J + 1) and a rotating
molecule then has its lowest angular
momentum.
May continue to calculate EJ with
increasing J values and in principle
there is no limit to the rotational
energy the molecule may have.
• In practice, reach a point where the centrifugal force of a
rapidly rotating diatomic molecule is greater than the
strength of the bond, and the molecule is disrupted, but this
point is not reached at normal temperatures.
Consider energy differences in order to discuss the spectrum.
Note that spectroscopists speak of term values, F(J), having
units of either frequency (by dividing by h):
EJ
h
F (J ) 
 BJ ( J  1); B  2
h
8 I
Or, units of wavenumbers, by dividing by hc:
To avoid confusion, I will continue to use
tildes to indicate wavenumbers.
FJ 1  FJ 0  2 B  0  2 B cm -1
Therefore absorption line will appear at 2B cm-1
Similarly
~
~
~ -1
J 1 J  2  6 B  2 B  4 B cm
~
~
  J J 1  2 B ( J  1) cm -1
~
Therefore step-wise raising of the rotational energy results in
an absorption spectrum consisting of lines at 2B, 4B, 6B …
cm-1 and separation 2B.
The Non-Rigid Rotor


The assumption of a rigid bond is only an approximation.
The separation between successive lines (and hence the
apparent B value) decreases steadily with increasing J

The reason for this decrease may be seen if we calculate
internuclear distances from the B values.

The bond length increases with J – this reflects the fact that
the more quickly a diatomic molecule rotates, the greater
the centrifugal force tending to move the atoms apart.
Two consequences of elasticity

1) When the bond is elastic, a molecule may have vibrational
energy. i.e., a bond will stretch and compress periodically with
a certain fundamental frequency dependent upon the masses of
the atoms and the elasticity (or force constant) of the bond.

If the motion is simple harmonic (which is a good
approximation) the force constant is given by:
k  4 2 2c 2 
Where  is the vibrational frequency (cm1)
- The variation of B with J is determined by the force constant –
the weaker the bond, the more readily will it distort under
centrifugal forces.
2) The quantities r and B vary during a vibration – when
these quantities are measured by microwave techniques many
hundreds of vibrations occur during a rotation, and hence the
measured value is an average.
However, from the equation for B we have
h
h
B 2  2
8 Ic 8 c r 2
Since all other quantities are independent of vibration.
The effect of rotation on a molecule. The centrifugal
force arising from rotation distorts the molecule,
opening out bond angles and stretching bonds slightly.
The effect is to increase the moment of inertia of the
molecule and hence to decrease its rotational constant.
The Spectrum of a Non-Rigid Rotator
The Schrodinger eqn may be set up for a non-rigid molecule
and the rotational energy levels are found to be for a simple
harmonic force field:
h2
h4
2
2


EJ  2 J J  1 
J
J

1
8 I
32 4 I 2 r 2 k
EJ ~
~ 2
2
or FJ 
 BJ J  1  DJ J  1 cm -1 where J  0, 1, 2.....
hc
~
D
h3
-1
cm
32 4 I 2 r 2 kc
which is a positive quantity
If the force field is anharmonic, the expression becomes
FJ  BJ J  1  DJ 2 J  1  HJ 3 J  1  KJ 4 J  1
2
3
4
Where H, K etc. are small constants dependent on the
geometry of the molecule, and negligible compared with D
Most spectroscopic data is adequately fitted by eqn for the
simple harmonic force field.
h
h
~
h3
~
B 2  2
D
2
4 2 2
8

Ic
8

c

r
32 I r kc
Solving for h in terms of B, and substituting into D gives:
~3 3 6 3 3 6 ~3
~3
2 2 3 4
B 16 c  r
B 16 2 c 2  3 r 4
~ B 8 c r
D


4 2 2
2 2
32 I r kc
I r k
 2r 4k
~3 2 2
~3
4B
~ 16 B  c
D
 2
where k  4 2 2 c 2 
k

cm-1
Note: vibrational frequencies are usually of the order of 103
cm-1
B is of the order 10 cm-1
D is of the order 10-3 cm-1 i.e. is very small compared with
B.
For small J, the correction term DJ2(J + 1)2 is almost
negligible
For J values of 10 or more, it may become appreciable.
Selection rules remain the same J =  1
~
~ ~ ~
FJ 1  FJ   J  B J  1J  2  J J  1
~
2
2
2
 D J  1 J  2  J 2 J  1
~
~
3
 2 B J  1  4 DJ  1 cm -1


Knowing D one can (rather inaccurately) calculate the
vibrational frequency of a diatomic molecule
~3
4B
2
  ~
D
The force constant is then calculated:
k  4 2 2c 2 
Raman Spectroscopy

Molecular energy levels are explored by examining
the frequencies present in radiation scattered by
molecules.

Most of the radiation is scattered without change of
frequency, giving the Rayleigh line.

About 1 in 107 of the incident photons give up some
energy in collision with the sample molecules,
emerging with lower energy, giving lower frequency
Stokes lines.

Other incident photons may collect energy from
excited sample molecules, emerging as higher
frequency anti-Stokes lines.

For Raman spectroscopy to work, lasers are required:
1) frequency shifts in the scattered radiation are small,
so highly monochromatic radiation is required in order
to observe the shifts.
2) The intensity of the scattered radiation is so low
that an intense incident beam is needed.


Rotational Raman Spectra
Gross selection rule for rotational Raman transitions: molecule
must be anisotropically polarizable
An electric field applied to a
molecule results in its distortion,
and the distorted molecule
acquires a contribution to its dipole
moment (even if it is nonpolar
initially). The polarizability may be
different when the field is applied
(a) parallel or (b) perpendicular to
the molecular axis (or, in general,
in different directions relative to the
molecule); if that is so, then the
molecule has an anisotropic
polarizability.
The distortion of a molecule in an electric field is determined
by its polarizability .
In addition to any permanent dipole moment a molecule may
have, the molecule acquires an induced dipole moment , if
the strength of the field is :
An atom is isotropically polarizable.
All linear molecules and diatomics (whether homonuclear or
heteronuclear) have anisotropic polarizabilities rotationally
Raman active
 allows study of many molecules that are inaccessible to
microwave spectroscopy.
but some types of rotors both rotationally Raman &
microwave inactive
Specific rotational Raman selection rules:
Linear rotors: J = 0,  2
The distortion
induced in a
molecule by an
applied electric
field returns to its
initial value after
a rotation of only
180 (that is,
twice a
revolution). This
is the origin of the
J = 2 selection
rule in rotational
Raman
spectroscopy.
Note: the J = 0 transitions do not
lead to a shift of the scattered
photon’s frequency in pure
rotational Raman Spectroscopy
contribute to unshifted Rayleigh
radiation in the forward direction
The rotational energy levels of a
linear rotor and the transitions
allowed by the J = 2 Raman
selection rules. The form of a
typical rotational Raman
spectrum is also shown.
Transition J = +2, scattered
radiation leaves it in a higher
rotational state, so the wavenumber
of the incident radiation is decreased
Stokes lines in the spectrum
where  i is incident radiation
Stokes lines appear to low
frequency of incident radiation and at
displacements 6B, 10B, 14B…from
the incident radiation (Rayleigh line)
for J = 0, 1, 2, ….
Transition J = 2, scattered
radiation leaves it in a lower
rotational state, so the wavenumber
of the incident radiation is increased
anti-Stokes lines in the spectrum
where  i is incident radiation
anti-Stokes lines appear to high
frequency of incident radiation and at
displacements 6B, 10B, 14B…from
the incident radiation (Rayleigh line)
for J = 2, 3, 4, …. as J = 2 is the
lowest state that can contribute under
the selection rule J = 2
Pure rotation spectra of polyatomics
The rotation of a 3-dimensional body may be quite
complex.
It is convenient to resolve it into rotational
components about 3 mutually perpendicular
directions through the centre of gravity – the
principal axis of rotation.
Therefore a body has three principal moments of
inertia, one about each axis, usually designated Ia, Ib,
and Ic
For linear
molecules, the
moment of inertia
around the
internuclear axis is
zero.
By
convention:
An asymmetric rotor has three different moments of
inertia; all three rotation axes coincide at the centre of
mass of the molecule.
The definition of moment of
inertia. In this molecule
there are three identical
atoms attached to the B
atom and three different but
mutually identical atoms
attached to the C atom. In
this example, the centre of
mass lies on the C3 axis, and
the perpendicular distances
are measured from the axis
passing through the B and C
atoms.
Where ri is the perpendicular distance of
the atom i from the axis of rotation of
interest.
Molecules may be classified into groups according to the
relative values of their three principal moments of
inertia:
1) Linear Molecules
2) Symmetric Tops
3) Spherical Tops
4) Asymmetric Tops
1) Linear Molecules (e.g., HCl, OCS (carbon oxy-sulfide))
3 directions of rotation:
(a) about the bond axis
(b) end-over-end rotation in the plane of the screen
(c) end-over-end rotation at right angles to the plane
(b) and (c) are the same (i.e., Ib = Ic)
(a) must be zero, as the distances of all the atoms to
the bond axis must be zero (i.e., Ia = 0)
2) Symmetric Tops (e.g., methyl fluoride)
(a) about the bond axis (e.g., C-F bond axis chosen as
the main rotational axis since the centre of gravity lies
along it is now not negligible because it involves the
rotation of 3 comparatively massive H atoms off this
axis - like a top
(b) end-over-end rotation in the plane of the screen
(c) end-over-end rotation at right angles to the plane
(b) and (c) are the still the same (i.e., Ib = Ic)
Symmetric Tops
2 subdivisions of this class:
• Prolate symmetric top (eg., methyl fluoride) Ib = Ic > Ia
revolution about longer axis
• Oblate symmetric top (eg., boron trifluoride) Ib = Ic < Ia
revolution of ellipse about shorter axis
3) Spherical Tops (eg., methane)
when a molecule has all 3 moments of inertia
identical
However, these molecules have no dipole moment
because of their symmetry, rotation alone can produce no
dipole change and hence no rotational spectrum is
observable.
4) Asymmetric Tops (eg., water, vinyl chloride)
these molecules to which the majority of
substances belong, have all 3 moments of inertia different
A schematic illustration of the classification of rigid rotors.
Classification based on point group

Linear rotors must, obviously, be Cv or Dh.

A symmetric rotor must have either a Cn axis with
n>2, or an S4 axis. Point groups include Sn (n ≥4),
Cn, Cnv, Cnh, Dn, Dnh (all with n≥3) and all Dnd
(note that D2d has an S4 axis).

Molecules in the higher symmetry groups, Td , Oh,
Ih, are spherical rotors.
Linear Molecules
Consider first molecules such as OCS or HC  CCl , where
all the atoms lie on a straight line  gives rise to
particularly simple spectra in the microwave region.
Since Ib = Ic , Ia = 0 as for diatomic molecules, the energy
levels are given by
EJ
2
2
FJ 
 BJ  J  1  DJ  J  1
hc
cm -1
And the spectrum will show the same 2B separation modified
by the distortion constant. i.e., discussion on diatomic
molecules applies equally to all linear molecules
However, consider following 3 points:
1) Since the moment of inertia for the end-over-end
rotation of a polyatomic linear molecule is
considerably greater than that of a diatomic
molecule, the B value will be much smaller, and
the spectral lines more closely spaced.
B values for diatomic molecules are ~ 10 cm1
while for triatomic molecules they can be 1 cm1
or less, and for larger molecules smaller still.





2) The molecule must as usual possess a dipole moment
if it is to exhibit a rotational spectrum: i.e., OCS will be
microwave active, while CO2 (OCO) will not. Note
also: isotopic substitution does not lead to a dipole
moment since the bond lengths and atomic charges are
unaltered by the substitution
16OC18O is microwave inactive
3) A non cyclic polyatomic molecule containing N
atoms has altogether N-1 individual bond lengths to be
determined.
 in OCS  rCO and rCS
But there is only 1 moment of inertia for the end-overend rotation of OCS, and only this one value can be
determined from the spectra.
Symmetric Top Molecules
Although the rotational energy levels of this type of
molecule are more complicated than those of the linear
molecules, because of their symmetry their pure rotational
spectra are still relatively simple.
e.g., CH3F where Ib = Ic  Ia
Ia  0
 2 directions of rotation in which the molecule can absorb
or emit energy: along main symmetry axis (C-F bond) and 
to it.
Need to have 2 quantum numbers to describe the degree
of rotation:
- one for Ia
- one for Ib or Ic
It is convenient mathematically to have a quantum number
to represent the total angular momentum of the molecule
 which is the sum of the separate angular momenta about
the two different axes
Use J – where J is the total angular momentum
(note that in linear molecules J also represents the angular
momentum
Use K – to represent the angular momentum about the top
axis,
i.e., C-F bond (spinning axis)
Allowed values for K and J: both must, by conditions of
quantum mechanics, be integral or zero.
The total angular momentum can be as large as we like –
i.e.,
(except that a real molecule will be disrupted at very high
rotational speeds)
Once we have chosen J, however, K is more limited.
e.g., J = 3
- the rotational energy can be divided in several ways
between motion about the main symmetry axis and motion
 to it.
- if all the rotation is about the top axis, then K = 3,
since J is total angular momentum (note: K can not be
greater than J)
-or could have K = 2, 1, 0 in which case the momentum 
to the axis would have 1, 2, 3 respectively
-additionally K can be negative i.e., imagine positive and
negative values of K to correspond with clockwise and
anticlockwise rotation about the symmetry axis
and  can have values of 1, 2, 3
K = J, J – 1, J – 2,…. 0 ….(J – 1), J
which is a total of 2J + 1 values altogether
The significance of
the quantum
number K. (a)
When |K| is close
to its maximum
value, J, most of
the molecular
rotation is around
the principal axis.
(b) When K = 0 the
molecule has no
angular momentum
about its principal
axis: it is
undergoing endover-end rotation.
Consider the rigid symmetric top: i.e., one in which the
bonds are not supposed to stretch under centrifugal forces)
Solving Schrodinger equation gives
EJ , K ~
~ ~ 2
~
FJ , K 
 BJ J  1  ( A  B ) K cm-1
hc
~
where B 
h
h
~
and A  2
2
8 I b c
8 I a c
Note: energy depends on K2 , so it is immaterial whether
the top spins clockwise or anticlockwise; the energy is the
same for a given angular momentum.
 for all K > 0, the rotational energy levels are doubly
degenerate.
- the selection rules for this molecule are:
Applying these to the above equation, the spectrum for
rigid symmetric top is given by:


FJ 1, K  FJ , K   J , K  B  J  1 J  2   A  B K 2


  BJ  J  1  A  B K 2 



Thus the spectrum is independent of K, and hence rotational
changes about the symmetry axis do not give rise to a
rotational spectrum.
- i.e., rotation about the symmetry axis does not change the
dipole moment  to the axis (which always remains zero),
and hence the rotation cannot interact with radiation.
the spectrum is the same as for a linear molecule and only
one moment of inertia, i.e., that for end-over-end rotation can
be measured.
Non-rigid rotor
Consider the case where centrifugal stretching is taken into
account.
The effect of rotation
on a molecule. The
centrifugal force
arising from rotation
distorts the molecule,
opening out bond
angles and stretching
bonds slightly. The
effect is to increase
the moment of inertia
of the molecule and
hence to decrease its
rotational constant.
The energy levels become:
EJ ,K ~
~ ~ 2 ~ 2
~
2
FJ , K 
 B J  J  1  ( A  B ) K  DJ J  J  1
hc
~
~ 4
2
 DJ , K J  J  1K  DK K cm -1
~
~
~
where DJ , DJ , K , and DK are small correction terms
for non - rigidity
The selection rules are unchanged and so the spectrum is:
~
~
~
~
3
~
FJ 1, K  FJ , K   J , K  2 B J  1  4 DJ J  1
~
 2 DJ , K J  1K 2 cm -1
 The spectrum will be basically that of a linear molecule
(including centrifugal stretching), with an additional term
dependent on K2
It is easy to see why this spectrum now depends on the
axial rotation (i.e., depends on K) although such rotation
produces no dipole change.
When K = 0  no axial rotation
When K > 0  i.e., K = 1, molecule is rotating about the
symmetry axis.  axial rotation widens the HCH angles
and stretches the C-H bonds.
The distorted molecule has a different moment of
inertia for end-over-end rotation than the undistorted
molecule.
Rewriting the above equation gives:


~
~
~ ~
~
2
~
FJ 1, K  FJ , K   J , K  2J  1 B  2DJ J  1  DJ , K K 2 cm-1
The centrifugal distortion constants DJ and DJ,K can be
considered as correction terms to the rotational constant B,
and hence as perturbing the momentum Ib.
Since each value of J is associated with 2J + 1 values of
K, each line characterized by a certain J value must have
2J + 1 components.
However, since K only appears as K2 in the equation,
there will be only J + 1 different frequencies
 all those with K > 0 being doubly degenerate.
Note: each spectrum yields only one value of B, but the
spectra of isotopic molecules can in principle give
sufficient information for the calculation of all bond
lengths and angles of symmetric top molecules, together
with estimates of the force constant of each bond.
Asymmetric Top Molecules
Since spherical tops show no microwave spectrum, the only
other class of molecules of interest here is the asymmetric
top which has all three moments of inertia different.
Their rotational energy levels and spectra are very
complex
 each molecule and spectrum must be treated as a
separate case and much tedious computation is necessary
before structural parameters can be determined.
The best method of attack so far has been to consider the
asymmetric top as falling somewhere between the oblate
and prolate symmetric top, interpolation between the two
sets of energy levels of the latter leads to approximation of
the energy level – and hence spectrum – of the asymmetric
molecule.
Methods such as this have been quite successful, and
much very precise structural data has been published.
Specific rotational Raman selection rules
Linear rotors: J = 0,  2
Symmetric rotors: J = 0,  1,  2; K = 0
Asymmetric rotors:
For the latter, K is not a good quantum number,
so additional selection rules become too
complex.
A good quantum number is one which is
conserved in the presence of an external
interaction.
Structure Determination from Rotational
Constants

First, we need to examine what
exactly we are measuring
when we determine the
rotational constant B, and
hence the moment of inertia I
and bond length r.

From the Uncertainty
Principle, it is impossible to
get to the bottom of the
potential energy well, only to
the zero-point vibrational
level.
A way around the problem

In general, we are measuring B0. However, it is
possible to obtain Be, relating to the unattainable
equilibrium position, at least for diatomic
molecules.

IF measurements can be made on excited vibrational
states, the rotational constants are both slightly
vibrationally dependent, so the term values should
be written as
FJ  B J  J  1  D J
2
 J  1
2
cm
-1
Needed equations

Spectral lines are now given by
FJ 1  FJ   J  2 B

 J  1  4 D  J  1
3
cm -1
The vibrational dependence of B is, to a good
approximation, given by:

B  Be     1
2

where  is a vibration-rotation interaction constant and  is
the vibrational quantum number.
Limitations

The vibrational dependence of the centrifugal
distortion constant is too small to be of concern.

In order to obtain Be and hence the equilibrium bond
length re, B must be obtained in at least two
vibrational states.

Rotational transitions in excited vibrational states
are in general very weak.
Does it really matter?

Unless we require a high degree of accuracy, it does
not matter much whether we are determining re or r0.
Molecule
14N
2
15N
2
r0 (Å)
1.100105 ±
0.000010
1.099985 ±
0.000010
re (Å)
1.097651 ±
0.000030
1.097614 ±
0.000030
Note that re is independent of isotopic
substitution but r0 is not.
Structure Determination from Rotational
Constants

Re refers to the bond length at
the minimum of the potential
energy curve. This does not
change with isotopic
substitution, whether in the
harmonic or anharmonic
oscillator.

On the other hand, vibrational
energy levels, and therefore r0,
are affected by isotopes, via the
mass dependence of the
vibration frequency.
Determination from
Raman spectrum
The separation of adjacent lines
in both the Stokes and antiStokes region is 4B, so from its
measurement I can be
determined and then used to
find the bond lengths exactly as
in the case of microwave
spectroscopy.
Finding most populated level of rotational
states

We can find the most populated level by
differentiating the population function wrt J,
setting the derivative to zero, finding the
maximum, and rounding to the nearest integer.

The population function, after substituting for E,
and using B in cm1, is:
 hcBJ ( J  1) 
(2 J  1) exp 

kBT


Example problem (populations)

Find the ratio of the population of one of the J = 2
states to that of the J = 0 state for CO, for which
the rotational constant is 5.7898 × 1010 s1.

Find the ratio of the population of the J = 2 level
to the J = 0 level at this temperature.

Find the level with the largest population at this
temperature.
Example problem (structures)


From the value of B0 of 1.923604 cm1, obtained
from the rotational Raman spectra of 14N15N,
calculate the bond length r0.
Note that this differs from the value given above
for 14N2, because the zero-point level is lower for
the heavier isotopic species. Therefore the midpoint of the zero-point level is at a smaller value
of the bond length. We needed at least 5 sig figs
in order to see this difference.
Spin Angular Momentum

In addition to orbital angular momentum, there is
also angular momentum arising from the motion
of the electron about its own axis. This is called
spin angular momentum or intrinsic angular
momentum, and has no classical analogue.

For spin angular momentum we use the quantum
number s. In general, the magnitude is given
by
Quantization of Spin

Particles with half-integral spins are called
fermions (includes electrons, protons and
neutrons).

Particles with integral spins (includes photons and
some atomic nuclei) are called bosons.

Atoms or molecules with an even number of
fermions are bosons; if they contain an odd
number of fermions they are also fermions.
Spin magnetic quantum number



For an electron it turns out
that only one value of s is
allowed, s =1/2.
The projection on the zaxis is denoted by the spin
magnetic quantum
number, ms, which may
have 2s +1 orientations, or
two for an electron, ms
=1/2.
The magnitude of this
component is ms 
Inclusion of Spin in Wavefunction



The spin can be included by multiplying by a spin
function.
The convention is to use  for a clockwise spin
function (also denoted by ) and  for a
counterclockwise spin (also denoted by ).
The spin function is thought of as a function of
some unspecified spin coordinates, which we do
not explicitly represent.
Two-Electron Spin Functions

To be fully correct, electron spin must be included
in the wavefunction. Our first impulse might be to
write the following four spin functions for twoelectron systems:
 

 
Uncertainty yet again

However, the last two functions are invalid
because they distinguish between the electrons.
Electrons are identical to one another, and there is
no way of experimentally determining which
electron has ms = +1/2 and which has ms = 1/2.

The Heisenberg uncertainty principle makes it
impossible to follow the path of a particle in
quantum mechanics. Therefore, the wavefunction
must not distinguish between the two electrons.
Probability Densities for Two Particles

The principle of indistinguishability means the
probability density must remain unchanged if the
locations of the two particles are interchanged:
–
With real functions, there are only two ways to
satisfy this condition.
The Two Ways



– In this case the wavefunction is symmetric with
respect to interchange. Bosons obey the
symmetric case.

– In this case the wavefunction is antisymmetric
with respect to interchange. Fermions obey the
antisymmetric case.
Proper Spin Functions


The two-electron spin functions
 are still valid. The other valid
spin functions are
C  (1)  (2)   (1) (2)
C  (1)  (2)   (1) (2)
where C is a normalization constant, in this case
equal to 2/.
These spin functions are combined with
symmetric or antisymmetric spatial functions to
give an antisymmetric total wavefunction.
Wavefunction must be Antisymmetric
 (1,2)  C  1 (1) 2 (2)   2 (1) 1 (2)


If the orbitals  and  are the same, the twoparticle wavefunction is the difference of two
identical terms and vanishes.
Any physically meaningful two-electron orbital
wavefunction cannot contain the same spin orbital
more than once in any term.
Pauli Exclusion Principle

Generalizating of our observation for two
electrons: In an orbital wavefunction, no two
electrons can occupy the same spin orbital.

This requires that if there are two electrons in one
spatial orbital, their spins must be paired, i.e., one
electron has spin  (ms = ½)
and the other electron has spin 
Nuclear spin

Exactly analogous to electron spin is nuclear spin,
which must also be accounted for. The total
wavefunction can be written as
   e v r ns
where the electronic wavefunction  e includes the
electron spin function, and the nuclear wave
function now includes a nuclear spin function  ns
in addition to the vibrational and rotational
wavefunctions.
Nuclear spin quantum numbers

Nuclei which are fermions (odd total number of
protons and neutrons) will be antisymmetric to nuclear
exchange, and have nuclear spin quantum number I =
n + ½, where n is zero or a positive integer.

Nuclei which are bosons will be symmetric to nuclear
exchange, and have nuclear spin quantum number I =
n.

The projection on the z-axis is denoted by the nuclear
spin magnetic quantum number, MI, which may have
2I +1 orientations.
Nuclear spin functions

For a diatomic molecule with I = 1/2, the possible
nuclear spin functions are:
 (1) (2)
 (1)  (2)
 (1)  (2)   (1) (2)
1
2 2  (1)  (2)   (1) (2) 
2
1
2
where the first 3 are symmetric (s) to nuclear
exchange and the 4th is antisymmetric (a).

In general for a homonuclear diatomic molecule
there are (2I+1)(I+1) symmetric and (2I+1)(I)
antisymmetric nuclear spin wave functions, so we
have
number of (s) functions I  1

number of (a) functions
I
which gives rise to nuclear spin statistical weights.
Symmetry properties

In the simple case of 1H2, the electronic and
vibrational wavefunctions are symmetric to
nuclear exchange, so we need only consider the
behaviour of  r ns

Since I=1/2 for 1H, we are dealing with fermions,
and , and thus  r ns must be antisymmetric to
nuclear exchange.
Rotational symmetry


For even values of the rotational quantum number
J, r is symmetric to exchange, and for odd values
of J it is antisymmetric.
In order that  r ns is always antisymmetric for
1H the antisymmetric  are associated with even
2
ns
J states and the symmetric ns are associated with
odd J states. These give rise to para- and orthohydrogen, respectively.
The strange case of oxygen

For 16O, I=0. Since each 16O is a boson, the total
wavefunction must be symmetric to nuclear
exchange.

In the case of 16O2, which has 2 electrons with
unpaired spins in the ground state, e is
antisymmetric, and there are no antisymmetric
nuclear spin functions (only ns=1). To obtain a
symmetric total wavefunction, r must be
antisymmetric, so only odd values of J are
allowed.