24Jan2011
Chemistry 21b – Spectroscopy
Lecture # 9 – Rotation of Polyatomic Molecules
The rotational spectra of molecules can be classified according to their “principal
moments of inertia”.
Assume that the molecule rotates as a rigid body, that is, the
relative nuclear positions are fixed. The moment of inertia
I
of any molecule about any
axis through the centerofmass is then given by
I
=
summationdisplay
i
M
i
R
2
i
(9
.
1)
where
M
i
and
R
i
are the mass and perpendicular distance of atom
i
from the axis. Recall
that the centerofmass position for a given axis may be found from the equation:
R
α
(
C.M.
) =
summationdisplay
i
M
i
R
i
/
summationdisplay
i
M
i
.
In general, for any rigid three dimensional body, there exists a 3
×
3 moment of inertia tensor
(that is, matrix). If we use any three axes defined by (
xyz
), the moment of inertia tensor
has diagonal elements defined as
I
xx
=
∑
i
M
i
(
y
2
i
+
z
2
i
) (using the
x
axis, for example),
and offdiagonal elements given by
I
xy
=
−
∑
i
M
i
x
i
y
i
, etc. (Note the minus sign in the
offdiagonal terms).
Clearly, the momentofinertia tensor is Hermetian, and so can be diagonalized to
yield three (potentially) distinct eigenvalues. These eigenvalues are called the “principal
moments of inertia,” and the eigenvectors corresponding to the diagonalized coordinate
system are referred to as the “principal axes” of the molecule.
Mathematically, the
eigenvectors of the matrix in any coordinate system form the
direction cosines
that relate
the arbitrary coordinate system to the principal axis coordinate system, that latter of
which must rotate with the molecule.
One can always find one axis, called the
c
axis, about which the moment of inertia has
its
maximum
value, and another axis, labeled the
a
axis, about which
I
has its
minimum
value.
It can be shown that the
a
and
c
axes must be mutually perpendicular.
Thus,
according to convention, the principal axes are ordered:
I
c
≥
I
b
≥
I
a
.
(9
.
2)
Why do we care? If we use these principal axes, then the components of the rotational
angular momentum
P
along these axes can be shown to be
P
a
=
I
a
ω
a
,
P
b
=
I
b
ω
b
,
P
c
=
I
c
ω
c
,
(9
.
3)
and the kinetic energy operator for a rigidrotor simply becomes
T
rot
=
P
2
a
2
I
a
+
P
2
b
2
I
b
+
P
2
c
2
I
c
71
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B
F
F
F
(c)
(a)
(b)
r
BF
0.13 nm
60
ο
Figure 9.1: The structure and principal axes of boron trifluoride (BF
3
).
In all other coordinate systems things are much more complicated!
As a simple example, let us calculate the moments of inertia for the molecule BF
3
.
This molecule is planar, with the three F atoms arranged around a central B atom at
angles of 120
◦
and a BF distance of 1.30
˚
A. (See Figure 9.1). The center of mass is clearly
at the B atom. One principal axis is perpendicular to the plane of the molecule, and since
this axis will possess the largest moment of inertia, it is the caxis. A second axis will pass
through one of the BF bonds, and the third will be mutually perpendicular to the first
two.
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 Fall '10
 list
 Atom, Mole, Angular Momentum, Moment Of Inertia, Rotation, principal axes

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