Week 6 W - Vibra&onal
Spectroscopy
 • 

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Unformatted text preview: Vibra&onal
Spectroscopy
 •  Vibra&onal
spectroscopy
is
concerned
with
the
 observa&on
of
the
degrees
of
vibra&onal
 freedom.
 •  A
molecule
with
n
atoms
have
3n
degrees
of
 freedom
describing
the
transla&onal
(3),
 vibra&onal
and
rota&onal
(3)
mo&ons
of
the
 molecule.
 •  How
many
vibra&onal
modes
are
there
for
a
 given
molecular
species?
 Two
classes
of
molecular
vibra&ons
 •  Two distinct classes of molecular vibrations, stretching and bending. •  Stretching vibrations change the bond length while bending vibrations change the bond angle. •  These two classes may be subdivided into different types depending on how the atoms move relative to each other. –  stretching (symmetric and asymmetric) –  bending (scissoring, rocking, wagging and twisting)
 Stretching
 symmetric
stretch
 asymmetric
stretch
 Bending
 scissoring
 rocking
 wagging
 twis&ng
 A
normal
mode
of
an
oscilla&ng
system
is
a
paCern
of
mo&on
in
which
all
parts
of
the
system
 move
sinusoidally
with
the
same
frequency.
The
modes
are
normal
in
the
sense
that
they
can
 move
independently,
that
is
to
say
that
an
excita&on
of
one
mode
will
never
cause
mo&on
of
a
 different
mode.

 Use
of
character
tables
to
iden&fy
all
the
 mo&ons
of
a
molecule
 Example
SiH2Cl2

 z
 y
 Character
table
 Si
 Cl1
 Cl2
 x
 
C 2 
σv(xz) 
σv(yz) 


h
=
4
 A1 
+1 
+1 
+1 
+1 


z 



x2,
y2,
z2
 
+1 
+1 
‐1 
‐1 


Rz 




xy
 B1 H2
 
E A2 H1
 C2v 
+1 
‐1 
+1 
‐1 


x,
Ry 



xz
 B2 
+1 
‐1 
‐1 
+1 


y,
Rx 



yz
 Draw
x,
y
and
z
vectors
on
all
atoms
 How
the
vectors
are
affected
by
symmetry?
 How
are
the
vectors
affected
by
 symmetry
 •  Count
0
if
atom
change
posi&on
during
the
 symmetry
opera&on
 •  If
the
atom
remains
in
its
original
posi&on
(in
the
 case
of
the
C2v
point
group)
 –  Count
+1

if
the
vector
direc&on
is
unchanged
 –  Count
‐1
if
the
vector
direc&on
is
changed
 Opera&on
E
 z
 H1
 y
 H2
 x
 Si
 Cl1
 Cl2
 
 
 
Si
atom 
x
transforms
into
Si
x 
count
+1
 
 
 
 
 
y
transforms
into
Si
y 
count
+1
 
 
 
 
 
z
transforms
into
Si
z 
count
+1
 
 
 
 
 
 
 
 
 
total
+3
 Same
for
other
4
atoms 
 
 
 
 
grand
total
+15
 z
 H1
 y
 Opera&on
C2
 H2
 x
 
Si
atom 
 
 
x
transforms
into
Si
‐x 
 
 
 
 
y
transforms
into
Si
‐y 
count
‐1
 
 
 
 
 
z
transforms
into
Si
z 
 
 
 
 
 
 
 
total
 
 H1
and
H2
move
‐
swap
places 
 
 
 
 
count
0
 Cl1
and
Cl2
swap
places 
 
 
 
 
 
count
0
 
 
 
grand
total
 
 Si
 Cl1
 
 Cl2
 
 
 
 
count
‐1
 
count
+1
 
‐1
 
‐1
 z
 H1
 H2
 y
 Opera&on
σv(xz)

 x
 Si
 Cl1
 Cl2
 
 
 
Si
atom 
x
transforms
into
Si
x 
 
 
 
 
y
transforms
into
Si
‐y 
count
‐1
 
 
 
 
 
z
transforms
into
Si
z 
 
 
 
 
 
 
 
count
+1
 
count
+1
 
total
 
 
+1
 H1
and
H2
also
lie
in
xz
plane,
and
behave
as
Si 
 
count
+1
each
 Cl1
and
Cl2
swap
places 
 
count
0
 
 
 
 
 
 
 
 
 
 
 
grand
total
 
+3
 z
 H1
 H2
 y
 Opera&on
σv(yz)

 x
 Si
 Cl1
 Cl2
 
 
 
Si
atom 
x
transforms
into
Si
‐x 
count
‐1
 
 
 
 
 
y
transforms
into
Si
y 
count
+1
 
 
 
 
 
z
transforms
into
Si
z 
count
+1
 
 
 
 
 
 
 
 
total
 
 H1
and
H2
swap
places 
 
 
 
 
 
+1
 
count
0
 Cl1
and
Cl2
also
lie
in
yz
plane,
and
behave
as
Si 
count
+1
each
 
 
 
 
 
 
 
grand
total
 
 
+3
 Reducible
representa&on
 Overall
we
have:
 E 
 
C 2 
 
σv(xz) 
 +15 
 
‐1 
 
+3 
 
σv(yz)
 
+3
 This
is
the
reducible
representa&on
of
the
set
of
3N
(=15)
 atomic
displacement
vectors
 15
degrees
of
freedom
 How
to
reduce
it
to
the
irreducible
 representa&on?
 Reduce
the
reducible
representa&on
 
Formula
is
 Reducible
representa&on
 
15 
‐1 
3 
3
 
 

 Character
table
 C2v 
1E 
1C2 
1σv(xz)
1σv(yz)


h
=
4
 A1 
+1 
+1 
+1 


+1 


z 



x2,
y2,
z2
 A2 
+1 
+1 
‐1 


‐1 


Rz 




xy
 B1 
+1 
‐1 
+1 


‐1 


x,
Ry 



xz
 B2 
+1 
‐1 
‐1 


+1 


y,
Rx 



yz
 No.
of
A1
mo&ons
=
1/4
[1.15.1
+
1.(‐1).1
+
1.3.1
+
1.3.1] 
=
5
 
Formula
is
 Reducible
representa&on
 
15 
‐1 
3 
3
 
 

 Character
table
 C2v 
1E 
1C2 
1σv(xz)
1σv(yz)


h
=
4
 A1 
+1 
+1 
+1 


+1 


z 



x2,
y2,
z2
 A2 
+1 
+1 
‐1 


‐1 


Rz 




xy
 B1 
+1 
‐1 
+1 


‐1 


x,
Ry 



xz
 B2 
+1 
‐1 
‐1 


+1 


y,
Rx 



yz
 No.
of
A1
mo&ons
=
1/4
[1.15.1
+
1.(‐1).1
+
1.3.1
+
1.3.1] 
 No.
of
A2
mo&ons
=
1/4
[1.15.1
+
1.(‐1).1
+
1.3.(‐1)
+
1.3.(‐1)] 
=
5
 
=
2
 
Formula
is
 Reducible
representa&on
 
15 
‐1 
3 
3
 
 

 Character
table
 C2v 
1E 
1C2 
1σv(xz)
1σv(yz)


h
=
4
 A1 
+1 
+1 
+1 


+1 


z 



x2,
y2,
z2
 A2 
+1 
+1 
‐1 


‐1 


Rz 




xy
 B1 
+1 
‐1 
+1 


‐1 


x,
Ry 



xz
 B2 
+1 
‐1 
‐1 


+1 


y,
Rx 



yz
 No.
of
A1
mo&ons
=
1/4
[1.15.1
+
1.(‐1).1
+
1.3.1
+
1.3.1] 
 
=
5
 No.
of
A2
mo&ons
=
1/4
[1.15.1
+
1.(‐1).1
+
1.3.(‐1)
+
1.3.(‐1)] 
=
2
 No.
of
B1
mo&ons
=
1/4
[1.15.1
+
1.(‐1).(‐1)
+
1.3.1
+
1.3.(‐1)] 
=
4
 No.
of
B2
mo&ons
=
1/4
[1.15.1
+
1.(‐1).(‐1)
+
1.3.(‐1)
+
1.3.1] 
=
4
 Degree
of
freedom
for
SiH2Cl2
 Transla&ons,
rota&ons,
vibra&ons
and
 symmetry
associated
with
them
 Symmetry
species
of
all
mo&ons
are:‐
 5A1
+
2A2
+
4B1
+
4B2

 3
of
these
are
transla&ons
of
the
whole
molecule
 3
are
rota&ons

 Symmetry
species
of
transla&ons
are
given
by
vectors
(x,
y,
z)
in
the
character
table
 Symmetry
species
of
rota&ons
are
given
by
Rx,
Ry
and
Rz
in
the
character
table
 Transla&ons,
rota&ons,
vibra&ons
and
 associated
symmetries
for
SiH2Cl2
 Symmetry
species
of
all
mo&ons
are:‐
 
 
 
5A1 
+
2A2



+
4B1


+
4B2
 Transla&ons
are:‐ 
 
A1
 
 


+
B1




+
B2
 Rota&ons
are:‐ 
 
 




A2 



+
B1




+
B2
 ‐
so
vibra&ons
are:‐ 

4A1 
+

A2



+
2B1


+
2B2
 
 
 Character
table
 C2v 
1E 
1C2 
1σv(xz)
1σv(yz)


h
=
4
 A1 
+1 
+1 
+1 


+1 


z 



x2,
y2,
z2
 A2 
+1 
+1 
‐1 


‐1 


Rz 




xy
 B1 
+1 
‐1 
+1 


‐1 


x,
Ry 



xz
 B2 
+1 
‐1 
‐1 


+1 


y,
Rx 



yz
 Vibra&onal
modes
of
SiH2Cl2
 Symmetry
species
of
vibra&ons
are:‐
 
4A1
+
A2
+
2B1
+
2B2
 What
does
each
of
these
modes
look
like?
 2
rules
 (i) 
there
is
1
stretching
vibra&on
per
bond
 (ii) 
must
treat
symmetry‐related
atoms
together
 ...
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This note was uploaded on 10/03/2011 for the course CHEM 113A taught by Professor Professornotknown during the Spring '09 term at San Jose State University .

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