Problem Set 5 Solutions - Department of Chemistry

Chemistry 362
Dr. Jean M. Standard
Problem Set 5 Solutions
1.
How many vibrational modes do the following molecules or ions possess? [Hint: Drawing Lewis
structures may be useful in some cases.]
In all of the cases, there will be 3N–6 vibrations for nonlinear molecules (or ions) and 3N–5 vibrations for linear
molecules (or ions).
a.) CH4
CH4 is a nonlinear molecule; therefore, it has 3*5–6 = 9 vibrational modes.
b.) HCN
HCN is a linear molecule from the Lewis structure shown below.
H
C
N
Therefore, HCN possesses 3*3–5 = 4 vibrational modes.
c.) NH3
NH3 is a nonlinear molecule; therefore, it has 3*4–6 = 6 vibrational modes.
d.) NO3
–
–
–
NO3 is a nonlinear molecular ion; therefore, it has 3*4–6 = 6 vibrational modes. [Note that NO3 is a
planar molecule, not a linear molecule.]
2.
2
Consider the molecule N2O, with connectivity N-N-O.
a.) Is the molecule linear or nonlinear? How many vibrational modes does N2O possess?
N2O is a linear molecule (you can predict that from the Lewis dot structure, shown below).
N
N
O
Since N2O is linear, the number of vibrational modes is 3N–5, where N is the number of atoms.
Thus, # modes = 3*3–5 = 4 modes.
b.) Describe the vibrational modes of N2O.
(1) N-N stretch
(2) N-O stretch
(3, 4) degenerate N-N-O bends
Note here that the two stretching vibrations are not labeled as 'symmetric' and 'antisymmetric'. This is
because for a molecule such as N2O, the two bonds are not identical and therefore no symmetry
combinations exist. In contrast, the two bonds in molecules such as H2O or SO2 are identical, and so the
stretching vibrations occur as symmetric and antisymmetric combinations.
c.) For N2O, determine which of the vibrational modes are IR active.
(1) N-N stretch - IR active
(2) N-O stretch - IR active
(3, 4) degenerate N-N-O bends - IR active
The dipole moment changes in all three of the vibrations; therefore, all are IR active. A gas phase IR
spectrum is available from NIST, showing the IR absorption peaks for the three IR active modes (at about
590, 1290, and 2220 cm–1, for the bends, N-O stretch, and N-N stretch, respectively), as well as several
overtone and combination bands. The spectrum is reproduced below for your information.
3.
3
Consider the molecule NO2, with connectivity O-N-O.
a.) Is the molecule linear or nonlinear? How many vibrational modes does NO2 possess?
NO2 is a nonlinear molecule (you again can predict that from the Lewis dot structure for one resonance
contributor, shown below).
N
O
O
Since NO2 is nonlinear, the number of vibrational modes is 3N–6, where N is the number of atoms.
Thus, # modes = 3*3–6 = 3 modes.
b.) Describe the vibrational modes of NO2.
(1) N-O symmetric stretch
(2) N-O antisymmetric stretch
(3) O-N-O bend
c.) For NO2, determine which of the vibrational modes are IR active.
(1) N-O symmetric stretch - IR active
(2) N-O antisymmetric stretch - IR active
(3) O-N-O bend - IR active
As for water (which has a similar bent structure), the dipole moment changes in all three of the vibrations;
therefore, all are IR active.
4.
Consider the formaldehyde molecule, H2CO.
a.) How many vibrational modes does formaldehyde possess?
Since formaldehyde is nonlinear, the number of vibrational modes is 3N–6, where N is the number of
atoms.
Thus, # modes = 3*4–6 = 6 modes.
b.) Describe the vibrational modes of formaldehyde (simple sketches may help).
(1) symmetric C-H stretch
(2) antisymmetric C-H stretch
(3) C-O stretch
(4) H-C-H bend (also referred to as the CH2 scissors mode)
(5) CH2 in-plane rock
(6) CH2 out-of-plane bend (also referred to as the CH2 wagging mode)
4 b).
4
continued
The first three modes listed above are similar to those we have seen before. Many times sketches of the
vibrational modes use arrows to indicate the direction of motion of the atoms. Typical sketches of this type
are included below for each of the vibrational modes of formaldehyde.
O
O
C
C
H
H
symmetric C-H stretch
H
H
antisymmetric C-H stretch
O
O
C
C
H
H
C-O stretch
CH2 scissors
O
O
C
H
H
H
C
H
CH2 in-plane rock
H
H
CH2 out-of-plane wag
Note that for the last sketch, corresponding to the CH2 out-of-plane wag, the plus and minus signs indicate
that the atom comes out of the plane of the molecule (+) or goes back into the plane (–).
c.) For formaldehyde, list the vibrational modes that are IR active.
All the modes are IR active. Each one leads to a change in the dipole moment (either a change in
magnitude or a change in direction).
5.) The molecule hydrogen peroxide, H2O2, is a nonplanar molecule with an H-O-O-H torsional angle of
about 120 degrees.
a.) How many vibrational modes does H2O2 have?
Since H2O2 is nonlinear, the number of vibrational modes is 3N–6, where N is the number of atoms.
Thus, # modes = 3*4–6 = 6 modes.
b.) Describe the vibrational modes of H2O2 (simple sketches may help).
(1) symmetric O-H stretch
(2) antisymmetric O-H stretch
(3) O-O stretch
(4) symmetric O-O-H bend
(5) antisymmetric O-O-H bend
(6) H-O-O-H torsional vibration
Sketches of the vibrational modes are shown below.
O
O
O
H
H
symmetric OH stretch
O
H
antisymmetric OH stretch
O
H
O-O stretch
O
H
O
H
O
H
H
symmetric OOH bend
O
O
H
O
O
H
H
antisymmetric OOH bend
H
torsion
c.) List which vibrational modes are IR active.
(1) symmetric O-H stretch - IR active
(2) antisymmetric O-H stretch - IR active
(3) O-O stretch - IR active (There is only a very small change in the dipole moment, so it is okay if you
stated that this mode was not IR active.)
(4) symmetric O-O-H bend - IR active
(5) antisymmetric O-O-H bend - IR active
(6) H-O-O-H torsional vibration - IR active
5
6.
−α x
The ground state wavefunction of the harmonic oscillator is ψ 0 ( x ) = N e
normalization constant N.
The normalization constant N is defined by the equation
€
N =
1
)1/ 2
*
ψ ( x ) ψ ( x ) dx +
+*
−∞
&
(
('
.
∞
∫
Substituting the form of the wavefunction,
€
∞
∞
∫
∫e
ψ * ( x ) ψ ( x ) dx =
−∞
−α x 2
dx .
−∞
From the handout on integrals,
€
1
∞
∫
e
−bx 2
0
1 & π )2
dx =
( + .
2 'b*
This is for only half the range. Since the function is an even function,
€
∞
∫
−∞
2
& π )1/ 2
e −bx dx = ( + .
'b*
Then, making the substitution that b = α , the integral becomes
€
∞
€
∫
∞
ψ * ( x) ψ ( x) dx =
−∞
∫
−∞
2
( π +1/ 2
e −α x dx = * - .
)α ,
Substitution into the normalization equation becomes
€
N =
1
&
(
('
1/ 2
)
ψ ( x ) ψ ( x ) dx +
+*
−∞
∞
∫
*
. α 11/ 4
N = 0 3 .
/π 2
€
=
1
& . π 11/ 2 )1/ 2
(0 3 +
(' / α 2 +*
2
/2
6
. Determine the
7.
The first excited state wavefunction of the harmonic oscillator is given by ψ1( x ) = N ⋅ 2 α ⋅ x e
−α x 2 / 2
7
,
1/ 2
# µk &
where α = % 2 (
$! '
. Show that this function is an eigenfunction of the harmonic oscillator Hamiltonian
€ and µ is the reduced mass.
operator and determine the energy eigenvalue. Here, N is a constant
€ In order to show that the function is an eigenfunction, we start by operating the Hamiltonian operator on it,
$ !2 d 2
'
2
1
Hˆ ψ1 (x) = % −
+ k x 2 ( N ⋅ 2 α x e −α x / 2
2
2
µ
2
dx
&
)
= −
2
2N α ! 2 d 2
x e −α x / 2
2
2µ
dx
(
)
2
k
2N α x 3 e −α x / 2 .
2
+
Evaluating the first derivative,
€
2
2
d $ −α x 2 / 2 &
xe
= e −α x / 2 − α x 2 e −α x / 2
'
dx %
= e −α x
2 /2
( 1− α x 2) .
Evaluating the second derivative,
2
d2
x e −α x / 2
2
dx
(
)
=
d −α x 2 / 2
e
1− α x 2
dx
{
)}
− α x ( 1− α x ) e
( 3 − α x ).
(
= − 2α x e −α x
= − α x e −α x
2
2
/2
/2
2
−α x 2 / 2
2
Substituting into the Hamiltonian equation,
€
2
2
2N α ! 2
k
Hˆ ψ1 (x) = −
− α x e −α x / 2 3 − α x 2
+
2N α x 3 e −α x / 2
2µ
2
+-% ! 2α (
2
k 2 /= ,'
x 0 2N α x e −α x / 2
* 3 − α x2 +
-.& 2µ )
2 -1
{
(
+-% ! 2α (
= ,'
* 3 − α x2
-.& 2µ )
(
€
(
)}
)
)
+
k 2 /x 0 ψ1( x ) .
2 -1
8
7. continued
Using the definition of α,
+-$ ! 2α '
Hˆ ψ1 (x) = ,&
) 3 − α x2
-.% 2µ (
+- $ ! 2 ' $ µ k '1/ 2
= ,3 & ) & 2 ) −
-. % 2µ ( % ! (
(
)
1/ 2
+3 $ k'
= , !& )
-. 2 % µ (
=
+
k 2 /x 0 ψ1( x )
2 -1
$ !2 ' $ µ k '
k 2 /x 0 ψ1( x )
& )& 2 ) x 2 +
2 -1
% 2µ ( % ! (
/
k 2
k 2−
x +
x 0 ψ1( x )
2
2 -1
1/ 2
3 $ k'
! & ) ψ1( x )
2 %µ (
1/ 2
3 h $ k'
Hˆ ψ1 (x) =
& ) ψ1 ( x ) .
2 2π % µ (
Next, the definition of the harmonic frequency can be used,
€
ν0 =
1/ 2
1 $ k'
& ) .
2π % µ (
Substituting,
€
1/ 2
3 h $ k'
Hˆ ψ1 (x) =
& ) ψ1 ( x ) ,
2 2π % µ (
3
or Hˆ ψ1 (x) =
hν 0 ψ1 ( x ) .
2
Thus, we see that the function ψ1( x ) is an eigenfunction of the Schrödinger equation for the harmonic oscillator.
ˆ
€ energy eigenvalue must be given by
Since H ψ1( x ) = E1 ψ1 ( x ) , the
€
E1 =
€
€
3
hν 0 .
2
8.
9
Two functions f(x) and g(x) are said to be orthogonal if
∞
∫f
*
( x) g( x) dx
= 0.
−∞
Show that the ground state (v=0) and first excited state (v=1) wavefunctions of the harmonic oscillator
are orthogonal.
€
For the two functions to be orthogonal, we must show that
∞
∫ψ
*
0
( x) ψ1( x) dx
= 0.
−∞
The harmonic oscillator functions for the ground and first excited states are
€
2
ψ 0 (x) = N 0 e −α x / 2
and ψ1 (x) = 2 N1 α x e −α x
Substituting,
€
∞
∫
∞
∫
ψ 0* ( x) ψ1 ( x) dx = 2 N 0 N1 α
−∞
2
x e −αx dx .
−∞
From the handout on integrals,
€
∞
∫
2
x e −bx dx =
0
1
.
2b
This is for only half the range. Since the function is an odd function,
€
∞
∫
2
x e −bx dx = 0.
−∞
Then, the orthogonality integral is
∞
∫
€
ψ *0 ( x ) ψ1 ( x ) dx = 2 N 0 N1 α
−∞
€
∫
−∞
= 0.
Therefore, the two functions are orthogonal.
∞
2
x e −α x dx
2
/2
.
9.
Determine the average value of the position for the ground state of the harmonic oscillator.
The expectation value of position is defined as
∞
x
∫ ψ ( x) xˆ ψ( x) dx .
*
=
−∞
The wavefunction for the ground state of the harmonic oscillator is
€
ψ 0 (x) = N 0 e −αx
2
/2
,
where the normalization constant N 0 is given by
€
$ α '1/ 4
N0 = & ) .
%π (
€
Substituting into the expression for the expectation value,
€
x
∞
=
∫ ψ ( x) xˆ ψ( x) dx
*
−∞
( α +1/ 2
= * )π ,
∞
∫e
−α x 2 / 2
−∞
1/ 2 ∞
x
(α +
= * )π ,
∫ xe
−α x 2
x e −α x
2
/2
dx
dx .
−∞
This integral is the same one evaluated in Problem 8, and it was shown to be 0. Therefore,
€
x
€
= 0.
10
10. Determine the average value of the momentum for the ground state of the harmonic oscillator.
The expectation value of momentum for normalized wavefunctions is defined as
∞
px
∫ ψ ( x) pˆ
*
=
x
ψ ( x ) dx .
−∞
The wavefunction for the ground state of the harmonic oscillator is
€
ψ 0 (x) = N 0 e −α x
2
/2
,
where the normalization constant N 0 is given by
€
$ α '1/ 4
N0 = & ) .
%π (
€
Substituting into the expression for the expectation value,
€
px
∞
∫ ψ ( x) pˆ
*
=
ψ ( x ) dx
x
−∞
( α +1/ 2
= − i !* )π ,
∞
∫e
−α x 2 / 2
−∞
d −α x 2 / 2
e
dx .
dx
In the last expression above, the definition of the momentum operator,
€
pˆ x = − i !
d
,
dx
has been used. Now, evaluating the derivative,
€
2
d −α x 2 / 2
e
= − α x e −α x / 2 .
dx
Substituting, the expectation value becomes
€
px
$ α '1/ 2
= iα !& )
%π (
∞
∫
−∞
This integral has already been shown to be 0. Therefore,
€
px
€
= 0.
2
x e −α x dx .
11
11. What are the most probable positions for the ground and first excited states of the harmonic oscillator?
To determine the most probable value, the maximum in the probability density must be determined. A
maximum in a function can be found by setting the derivative equal to zero.
Ground State
The wavefunction for the ground state of the harmonic oscillator is
2
% α (1/ 4
ψ 0 (x) = ' * e −α x / 2 ,
&π )
and so the probability density is
€
2
% α (1/ 2
ψ 0* ( x ) ψ 0 ( x ) = ψ 02 ( x ) = ' * e −α x .
&π )
Setting the derivative equal to zero,
€
d *
ψ 0 ( x) ψ 0 ( x) = 0
dx
% α (1/ 2 d −α x 2
e
= 0
' *
& π ) dx
% α (1/ 2
−α x 2
' * −2α x e
&π )
(
)
= 0.
Dividing both sides by the constant terms,
€
x e −α x
2
= 0.
Since the exponential term is never zero except if x → ∞ , the solution is
€
x = 0.
€
Thus, the most probable position for the ground state of the harmonic oscillator occurs at x = 0.
€
First Excited State
The wavefunction for the first excited state of the harmonic oscillator is
ψ1 (x) = 2N1 α x e −α x
2
/2
.
The probability density for this function is
€
€
2
ψ1* ( x ) ψ1 ( x ) = ψ12 ( x ) = 4 N12α x 2 e −α x .
12
13
11. continued
Setting the derivative equal to zero,
d
ψ1* ( x )ψ1 ( x ) = 0
dx
2
d
4 N12 α
x 2 e −α x
= 0
dx
{
}
{
(
}
2
4 N12 α 2x e −α x − 2 α x 3 e −α x
(
)
8 N12 α x 1 − α x 2 e −α x
2
2
)
= 0
= 0 .
Dividing both sides by the constant term 8 N12 α , the equation becomes
€
(
2
)
x 1 − α x 2 e −α x
= 0.
€
Since the exponential is never zero unless x → ∞ , the equation can be simplified to
€
(
x 1 − α x2
)
= 0.
€
This equation has solutions when
€
x=0
or
1− α x 2 = 0 .
It is easy to see from a plot of the probability density for the first excited state that the solution x = 0
corresponds to a node in the wavefunction and therefore a minimum in the probability distribution. The other
€
solution is
1 − α x2 = 0
α x2 = 1
1
x2 =
α
x = ±
1
.
α
Thus, there are two most probable positions for the first excited state of the harmonic oscillator. The most
1
1
probable positions are x =
and
.
€ x = −
α
α
€
€