Chemistry 362
Dr. Jean M. Standard
Problem Set 13 Solutions
1.
2
+
The molecular term symbol of the ground electronic configuration of B2 is Π u . Two of the excited
2
2
2
2
1
( ) (1σ *u ) (2σ g ) (2σ *u ) (3σ g )
+
states of B2 have electron configurations given by 1σ g
2
2
1σ *u
2
2
2σ *u
1
1π *g .
(1σ g ) ( ) (2σ g ) ( ) ( )
and
€
€
+
a.) Determine the molecular term symbols for the two excited electronic configurations of B2 .
€
2
2
2
2
1
( ) (1σ *u ) (2σ g ) (2σ *u ) (3σ g )
For electron configuration 1σ g
The closed shells in the electron configuration do not contribute to the term symbol. Therefore, only the
1
(3σ g )
€
portion of the electron configuration needs to be included in the determination of the term symbol.
€
Since there is only a single electron to consider, we have s1 =
multiplicity of 2S +1 = 2 , or a doublet state.
1
2
and therefore S = 12 . This gives a
For the molecular orbital angular momentum, a€σ molecular orbital €
corresponds to λ1 = 0 which leads to
Λ=€
0 , or a Σ state.
Since the electron that contributes to the
€ term symbol is in a σ g molecular orbital, the parity of the
€
€
molecular
state is the same, gerade.
€
2
2
€2
2
1
( ) (1σ *u ) (2σ g ) (2σ *u ) (3σ g )
The term symbol for the 1σ g
2
2
2
2
2
configuration is therefore Σ g .
1
( ) (1σ *u ) (2σ g ) (2σ *u ) (1π *g )
€
1σ g
For electron configuration
1
( )
*
Only the 1π g
symbol.
€
portion of the electron configuration needs to be included in the determination of the term
€
€Since there is only a single electron to consider, we have a multiplicity of 2S +1 = 2 , or a doublet state just
like for the previous case.
For the molecular orbital angular momentum, a π molecular orbital
€ corresponds to λ1 = ±1 which leads to
Λ = 1 , or a Π state.
€
Since the electron that contributes to the
€ term symbol is in a π g molecular€orbital, the parity of the
molecular
state is the same, gerade.
€
2
2
2
€2
1
( ) (1σ *u ) (2σ g ) (2σ *u ) (1π *g )
The term symbol for the 1σ g
€
2
configuration is therefore Π g .
€
2
1.
continued
+
b.) Will the equilibrium bond lengths of the two excited configurations of B2 be approximately the same
as the bond length of the ground state, or will they be significantly different? Explain.
+
The relative bond lengths of the electronic states of B2 can be estimated from the bond orders.
+
2
2
2
2
2
2
( ) (1σ *u ) (2σ g ) (2σ *u )
The bond order of the ground state of B2 , with electron configuration 1σ g
1
(1π u ) ,
is given by:
# Bonding electrons −€# Antibonding electrons
2
5− 4
=
2
BondOrder = 12 .
BondOrder =
+
2
2
1
( ) (1σ *u ) (2σ g ) (2σ *u ) (3σ g )
The bond order of the excited state of B2 with electron configuration 1σ g
€
given by:
is
# Bonding electrons −€ # Antibonding electrons
2
5− 4
=
2
BondOrder = 12 .
BondOrder =
The bond order of this excited state is the same as that of the ground state. Thus, we would expect the bond
+
length of this
€ particular excited state to be approximately the same as that of the ground state of B2 .
+
2
2
2
2
1
( ) (1σ *u ) (2σ g ) (2σ *u ) (1π *g )
The bond order of the excited state of B2 with electron configuration 1σ g
is
given by:
# Bonding electrons −€ # Antibonding electrons
2
4−5
=
2
1
BondOrder = − 2 .
BondOrder =
The bond order of this excited state is less than zero, implying that the state may not even be bound. If it is
bound, we would
expect a very weak bond strength and a very long bond.
€
3
1.
continued
c.) On the basis of selection rules, what are the allowed electronic transitions from the ground state of
+
B2 to the excited states?
+
The ground state of B2 has term symbol 2 Π u . Selection rules require ΔS = 0 , ΔΛ = 0, ± 1, and there must
be a change in parity ( g → u or u → g ).
Because of the ΔS = 0 spin selection
rule, since the ground electronic
state is a doublet state, the only
€
€
€
allowed transitions
occur
€
€ to excited doublet states. Both the excited states given in this problem are
doublet states, so transitions from the ground state to those excited states would be allowed by the spin
selection
€ rule.
Since the ground state is a Π state, it has Λ = 1 . Therefore, the ΔΛ = 0, ± 1 selection rule would allow
transitions to states with Λ = 0, 1, or 2 ( Σ , Π , or Δ states).
From the ground€state, 2 Π u , only
€ transitions from u →€g are possible because of the parity selection rule.
€ term symbols 2 Σ g , 2 Π g , and 2Δ g are the allowed transitions
€
€
€ with
Therefore, transitions to excited states
from the ground state.
€
€
2
2
2
2
To summarize, the transitions possible based on€selection
€ rules are
€ Π u → Σ g , Π u → Π g , and
€
€
2
2
Π u → 2Δ g . Therefore, both the excited states mentioned in the problem, with term symbols Σ g and
2
Π g , correspond to allowed transitions from the ground state of B2+.
€
€
€
4
2.
Assume that an electron and a hydrogen nucleus (a proton) are both given the same kinetic energy, 13.6
eV. (This is the energy of the electron in the ground state of the hydrogen atom.) Approximate the
kinetic energy using the classical expression, T = 12 mv 2 as an estimate. Determine the velocity of the
electron and the velocity of the proton. How do they differ? Note that this difference forms the
foundation of the Born-Oppenheimer approximation.
Solving the kinetic energy expression for the velocity yields
! 2T $1/2
v = # & .
"m%
For the electron, substituting 13.6 eV for the kinetic energy and using the electron mass gives
1/2
)
# 1.60217 ×10 −19 kg & ,
+ 2 (13.6 eV) %
(.
1 eV
$
'.
v electron = ++
.
9.10938 ×10 −31 kg
+
.
.*+
(
)
v electron = 2.19 ×10 6 m/s.
For the proton, substituting 13.6 eV for the kinetic energy yields
v proton
1/2
)
# 1.60217 ×10 −19 kg & ,
+ 2 (13.6 eV) %
(.
1 eV
$
'.
= ++
−27
.
1.67262 ×10 kg
+
.
+*
.-
(
)
v proton = 5.10 ×10 4 m/s.
Thus, we see that the velocity of the electron is much larger than the velocity of the proton. This is why it is
okay to assume that the nuclei are fixed when solving the Schrödinger equation for the motion of the electrons
in a molecule. In this particular case, the velocity of the electron is about 43 times larger than the velocity of
the proton; for all other cases, the nucleus heavier and the difference is even more pronounced.
5
3.
Give the form of the Hamiltonian operator for the H2 molecule in atomic units. Make sure to write out
all the terms explicitly.
The H2 molecule includes two electrons and two nuclei (with Z=1). Within the Born-Oppenheimer
approximation, the nuclei are considered to be fixed, so there will be no kinetic energy terms for the nuclei.
The Hamiltonian operator in general is the sum of the kinetic and potential energy operators, Ĥ = T̂ + V̂.
For H2, there are two electrons, so the kinetic energy operator will be the sum of the kinetic energy terms for
each electron,
T̂ = −
1 ˆ2
1 ˆ2
∇1 − ∇
2 .
2
2
Note that the electrons are labeled 1 and 2 in the expression above.
The potential energy operator V̂ will consist of the Coulomb interactions between the electrons and protons. In
particular, electron 1 will feel an attraction to both nuclei, as will electron 2. The positively charged nuclei will
feel a repulsive interaction between them, as will the two negatively charged electrons.
Labeling the nuclei A and B, the electron-nuclear attractions can be written in atomic units as:
Vel−nuc = −
1
1
1
1
−
−
−
,
r1A
r1B
r2 A
r2 B
where, for example, r1A is the distance between electron 1 and nucleus A. The nuclear-nuclear repulsion in
atomic units is:
Vnuc−nuc = 1
,
rAB
where rAB is the distance between the two nuclei, A and B. The electron-electron repulsion in atomic units is:
Vel−el = 1
,
r12
where r12 is the distance between the two electrons, 1 and 2.
Putting it all together and writing out the Hamiltonian operator explicitly, we have
1 ˆ2
1
∇1 − ∇ˆ 22
2
2
1
1
1
1
−
−
−
−
r1A
r1B
r2 A
r2 B
1
1
+
+ .
r12
rAB
Ĥ = −
The first two terms are the kinetic energy operators for electrons 1 and 2, while the terms in the second row are
the electron-nuclear attractions, and the terms in the last row are the electron-electron repulsion and the nuclearnuclear repulsion, respectively.
6
4.
Write down the set of secular equations for an LCAO-MO function made up of 3 atomic orbitals,
ψ MO = c1 χ1 + c2 χ 2 + c3 χ 3 . Also give the corresponding secular determinant.
The general form of the secular equations for the LCAO-MO method is
n
∑ ci ( H ki − E Ski )
= 0,
k = 1, 2, … n.
i=1
Here, H ki are the Hamiltonian integrals, Ski are the overlap integrals, ci are the linear coefficients, E is the
approximate energy eigenvalue, and n is the number of AOs which comprise the MO.
For 3 atomic orbitals, n=3, and so the secular equations are
3
∑ ci ( H ki − E Ski )
= 0,
k = 1, 2, 3.
i=1
Writing these out, each secular equation includes three terms,
c1 ( H11 − E S11 ) + c2 ( H12 − E S12 ) + c3 ( H13 − E S13 ) = 0
c1 ( H 21 − E S21 ) + c2 ( H 22 − E S22 ) + c3 ( H 23 − E S23 ) = 0
c1 ( H 31 − E S31 ) + c2 ( H 32 − E S32 ) + c3 ( H 33 − E S33 ) = 0 .
If the atomic orbitals are normalized, then Sii = 1 and the secular equations simplify to
c1 ( H11 − E ) + c2 ( H12 − E S12 ) + c3 ( H13 − E S13 ) = 0
c1 ( H 21 − E S21 ) + c2 ( H 22 − E ) + c3 ( H 23 − E S23 ) = 0
c1 ( H 31 − E S31 ) + c2 ( H 32 − E S32 ) + c3 ( H 33 − E ) = 0 .
The secular determinant associated with these secular equations must equal 0 for there to be a solution to the
equations. This equation is given by
H11 − E
H12 − E S12
H13 − E S13
H 21 − E S21
H 22 − E
H 23 − E S23
H 31 − E S31
H 32 − E S32
H 33 − E
= 0 .
7
8
5.
(
+
The bonding orbital for H2 has the form ψ MO1 = N1 χ1sA + χ1sB
(
form ψ MO2 = N 2 χ1sA − χ1sB
)
)
and the antibonding orbital has the
, where χ1sA and χ1sB are 1s atomic orbitals on atoms A and B,
respectively. Determine the normalization constants N1 and N 2 for the bonding and antibonding
€ will depend on the overlap integral, SAB , which is defined as
molecular orbitals. Your answers
€
€
€
SAB = χ1s* A €
χ 1sB dτ €
= χ1s* B χ 1sA dτ .
∫
∫
€
Bonding MO
To find the normalization constant N1 , we use the normalization condition,
€
∫ψ
*
MO1 ψ MO1 dτ
= 1.
€
Substituting, and assuming that N1 is real,
€
N12
€
∫ (χ
+ χ1sB
1sA
*
) ( χ1s
)
+ χ1sB dτ = 1 .
A
Expanding,
N12
€
{∫χ
*
1sA
χ1sA dτ +
∫χ
*
1sA χ1sB dτ
+
∫χ
*
1sB χ1sA dτ
+
∫χ
*
1sB χ1sB dτ
From AO normalization,
€
∫χ
*
1sA χ1sA dτ
=
∫χ
*
1sB
χ1sB dτ = 1,
and by definition of the overlap intergral,
€
SAB =
∫χ
*
1sA
χ1sB dτ =
∫χ
*
1sB
χ1sA dτ .
Therefore, the normalization requirement becomes
€
N12 + 2N12 SAB + N12 = 1
2N12 (1 + SAB ) = 1.
Solving for N1 leads to
€
1
N1 =
2(1+ SAB )
€
.
Thus, the best approximation for the ground state MO is
€
ψ MO1 =
€
1
2(1+ SAB )
( χ1s
A
)
+ χ1sB .
}
= 1.
9
5. continued
Antibonding MO
To find the normalization constant N 2 , we use the normalization condition,
∫ψ
*
MO2 ψ MO 2
dτ = 1.
€
Substituting, and assuming that N 2 is real,
€
N 22
€
∫ (χ
1sA
− χ1sB
*
) ( χ1s
)
− χ1sB dτ = 1 .
A
Expanding,
N 22
€
{∫χ
*
1sA χ1sA dτ
−
∫χ
*
1sA χ1sB dτ
−
∫χ
*
1sB χ1sA dτ
+
∫χ
*
1sB
χ1sB dτ
}
= 1.
The first and last integrals are again 1 from atomic orbital normalization, and the second and third integrals
correspond to the overlap SAB .
€
Substituting,
€
N 22 − 2N 22 SAB + N 22 = 1
2N 22 (1 − SAB ) = 1.
Solving for N 2 leads to
€
1
N2 =
2(1− SAB )
€
.
Thus, the excited state MO is given by,
€
ψ MO2 =
€
1
2(1− SAB )
( χ1s
A
)
− χ1sB .
10
6.
+
Use the expression for the LCAO-MO energy expectation value for H2 ,
E =
c A2 H AA + 2c A c B H AB + c B2 H BB
,
c A2 + 2c A c B SAB + c B2
# ∂E &
along with the condition %
( = 0 , to obtain the first secular equation,
€ $ ∂c A 'c B
c A ( H AA − E) + c B ( H AB − E SAB ) = 0 .
€
# ∂E &
( = 0 . Since the expression for E involves a quotient, we
To get the first secular equation, we require %
$ ∂c A 'c B
€
can represent it as
€
E =
num
-1
= num ⋅ ( denom) ,
denom
with num = c A2 H AA + 2c A c B H AB + c B2 H BB and denom = c A2 + 2c A c B SAB + c B2 . Then, using the product rule,
the partial derivative becomes
€
# ∂E &
%
(
$ ∂c A 'c
€
−1
= ( denom)€ ⋅
B
∂
−2 ∂
( num) + num ⋅ (−1) ⋅ ( denom)
( denom) .
∂c A
∂c A
Substituting and setting the partial derivative equal to zero,
€
# ∂E &
%
(
$ ∂c A 'c
−1
= ( denom) ⋅
B
∂
c A2 H AA + 2c A c B H AB + c B2 H BB
∂c A
(
− num ⋅ ( denom)
−2
∂
c A2 + 2c A c B SAB + c B2
∂c A
(
)
)
= 0.
Evaluating the partial derivatives yields
€
−1
( denom) ⋅ ( 2c A H AA
+ 2c B H AB ) − num ⋅ ( denom)
−2
( 2c A
+ 2c B SAB ) = 0.
Since the right hand side of the equation equals 0, we can multiply both sides of the equation by "denom" to
simplify,
€
( 2c A H AA
# num &
+ 2c B H AB ) − %
(( 2c A + 2c B SAB ) = 0.
$ denom '
num
From the definition, we know that E =
. Substituting this expression and dividing both sides of the
denom
€
equation by a factor of 2 gives
€
c A H AA + c B H AB − E(c A + c B SAB ) = 0.
Rearranging by collecting terms involving c A and terms involving c B leads to the first secular equation,
€
c A ( H AA − E) + c B ( H AB − ESAB ) = 0.
€
€
€
11
7.
+
In class, we solved the secular determinant for H2 and found the ground state energy to be
E1 =
H AA + H AB
.
1+ SAB
We then substituted this energy back into the secular equations in order to determine the coefficients ( c A
and c B ) for the linear variational€approximation to the ground state wavefunction. Use the expression
that we found in class for the first excited state energy, along with the secular equations, to determine the
coefficients ( c A and c B ) for the approximation to the first excited state wavefunction,
€
€
ψ2 = cA χ A + cB χ B .
€
€
The first excited state energy is given
€ by
E2 =
H AA − H AB
.
1− SAB
The secular equations are
€
(1) c A ( H AA − E) + c B ( H AB − ESAB ) = 0
c A ( H AB − ESAB ) + c B ( H BB − E) = 0.
(2)
Substituting E2 into secular equation (1) leads to
€
#
#
SAB ( H AA − H AB ) &
H AA − H AB &
c A % H AA −
( = 0.
( + c B % H AB −
1− SAB '
1− SAB
$
$
'
€
Multiplying both sides of the equation by 1− SAB gives
€
c A [ H AA (1− SAB ) − ( H AA − H AB )]
+ c B [ H AB (1− SAB ) − SAB ( H AA − H AB )] = 0.
€
Simplifying by canceling terms yields
€
c A ( H AB − H AA SAB ) + c B ( H AB − H AA SAB ) = 0.
By dividing both sides by the factor H AB − H AA SAB , the expression becomes
€
c A + c B = 0,
€
or c A = − c B .
€
Note that substitution of E2 into secular equation (2) leads to exactly the same result, so it will not be shown
here (though you should check this). €
€
12
7. continued
+
The approximate first excited state wavefunction for H 2 is so far determined to be:
(
)
ψ 2 = c A χ1sA − χ1sB .
€
To complete the solution and determine the coefficient c A , we use the normalization condition,
€
∫ψ ψ
*
2 2
dτ = 1 .
€
Substituting, and assuming that the coefficient c A is real gives,
€
c A2
∫ (χ
− χ1sB
1sA
€
*
) (χ1s
)
− χ1sB dτ = 1 .
A
Expanding,
c A2
∫
€
χ1s* A χ1sA dτ − c A2
∫χ
*
1sA χ1sB dτ
− c A2
∫χ
*
1sB χ1sA dτ
+ c A2
∫χ
The atomic orbitals are assumed to be normalized,
€
∫χ
*
1sA χ1sA dτ
=
∫χ
*
1sB χ1sB dτ
= 1.
Also, the overlap integrals are defined,
€
SAB =
∫χ
*
1sA χ1sB dτ
∫χ
= SBA =
*
1sB χ1sA dτ
Therefore, the normalization requirement becomes
€
2c A2 − 2c A2 SAB = 1 ,
or
2c A2 (1− SAB ) = 1 .
€
Solving for c A leads to
€
1
cA =
2(1− SAB )
€
.
Thus, the best approximation for the first excited state wavefunction is
€
ψ2 =
€
1
2(1− SAB )
(χ1s
A
)
− χ1sB .
.
*
1sB χ1sB dτ
= 1.