Chemistry 362
Dr. Jean M. Standard
Problem Set 12 Solutions
1.
+
Using the correlation diagram, what is the molecular electronic configuration of B2 ? Determine the
bond order.
+
Using the correlation diagram, the electron configuration of B 2 is
2
2
2
(1σ g ) (1σ *u ) (2σ g ) (2σ *u )
2
1
(1π u )
.
€
The bond order is
€
# Bonding electrons − # Antibonding electrons
BondOrder =
2
5− 4
=
2
1
BondOrder = 2 .
€
2.
Using the correlation diagram, what is the molecular electronic configuration of N2? Determine the bond
order.
Using the correlation diagram, the electron configuration of N2 is
2
2
2
(1σ g ) (1σ *u ) (2σ g ) (2σ *u )
2
(1π u )
4
(3σ g )
2
.
The bond order is
€
# Bonding electrons − # Antibonding electrons
BondOrder =
2
10 − 4
=
2
BondOrder = 3.
€
2
3.
–
Determine the molecular electronic configurations and bond orders of the following compounds: C2, C2 ,
2–
C2 using the correlation diagram. Which would be expected to have the longest bond length?
Using the correlation diagram, the electron configuration of C2 is
2
2
2
(1σ g ) (1σ *u ) (2σ g ) (2σ *u )
2
(1π u )
4
.
The bond order of C2 is
€
# Bonding electrons − # Antibonding electrons
2
8− 4
=
2
BondOrder = 2 .
BondOrder =
–
Similarly, the electron configuration of C 2 is
€
2
€
2
2
(1σ g ) (1σ *u ) (2σ g ) (2σ *u )
2
(1π u )
4
1
(3σ g ) .
–
The bond order of C 2 is
€
# Bonding electrons − # Antibonding electrons
2
9− 4
=
= 2.5.
2
BondOrder =
€
2–
Finally, the electron configuration of C 2 is
€
2
€
2
2
(1σ g ) (1σ *u ) (2σ g ) (2σ *u )
2
(1π u )
4
(3σ g )
2
.
–
The bond order of C 2 is
€
# Bonding electrons − # Antibonding electrons
2
10 − 4
=
= 3.
2
BondOrder =
€
We would expect the molecule with the lowest bond order to have the longest bond length. Therefore, we
would predict€that the C2 molecule has the longest bond length.
3
4.
1
1
( ) (σ *u )
Determine all the possible molecular term symbols arising from the σ g
The total spin angular momentum is calculated using s1 =
ranges from s1 − s2 to s1 + s2 in steps of 1. Thus,
€S =
€
1
2
1
2
electron configuration.
and s2 = 12 . The total spin angular momentum S
€
− 12€, …,
1
2
+ 12 ,
€
or S = 0, 1 .
€
The multiplicity 2S+1 is therefore 1 or 3 (singlet or triplet).
€
The total molecular orbital angular momentum is calculated using λ1 = 0 and λ 2 = 0 , since both electrons are
in σ orbitals. The total molecular orbital angular momentum Λ takes the values λ1 − λ 2 and λ1 + λ 2 .
Therefore,
€
€
Λ = 0 − 0 and 0 + 0 , €
€
or Λ=0.
€
The state is therefore a Σ state, and the possible term symbols are 1 Σ and 3 Σ .
Finally, to determine the parity, we note that one electron is in a molecular orbital with g parity, while the other
electron occupies a molecular orbital with u parity. Thus,
parity of the state is g×u=u. Therefore,
€ the overall
€
the complete molecular term symbols are 1 Σ u and 3 Σ u .
€
€
4
5.
1 1
Determine all the possible molecular term symbols arising from the σ π electron configuration.
and s2 = 12 (the result will be the same as that for
problem 1). The total spin angular momentum S ranges from s1 − s2 to s1 + s2 in steps of 1. Thus,
The total spin angular momentum is calculated using s1 =
€S =
1
2
1
2
− 12€, …,
1
2
+ 12 ,
€
€
or S = 0, 1 .
€
The multiplicity 2S+1 is therefore 1 or 3 (singlet or triplet).
€
The total molecular orbital angular momentum is calculated using λ1 = 0 and λ 2 = ±1 , since one electron is in
a σ molecular orbital and the other is in a π molecular orbital. The total molecular orbital angular momentum Λ
takes the values λ1 − λ 2 and λ1 + λ 2 . Therefore,
€
€
Λ = 0 −1 and 0 +1 ,
€
€
or Λ=1.
€
The state is therefore a Π state, and the possible term symbols are 1 Π and 3 Π .
In this particular case, the parities of the individual molecular orbitals are not given, so we cannot determine the
€
€
overall parity of the state.
6.
+
The simplest molecular ion is H2 . Determine the electron configuration and molecular term symbol of
+
the ground state of H2 .
1
( )
+
There is only one electron in H2 ; therefore, the electron configuration is 1σ g . Since there is only a single
electron to consider, we have s1 =
state.
1
2
and therefore S = 12 . This implies a multiplicity of 2S +1 = 2 , or a doublet
€
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 1σ g molecular orbital, the parity of€the molecular
€
+
2
state is the same, gerade. The term symbol for H 2 is therefore Σ g .
€
€
€
€
5
7.
+
For the ground electronic configuration of B2 , determine the bond order and molecular term symbols.
2
2
2
( ) (1σ *u ) (2σ g ) (2σ *u )
+
From the correlation diagram, the electron configuration of B 2 is 1σ g
2
1
(1π u ) .
The bond order is (this is a repeat of problem 1):
€
€
# Bonding electrons − # Antibonding electrons
BondOrder =
2
5− 4
=
2
1
BondOrder = 2 .
For the molecular term symbol, note that the closed shells in the electron configuration do not contribute to the
1
€
term symbol. Therefore, only the (1π u ) 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 2S +1 = 2 , or a doublet state.
1
2
and therefore S = 12 . This implies a
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 u molecular€orbital, the parity of the molecular
state
€ is the same, ungerade.
+
2
The term symbol for B 2 is therefore Π u .
€
€
€
6
8.
1
Diatomic halogen molecules, X2, with X=F, Cl, Br, I, have a ground electronic state with term symbol Σ g.
1
1
3
3
3
The excited electronic states of these molecules have Σ u, Π u, Σ g, Σ u, and Π u term symbols. What are
the allowed electronic transitions from the ground state?
The selection rules for molecular electronic transitions are 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 singlet state, the only allowed
1
€ transitions
€ occur to excited singlet states. Therefore, the only possible allowed transitions occur to the Σ u and
1
Π u excited states.
€
€
Since the ground state is a Σ state, it has Λ = 0 . Therefore, the ΔΛ = 0, ± 1 selection rule would
€ allow
1
1
Λ
Σ
Π
transitions to states with
= 0 or 1 ( or
states). Thus, the Σ u and Π u excited states are still potential
candidates for allowed transitions.
€
€
€
€ state, 1 Σ g€
From the ground
, only€transitions from g →€u are possible
because of the parity selection rule.
€
Therefore, the excited states 1 Σ u and 1 Π u are left as the allowed transitions from the ground state. To
1
1
1
1
summarize, the transitions possible based on selection rules are Σ g → Π u and Σ g → Π u .
€
€
€
€
€
€
7
9.
+
The attached figure shows a potential energy diagram for some of the electronic states of O2 (and O2 ).
3
a.) From the ground state of O2 (X Σ), what is the allowed electronic transition?
The only possible transition that is allowed by the selection rules for the electronic states shown in the
figure is
X 3Σg → B 3Σu .
This transition has ΔΛ = 0 and ΔS = 0 (and is g → u ), which correspond to allowed transitions. The other
3
1
3
1
possible transitions, X Σ g → a€Δ g and X Σ g → b Σ g , both violate the ΔS = 0 selection rule (and
3
1
both violate the g → u or u → g selection rule). The X Σ g → a Δ g transition also violates the
€
ΔΛ = 0, ±€1 selection rule.
€
€
€
€
€
€
€
€
b.) From the figure, estimate the vertical transition energy in eV for the allowed transition.
3
The X Σ g ground state is shown at zero energy in the figure. From the graph, the vertical excitation
3
energy to the B Σ u state can be roughly estimated to be about 9 eV.
€
c.) Will€the molecule in the excited electronic state be vibrationally excited? Explain.
3
From the graph the vertical excitation occurs high up on the B Σ u potential energy curve. Therefore, the
Franck-Condon principle would predict that the excited electronic species would be produced with
significant vibrational excitation.
€
Another way to come to this same conclusion is to note from the graph that the equilibrium bond distance
of ground state O2 is approximately 120 pm, while the excited state equilibrium bond distance is
3
significantly longer at about 170 pm. This implies that upon excitation to the B Σ u state the molecule will
have a bond distance (120 pm) that is way too short, and will begin to vibrate to alleviate that stress.
€