Chemistry 362
Fall 2015
Dr. Jean M. Standard
Solutions: Sample Questions for Non-Cumulative Portion of Final Exam
1.)
Consider an atom with excited electron configuration 3p13d1. Determine the possible
atomic term symbols for this electron configuration.
1
1
The spin angular momentum is calculated using s1 = 2 and s2 = 2 . The total spin angular
momentum S ranges from s1 − s2 to s1 + s2 in steps of 1. Thus,
S = 12 − 12 ,€…, 12 + 12 ,€ or S = 0, 1 .
€
€
The multiplicity 2S+1 is therefore 1 or 3 (singlet or triplet).
€
€
The orbital angular momentum is calculated using ℓ 1 = 1 and ℓ 2 = 2 . The total orbital
angular momentum L ranges from ℓ 1 − ℓ 2 to ℓ 1 + ℓ 2 in steps of 1. Therefore,
L = 1€− 2, …, 1+
€ 2,
€
or L = 1, 2, 3 .
€
€
1
1
The state possible therefore include P, D, and F, and the possible terms are P , D ,
1
F , 3P , 3D , and 3F .
€
€ €
€ be€determined. The
To complete the term symbols, the total angular momentum J must
€total angular
€
momentum J ranges from L − S to L + S .
For
€
€
€
1
P : L = 1 and S = 0 leads to J = 1 and the term symbol 1P1 .
€
€
1
For D : L = 2 and S = 0 leads to J = 2 and the term symbol 1D2 .
€
€
€
€
1
For F : L = 3 and S = 0 leads to J = 3 and the term symbol 1F3 .
€
€
€
€
For
€
3
P : L = 1 and S = 1 leads to J = 0, 1, 2 and the term symbols 3 P0 , 3P1, 3P2 .
€
€
€
3
€
D : L = 2 and S = 1 leads to J = 1, 2, 3 and the term symbols 3 D1, 3D2 , 3D3 .
€
€
€
3
For F : L = 3 and S = 1 leads to J = 2, 3, 4 and the term symbols 3F2 , 3F3, 3F4 .
€
€
€
€
€
€
€
For
€
€
€
€
2
1. continued
The possible term symbols are therefore:
1
P1, 1D2 , 1F3, 3P0 , 3P1, 3P2 , 3D1, 3D2 , 3D3, 3F2 , 3F3, and 3F4 .
€
2.)
Provide a statement of the Pauli Principle. Explain how the Pauli Principle impacts
the filling of atomic and molecular orbitals and the symmetry of atomic and
molecular wavefunctions.
One statement of the Pauli Principle is that no two electrons in a many-electron atom or
molecule may have the same set of quantum numbers.
As a result of the Pauli Principle, the filling of atomic and molecular orbitals is affected in
that only two electrons may be placed in any one orbital. If an attempt is made to place a
third electron in an orbital, it would result in that electron possessing the same set of
quantum numbers as one of the others.
Another statement of the Pauli Principle is that the wavefunction of a many-electron system
must be antisymmetric with respect to interchange of any two electrons. This statement
places a strict criterion on the form of the wavefunction for atoms and molecules.
3
3
a.)
Determine the molecular electronic configuration of the ground state of the O2+
molecular cation.
Using the correlation diagram, the electronic configuration of O2+, with 15 electrons,
is
2
2
2
2
2
1
(1σg ) (1σ*u ) (2σg ) (2σ*u ) (3σg ) (1π u )4 (1π g* ) .
€
b.)
Determine the bond order of the ground state of the O2+ molecular cation.
The bond order is
# Bonding electrons − # Antibonding electrons
2
10 − 5
=
2
BondOrder = 2.5.
BondOrder =
c.)
€
Draw a sketch of the highest occupied molecular orbital of the O2+ molecular
cation. Does this molecular orbital have bonding or antibonding character?
The 1πg* MO is formed from a linear combination of the 2p atomic orbitals, as shown
below:
It is an ANTIBONDING MO.
4.)
The following terms arise from a d8 electron configuration: 1S, 3P, 1D, 3F, 1G. Use
Hund's rules to properly order these terms according to increasing energy and sketch
an appropriate energy diagram including J values.
3
Hund’s Rule #1 tells us that both the triplet terms ( P and 3 F ) will be lower in energy
than the singlet terms ( 1 S , 1 D , and 1 G ), since the terms with higher multiplicity are lower
in energy.
€
€
€ among the triplets, the one with the highest L value will be
Hund’s Rule €
#2 €
tells us that
3
3
lowest energy ( F < P ), and similarly among the singlets, the one with the highest L will
1
1
1
have the lowest energy ( G < D < S ).
Thus,€the ordering of the terms according to increasing energy is:
€
3
F < 3P < 1G < 1D < 1S .
To complete the term symbols and energy ordering, the total angular momentum J must be
determined. The total angular
momentum J ranges from L − S to L + S .
€
1
S : L = 0 and S = 0 leads to J = 0 and the term symbol 1S0 .
€
€
1
For D : L = 2 and S = 0 leads to J = 2 and the term symbol 1D2 .
€
€
€
€
For
€
1
For G : L = 4 and S = 0 leads to J = 4 and the term symbol 1G4 .
€
€
€
€
€
3
For
€
€
For
€
3
P : L = 1 and S = 1 leads to J = 0, 1, 2 and the term symbols 3 P0 , 3P1, 3P2 .
€
€
€
€
F : L = 3 and S = 1 leads to J = 2, 3, 4 and the term symbols 3 F2 , 3F3 , 3F4 .
€
€
€
Since the shell (d8) is greater than half filled, for term symbols with the same L and S but
€ J, the one
€ with higher€J value is lower in energy. Thus,
€different
€ for the term symbols
3
3
3 , the lowest in energy is 3 . Similarly, for the term symbols 3
P0 , P1, P2
P2
F2 , 3F3 , 3F4 ,
the lowest in energy is 3 F4 .
€
Putting this all together, the€ordering of the term symbols according
€ to increasing energy is:
€
€
3
F4 < 3F3 < 3F2 <
3
P2 < 3P1 < 3P0 <
1
G4 <
1
D2 <
1
S0 .
4
4.)
5
continued
Finally, an approximate energy diagram is sketched below.
Energies of Atomic Terms for d8 Configuration
1
S0
1
D2
1
G4
E
3
P0
3
P1
3
P2
3
F2
3
F3
3
F4
5
a.)
6
Explain what is meant by LCAO-MO.
The term LCAO-MO refers to a method used to construct an approximate wavefunction
for a molecular orbital. It means that a linear combination of atomic orbitals is used to
form the molecular orbital. It is described by the equation
K
ψ MO = ∑ ci χ i,AO ,
i=1
where ψ MO is the molecular orbital, χ i,AO corresponds to the set of atomic orbitals, and
ci corresponds to the set of linear coefficients.
b.) Show, using sketches, how two p-type atomic orbitals may be combined to form σ
bonding and σ* antibonding orbitals.
Two p-type orbitals, such as the py orbitals shown below, can be put together in +/–
combinations to form σ and σ* molecular orbitals.
AOs
MOs
The top sketch shows how the two p-type AOs may be subtracted to form a σ bonding
MO. In this case, the two lobes in the center have the same sign so positive
reinforcement occurs, and the electron density is enhanced between the two centers.
The bottom sketch shows how the two p-type AOs add to form a σ* antibonding MO.
In this case, the two lobes in the center have the opposite sign so cancellation occurs,
the electron density is depleted between the two centers and a node is formed.
6.)
Emission lines from the diatomic carbon molecule, C2, are often observed in spectra
of interstellar gas clouds. These emission lines are known as the Swan bands.
a.)
What is the molecular electronic configuration of the ground state of the C2
molecule? Determine the molecular term symbol of the ground state.
Using the correlation diagram, the electron configuration of C2 is
2
2
2
2
(1σ g ) (1σ *u ) (2σ g ) (2σ *u ) (1π u )4 .
Since all the electrons are in filled shells, we have S = 0 and Λ = 0 , and the term
1
Σg .
symbol is automatically
€
€
b.)
€
€
The electron configuration of two important excited states of C2 have electron
1
( )
configurations (1π u ) 3σ g
1
1
( )
1
*
and (1π u ) 1π g . Determine the molecular term
symbols of the two excited state configurations.
Both of
€ the excited states€have two electrons in unfilled shells. Thus, for both states,
the total spin angular momentum is calculated using s1 = 12 and s2 = 12 . The total spin
angular momentum S ranges from s1 − s2 to s1 + s2 in steps of 1. Thus,
S=
€
1
2
− 12 ,€…,
1
2
+ 12 ,€
€
or S = 0, 1.
€
The possible multiplicities for both states, 2S+1, are therefore 1 or 3 (singlet or
triplet).
€
7
6
b.)
8
continued
1
( )
1
For electron configuration (1π u ) 3σ g :
The total molecular orbital angular momentum is calculated using λ1 = ±1 and
λ2 = 0 , since one€electron is in a π molecular orbital and the other is in a σ molecular
orbital. The total molecular orbital angular momentum Λ takes the values λ1 − λ2
and λ1 + λ2 . Therefore,
€
€
Λ = 1 − 0 and 1+ 0 ,
€
€
or Λ=1.
€
The state is therefore a Π state, and the possible term symbols are 1 Π and 3 Π .
Finally, to determine the parity, we note that one electron is in a molecular orbital
with u parity, while the other electron occupies a molecular
€ orbital
€ with g parity.
Thus, the overall parity of the state is u×g=u. Therefore, the complete molecular
term symbols are 1 Πu and 3 Πu .
1
1
€ configuration
€
(1π u ) 1π *g :
For electron
( )
The total molecular orbital angular momentum is calculated using λ1 = ±1 and
λ2 = ±1, since both
€ electrons are in π molecular orbitals. The total molecular orbital
angular momentum Λ takes the values λ1 − λ2 and λ1 + λ2 . Therefore,
Λ = 1 −1 and 1+1 ,
€
€
€
€
or Λ = 0, 2.
€
The state are therefore Σ and Δ states, and the possible term symbols are 1 Σ , 3 Σ , 1Δ ,
and 3Δ .
€
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, the overall parity of the state is g×u=u.
Therefore, the complete molecular term symbols are 1 Σ u , 3 Σ u , 1Δ u , and 3Δ u .
€
€
€
€
6.)
9
continued
c.)
Determine the allowed spectroscopic transitions from the ground state to the
excited states.
1
The ground state of C2 has term symbol Σ g . 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 singlet
€
state, the only allowed transitions occur €
to excited€singlet states.
€
Since€the ground state is a Σ state, it has Λ = 0 . Therefore, the ΔΛ = 0, ± 1 selection
rule would allow transitions to states with Λ = 0 or 1 ( Σ or Π states).
€only transitions
€ from 1 Σ → 1Σ or 1€Σ → 1Π would be allowed.
Thus, in this case,
€
€ €
1
€ transitions€from g → u are possible because of the
From the ground state, Σ g , only
parity selection rule. Therefore, transitions to excited states with term symbols 1 Σ u
and 1 Πu are the allowed transitions from the ground state.
€
€
€
1
1
To summarize, the transitions possible based on selection rules are Σ€
g → Σ u and
1
Σ g → 1Πu .
€
€