Chemistry 362
Dr. Jean M. Standard
Problem Set 11 Solutions
1.
Use the Pauli and Aufbau principles, along with Hund's Rule of maximum multiplicity, to draw orbital
energy diagrams for the ground states of the Si and P atoms. Classify the multiplicity of the ground
states of the Si and P atoms as singlet, doublet, triplet, or quartet.
2
2
6 2
2
Using the Pauli and Aufbau Principles, the electron configuration of Si is 1s 2s 2 p 3s 3p .
2
2
6 2
3
The electron configuration of P is 1s 2s 2 p 3s 3p .
€
Orbital energy diagrams for Si and P are shown below. Note that using Hund's rule of maximum multiplicity,
the electrons in the 3p orbitals
are placed in different orbitals and have the same electron spin.
€
Atomic Energy Levels of Si
Atomic Energy Levels of P
3p
3p
3s
3s
2p
E
2p
E
2s
2s
1s
1s
For Si, placing two electrons in the 3p set of orbitals leads to two unpaired electrons. Thus, the total spin of
the two unpaired electrons is S=1, and therefore the multiplicity is 2S+1 = 2·1+1 = 3. Therefore, the
multiplicity of the ground state of Si is a triplet.
For P, placing three electrons in the 3p set of orbitals leads to three unpaired electrons. Thus, the total spin
of the three unpaired electrons is S=3/2, and therefore the multiplicity is 2S+1 = 2·(3/2)+1 = 4. Therefore,
the multiplicity of the ground state of P is a quartet.
2
2.
A researcher employs the following function to represent the spatial part of the ground state
wavefunction of the Li atom,
ψ (1,2,3) = χ1s (1) χ1s ( 2) χ 2s ( 3) .
Here, χ1s or χ 2s is shorthand notation used to refer to the spatial form of the atomic orbital; for
example,
€
€
$ 1 '1/ 2
χ1s (1) = & 3 )
e − r1 / a o .
% πa o (
€
Does the form given above for the Li atom correspond to an acceptable spatial wavefunction? Explain.
€
2 1
No, it is not an acceptable function. The electron configuration of the ground state of Li is 1s 2s , so the atomic
orbitals used to construct the function are correct. However, for example, the function specifically places
electron 3 in the 2s orbital; thus, the function distinguishes between the electrons.
An acceptable function would allow for electron1, 2, or 3 being placed in the 2s orbital. Shown below is an
example of an acceptable spatial wavefunction that does not distinguish between the electrons:
ψ (1, 2, 3) =
1
[ χ1s (2) χ1s (3) χ 2s (1) + χ1s (1) χ1s (3) χ 2s (2) + χ1s (1) χ1s (2) χ 2s (3)] .
3
This function places each of the three electrons in the 2s orbital, so it does not distinguish. This example spatial
wavefunction
happens to be symmetric with respect to interchange of electrons. When constructing the full
€
wavefunction for Li, including spatial and spin components, this symmetric spatial wavefunction would have to
be combined with an antisymmetric spin wavefunction so that the overall wavefunction was antisymmetric and
satisfied the Pauli Principle.
3.
Determine the possible atomic term symbols arising from a d1 electron configuration.
1
Since the d electron configuration only involves one outer electron, the total orbital angular momentum
quantum number L and total spin angular momentum quantum number S are identical to the orbital and spin
angular momentum quantum numbers of the outer electron. Thus,
S = s1 =
1
2
and L = ℓ1 = 2.
2
The multiplicity 2S+1 is therefore 2 (a doublet) and the term corresponds to a D term. Then, all that is needed
is to determine the total angular momentum
quantum
€ number J. The total angular momentum quantum number
€
2
L
−
S
L
+
S
D
J ranges from
to
. For the
term, L=2 and S = 12 leads to J = 32 , 52 .
€
1
Therefore, the possible term symbols for the d electron configuration are 2 D3/ 2 and 2 D5 / 2 .
€
€
€
€
€
€
€
3
4.
8
1
1
1
3
3
The following terms arise from a d electron configuration: S, D, G, P, F. Use the first two of Hund's
rules to properly order these terms according to energy. Then, determine the term symbols arising from
each term, and use the third Hund's rule to complete the energy ordering.
Hund's Rules #1 and 2 allow us to order the terms according to increasing energy:
3
€ €
F < 3P < 1G < 1D < 1S .
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
1
1
S , D , and G ). Hund's Rule #2€tells us that among the triplets, the one with the highest L value will be
3
3
lowest energy ( F < P ), and similarly among the singlets, the one with the highest L will have the lowest
€
1
1
1
€
energy ( G < D < S ).
€
Next,€the complete term symbols for each term may be determined:
€
3
3
3
For 3 F : L = 3 and S = 1 leads to J = 2, 3, 4 and term symbols F2 , F3 , F4 .
€
For
3
€
P : L = 1 and S = 1 leads to J = 0, 1, 2 and term symbols 3 P0 , 3P1, 3P2 .
€
€
€
1
For 1G : L = 4 and S = 0 leads to J = 4 and the term symbol G 4 .
€
€
€
€
For
1
D:
€
€
L = 2 and S = 0 leads to J = 2 and the term symbol 1D2 .
€
€
€
€
1
For S : L = 0 and S = 0 leads to J = 0 and the term symbol S0 .
€
€
€
€
1
€
8
Finally, Hund's Rule #3 may be used to order the term symbols with each term. Since the d electron
€
configuration
corresponds
to a shell
the term symbol with the highest J value will
€
€
€ that is greater than 1/2 filled,
€
3
3
3
3
3
be lowest in energy. Therefore, for the F term, the energy ordering is F4 < F3 < F2 , and for the P term,
3
3
3
the energy ordering is P2 < P1 < P0 .
€
To summarize, the overall energy ordering is therefore:
€
€
3
F4 < 3F3 < 3F2
<
3
P2 < 3P1 < 3P0
€
€
<
1
G4
<
1
D2
<
1
S0 .
4
5.
1
1
Determine the possible atomic term symbols arising from a 2s 2p electron configuration. Give the
degeneracy of each term symbol. Use Hund's rules to order the term symbols with respect to increasing
energy. Show an energy diagram that includes splittings due to spin-orbit coupling.
The spin angular momentum is calculated using s1 =
from s1 − s2 to s1 + s2 in steps of 1. Thus,
1
2
1
and s2 = 2 . The total spin angular momentum S ranges
1
S= €
− 12 , …,
2
€
1
2
+ 12 ,
€
€
or S = 0, 1 .
€
The multiplicity 2S+1 is therefore 1 or 3 (singlet or triplet).
€
The orbital angular momentum is calculated using ℓ1 = 0 and ℓ 2 = 1 . The total orbital angular momentum L
ranges from ℓ1 − ℓ 2 to ℓ1 + ℓ 2 in steps of 1. Therefore,
€
€
L = 0€
−1, …, 0 +1,
€
or L=1.
€
1
3
The state is therefore a P state, and the possible terms are P and P .
To complete the term symbols, the total angular momentum J must be determined. The total angular
€
€
momentum J ranges from L − S to L + S .
1
For P : L = 1 and S = 0 leads to J = 1 .
€
€
3
P : L = 1 and S = 1 leads to J = 0, 1, 2 .
€ €
€
€
The possible term symbols (and their degeneracies) are therefore
€ €
€
€
1
3
3
Term Symbol
P1
P0
P1
For
Degeneracy (2J+1)
€
3
1
€
3
3
€
P2
5
€
5
5.
continued
1
3
3
1 1
The terms for the 2s 2p electron configuration are P and P . Using Hund's rule 1, the P term is lower in
3
3
3
energy because it has a higher multiplicity. Spin-orbit coupling will split the P0 , P1 , and P2 term symbols
by a small amount. A sketch of the energy level diagram is shown below.
€
€
€
1
P
€
€
€
1
P1
E
3
P
3
Po , 3P1 , 3P2
2s12 p1
elec-elec
repulsions
spin-orbit
coupling
6
6.
1
1
Determine the possible atomic term symbols arising from a 3s 3d electron configuration. Give the
degeneracy of each term symbol.
The spin angular momentum is calculated using s1 =
from s1 − s2 to s1 + s2 in steps of 1. Thus,
1
2
1
and s2 = 2 . The total spin angular momentum S ranges
1
S= €
− 12 , …,
2
€
1
2
+ 12 ,
€
€
or S = 0, 1 .
€
The multiplicity 2S+1 is therefore 1 or 3 (singlet or triplet).
€
The orbital angular momentum is calculated using ℓ1 = 0 and ℓ 2 = 2 . The total orbital angular momentum L
ranges from ℓ1 − ℓ 2 to ℓ1 + ℓ 2 in steps of 1. Therefore,
€
€
L = 0−
€2 , …, 0 + 2,
€
or L = 2 .
€
1
3
The state is therefore a D state, and the possible terms are D and D .
€
To complete the term symbols, the total angular momentum J must be determined. The total angular
€
€
momentum J ranges from L − S to L + S .
1
For D : L = 2 and S = 0 leads to J = 2 .
€
€
3
D : L = 2 and S = 1 leads to J = 1, 2, 3 .
€ €
€
€
The possible term symbols (and their degeneracies) are therefore
€
€
€
€
1
3
3
Term Symbol
D2
D1
D2
For
Degeneracy (2J+1)
€
5
3
€
3
5
€
D3
7
€
7
7.
1
1
Determine the possible atomic term symbols arising from a 3p 3d electron configuration. Give the
degeneracy of each term symbol.
The spin angular momentum is again calculated using s1 =
ranges from s1 − s2 to s1 + s2 in steps of 1. Thus,
1
2
S=
€
€
1
2
− 12 ,€…,
1
and s2 = 2 . The total spin angular momentum S
1
2
+ 12 ,
€
or S = 0, 1 .
€
The multiplicity 2S+1 is therefore 1 or 3 (singlet or triplet).
€
The orbital angular momentum is calculated using ℓ1 = 1 and ℓ 2 = 2 . The total orbital angular momentum L
ranges from ℓ1 − ℓ 2 to ℓ1 + ℓ 2 in steps of 1. Therefore,
€
€
€
L = 1−
€ 2 , …, 1+ 2,
or L = 1, 2, 3.
€
1
3
1
3
1
3
The states possible are therefore P, D, and F states, and the possible terms are P , P , D , D , F , and F .
€
To complete the term symbols, the total angular momentum J must be determined. The total angular
€ € € € €
€
momentum J ranges from L − S to L + S .
1
For P : L = 1 and S = 0 leads to J = 1 .
€
€
For
€
€
For
€
€
For
€
€
For
€
€
For
€
€
€
€
3
P : L = 1 and S = 1 leads to J = 0, 1, 2 .
€
€
1
D : L = 2 and S = 0 leads to J = 2 .
€
€
3
D : L = 2 and S = 1 leads to J = 1, 2, 3 .
€
€
1
F : L = 3 and S = 0 leads to J = 3 .
€
€
3
F : L = 3 and S = 1 leads to J = 2, 3, 4 .
€
€
€
€
8
7.
continued
The possible term symbols (and their degeneracies) are therefore
Term Symbol
P1
Degeneracy (2J+1)
3
P0
1
1
3
3
P1
3
P2
5
D2
5
€
3
€
1
€
3
D1
3
€
3
D2
5
€
3
D3
7
€
1
€
3
F2
5
€
3
F3
7
€
3
F4
9
F3
7
€
€
€
8.
Classify the following atomic transitions as allowed or not. If a transition is not allowed, indicate which
selection rule is violated.
a.) 3P2 →
1
D2
Not allowed. This transition corresponds to ΔS= –1, which violates the ΔS=0 selection rule. Otherwise, it
corresponds to ΔL=1 and ΔJ=0, which are both OK.
b.) 2P3/2 →
2
S1/2
Allowed. This transition has ΔS=0, ΔL= –1, and ΔJ= –1, which are all allowed by the selection rules.
c.)
3
D2 →
3
F2
Allowed. This transition has ΔS=0, ΔL=1, and ΔJ=0, which are all allowed by the selection rules.