CHEM 120A Problem Set 7 Solutions
David Hoﬀman
,
Doran Bennett
,
Tara Yacovitch
4/2/2010
1. The solutions can be found in the McQuarrie and Simon Solutions manual.
2. The solutions can be found in the McQuarrie and Simon Solutions manual.
3. The solutions can be found in the McQuarrie and Simon Solutions manual.
4. We will use McQuarrie and Simon’s method of counting microstates where the mi
crostates are listed in a table with the possible
M
L
’s in the rows and the possible
M
S
’s
in columns. We will consider the following quantum numbers:
m
l
,l,M
L
,L
;
m
s
,M
S
and
S
(a)
np
5
n
0
p
Here we have two diﬀerent
p
orbitals. For the ﬁrst orbital,
l
= 1 (p orbital) such
that
m
l
= 1
,
0
,

1
We also ﬁnd that the spins for paired electrons cancel such that
m
s
=
±
1
2
For the second orbital,
l
0
= 1 and
m
l
0
= 1
,
0
,

1
m
s
0
=
±
1
2
We add the maximum values for the individual orbital quantum numbers to ﬁnd
M
L
= 2
,
1
,
0
,

1
,
2
M
S
= 1
,
0
,

1
We construct a table and write in the possible microstates,
m
l
m
s
m
l
0
m
s
0
. For
simplicity, we write
α
for
m
s
= +
1
2
and
β
for
m
s
=

1
2
. Note that we include
both 1
α
0
β
and 0
β
1
α
(and similar states) because the orbitals are nonequivalent,
n
6
=
n
0
.
1
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P
P
P
P
P
P
P
P
M
L
M
S
1
0
1
2
1
α
1
α
1
α
1
β
1
β
1
β
1
β
1
α
1
1
α
0
α
1
α
0
β
, 1
β
0
α
1
β
0
β
0
α
1
α
0
α
1
β
, 0
β
1
α
0
β
1
β
0
0
α
0
α
0
α
0
β
, 0
β
0
α
0
β
0
β
1
α
(1)
α
1
α
(1)
β
, 1
β
(1)
α
1
β
(1)
β
(1)
α
1
α
(1)
α
1
β
, (1)
β
1
α
(1)
β
1
β
1
(1)
α
0
α
(1)
α
0
β
, (1)
β
0
α
(1)
β
0
β
0
α
(1)
α
0
α
(1)
β
, 0
β
(1)
α
0
β
(1)
β
2
(1)
α
(1)
α
(1)
α
(1)
β
(1)
β
(1)
β
(1)
β
(1)
α
From this table we see that there are a total of 36 microstates. The given term
symbols
3
P
,
3
D
,
1
P
,
1
D
account for 32 of these states.
Next, we start collecting microstates for each term symbol. Within a given cell
of the table, it doesn’t matter which microstate you cross out.
i. Consider microstates with
L
= 2 and
S
= 1:
M
L
= 2
,
1
,
0
,

1
,

2, a D state.
M
S
= 1
,
0
,

1, a spin multiplicity of 2
S
+ 1 = 3
This is a
3
D
state. Cross out one microstate for each of the 15 possible com
binations of the above
M
L
and
M
S
. You end up crossing out one microstate
in each cell of the table.
ii. Consider microstates with
L
= 2 and
S
= 0:
M
L
= 2
,
1
,
0
,

1
,

2, a D state.
M
S
= 0, a spin multiplicity of 2
S
+ 1 = 1
This is a
1
D
state. Cross out one microstate for each of the 5 possible com
binations of the above
M
L
and
M
S
.
iii. Consider microstates with
L
= 1 and
S
= 1:
M
L
= 1
,
0
,

1, a P state.
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 2S, Spin multiplicity, Simon Solutions

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