Midterm Exam I
CHEM 181: Introduction to Chemical Principles
September 24, 2015
Key
1. A Li2+ ion in an unknown, highly excited electronic state, first emits a photon
at a wavelength of 4.316 µm, and following that, emits a second photon at a
wavelength of 10.25 nm. What is the initial state and final state of this Li2+
ion?
You can’t use the Rydberg equation from your equation sheet as-is:
1
1
1
= −R
−
λ
n21 n22
The +3 nuclear charge of the lithium nucleus also has to be included, so:
1
1
1
2
= −R · Z
−
λ
n21 n22
2 1
1
7
−1
= 1.09737 × 10 m
3
−
n21 n22
1
1
7
−1
−
= 9.876 × 10 m
n21 n22
1
1.012 × 10−8 m
1
−
=
n21 n22
λ
For labeling, say we start at na , go to nb , and end at nc . It’s easier to solve the
nb → nc step first:
1
1
1.012 × 10−8 m
−
=
n2c
n2b
10.25 × 10−9 m
= 0.9878
nc (the final state) must be 1 for this to work at all, so
1−
1
n2b
1
n2b
n2b
nb
1
= 0.9878
= 0.0122
= 82.14
≈ 9
It’s true that 82 6= 81, but this is from round-off errors (the wavelength of the
photon to more significant figures is 10.25178 nm.) Now that we’ve got nb , we
can solve for na
1
1
1.012 × 10−8 m
−
=
n2b
n2a
4.316 × 10−5 m
= 2.35 × 10−3
1
1
− 2 = 2.35 × 10−3
81 na
1
= 1.00 × 10−2
n2a
na = 10
The initial state is n = 10, and we end up at n = 1.
2. Write down the ground-state electron configurations for the following atoms or
ions. You can either include all electrons or use core/valence notation; e.g.,
sodium could be written as either
Na: 1s2 2s2 2p6 3s1
or
Na: [Ne] 3s1
In cases where it is difficult to predict the exact electron configuration, you can
write two configurations along with a brief (one sentence) explanation. (If there
is only one reasonable configuration, do not include extras—you will lose points
for doing so.)
(a) Cl:
[Ne]3s2 3p5
(b) Ni+ :
[Ar]4s1 3d8
or
[Ar]4s0 3d9
It’s ok to miss the second of these but not the first (though my bet is that
the second is what actually happens).
(c) V− :
[Ar]4s2 3d4
or
[Ar]4s1 3d5
I think the first is much better, because you know that if you ionize V− ,
it’s the 3d electron that leaves. We’ve got the opposite of the behavior we
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see for positive ions—the effective nuclear charge is low for the number of
electrons—and it seems unlikely that V− is isoelectronic with Cr.
(d) Os2+ :
[Xe]6s0 5d6
or
[Xe]6s1 5d5
(FYI: Ni+ and V− are atypical states found in relatively few compounds, though
both can be produced in the laboratory.)
3. N2 and N+
2 both have multiple stable excited states with electron configurations that are different from the ground-state configuration. Use the following
information about energies and bond lengths to figure out these electron configurations. (All answers that are logical and self-consistent will be marked as
correct. You can use the MO diagram for N2 on the next page.)
Starting points:
• σ2pz is the highest-energy orbital with electrons in it, so it must be a σ2pz
electron that leaves to make N+
2 , so its ground state must be (224100).
• The 0.18 aJ state in N+
2 is significantly lower in energy than anything that
N2 does, so it must correspond to a configuration with no close analogue
in N2 : (223200).
• This also tells us that quantitatively, the π bonding orbitals are ∼0.18 aJ
higher than the σ2pz . (This is a good approximation, though not exact, as
orbital energies will be slightly different between N2 and N+
2.
• The 1.77 aJ state in N2 keeps bond length the same, so we’re looking for
a configuration with the same bond order. This has to happen by moving
∗
a σ2s
anti-bonding electron into another anti-bonding orbital: (214210) or
(214201).
• The 1.00 and 1.18 states in N2 have reduced bond order (longer bonds),
so will have one fewer bonding and one more anti-bonding electron. If we
say the lower-energy of these has one fewer σ2pz electron and one more
π ∗ electron, then the higher-energy one would have the same π ∗ electron,
but will have lost a π bonding electron instead—this fits with the 0.18 aJ
spacing we’d think these two levels would have.
There are other self-consistent answers that will be marked as correct. The
actual, measured electron configurations are shown in the table.
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N2
σ2s
∗
σ2s
π2px , π2py
σ2pz
∗
∗
, π2p
π2p
y
x
∗
σ2p
z
1.10
2
2
4
2
0
0
1.00
1.29
2
2
4
1
1
0
1.18
1.21
2
2
3
2
1
0
1.77
1.15
2
1
4
2
1
0
energy (aJ)
bond length (nm)
0.0
N+
2
σ2s
∗
σ2s
π2px , π2py
σ2pz
∗
∗
, π2p
π2p
y
x
∗
σ2p
z
1.12
2
2
4
1
0
0
0.18
1.17
2
2
3
2
0
0
1.04
1.47
2
2
4
0
1
0
1.28
1.26
2
2
3
1
1
0
energy (aJ)
bond length (nm)
0.0
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4. The following diagram shows the shapes and energies of the molecular orbitals
for BeH2 , which can be described as combinations of the Be 2s and 2p orbitals
with the H atoms’ 1s orbitals:
H
Be
H
a
a
nb
Be 2p
Energy
Be 2s
H 1s
b
b
Do the following:
(a) Label each orbital (b for bonding, a for antibonding, and nb for nonbonding).
The two molecular orbitals with the same energy look exactly like beryllium
2p orbitals, and they must have the same energy as these orbitals (which
allows us to locate the energy of Be 2p as well.) They are non-bonding—
the electrons do not care whether the Be is part of a BeH2 molecular or
not.
(b) Place the appropriate number of electrons into the molecular orbitals (draw
them on top of the horizontal lines used for energy levels.)
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The molecule has six electrons total, but two of these are core 1s electrons
on the beryllium, and not involved in molecular orbitals at all. This diagram needs to be filled with the four valence electrons (2 from Be, 1 from
each H).
(c) On the same energy scale as the molecular orbitals, indicate (use horizontal
lines) the energies of
• the 2s and 2p orbitals for atomic beryllium and
• the 1s orbital for atomic hydrogen.
Explain below how you determined these energies:
The Be 2p must be at the same energy as the non-bonding molecular
orbitals. The 2s must be lower in energy than the 2p, but must be higher in
energy than the lowest-energy bonding orbital, which involves constructive
interference between (and thus a drop in energy from) Be 2s and H 1s.
The 1s orbitals of the H atoms must be lower in energy than the first antibonding orbital (destructive between 1s and 2s) and higher than the second
bonding orbital (constructive between 1s and 2p.) However, because H is
more electronegative than Be, we expect the valence electron of H to be
lower in energy than the valence electrons of Be—that is, H 1s should be
lower in energy than Be 2s.
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