Chemistry 181
Professor S. Alex Kandel
Fall 2015
Due 9/2/2015
Problem Set 1
Discussion problems:
1. Use the periodic table on page 159 and find each instance where elements in
the same column do not have analogous ground-state electronic structures. Can
you come up with any general rules as to when this occurs, and any general
ideas as to why? (You can ignore the lanthanides and actinides.)
The fill order for the d orbitals is not so regular, and you can (for example)
have s2 d4 in one row, and s1 d5 below it. We will discuss this in detail in the
next week.
2. Study the plots and diagrams of hydrogen orbitals on pages 141 and 143.
The important thing here is to understand the three-dimensional structure of
each orbital. A useful web site is:
http://www.falstad.com/qmatom/
3. The first through sixth ionization energies of carbon, in attojoules (1 aJ =
10−18 J) are:
Z
Element
I1
I2
I3
I4
I5
I6
6
C
1.81
3.90
7.67
10.3
62.8
78.5
In photoelectron spectroscopy, a single photon detaches an electron from an
atom A:
A + photon −→ A+ + e− (with kinetic energy)
Don’t consider processes where one photon comes in and two or more electrons come out (this is possible, but quite uncommon) and assume none of the
photon’s energy remains in the atom (also mostly true).
The ionization energies given above are useful and relevant, but they
only indirectly give you the information you need to solve the problem. Thinking and (educated) guessing will be necessary.
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(a) Carbon has 1s, 2s, and 2p electrons. Which of these can be detached using
light with wavelength λ = 100 nm? Explain your answer.
First calculate the photon energy:
hc
λ
6.626 × 10−34 J s · 2.998 × 108 m s−1
=
100 × 10−9 m
−18
= 1.99 × 10
J
= 1.99 aJ
E =
The definition of I1 is the minimum energy required to ionize the atom—
this means it is the energy to detach the most weakly bound electron, which
is a 2p electron. The 2s orbital is lower in energy than the 2p orbital; the
question is, how much lower?
• It’s not going to take a full 7.67 aJ to pull off an electron, as that’s
the energy needed to remove a 2s electron after both 2p electrons are
gone. (Definition of I3 .)
• On the other hand, you can look at the difference between I1 and I2 :
in both cases you’re pulling off a 2p electron, but the energy goes up
when you are pulling the electron off of C+ instead of C.
• If the 2p and 2s orbitals were exactly the same energy, you’d expect
I3 to be bigger than I2 , but a more reasonable value might be ∼6 aJ;
i.e., the increase from I2 to I3 would be about as large as the increase
from I1 to I2 .
• You can eyeball the large jump between I2 and I3 , compared to the
smaller jumps between I1 and I2 , and I3 and I4 , and guess that the 2s
orbital is approximately (or at least) 1 aJ more stable than the 2p.
So 100 nm can only remove a 2p electron, as the 2s is too strongly bound.
(b) What is the maximum speed for an electron detached with 100 nm light?
The kinetic energy released is
KE =
hc
− Ebinding = 1.99 aJ − 1.81 aJ = 0.18 aJ
λ
2
Plugging this in with the mass of the electron:
1
me v 2
2
s
2E
v =
me
s
2 · 0.18 × 10−18 J
=
9.11 × 10−31 kg
s
J
3.95 × 1011
=
kg
KE =
=
v
u
u
5t
6.3 × 10
J 1 kg m2 s−2
·
kg
1J
= 6.3 × 105 m s−1
(c) Which electrons (1s, 2s, and/or 2p) can be detached using 28 nm light?
Explain your answer.
You can do another calculation, or go ahead and say that this light has
100/28 times the energy:
100
· 1.99 aJ = 7.09 aJ
28
Based on the previous discussion, this will be enough to detach either a 2s
or 2p electron. It is almost enough to detach a 2s electron even after both
2p electrons are gone.
The 1s orbital energy is going to be less than 62.8 aJ, but it’s not going
to be too much less: the 2s and 2p electrons don’t shield the 1s electrons
from the nucleus much. (Or make a similar argument to that made in part
(a).)
(d) Assume no energy is lost as heat or remains in the atom; that is, all excess
energy goes into kinetic energy. If light with a wavelength of 2 nm is
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used, how many different electron speeds will be observed? (Explain your
answer.)
The energy is 50 times higher than the 1.99 aJ you calculated in part (a),
so almost 100 aJ: plenty of energy to knock out any one of the electrons.
(Again, we’re not interested in the cases where more than one electron is
ejected.) The possible processes are:
C(1s2 2s2 2p2 ) + photon −→ C+ (1s2 2s2 2p1 ) + energy
C(1s2 2s2 2p2 ) + photon −→ C+ (1s2 2s1 2p2 ) + energy
C(1s2 2s2 2p2 ) + photon −→ C+ (1s1 2s2 2p2 ) + energy
(2p is removed)
(2s is removed)
(1s is removed)
There are 3 possible electron speeds: it doesn’t matter which of the two
2p electrons you ionize, as they have the same energy, and so on.
Graded problem:
This should be written up and handed in before class begins on Wednesday the 2nd.
This problem may also be discussed in class.
1. Look at the atomic emission spectra on page 117 of your text. The wavelength
of every bright line in a given spectrum is determined by the energy difference
between an initial (higher energy) state and a final (lower energy) state.
(a) Which initial and final states are responsible for each of the bright lines in
the spectrum of hydrogen shown? Also explain why there are no transitions
with energies that would put them in the dark regions of the hydrogen
spectrum.
Table 4.3 actually lists these out for you; they are all lines that end at a
final state with n = 2 (the Balmer series of hydrogen). The highest-energy
(lowest-wavelength) line at 4100 Å is from n = 6 → n = 2, and the others
are 5 → 2, 4 → 2, and 3 → 2.
The next question is, how do we know that some other random transition,
like n = 17 → n = 4, doesn’t end up smack in the middle of this area
of the spectrum? This is entirely possible with multiple-electron atoms
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(which is why their spectra can be so much more complicated), but not
with hydrogen. In hydrogen, because the levels get so much closer together with increasing n, the first several spectral series are completely
non-overlapping. The Lyman spectrum (ending in n = 1) has wavelengths
given by
1
1 = 1.097 × 107 m−1 1 − 2
λ
n
with
n = 2, 3, 4, 5, . . .
The second term in this equation hits a minimum when n = 2, and a
maximum as n → ∞, so
1
0.75 ≤ 1 − 2
n
<1
For the Balmer series (ending at n = 2), this becomes
1
0.139 ≤ 1 − 2
n
< 0.25
For a final n = 3 (Paschen):
0.049 ≤ 1 −
1
n2
< 0.11
and so on.
(b) Not every combination of initial and final energy levels will result in emission, however: the process is subject to the selection rule:
∆` = ±1
This means that transitions can occur where an electron moves between p
(` = 1) and s (` = 0) orbitals, or between p (` = 1) and d (` = 2) orbitals,
but not between p and p (∆` = 0) or s and d (∆` = 2).
Why do we not need to consider selection rules to understand the hydrogen
spectrum?
To clarify: the selection rule is 100% applicable for hydrogen; however,
you don’t need it to understand the spectrum. This is because only n
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determines the energy of an orbital, so all of the following transitions would
have exactly the same energy:
3p
3s
3s
3p
−→
−→
−→
−→
2s
2p
2s
2p
3
3
7
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The fact that the selection rule prohibits the final two from occurring does
not change the spectrum at all—or, you still predict the right spectrum
even if you don’t know the selection rule. Since all of the above transitions
would be at different energies in a multi-electron atom, the selection rule
is needed to explain why the final two don’t show up in an experimental
spectrum.
(c) Come up with a plausible explanation for the energy levels involved for
each of the bright lines in the spectrum for lithium and for sodium. Again,
also explain why regions of the spectrum that are dark have no transitions.
The cartoon diagram on page 150 may be helpful in guiding your intuition.
This is the actual answer for lithium:
2p → 2s
3d → 2p
4s → 2p
4d → 2p
3p → 3s (closer than 2p and 2s)
3s → 2p (closer than 2p and 2s)
too high energy 3p → 2s
6710 Å
6100 Å
4970 Å
4600 Å
too low energy
though other answers are certainly plausible: call the 6710 line 3s → 2p,
and shift the other assignments up, for example. Note that for excited
lithium, the 3d orbital is lower in energy than the 4s–why?
For sodium, the bright, thick line at ∼5900 Å (the only line you see in
the diagram from your textbook) is from 3p → 3s. (It’s the spin of the
electron that makes there be two lines there, which is what makes for the
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thickness.) 4s → 3p and 3d → 3p are too low in energy to show on this
scale, and 4p → 3s is too high. You can see (many) other lines in the
spectrum included on the problem set. The next brightest one, just left of
the monster, is 4d → 3p. 5s → 3p gives one of the brighter red lines, and
the dim blue-greenish lines are 6s → 3p and 5d → 3p. Again, there are
lots of other plausible explanations, and from a grading perspective, any
of them (any that are plausible) would be fine here.
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