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Exam 2 Solutions
Chem 6, 9 Section, Spring 2002
1. Dartmouth’s FM radio station, WDCR, broadcasts by emitting from its antenna
photons of frequency 99.3 MHz (99.3 × 106 Hz).
(a) What is the energy of a single WDCR photon?
The photon energy is simply Planck’s constant times the photon frequency:
Ephot = hν = (6.62 × 10–34 J) (99.3 × 106 Hz) = 6.58 × 10–26 J
(b) Suppose one of these photons strikes a hydrogen atom in a particular quantum
state with a principal quantum number n. What is the minimum value for n that would
allow this photon to ionize the atom?
The energy of the photon must equal or exceed the energy difference between the
energy of the state with the n value we seek and the energy of the ionized atom (which is at
our chosen zero of energy, or, equivalently, at the state with n = ∞):
Ephot = 6.58 × 10–26 J = E n = ∞ – En = 0 – –
2.18 × 10–18 J
n2
or
n=
2.18 × 10–18 J
6.58 × 10–26 J
= 5756
(c) You now know n, but not l or m for this atom. Calculate the maximum average
radius this atom could have, and in doing so, tell me what l and m values lead to this
maximum value. (If you didn’t get a value for n in part (b), assume n = 100, which isn’t
the answer to part (b).)
If n is this huge number, then there are many, many possible l and m values for atoms
with the same energy. But if we look at the expression for the average radius, we can see
that as l increases from its minimum value of 0, the average radius decreases, which tells us
that l = 0 has the largest radius, and if l = 0, then m = 0 as well:
5.29 × 10–11 m
a
a
rnl = 0 3n 2 – l(l +1) = 0 3(5756)2 – 0 (0 +1) =
3(5756)2 = 2.63 × 10–3 m
2
2
2
This is one HUGE atom, with a 2.63 mm radius!
2. (a) Given the H atom orbital picture below, specify those quantum numbers that are
determined by the picture, and give possible values for the quantum numbers not
specifically determined.
First, note that three views are needed to see the entire symmetry of the wavefunction.
That tells us that it is NOT cylindrically symmetric about the z axis, meaning m ≠ 0. Next,
note that there are three planar nodes: the xy plane, the xz plane, and the yz plane. This
tells us that l = 3 (it is an f orbital). There are no radial nodes, so n = l + 1 = 4. This is one
of the seven 4f orbitals. Finally, m can be anything allowed for l = 3 except zero:
n=4
l=3
m = +3, +2, +1, –1, –2, –3
(b) One of the H atom s orbital radial probability distribution functions is
Exam 2 Solutions
Chem 6, 9 Section, Spring 2002
rpd ∝ r 2 2 – ar 2 e –r/a0
For this orbital, n = 2.
0
This is an s orbital, and thus there are only radial nodes. From the mathematical form
of the rpd, we see that this function is zero at r = 0 (which doesn’t count as a node), at r =
∞ (which also doesn’t count), and when the factor in parentheses, (2 – r/a0), equals zero.
There is only one value of r that meets this criterion, r = 2a0, and thus there is only one
radial node. Now we know n = 2, and this is the rpd for the 2s orbital.
(c) The ratio of the wavelength of the n = 2 to n = 1 emission photon in H to the
wavelength of the n = 4 to n = 2 emission photon in the one-electron ion He+ is 1.
In general, the wavelength λ of a one-electron atom emission is given by
hc = (2.18 × 10–18 J) Z2 – Z 2
λ
nf2 ni2
If we write this equation once for the H atom case and once for the He+ case, take the ratio
of these two equations, we find that the ratio is 1:
22 – 2 2
λ(H) 22 42
=
=1
λ(He) 12 – 12
12 22
3. (a) If I assign ms values to all the electrons in the ground state of the N atom, I find
that 2 of them have one value while 5 of them have the other value.
N has 7 electrons: 2 are paired in the 1s orbital, 2 are paired in the 2s orbital, and the
other three are unpaired in separate 2p orbitals.
(b) The electrons lost by Sn when it forms the Sn2+ cation come from the subshell(s)
5p.
Sn has a Kr core of electrons plus the 5s2 4d10 5p2 valence configuration. On
ionization, the two 5p electrons are lost.
(c) A Rb atom is in an excited state with the electron configuration [Kr] 9d1. The
effective nuclear charge seen by the 9d electron is closest to (circle one)
3e
2e
1e
0
–1e
–2e
–3e
The 9d orbital holds the electron far from the inner core of electrons. These core
electrons shield all but one proton very well, making the effective nuclear charge close to
the charge on a proton, which is just e, the elementary charge. (Electrons have the charge
–e.)
Exam 2 Solutions
Chem 6, 9 Section, Spring 2002
(d) The first four ionization energies of a particular element are IE1 = 9.6 × 10–19 J,
IE2 = 30.2 × 10–19 J, IE3 = 45.6 × 10–19 J, and IE4 = 192.2 × 10–19 J. The element is
F
Ar
Al
C
Na
Li
Ca
The ionization energies are, in turn, one small amount, two larger but close to the
same, and one last one quite large. This trend reveals the electron configuration shell
sequence: the first electron to go is the Al 3p electron, then come two 3s electrons, and
finally a 2p electron, requiring quite a large amount of energy due to the change in principal
quantum number from 3 to 2. None of the other choices has the same shell sequence
through the first four ionization steps.
(e) Two elements in the range between Mg and Ar form stable anions that are
spherical. These elements are: (circle them)
Mg
Al
Si
P
S
Cl
Ar
Si– is isoelectronic to P and thus has a half-filled p subshell outside a closed core of
electrons. Cl– is isoelectronic to Ar with closed subshells. (In addition, both Al2– and S2–
would be spherical, and if you circled these, you were not penalized as long as you also
circled Si and Cl.)
(f) The Cr4+ ion in the compound CrO2 is exploited in magnetic audio and video
recording tape. The electron configuration of Cr4+ is [Ar] 3d2.
The question here is simple: which four electrons does Cr with the [Ar] 4s1 3d5
configuration lose? There are two obvious possibilities for Cr4+ that make sense following
Hund’s rules: [Ar] 4s2 and [Ar] 3d2. In the first, however, the two electrons would be
spin-paired and the ion would be diamagnetic, a lousy choice for a component of a
magnetic tape! In the other, however, the two electrons are not spin paired, and the ion is
paramagnetic.
(g) The smallest of the following atoms or ions is (circle one)
Sc3+
Ti
K+
S2–
Fe2+
Zn
Ar
Highly positively charged ions that have lost all their valence electrons will be smaller
than isoelectronic neutrals or isoelectronic positive ions with smaller nuclear charge.
4. The Li atom in its ground state can absorb many different photons, producing many
different excited states. This question is concerned with the photon of longest wavelength
that Li can absorb. This photon has an energy of 2.96 × 10–19 J.
Exam 2 Solutions
Chem 6, 9 Section, Spring 2002
(a) What is the electron configuration of the Li atom produced when ground state Li
absorbs this photon?
Just as in the Na atom you studied in lab and we discussed in class, the longest
wavelength excitation (smallest energy excitation) promotes the s valence electron to the p
orbital of the same principal quantum number. For Li, this means the 1s2 2s1 ground state
becomes the 1s2 2p1 excited configuration.
(b) The first ionization energy of Li is 8.64 × 10–19 J. Draw an energy level diagram
that locates the Li ground state, the excited state, and the state corresponding to
Li+ + e– below. Label each of your energy levels.
The ionization limit is at the zero of energy, and the ground state is 8.64 × 10–19 J
lower (i.e., at –8.64 × 10–19 J). The excited state is above the ground state by an amount
equal to the photon energy, placing it at –8.64 × 10–19 J + 2.96 × 10–19 J = –5.68 × 10–19
J. The diagram below summarizes these energies.
Li+ + e–
0
Energy/10–19 J
–2
–4
Excited State
–6
–8
Ground State
–10
(c) Calculate the effective nuclear charge felt by the valence electron in the excited
state.
We note that the excited state valence electron has n = 2, and we equate the energy of
this state, –5.68 × 10–19 J, to the expression for the energy in terms of the effective nuclear
charge:
En,l = – 5.68 ×
10–19
J = –(2.18 ×
2
Zeff,
n, l
–18
10 J)
22
Zeff, n, l =
(4) (– 5.68 × 10–19 J)
– 2.18 × 10–18 J
= 1.02
5. (a) Which of the following factors appears in the complete wavefunction for an H atom
in the 2pz state? Circle the correct choice and explain your answer.
cos θ
sin θ sin φ
sin θ sin φ
sin θ
Exam 2 Solutions
Chem 6, 9 Section, Spring 2002
The 2pz wavefunction has a nodal plane (i.e., is zero) everywhere in the xy plane.
This is the plane for which the spherical polar angle θ = 90° for any value of the spherical
polar angle φ. When θ = 90°, cos θ = 0.
(b) A particle in a box of length l is in a state in which the probability of finding the
particle in the region 0 ≤ x ≤ L/4 is equal to the probability of finding the particle in the
region L/2 ≤ x ≤ 3L/4. The quantum number for the particle could be (circle all that are
possible):
0
1
2
3
4
First, n = 0 is not an allowed quantum number value for the particle in a box system.
Next, we look for a wavefunction with the correct symmetry to satisfy the specified
criterion. This symmetry shows up for any state with an even quantum number.
(c) An electron is in a one-dimensional box of length 10.0 Å in the quantum state n =
2. What is the electron’s de Broglie wavelength?
Embarrassingly easy way to find the answer: for n = 2, the wavefunction is a single,
full wave spanning the full length of the box. In other words, the wavelength of the
wavefunction equals the box length. But the wavelength of the wavefunction is the de
Broglie wavelength! So λ = 10.0 Å!
Equally good, straightforward way to find the answer: equate the energy of the state to
the expression for kinetic energy in terms of momentum, express the momentum in terms
of the de Broglie wavelength, and solve for the de Broglie wavelength:
2
2 2
h/λ
p2
E= h n =
=
8meL 2 2me 2me
so
λ = 2L
n = L = 10.0 Å