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Chem 6, 10 AM Section
Exam 2 Solutions
Spring 2012
1. (a) There is a molecule with empirical formula CN2O. Shown below are several possible
ways these atoms could be connected. The single, double, or triple bonds are shown for each,
but any possible lone pairs are not shown. Enter in the boxes any nonzero formal charges,
and circle the structure that you believe is the correct one.
+1
–1
O = N = C = N
+1
+1
–1
–1
O = N – C
N
O
N – C = N
–2
+1
–1
+1
+1
O = N = N = C
O = N – N
C
O
N – N = C
+1
+1
–2
The lone pairs are shown along with the non-zero formal charges. The circled structure
with no formal charges on any atom is the best, and it is in fact the correct, observed structure.
Note that this molecule is a simple coupling between NO, with an unpaired electron on the N,
and CN (which is in part (c) below), with an unpaired electron on the C.
(b) Of LiF, ClF, and F2, the one with the largest dipole moment is LiF because:
the bond in LiF is strongly ionic: a large electronegativity atom (f) bonded to a small
electronegativity atom (Li). Cl and F have similar electronegativities and thus a small dipole
moment for ClF, and F2, being homonuclear, has zero dipole moment because of its symmetry.
(c) Of CO, CN, and CN–, the one that is paramagnetic is CN because:
CO and CN– have an even number of electrons, and all of them can be assigned to bonding
pairs or to lone pairs. CN, however, has an odd number of total electrons (6 + 7 = 13); one
must be unpaired, which ensures paramagnetism.
.
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.
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Chem 6, 10 AM Section
Exam 2 Solutions
n = ______4________
Spring 2012
l = ______3_______
m = –3, –2, –1, 1, 2, 3
The three views each show two nodal planes for three total: xy, xz, and yz. Thus, l must
equal 3 (i.e., this is an f orbital). There are no spherical nodes present; thus, n = 3 + 1 = 4 (i.e.,
a 4f orbital). The orbital is not cylindrically symmetric about the z (or any) axis; thus m ≠ 0,
but m could be any of the other possibilities for l = 3.
(b) Ignoring electron spin, how many states in all have the same energy as (i.e., are
degenerate with) the state shown in part (a)? (Include the state in (a) in your count.)
The energy of an H atom state depends only on the n quantum number. For n = 4, the
possibilities are 4s (only 1, m = 0), 4p (3 of them, m = –1, 0, +1), 4d (five of them, m = –2, –1,
0, +1, +2), and 4f (seven of them, m = –3, –2, –1, 0, +1, +2, +3). That’s 16 in all (and in
general, the degeneracy of a given n value, neglecting spin, is n2).
(c) One of the H atom s orbital radial probability distribution functions (rdf) is
rdf ∝ r2 2 – ar 2 e –r/a0
0
For this orbital, n = 2 because:
an s orbital has no planar nodes (l = 0), and here, we see that the radial part of the
2 (r). This radial
wavefunction is Rn, l(r) = 2 – r/a0 e –r/2a0 because we know rdf = r2 Rn,
l
wavefunction has one zero (one node) at r = 2a0:
Rn, l(r) = 2 – r/a0 e –r/2a0 = 0 or 2 – r/a0 = 0 so that r = 2a0 locates a node.
The total number of nodes is n – 1. Thus, with one node here, n = 2.
3.
(a) If I assign ms values to all the electrons in the ground state of the P atom, I find that
6 of them have one value while 9 of them have the other value.
The P ground-state configuration is 1s2 2s2 2p6 3s2 3p3 for a total of 15. Those three 3p
electrons are not spin-paired. Each is in a unique 3p orbital with the same spin. The remaining
twelve electrons are spin-paired. Thus, 6 have one spin quantum number value, and 6 + 3 = 9
have the other value.
(b) The electrons lost by Sn when it forms the Sn2+ cation come from the subshell(s)
5p.
The Sn ground-state configuration is [Kr] 5s2 4d10 5p2. The easiest electrons to remove
are the outer-most, highest energy, 5p electrons.
(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
As the outer electron moves farther and farther from the rest of the atom, it sees an object
(the rest of the atom) with a net charge of 1e, i.e., the total nuclear positive charge is reduced to
the charge of a single proton by the presence of the other electrons left in the core. Here, that
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Chem 6, 10 AM Section
Exam 2 Solutions
Spring 2012
core therefore looks like Rb+. The effective charge can never be less that this. Normally, the
outermost Rb electron is in the 5s orbital. Moving it from 5s to 9d moves it a significant
distance away from the Rb+ core.
(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 numbers indicate that the first electron is easy to remove, the next two are a bit harder,
but comparable, and the fourth is significantly harder to remove. This suggests that the first
three have the same principal quantum number, the second and third have a different angular
momentum quantum numbers from the first, and the fourth has a principle quantum number
one less than that of the first three electrons. That the case for Al: 1s2 2s2 2p6 3s2 3p1. In
order, we remove the 3p electron, then the two 3s electrons, and finally one of the 2p electrons.
(e) Two elements in the range between Mg and Ar form stable singly charged anions that
are spherical.
These elements are: (circle them)
Mg
Al
Si
P
S
Cl
Ar
We want anions with full or half-full subshells. These are Si– and Cl–.
(f) Cobalt, as Co3+, binds six F– ions to form the paramagnetic complex ion CoF63–.
Show, using “spin arrows” (↑ and/or ↓), the valence electron configuration of Co in this ion.
3d
4s
2
7
The Co atom’s configuration is [Ar] 4s 3d , but we note that we are well past the midpoint
of the first transition metal row, which means the 3d electrons are lower in energy than the 4s
electrons. When Co is oxidized (loses electrons) to Co3+, we will lose those two 4s electrons
first, and then lose one of the 3d electrons: Co3+ [Ar] 3d6. Because the ion is paramagnetic,
those six 3d electrons cannot be spin-paired. In fact, the must be as unpaired as possible,
leading to the spin assignments shown.
(g) The smallest of the following atoms or ions is (circle one)
Sc3+
Ti
K+
S2–
Fe2+
Zn
Ar
We note that the ions S2–, Sc3+, and K+ are isoelectronic to Ar. The others have electrons
outside the [Ar] core, and we can rule them out. Of these three, the smallest will be the one with
the largest nuclear charge. That’s Sc3+.
4. The Li atom in its ground state can absorb many different photons, producing many
different excited states. The photon of longest wavelength that ground-state Li can absorb has
an energy of 2.96 × 10–19 J.
(a) What is the electron configuration of the excited Li atom produced when ground state
Li absorbs this photon?
The longest wavelength photon is the smallest energy photon Li can absorb. That means
the ground state (1s2 2s1) is excited to the first excited state. Because Li is a many-electron
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Chem 6, 10 AM Section
Exam 2 Solutions
Spring 2012
Energy/10–19 J
atom, the 2s and 2p orbitals are not degenerate; 2p is higher in energy than 2s. Thus the excited
atom is 1s2 2p1.
(b) The first ionization energy of Li is 8.64 × 10–19 J. Draw an energy level diagram
below that locates the Li ground state, the excited state of part (a), and the state
corresponding to Li+ + e– below. Label each of your energy levels carefully.
10
8
6
4
2
Li+ + e–
0
–2
First ionization
energy
–4
Excited 1s2 2p1 state
–6
Photon
energy
–8
Ground 1s2 2s1 state
–10
We locate the ionized Li+ + e– energy at zero; that is the physical situation our effective
nuclear charge energy expression yields for n = ∞, which represents ionization. This choice
places the ground state at a lower energy, lower by the first ionization energy. The excited state
is above the grounds state by an amount equal to the photon energy that the ground state
absorbs.
(c) Calculate the effective nuclear charge felt by the Li valence electron in its ground state.
We use the following equation with n = 2, the principal quantum number of the Li valence
electron:
Z2
En,l = –(2.18 × 10–18 J) eff, n, l = –IE1 = –8.64 × 10–19 J
n2
Zeff, 2, 0 =
22 –8.64 × 10–19 J
= 1.26
–2.18 × 10–18 J
(d) A Li atom in an excited state that is not the state in parts (a) and (b) can emit two and
only two distinct photons. One of these photons, the one with the larger wavelength, has λ =
813 nm. What are the Li atom electron configurations of the two states involved in this
emission?
Initial state configuration: 1s2 3s1
Final state configuration: 1s2 2p1
An excited state that can emit only two unique photons must be the second excited state: it
can emit a photon to the first excited state and a different one to the ground state. Of these two,
the larger wavelength (smaller energy) photon would be between the third excited state to the
second excited state. The 1s2 3s1 configuration describes the third excited state, and the
seconds excited state is the 1s2 2p1 state of part (a).
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Chem 6, 10 AM Section
Exam 2 Solutions
Spring 2012
5.
Here are some questions about wavefunctions and their interpretation.
(a) Which of these H atom angular wavefunctions must be for a state with m = 0? Circle
your choice or choices (there may be more than one!), but also explain your reasoning briefly.
5 cos3 θ – cos θ
sin2 θ sin 2φ
cos θ
sin θ cos θ sin φ
3
States with m = 0 are cylindrically symmetric about the z axis; that means they must be
independent of φ. The first and third of these functions fit this requirement.
(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):
2
3
4
0
1
First, the particle in a box system does not allow n = 0; so, we can rule that out right away.
For n = 1, the wavefunction is half a sine wave, extending from x = 0 to x = L. To have the
probability of finding the particle in the region 0 ≤ x ≤ L/4 be equal to the probability of finding
the particle in the region L/2 ≤ x ≤ 3L/4, the square of the wavefunction must be identical over
these two spans. The n = 1 wavefunction lacks this symmetry. But for n = 2, the square of the
wavefunction is identical over these spans (look at a graph of Ψ2(x) = (2/L) sin2 (2πx/L) to see
this). Likewise, n = 4 works, but n = 3 doesn’t.
(c) An electron is in a one-dimensional box of length 10.0 Å in the quantum state n = 4.
What is the electron’s de Broglie wavelength?
There are three ways to approach this question. The first is the easiest: a picture of the n =
4 wavefunction shows that it is two full sine waves extending from x = 0 to x = 10 Å. But the
de Broglie wavelength is the length of one full sine wave in the full wavefunction. Thus, λde
Broglie = L/2 = 5.0 Å. The second way is symbolic:
2 2
p2
h
h2
=
p=
, E4 = h 4 =
λde Broglie
8mL 2 2m 2m λ 2
de Broglie
8L 2
and λde Broglie = L
2(16)
2
The third way would be numeric: calculate the n = 4 energy (9.64 × 10–19 J), then calculate the
momentum (from E4 = p2/2me or p = 1.34 × 10–24 kg m s–1), and finally calculate the
wavelength (from h/p = 5.0 × 10–10 m = 5.0 Å).
or
2
λde
Broglie =
6.
Some dipole moment questions:
(a) The CsF molecule has a dipole moment of 7.88 D and a bond length of 2.55 Å. What
would its dipole moment be if its bond were purely ionic? What is its actual percent ionicity?
A “purely ionic” bond has a full electron charge transferred (from Cs to F here) across the
bond: μ = QR with Q = e and R = 2.55 Å. This gives μ = 4.09 C m = 12.2 D. But the
observed dipole moment is only 7.88 D; thus, the percent ionicity is (7.88 D)/(12.2 D)×100 %
= 64.3 %. (If you relied on remembering the quickie little formula in the text involving δ, the
fractional ionicity, it will certainly work, assuming you remembered it correctly. But
remembering such a formula is never a substitute for really understanding a physical situation.)
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Chem 6, 10 AM Section
Exam 2 Solutions
Spring 2012
(b) Which of the following triatomics, all with the N atom in the central position, would
you predict to have a dipole moment and why? (Note: molecular ions can be polar or non-polar
just like neutral molecules!)
NO2+
NO2
NO2–
Here are the correct Lewis dot structures. Note that NO2+ is isoelectronic to CO2, which is
non-polar (linear molecular geometry and linear electron pair geometry), while NO2– is
isoelectronic to ozone, O3, which is polar and has a bent molecular geometry (as does NO2).
The bent molecules have trigonal planar electron pair geometries.
[O=N=O]
+
O = N – O
[O=N–O]
–
Both NO2 and NO2– have one other resonance structure that could be drawn as well, and the
Lewis structures could be drawn as “bent” for the final two molecules.
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