Chemistry 312
Sample Midterm Exam I
Autumn 2013
Your NAME:_ANSWER KEY___________________ Student Number:___________________
Your Signature:_____________________________________
Your Seat Number:_________
Notes: (i) The total points for the exam is 107.
(ii) Many of the questions contain multiple parts. Be sure to read the
questions carefully and answer all of the parts.
(iii) Some questions ask for explanations while others don't.
1. (9 points) (a)[3p] Which orbital has two radial nodes, two angular nodes, and is
large along the z axis?
!
!
(b)[4p] On the graph at right, plot Ψ vs. r for this orbital
along the z axis. Mark the position of the nucleus on
the graph. The nucleus is at the origin (r = 0). The plot
5dz2
r (zr axis)
is zero at the nucleus and has two minima or maxima,
crossing the r axis twice and changing sign each time.
(c)[2p] On the graph at right, plot Ψ vs. r for this orbital
along the x axis. This is the same as part (b) (actually
!
!
r (xr axis)
it’s of opposite sign but we’ll give credit for either sign).
2. (4 points) Order the following elements in decreasing terrestrial abundance, and briefly explain your
reasoning: Titanium (#22), Rhodium (#45), Palladium (#46)
Titanium is the most abundant, then palladium, then rhodium. The general trend is that
elements get less abundant with increasing atomic number, which would predict Ti > Rh
> Pd (3 points for this answer). But there is also the trend that odd elements are less
abundant than even ones.
3. (6 points) (a)[3p] Aluminum has an atomic weight of 27. It could in principle be:
(i) 100% of a single isotope,
14
or (ii) a 1:1 mixture of two isotopes:
Al13
13
(fill in each of the boxes with a number)
Al13 +
15
Al13
[answers in part b below]
(b)[3p] Given your knowledge of isotopic abundances, is the isotopic abundance of aluminum best
described by:
(i) a single isotopeü
or
(ii) a mixture?
Circle your answer; no explanation is needed.
27
Aluminum is 100% Al13, with has 13 protons and 14 neutrons. The alternative would be a 1:1 mixture of
28
Al13, both of which would have odd numbers of both protons and neutrons. Such isotopes are very rare.
26
Al13 and
Chem312, Autumn 2013
Sample Midterm I, p. 2
4. (27 points). (a)[4p] Which has a higher ionization energy, K or Ca+? Explain your reasoning based
on the general patterns for ionization energies; don’t refer to Slater’s rules.
Ca+ has the higher ionization energy (I.E.) because it has the higher charge. Both ions have the
[Ar]4s1 configuration. All other things being equal, the ion with higher charge will have the higher
I.E. (In addition, Ca is to the right of K so even as a neutral atom it probably has a higher I.E.)
[4]
(b) Now write out the electronic configurations of K or Ca+ consistent as appropriate for Slater’s rules.
K: _________________________________________
(1s2)(2s22s6)(3s23s6)(4s1)
Ca+: ________________________________________
(1s2)(2s22s6)(3s23s6)(4s1)
(c)[4p] According to Slater’s rules, what is the difference in the shielding – the value of S – for the
outermost valence electron of K vs. Ca+? Explain. (You don’t need to calculate the shielding.)
According to Slater’s rules, there is no difference in the shielding for the outermost 4s1 valence
electron of K vs. Ca+. This is because they have exactly the same electronic configuration.
(d)[4p] Explain how your answer to (c) is consistent with your answer to (a).
According to Slater’s rules, the effective nuclear charge felt by an electron (Z*) is the nuclear
charge Z minus the shielding S. S is the same for the shielding for the outermost 4s1 valence
electrons of K vs. Ca+, but Z is one unit higher for Ca. Therefore the Z* will be one unit higher for
Ca+ and it’s ionization energy will be higher.
(e)[4p] Now consider Sc2+, which has a …d1 configuration. Use Slater’s rules to give the algebraic
equation for the difference in the shielding (S) for the outermost valence electron of Ca+ vs. Sc2+.
According to Slater’s rules, an outermost s electron is shielded 0.85 by each electrons to the left,
but an outermost d electron is shielded 1.0 by the electrons to the left. The Sc2+ is shielded by
0.15 more per 3s/3p electron and there are eight such electrons, so it is more shielded by 8 x 0.15
electrons (or 1.2 electrons [numerical answer not required]).
(f)[4p] Explain why the outermost valence electrons of Ca+ (…4s1) and Sc2+ (…3d1) have the same or
different shielding, in terms of the shapes of the orbitals involved.
An electron in a d orbital has very little density near the nucleus so it penetrates the core electrons
very little and therefore it is well shielded by those electrons (Slater 1.0). An s electron, however,
has some density close to the nucleus and therefore is not perfectly shielded by the core
electrons, so the Slater shielding constant is only 0.85.
(g)[3p] What are the charges on the cations Kn+, Cam+, and Scp+ in stable chemical compounds?
Fill in the boxes:
K
+
Ca
2+
Sc
3+
Chem312, Autumn 2013
Sample Midterm I, p. 3
5. (11 points) (a)[4p] Fill in the blanks at the top of the thermochemical cycle below.
Answers: Al3+ (g) and 3F– (g)
_______ (g) + ______ (g)
(–lattice energy) A
AlF3 (s)
B
C
Al3+ (aq) + 3 F- (aq)
D
(b)[4p] Which step has ∆H˚ closest to zero (not highly positive or negative)?
A
B
C
Dü
(c)[3p] The cycle for AlCl3 looks very similar. Which of the term(s) in the cycle are exactly the same
for AlF3 and AlCl3? (circle all that are correct):
A
Bü
C
D
6. [17 points] (a)[2p] Draw the d-orbital splitting pattern for an
octahedral complex ML6 in the box at right. What is the vertical
scale? (I’m looking for an answer like “temperature”). Energy
(b)[4p] Label each of the orbitals (dxy, etc.) in the diagram. Pick one
of the d orbitals and explain why you placed it higher or lower in
your diagram.
dz2
dxy
dx2-y2
dxz
dyz
dx2-y2 and dz2 are higher in the diagram because they point at the ligands, so electrons in these
orbitals are more repelled by the negative charge of the ligands. dxy, dxz and dyz point in between
the ligands.
(c)[2p] Consider the complex ion FeF64–. The d electron-count of this ion is: ____6_____.
(d)[5p] Fluoride is at the very low end of the spectrochemical series, so its complexes have very small
values of ∆o. This means that FeF64– is (circle one):
high spinü
low spin
In the orbital diagram you drew in part (a), fill in the electrons as appropriate for FeF64–.
(e)[2p] The octahedral iron complex with water ligands that has
the same electronic configuration as FeF64– has the formula: _________Fe(H2O)62+____________
(f)[2p] Water has a higher position on the spectrochemical series than fluoride. How will this change
the d-orbital splitting diagram for the answer to part (e), versus FeF64–? Remember that the two
complexes have the same electronic configuration (e.g., if one is low-spin, so is the other).
The diagram will look qualitatively the same, but the gap between the two sets of orbitals, ∆o, will
be larger for the water complex.
Chem312, Autumn 2013
Sample Midterm I, p. 4
7. (8 points). Complexes of ions of the 2nd and 3rd transition series (the 4d and 5d transition metals) are
always low spin. This is in part because the ∆o is larger for these ions. But there is a second
important issue that causes them to be always low-spin. Describe this issue and why it makes 4d
and 5d transition metal complexes low-spin.
The high spin/low spin dichotomy is a balance between the energetic cost of putting an electron in
a higher orbital (∆o) and the cost to pair the electron in a lower orbital. Thus the other issue is the
electron pairing energy. This paring energy is smaller for the larger metal ions because their
orbitals are larger. With larger orbitals, the electrons are on average farther apart and there is less
energetic cost to pair them. Since there is less cost to pair electrons, the low spin forms are
favored.
8. (12 points) (a)[3p] On the drawing at right, draw the
periodic trend for ionic radius, indicating the
small and large ends of the trend. answer below.
(b)[5p] Using this trend, order the cations below in terms of decreasing ionic radius, based on the
periodic trends discussed in class. The atomic numbers of the elements are given in parentheses. If
there are case(s) where you can’t predict the relative sizes based on the trend, state that.
Al3+ (#13), Sc3+ (#21), Cr3+ (#24), Zr4+ (#40).
r(Zr4+) ≅ r(Sc3+) > r(Cr3+) > r(Al3+)
Using the “similarly charged ions get smaller moving up and to the right” part of the trend, Al3+ is
smaller than Cr3+ which is smaller than Sc3+. But it’s hard to predict Sc3+ vs. Zr4+. Zirconium is a
step down the periodic table, which should make it larger, but it is a step to the right which should
make it smaller. And Zr4+ has a higher positive charge than Sc3+, which should make it smaller. So
it’s hard to predict.
(c)[4p] Thinking beyond the general trend in part (a), why is Sc3+ (#21) vs. Al3+ (#13) a simpler
prediction than Cr3+ (#27) vs. Al3+ (#13)?
While Al3+ and Sc3+ have the same valence electron configuration, Co3+ has a different valence
configuration, [Ar]3d6. And Co3+ is clearly smaller than Sc3+. Thus the Co3+ vs. Al3+ prediction is
less confident than the Sc3+ vs. Al3+ one.
smallest
ionic
radius
largest
ionic
radius
Chem312, Autumn 2013
Sample Midterm I, p. 5
9. (7 points) (a)[3p] What is the general trend for the heat of hydration, ∆Hhyd° ∝ ___Z2/r________
(b)[4p] There is roughly the same trend for the acidity of cations, and it uses the symbol Z. Does this Z
have the same meaning as the Z in Slater’s rules?
No, these Z’s are different. The Z in Slater’s rules is the charge on the nucleus. The Z in
the heat of hydration and acidity trends is the net charge on the cation. (For instance, for
Fe3+, Z(Slater) is +26 but Z(for ∆Hhyd) is +3.
Full credit if the answer to (a) is Z2/(r+50).
10. (6 points) You own a property in the mountains that is starting to get exposed to acid rain, because
the ski area upwind is burning more and more coal for heating. Your soil is a titanium aluminum
silicate – in other words it’s an oxide mineral in which the cations are Si4+, Al3+, and Ti4+. As the
rain gets more and more acidic, the oxide ions in the mineral can become protonated and the
mineral in part dissolves, liberating one of the possible aquo ions [M(H2O)6]n+. Which cation would
be expect to leach first?
The cation that leaches first is Al3+. This is because it is the least acidic cation, meaning
that oxide ligands bound to Al3+ are more basic. Al3+ is the least acidic because it has the
4+
4+
smallest charge, compared with Ti and Si . The differences in radii are smaller than
the differences in Z2 (16 vs. 9). The least acidic species has the most basic conjugate
base. So the first cation that dissolves from the solid oxide is aluminum, as [Al(OH2)6]3+.
In fact, a major cause of tree death from acid rain is aluminum poisoning.