On the Pages Herein One May Look Upon
A Fine Collection of
Rare And Exciting Questions
Exploring Various Facets and Aspects of
Numerous Worthy Subjects,
including :
Atomic Structure
Lewis Theory
Valence Bond Theory
and
Crystal Field Theory
As Suitable for Examination and Inquiry by
Adventurous Young Gentlemen and Ladies
Engaged in the Course of Study of
Chemistry Two Hundred and Twenty
1) Of the following mathematical functions, which correspond(s) to an orbital with  = 0 ?
(For all functions,  = Zr/a0)
 Z
 
a)
81 6  a0 
1
2  Z
 
b)
81   a0 
32
2e–/3(3cos2
– 1)
32
(6 –
2)e–/3sincos
 Z
 
c)
81 3  a0 
1
 Z
 
d)
4 2  a0 
1
32
(27 – 18 – 22)e–/3
32
e–/2cos
2) At right is shown the structure of methyl cyanoacrylate, an adhesive
commonly used for bonding plastics. Assume that the carbon atoms
labeled "c" "d" and "e" lie in the yz plane. Answer the following
questions about the valence bond structure of methyl cyanoacrylate.
H
O
H
c
H aC O b C C C e
H
dC
H
N
What is approximate H-C-O bond angle at carbon a ?
______
What is the hybridization of atomic orbitals at carbon e ?
______
Which orbital on oxygen b, atomic or hybrid, forms the Ca-Ob  bond?
______
Which orbital on carbon c, atomic or hybrid, forms the Cc-Ce  bond?
______
3) Use VSEPR theory to rationalize the observed geometries for the molecules [N3]– (azide,
linear) and O3 (ozone, bond angle 117°). For each, describe the (approximate) VBT
hybridization of valence atomic orbitals at the central atom required to account for the
observed geometry, and determine the specific orbitals each central atom uses to form each
bond.
4) In the solid and liquid phases, the compound with the empirical formula ClO3 is actually
chloryl perchlorate, [ClO2]+[ClO4]–. In the gas phase, it forms a neutral Lewis acid-base
adduct of those two ions.
a) Draw Lewis structures for both the chloryl cation and the perchlorate anion.
b) Use your Lewis structures to account for the fact that the cation is polar but the anion is
not.
c) Draw a Lewis structure for the adduct that forms between [ClO2]+ and [ClO4]–. What
qualitative geometries do you expect at the Cl atoms in the product? Do you expect
terminal Cl-O bonds to be stronger or weaker than the bridging Cl-O bonds?
5) Concisely rationalize the following observations.
a) The bond strength of iodine fluoride, IF (278 kJ/mol), is far greater than those of either I2
(149 kJ/mol) or F2 (155 kJ/mol).
b) The chloryl cation [ClO2]+ has a larger bond angle (122º) but shorter bonds (131 pm) than
the chlorite anion [ClO2]– (111º, 156 pm).
c) Nitrogen has a less favourable electron affinity (+0.07 eV) than carbon (–1.59 eV), even
though N is more electronegative than C.
d) The electron density in a hydrogen atom is greatest at the nucleus, but the most likely
distance from the nucleus at which to find the electron is 53 pm rather than zero.
e) A typical Si-F single bond (565 kJ/mol) is far stronger than an S-F single bond (284
kJ/mol), even though they have the same bond length (156 pm).
f) A typical S-Cl single bond (201 pm) is longer than a F-Cl single bond (163 pm), even
though they have the same bond strength (~250 kJ/mol).
6) Sulfur tetrafluoride reacts with boron trifluoride to yield an ionic salt [SF3][BF4]
a) For each reactant molecule and each product ion
 draw a valid Lewis structure, and
 describe the molecular geometry as predicted by VSEPR, and
 describe the valence bond hybridization of orbitals at the B or S, assuming a “perfect”
VSEPR-predicted geometry.
b) For SF4 and BF3, sketch a valence bond promotion / hybridization diagram for the central
atom. What orbital bears the sulfur lone pair? What boron orbital is empty? Be as specific
as possible.
c) The [SF3]+ cation deviates from that “perfect” geometry, with a bond angle of 97º. Offer
a rationale for this observation.
d) There are lots and lots of reasons that BF3 has the strongest bonds of any of those four
molecules. Name as many as you can think of.
7) For each complex below, use a crystal-field orbital splitting diagram to explain the observed
number of unpaired electrons and to calculate the crystal field stabilization energy (CFSE) in
Dq units. If the CFSE requires some number of pairing energy terms, include them
(e.g. –4Dq + 2P). In each case, describe the complex as either weak field or strong field, if
appropriate. (It’s appropriate only if HS vs LS would result in different configurations.)
a) [Mn(CN)6]4–, 1 upe
f) Mo(CO)6, 0 upe
b) [Mn(CN)6]2–, 3 upe
g) [V(H2O)6]2+, 3 upe
c) [Cr(NH3)6]2+, 4 upe
h) [ZnCl4]2– , 0 upe
d) [FeF6]3–, 5 upe
i) [Hg(NH3)2]2+, 0 upe
e) [CoCl4]2–, 3 upe
j) [CuCl4]2–, 1 upe
a) Explain why the overall trend is a decrease
in radius across the row.
b) Explain why the data exhibit the pattern
that they do, i.e. an increase contrary to the
overall trend for certain ions..
120
115
M2+ ionic radius /pm
8) At right is a graph of the measured ionic radii
for octahedral complexes of first-row
transition metals in their +2 oxidation state.
110
105
100
95
90
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
c) Tables of ionic radii list two values for
85
many of these ions, and the second value
80
always smaller than those plotted on this
graph. Explain the need for two values for
some (but not all) ions, explain why it will
be smaller, and explain why the other value for Fe2+ is the smallest radius in the series.
9) Complexes of Ni(II) can be either octahedral, tetrahedral, or low-spin square planar,
depending on the ligands. For each geometry, sketch a crystal-field orbital splitting diagram,
determine the number of unpaired electrons, and calculate the CFSE in Dq units. If the CFSE
requires some number of pairing energy terms, include them (e.g. –4Dq + 2P).
d-orbital energies (Dq)
geometry
z2
x2–y2
xy
xz
yz
octahedral
+6.00
+6.00
–4.00
–4.00
–4.00
tetrahedral
–2.67
–2.67
+1.78
+1.78
+1.78
square planar
–4.28
+12.28
+2.28
–5.14
–5.14
10) For each of the following pairs of complexes, indicate which one has the larger energy
separation of d-orbitals, and briefly explain your reasoning.
a) [Fe(NH3)6]3+ or [Fe(CN)6]3–
b) [CrCl6]3– or [MoCl6]3–
c) [Co(NH3)4]2+ or [Co(NH3)6]2+
d) [Fe(CN)6]3– or [Fe(CN)6]4–
e) [Cr(py)6]3+ or [CrF6]3–
(py = pyridine, C6H5N)