Final Exam - Fall 2000

Name_______________________________________
Chem 1311 Final Exam 15 December 2000
In taking this examination, you are expected to adhere to the GT academic honor code.
At a minimum this requires that you utilize only the materials supplied to you (except for
a non-programable calculator), and that you do not give help to, or accept help from,
others. Calculators cannot be shared. Numerical problems must show all equations
used to arrive at an answer; zero credit will be given for just the answer.
1. Shown to the right is the pH vs volume curve for
titration of 15 mL of a weak base with 0.10 M HCl.
a) (6) Calculate the concentration of the weak base in
the original solution.
0.010 L x 0.10 mol/L = 0.0010 mol H + = 0.0010 mol B
0.0010 mol B/0.015 L = 0.067 M B in original solution
b) (3) If the endpoint was to be determined using an
indicator, what would be the ideal value of the pK a of
the indicator?
5
c) (3) What is the pK b of the weak base?
pK b =pK w - pK a = 14 - 9 = 5 (see next part)
pH
12
11
10
9
8
7
6
5
4
3
2
1
0
0 2 4 6 8 101214161820
volum e 0.10 M acid added
d) (3) What is the pK a of the conjugate acid of the weak base?
pH = pK a + log [B]/[BH + ] at half-equivalence point pH = 9. Therefore pK a = 9
2. (12) Equimolar amounts of F (as the sodium salt), (CH 3 ) 2 S, Hg 2+ (as the nitrate salt) and BF 3
are combined in a suitable solvent. What are the four possible acid-base products? Circle the
two acid-base reaction products predicted to be present in greatest amount after equilibrium is
established. Assume a 1:1 stoichiometry for all acid-base reactions.
BF 4
HgF + (CH 3 ) 2 SBF 3 (CH 3 ) 2 SHg 2+
3. (6) Which of the compounds FeO, Fe 2 O 3 , FeS, or Fe 2 S 3 is most likely to have the largest
lattice energy? Briefly explain the basis for your choice.
Lattice energy is proportional to the product of the charge on the ions and inversely proportional
to the internuclear separation between cation and anion. Thus Fe 2 O 3 will have the largest
lattice energy since it has the smallest and most highly charged ions

4. (14) Draw all diastereoisomers of the complex [Co(H 2 NCH 2 CH 2 S)(H 2 NCH 2 CH 2 NH 2 )(Cl) 2 ].
Indicate those isomers that are optically active (have nonsuperimposible mirror image forms) by
drawing a box around the structure. You may abbreviate the chelating ligands as N ð N and
S ð N in making your drawings.
Cl
+ N
+ N
+
N
S
Co
N
N
N
S
Co
Cl
Cl
N
N
Co
Cl
Cl
Cl
N
S
5. (15) a) Draw structures of the following molecules and ions. Your structures should: i) depict
the geometry of the species, ii) show the location of all non-bonding electron pairs, if any, on all
terminal atoms and, iv) show formal charges on individual atoms consistent with the number of
valence shell electrons assigned to each and consistent with the overall charge on the species.
b) For each of the above species indicate what hybrid orbital set would be utilized for bonding
by the central atom.
Hybrid orbital set
SOCl 2
..
sp 3
..
Cl S ..
.. O
..
Cl
..
ICl 4
NO 3
..
- O
+
N
- O
..
..
O..
..
Cl ..
..
Cl
..
..
-
I
..
..
Cl
..
Cl
..
sp 3 d 2
7. (6) Give ground state electron configurations for the following species and indicate the
number of unpaired electrons for each.
Co 3+ [Ar]3d6 Sn + [Kr]5s 2 4d 10 5p 1
sp 2
unpaired electrons __4____
unpaired electrons __1____
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8. Shown below the molecular orbitals for CO 2 (O=C=O).
a) (6) Indicate which orbital (by number) is the HOMO (highest occupied molecular orbital) and
which is the LUMO (lowest unoccupied molecular orbital) and whether each is σ bonding (σ b ), σ
nonbonding (σ n ), σ antibonding (σ*), π bonding (π b ), π nonbonding (π n ), or π antibonding (π*).
HOMO 5 π n LUMO 4 π*
b) (8) Attack of hydride ion (H - ) at the LUMO of CO 2 will result in the formation of the formate
anion. Draw resonance structures of the formate ion that indicate that the carbon-oxygen
bonds are equivalent. Your Lewis structures must show all valence shell electron pairs and
formal charges on the appropriate atoms.
-
..
O
What is the bond order of the C-O bonds?__1.5__________
c) (5) Draw the bonding molecular orbital that would describe π bonding in the anion that is
equivalent to the resonance structures in part b). (Be careful to take into account the relative
contribution and phase of each atomic orbital involved in the molecular orbital.)
H
C
..
O..
..
O
H
C
..
O..
-
H
C
O
O
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9. (7) Iridium metal has a face-centered unit cell with a cell dimension
of 3.833 Å. Calculate the density of the metal in g cm -3 .
−1
4 atoms x 192.217 g mol
23 −1
d =
6.022 x10 atoms mol
−8 3
=
(3.833 x10 cm)
22.67 g cm
3
10. (6) Briefly explain why many coordination compounds absorb in the visible region of the
electromagnetic spectrum and are therefore colored.
Interaction between a metal ion and ligands results in a splitting of the d orbitals into two or
more groups, depending upon the number and geometry of the ligands, of different energy.
Often the energy difference in the groups of d orbitals is comparable to that of visible light and
excitation of electrons from the lower energy orbitals to the higher energy orbitals can occur
with absorption of photons which gives the appearance of color.
11. (12) Assume that the ions H 3 + and HFH + are both linear. Describe the bonding in each ion
in terms of the molecular orbitals formed and the number of electrons available to populate the
bonding orbitals. Be careful to explicitly describe differences in the nature of the bonding in the
two species. Simple orbital energy level diagrams could be useful in this regard, although not
absolutely necessary. In HFH + assume that the fluorine atom uses its s and one of the p atomic
orbitals to bond to the hydrogen atoms.
antibonding, no electrons
nonbonding, no electrons
bonding, 2 electrons
H + 3 is electron deficient
antibonding, no electrons
antibonding, no electrons
bonding, 2 electrons
bonding, 2 electrons
HFH + is not electron deficient
H
F p
F s
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12. (42) Circle the answer corresponding to the correct answer for each of the following.
Which species requires the most energy for ionization?
a) Cl 1- b) Ar « K 1+ d) all are the same
Which species has the largest radius?
« Cl 1- b) Ar c) K 1+ d) all are the same
Which of the following substances is most likely to be a solid at room temperature?
a) SO 2 b) ClO 2 « GeO 2 d) NO 2
Which of the following species has the lowest bond order?
1-
1+
a) O 2 b) O 2 c) O 2
« O 2
2-
Which of the following species has the smallest number of unpaired electrons?
1-
1+
2-
a) O 2 b) O 2 c) O 2 « O 2
Which should have the largest ionization energy?
1-
1+
a) O 2 b) O 2 « O 2
d) O 2
2-
Which of the following molecules will have an ionization energy that is less than that of the atom
of the element of which it is composed?
a) N 2 b) C 2 c) B 2 « none of these
Which coordination complex should have the largest number of unpaired electrons (fluorine is a
weak field ligand)?
a) [CrF 6 ] 3- b) [MnF 6 ] 3- « [FeF 6 ] 3- d) [CoF 6 ] 3-
Which should be the most acidic species
« [Al(OH 2 ) 6 ] 3+ b) [Ru(OH 2 ) 6 ] 3+ c) [Mg(OH 2 ) 6 ] 2+ d) none of these is acidic
Which of the following has the highest proton affinity (is the strongest base)?
« NH 2
+
b) NH 2 c) NH 3 d) NH 4
Band structure diagrams are shown below for three substances. Which is most likely to be an
insulator? The shaded area represents the fraction of the band that is filled.
a) « c) d) none of these
p
p
p
s
s
Which of the following compounds will generate the largest number of ions in solution?
« K 4 [Fe(CN) 6 ] b) [Co(NH 3 ) 6 ]Cl 3 c) Na 2 [PtCl 6 ] d) [Ni(NH 3 ) 6 ](PF 6 ) 2
s
Which of the following phosphorus-chlorine species is most likely to not be a stable moiety?
+
a) PCl 4 b) PCl 5 « PCl 2 d) PCl 6
page 5 of 7

13. (36) Answer six (ONLY) of the following. Letter each answer to match the question and
continue on the next page as necessary. Use balanced reactions and diagrams to illustrate
your answer whenever possible. Keep your rhetoric to a minimum (40 words or less per
answer).
a) [Fe(OH 2 ) 6 ] 2+ has four unpaired electrons, whereas [Fe(CN) 6 ] 4 is diamagnetic.
b) Doping silicon with a small amount of gallium greatly increases its conductivity.
c) Addition of SO 2 to aqueous base decreases the basicity of the solution.
d) The equilibrium constant for the reaction [Ni(en) 2 (OH 2 ) 2 ] 2+ + trien = [Ni(trien)(OH 2 ) 2 ] 2+ + 2 en
is >>1. en is H 2 NCH 2 CH 2 NH 2 ; trien is H 2 NCH 2 CH 2 NHCH 2 CH 2 NHCH 2 CH 2 NH 2
e) The pK a for unbound water is 14; however, when coordinated to certain metal ions its pK a
may drop to values as low as 2-3.
f) Compared to CO and O 2 , CS and S 2 are very unstable molecules and can be generated only
at very low concentrations in the gas phase.
g) A gas is evolved when CH 3 CH 2 CO 2 D is added to a suspension of NaH in in THF (an inert
solvent).
h) A gas is evolved when Ca 3 P 2 is treated with water.
i) Phosphine reacts with NaNH 2 in liquid ammonia or in organic solvents such as THF (a cyclic
ether) where NaPH 2 is produced by the reaction NaNH 2 + PH 3 S NaPH 2 + NH 3 . However,
phosphine does not react with NaNH 2 in water.
page 6 of 7

a) Both complexes contain d 6 Fe(II). Water is a weak field ligand whereas CN is a strong field
ligand. Thus the former complex is high spin and the latter is low spin.
b) Addition of gallium to silicon results in the creation of holes in the valence band which act as
low-lying acceptor levels for electrons and the formation of a p-type semiconductor.
c) SO 2 + OH = HOSO 2 or SO 2 + 2 OH = SO 3 2 + H 2 O
d) This is an example of the chelate effect, which is primarily entropy driven. Formation of the
products results in the formation of an additional free molecule.
e) Coordination of water to a metal ion results in electron transfer from the oxygen to the metal
and effectively an increased electronegativity for the oxygen atom. This in turn increases the
polarization of the O-H bond and its acidity.
f) CS and S 2 are multiply bonded species and contain third row elements. Pi bonds involving p
orbitals of heavier elements are weak relative to sigma bonds involving sp hybrid orbitals. Thus
such molecules are unstable to reaction with themselves or other molecules to form sigma
bonded frameworks.
g) CH 3 CH 2 CO 2 D + LiH = HD + Li + + CH 3 CH 2 CO 2
h) Ca 3 P 2 + 6 H 2 O = 3 Ca 2+ + 6 OH + 2 PH 3
i) NH 2 will react with water according to NH 2 + H 2 O = NH 3 + OH . Hydroxide is not a strong
enough base to deprotonate PH 3 . The intrinsic basicity of NH 2 has been leveled to that of OH .
page 7 of 7