Chem 400
Inorganic Chemistry
Practice Exam 2
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KEY
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(please print)
1. Using balanced chemical equations, give an explanation for the following
observations: AlF3 has only a low solubility in liquid HF, but a combination of NaF
and AlF3 leads to dissolution of the reagents (1st equation); when BF3 is added to
this solution, a precipitate forms (2nd equation). (10 points)
AlF3 is a Lewis acid and, in the presence of F-, forms the adduct [AlF4]-:
AlF3 + NaF → Na[AlF4]
(Equation 1)
and the ionic salt Na[AlF4] is soluble in liquid HF.
BF3 is a stronger fluoride acceptor than AlF3, and accepts F- from [AlF4]-:
Na[AlF4] + BF3 → AlF3 + Na[BF4]
(Equation 2)
In the absence of excess F-, AlF3 precipitates.
For acid-base chemistry, the self-ionization you need to consider is:
3HF ↔ [H2F]+ + [HF2]-
2. The σ bonding in the linear molecule XeF2 may be described as a 3-center, 4electron bond. If the z axis is assigned as the internuclear axis, use the pz orbitals
on each of the atoms to prepare a molecular orbital description of the sigma bonding
in XeF2. (10 points)
Energy
Two electrons are in the bonding orbital and two are in the non-bonding orbital.
Xe
XeF2
1
F- - - F
3. Suggest likely products for the following reactions (which are balanced on the lefthand side): (2 points each)
+
-
a.
2 XeF2 + SbF5 → [Xe2F3] [SbF6]
b.
Me3SiCl + Na[C5H5] → NaCl + Me3Si[η -C5H5]
c.
BCl3 + 3 C2H5OH → B(OC2H5)3 + 3 HCl
d.
n
e.
2 (CH3)3SnCl + Mg → (CH3)3SnSn(CH3)3 + MgCl2
1
BuLi + C6H6 → C6H5Li (PhLi) +
n
BuH (or n-C4H10)
f.
B2H6 + 6 CH2=CH2
1. THF
2. H2O2
2 B(OH)3 + 6 CH3CH2OH
g.
2PCl5 + N2H4 → Cl3P=N-N=PCl3 + 4 HCl
h.
2XeF6 + SiO2 → SiF4 + 2XeOF4
i.
AlCl3 + 3 CH3CH2MgCl → Al(CH2CH3) + 3 MgCl2
j.
2 NaOH + 2F2 → OF2 + 2 NaF + H2O
2
4. ClO2 and [ClO2]- are bent, each with equivalent Cl-O bond lengths: 147 pm in ClO2
and 157 pm in [ClO2]-. Give a description of bonding in ClO2 and [ClO2]- and
rationalize the difference in Cl-O bond lengths. Use both resonance structures and
MO theory arguments. Draw a representation of the π-molecular orbitals. (10
points)
Resonance structures can be drawn to rationalize equivalence of bonds:
O
O
Cl
Cl
O
O
O
O
Cl
Cl
O
O
O
O
Cl
Cl
O
O
Energy
Or, use MO theory to show delocalized bonding schemes analogous to π-allyltype bonding. There are 12 atomic orbitals in the basis set of ClO2 (4 valence
orbitals per atom) and therefore 12 MOs; these are occupied by 19 valence
electrons. In [ClO2]-, 10 MOs are fully occupied. The π-MOs are shown below.
In ClO2, the HOMO is the π* MO and is singly occupied. In [ClO2]-, this MO is
fully occupied, giving more antibonding character to the Cl-O interactions.
The MO scheme illustrates delocalization of electrons over the Cl-O
framework, and rationalizes the lengthening of bonds on going from ClO2 to
[ClO2]-.
π*
non-bonding
π
π-Molecular orbitals in ClO2.
3
5. Draw reasonable three-dimensional structures for each of the following inorganic
molecules (2 points each)
a.
(CH3Li)4
Li
CH3
H3C
Li
H3C
Li
Li
C
H3
b.
[B3O6]3OB
O
O
B
-
B
c.
O-
O
O
(HBNH)3
H
B
H
N
H
N
B
B
H
H
N
H
d.
[Si4O12]8-
-
-
-
O
O
-
O
Si
O
Si O-
Si
O
O
O
Si
-
e.
Bi2Cl8
O-
O
2-
O-
O
2-
Cl
Cl
Cl
Bi
Cl
Bi
Cl
Cl
Cl
4
Cl
f.
[H9O4]+
H
H
O
H
H
O
O
H
H
H
H
O
H
g.
S8
S
S
S
S
S
h.
S
[S4]2+
+
i.
S
S
S
S
S
S
S+
S
S
2+
S
(η5-Cp)BeH
H
H
Be
Be
or
j.
Al2(CH3)6
H3C
H3 C
H3
C
Al
Al
C
H3
5
CH3
CH3
6. The triiodide ion, I3-, is linear, but I3+ is bent. Explain. (5 points):
I3- is linear because there are three lone pairs in a trigonal geometry on the central
I. I3+ is bent because there are only two lone pairs on the central I, leaving a
structure similar to H2O.
7. The sulfur-sulfur distance in S2, the major component of sulfur vapor above ~720 ºC,
is 189 pm, significantly shorter than the sulfur-sulfur distance of 206 pm in S8.
Suggest an explanation for the shorter distance in S2. (5 points)
S2 is similar to O2 with a double bond. As a result, the bond order is shorter than
single bonds in S8.
8. Suggest a balanced chemical equation to describe the autodissociation of ICl (1st
equation). NaCl behaves as a base in this solvent, while AlCl3 behaves as an acid.
Write balanced chemical equations for the dissolution of NaCl in ICl (2nd equation)
and AlCl3 in ICl (3rd equation). (10 points)
Equation 1:
3 ICl ↔ I2Cl+ + ICl2Both solutes increase the concentration of ions:
NaCl + ICl → Na+ + ICl2NaCl acts as a base.
AlCl3 + 2ICl → I2Cl+ + AlCl4AlCl3 acts as an acid.
9. Describe the bonding in Ga2H6 and Ga2Cl6. (10 points)
Ga2H6 is isostructural with B2H6; analogous bonding pictures are appropriate. In
Ga2H6, 2c-2e localized terminal Ga-H bonds and 3c-2e Ga-H-Ga bridge bonds:
Overlap of H 1s
orbital with two B
H
Ga Ga
H
H
H
3
sp orbitals.
H
H
In Ga2Cl6 there are sufficient valence electrons to invoke all 2c-2e Ga-Cl
interactions; analogous to Al2Cl6:
Cl
Cl
Cl
Ga Ga
Cl
Cl
6
Cl