PSI AP Chemistry: Chemical Bonding
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Chapter Problems Part B Lewis Structures, Hybridization, and Bond Energy
1. Sulfur trioxide and the sulfite ion have the same number of atoms but
different structures and properties.
a. Draw the correct Lewis structure for both.
b. Compare the S-O bond lengths in both molecules. Justify your answer.
c. What is the O-S-O bond angle in both sulfur trioxide and in the sulfite
ion?
d. What is the hybridization of the S atom in both compounds?
e. Which of the two molecules would be planar?
f. What is the oxidation state of S in sulfur trioxide compared to in the
sulfite ion?
g. Are all of the bonds in sulfur trioxide the same length? Explain your
answer.
h. Why is the there a pi bond in the sulfite ion? Explain using the concept
of formal charge.
2. XeF2 and OF2 are two rare molecules.
a. Draw the Lewis structures for both molecules
b. Explain why the bond angles in XeF2 are larger than the bond angles
in OF2.
c. What is the formal charge on the central atom in both molecules?
Show all calculations.
3. Using Lewis structures, explain the following data:
a. The C-O bond enthalpy in carbon monoxide is 1072 kJ/mol while the
C-O bond enthalpy in carbon dioxide is 732 kJ/mol.
b. The C-F bond angles in CF4 are larger than they are in SF4.
c. The bond length in nitrate ions is longer than in nitrite ions.
d. The C-O bond lengths in the carbonate ion are all the same.
e. I3- is linear while H2O is bent.
4. Butanal (CH3CH2CH2CHO) really smells. Over 6 billion kg of it are produced
annually to be used in laboratories and plastics.
a. Draw a correct lewis structure for butanal.
b. What hybridizations exist on the carbon atoms within the molecule?
Show which carbons have which hybridizations.
c. What would be the O-C-H bond angle? Why is it larger than the H-C-H
bond angles found elsewhere in the molecule?
d. Draw the lewis structure for butanoic acid (CH3CH2CH2COOH).
e. What would be the C-O-H bond angle in butanoic acid?
f. Which of the two compounds (butanal or butanoic acid require
resonance structures?
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AP Chemistry
Chemical Bonding
1.
Answers
a.
b. The S-O bond lengths are the same due to the same bond order =
1.33 in each compound.
c. 120 degrees and roughly 107 degrees
d. sp2 and sp3 respectively
e. SO3
f. +6 and +4 respectively
g. Yes, the pi bond is spread amongst the three bonds making them
all slightly shorter than a single bond
h. Without the pi bond, sulfur would have a +1 formal charge, with it
the formal charge on S is 0, a more stable state.
2.
a.
b. The repulsions of the three unbonded pairs drive the bonded
electrons into a linear shape. In OF2, the 2 unbonded pairs drive
the bonded electrons into a bent shape.
c. XeF2 = 8-8 = 0 OF2 = 6-6 = 0
3.
a. The bond order in CO is three while the bond order is 2 in CO2.
b. The unbonded pair in SF4 drive the other atoms closer to each
other resulting in a smaller bond angle.
c. The bond order in nitrate (1.33) is less than that in nitrite (1.5)
resulting in a weaker and longer bond.
d. The pi bond is shared across all three bonds.
e. I3- has three unbonded pairs to waters two, causing repulsions
that bring the I atoms into a linear configuration.
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AP Chemistry
Chemical Bonding
4.
a.
b. From left to right, first three are sp3 while the rightmost carbon is
sp2
c. 120 degrees, the fewer sigma bonds make for greater distance
between bonds.
d.
e. 104.5
f. Butanoic acid due to the presence of the pi bond, that said, the
resonance structure with the pi bond on the non-hydrogen bearing
oxygen is preferred due to formal charge considerations.
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AP Chemistry
Chemical Bonding