AP Chemistry
Unit 3 Practice Problems
Key
1.
a) When two atoms approach each other, the electrons and nucleus of one atom interact with the e- and
nucleus of the other. We calculate the effect of these interactions on the energies of the electrons. As
atoms approach the overall energy of the system decreases (or becomes more negative) until the atoms
are bonded.
b) If the potential energy decreases and bonds form, then energy is released. If atoms are separated and
bonds are broken the potential energy increases and energy is released.
c) As the atoms approach and the optimal separation distance is obtained, their orbitals overlap and a bond
is formed. The system is at it’s lowest energy. If the atoms are forced to get closer the repulsion from the
nuclei causes an increase in potential energy.
d) Bond B shows that the potential energy is lower than bond A, therefore more energy released.
e) Bond A is shorter based on the graph.
2. 2(413) + 839= 1665 kJ/mol
3. 4(413) + 614 + 436 = 2702 kJ/mol must be absorbed to break all reactant bonds
6(413) + 348 = -2826 kJ/mol is released when product bonds are formed
∆𝐻𝑟𝑥𝑛 = 2702 − 2826 = −124 𝑘𝐽/𝑚𝑜𝑙
4. a) nonpolar
nonpolar)
5.
b) polar
c) ionic
d) polar e) nonpolar
6.
a
b
c
7.
1
1
2
2
13
3
1+
2+
3+
14
4
15
5
16
6
17
7
3-
2-
1-
18
8
8
8
8
d forms ionic bonds
What combinations of ions are impossible?
8
AB & CD
What would have the greatest lattice energy?
AC
What would have the least lattice energy?
BD
f) polar (technically, but it is often considered
Ionic Bond
8. a.
Relative Lattice Energy
E
LiF
(1+)(1-)/(0.68
+ 1.33) = 0.50
MgO
E  (2+)(2-)/(0.66 + 1.40) = 1.94
NaCl
E  (1+)(1-)/(0.97 + 1.81) = 0.36
Al2S3
E  (3+)(2-)/(0.51 + 1.84) = 2.55
b. NaCl < LiF < MgO < Al2S3
c. Al2O3: E  (3+)(2-)/(0.51+ 1.40) = 3.14
a. What is the range of electronegativities?
9.
metals
0.8 (Rb) - 1.8 (Sn)
metalloids
1.8 (Si) - 2.1 (Te)
nonmetals
2.1 (H) - 4.0 (F)
Rank the following bonds from most polar (1) to least polar (6). Place + next to the atom with the lower electronegativity.
2
1
N–S +
O–S +
F–S +
3.0 – 2.5 = 0.5
3.5 – 2.5 = 1.0
4.0 – 2.5 = 1.5
b.
4
P–S +
2.5 – 2.1 = 0.4
5
6
S–S
2.5 – 2.5 = 0
3
Cl–S +
3.2 – 2.5 = 0.7
10. Consider the following data for hydrogen halides.
Indicate whether the following correlate directly or inversely.
Electronegativity difference & dipole moment
Electronegativity difference & bond strength
Dipole moment & bond strength
Bond length & bond strength
Direct
x
x
x
Inverse
x
11. Consider the following data for C-C bonds.
Bond strength correlates directly with number of
shared electrons.
12. Complete the chart with the formula or name of the binary molecule.
Formula
N2O5
Name
dinitrogen pentoxide
CCl4
carbon tetrachloride
CO2
carbon dioxide
CuSO4
copert (II) sulfate
NO
nitrogen monoxide
OF2
oxygen difluoride
Al(PO4)3
aluminum phosphate
13.
1
1
2
2
13
3
14
4
15
3
16
2
17
1
18
0
14. In the Lewis structure shown below, A, D, E, Q, X and Z represent non-metal elements in the first two rows of the periodic table.
Identify the elements.
A
F
D
C
E
O
Q
N
X
H
Z
H
15. There are three ways to draw Lewis structures for NCO–
a.
Calculate the formal charge for each version
Structure [:::N–CO:]–
[::N=C=O::]–
[:NC–O:::]–
5 4 6
5 4 6
5 4 6
Formal
7 4 5
6 4 6
5 4 7
Charge
-2 0 +1
-1 0 0
0 0 -1
b.
Which is the preferred structure? Give two reasons.
[:NC–O:::]- is preferred because formal charges are closest to zero and negative charge is on the higher electronegativity atom—
oxygen.
16. afsd
CNO– (lowest formal charge)
POCl3 (lowest formal charge)
Cl:::
|
::: Cl – P = O::
|
Cl:::
[: N  C – O:::]-
SCN– (C is central atom)
IF4F:::
\ ..
[:::F – I – F:::].. \
F:::
[::S = C = N::]-
SF6
BrF3
:::F
.. ..
:::F – Br – F:::
|
F:::
F:::
\/
:::F – S – F:::
/\
:::F F:::
SO3
N2O (lowest formal charge)
:N  N – O:::
:::O – S = O::
|
O:::
IF5
NO2
.
:::O – N = O::
:::F F:::
\/
:::F – I – F:::
.. \
F:::
SF4
SO2Cl2 (lowest formal charge)
F:::
|
:::F – S – F:::
.. \
F:::
Cl:::
|
::O = S = O::
|
Cl:::
17. asdsd
Molecule
 Bonds
 Bonds
Lone Pairs
Hybridization
CH4
0
4
0
sp3
NH3
0
3
1
sp3
CO2
2
2
0
sp
CH2O
1
3
0
sp2
POCl3
1
4
0
sp3
CNO-
2
2
0
sp
SCN-
2
2
0
sp
BCl3
0
3
0
sp2
SF6
0
6
0
sp3d2
BrF3
0
3
2
sp3d
HF
0
1
3
sp3
SO3
1
3
0
sp2
N2O
2
2
0
sp
IF5
0
5
1
sp3d2
SO2Cl2
2
4
0
sp3
XeF4
0
4
2
sp3d2
18.
sp2
sp2
sp3
sp
sp
sp3
19.
Molecule
Resonance Structures
CO32-
SO2
..
..
O=S–OO–S=O
OCN-
20.
Shortest bond length is the CO triple bonds. The electron density between the two nuclei is the greatest as this is three
pairs of electrons being shared. Think of coulombs law, more charges interacting, therefore stronger force of attraction.