Practice for Unit III Test – Answer Key
AP Chemistry
Name_____________________________ Date_________________________
Part I—Free Response.
1)
a) IF3 Lewis structure. Iodine has two lone pairs and is singly bonded to three fluorines.
b) T-shaped
c) Draw the two equivalent resonance structures of the SO2 molecule. Bond lengths are equal
because the pi bonding electrons are spread over the entire molecule not localized in one bonding
area.
d) sp2
2)
a) Draw Lewis structures. AsF3 molecules are polar because there is a lone parir of electrons on
center atom pushing each As-F bond downward. The result is that the bond dipoles do not cancel.
The AsF5 molecules are trigonal bipyramidal. The axial As-F bond dipoles cancel. The equatorial
As-F bond dipoles cancel. Cancelation of bond dipoles results in non polar molecule.
b) Show the resonance structures of the NO2- ion. Explain equal bond lenth as in 1c) above. Once a
coordinate covalent bond is formed between a proton and an electron pair on one of the oxygen
atoms, the molecule no longer exhibits resonance: H-O-N=O (lone electron pairs not shown.)
c) Sulfur can have an expanded valence shell due to empty d orbitals in the n=3 energy level. Oxygen
cannot have an expanded valence shell b/c its valence electrons occupy the n=2 energy level which
does not contain empty d orbital.
3) Use principles of atomic structure and/or chemical bonding to answer each of the following.
a) To form the Ca2+ ion two electrons are removed from the n=4 energy level. The valence shell in the
ion is n=3 and there are two more protons then electrons. The result is less shielding and greater
electrostatic attraction between the protons and the valence electrons causing the radius of the ion
to be smaller than the radius of the atom.
K
Ca
1st Ionization Energy (kJ/mol)
419
590
2nd Ionization Energy (kJ/mol)
3,050
1,140
b) Explain the difference between Ca and K in regard to
(i)
Ca and K have the same valence shell (n=4). K contains one valence electron.
Ca contains 2 valence electrons. Ca valence elecrtons are held more tightly to
the nucleus b/c they experience a greater effective nuclear charge (due to more
protons in the nucleus, same shielding. Thus, it takes it more energy to remove
an electron from Ca then K (higher 1st IE).
(ii)
K has the higher second ionization energy because the second electron from K is
taken from the n=3 energy level while the second electron from Ca is taken from
the n=4 energy level. Electrons in the n=3 energy level experience a greater
ENC due to less shielding and greater proximity to the protons in the nucleus.
c) The first ionization energy of Mg is 738 kJ/mole and that of Al is 578 kJ/mole because is takes more
energy to remove an electron in the n=3 energy level from the spherically shaped s orbital then the
1
dumbbell shaped p orbital. The s orbital electrons feel a greater ENC because within a given
energy level s-orbital electrons are closer to the nucleus then p-orbital electrons.
d) Potassium has a lower first-ionization energy than lithium because the valence electron in K is in a
higher energy level (n=4) than the valence electron in Li (n=2). Thus the valence electron in K has
a greater ENC due to more shielding and is easier to remove.
4) Answer the following questions about the structures of ions that contain only sulfur and fluorine.
a) The compounds SF4 and BF3 react to form an ionic compound according to the following equation:
SF4 + BF3  SF3BF4
(i)
(ii)
(iii)
(iv)
SF3+ cation has a lone pair of electrons on the center S.
sp3.
trigonal pyramidal
The bond angle is less than 109.5° because the lone pair of electrons on S
pushes the bonding electrons downwards.
b) The compounds SF4 and CsF react to form an ionic compound according to the following equation:
SF4 + CsF  CsSF5
(i)
(ii)
(iii)
(iv)
Center S has one lone pair of electrons.
sp3d2
square pyramidal
+4 (Cs has +1, each F has -1)
5) The reaction represented below is one that contributes significantly to the formation of photochemical
smog.
∆Hrxn = -114.1 kJ/mol
2NO(g) + O2(g)  2NO2(g)
a) 73.1 g NO(g) x (1 mol NO / 30 g NO) x (114.1 kJ / 2 mol of NO) = 138.9 kJ
b) -114.1 = 4 x – [(2 x 607) + 495]
BE Nitrogen-oxygen bond in NO2 = 399 kJ/mol
c) NO2 has resonance; NO has a double bond. In each N has less then an octet of electrons.
d) The NO bond in NO has a bond order of 2. Each NO bond in NO2 has a BO of 3/2. It take more
energy to break a bond with BO = 2 versus BO = 3/2.
6)
a) All balloons contain equal moles and therefore equal particles. The balloon with CO2 has the
highest mass b/c CO2 has the highest molar mass of the gases shown.
b) All balloons are at the same temperature so the average kinetic energy is the same.
c) CO2, the largest gas, should deviate most from the ideal gas law because it has the greatest
intermolecular forces of attraction.
d) The balloon containing He should be the smallest. He has the lowest molar mass of the given
gases and therefor will effuse the fastest.
7)
a)
b)
c)
d)
mass of dry air: 0.843 L x (1.18 g / L) = 0.995 g
mass of empty flask: 157.70 g – 0.995 g = 156.70g
mass of unknown gas: 158.08 g – 156.70 g = 1.38 g
moles unknown gas: n = PV/RT = (750/760)(0.843)/(0.08206)/(296) = 0.0342 moles
Molar mass = g/mol = 1.39 g / 0.0342 moles = 40.3 g/mol
2
After the experiment was completed, the instructor informed the student that the unknown gas was
carbon dioxide.
e) percent error = (44.0 - 40.3)/44.0 x100 = 8.4%
f) Occurrence 1: Yes. The molar mass of air (apporx 29 g/mol) is less than the molar mass of CO 2.
So this would result in a lower mass of gas in the flask and through calculation a lower molar mass
of the unknown gas.
Occurrence 2: Yes. The calculated mole value (using PV=nRT) would be too high. This would
cause the calculated molar mass (grams/mol) to be too low as well. he temperature of the air was
23.0°C, but the temperature of the CO2(g) was lower than the reported 23.0°C.
g) Describe the steps of a laboratory method that the student could use to verify that the volume of the
rigid flask is 843 mL at 23.0°C. You need not include any calculations with your answer.
Part II—Multiple Choice.
1) B
8) D
2) D
9) E
3) C
10) E
4) D
11) A
5) E
12) B
6) A
13) D
7) A
14) C
15)
16)
17)
18)
19)
20)
21)
D
B
A
A
D
B
C
22)
23)
24)
25)
26)
27)
28)
3
B
C
D
B
D
A
B
29) B