CHEMISTRY 103 – Practice Problems #3
Chapters 8 – 10
http://www.chem.wisc.edu/areas/clc (Resource page)
Prepared by Dr. Tony Jacob
Suggestions on preparing for a chemistry exam:
1. Organize your materials (quizzes, notes, etc.).
2. Usually, a good method to prepare for a chem exam is by doing lots of problems. Re-reading a section of a chapter is fine,
but re-reading entire chapters takes up large amounts of time that generally is better spent doing problems.
3. Old exams posted by your instructor should be completely worked though. Old exams give you a sense of how long the exam
will be, the difficulty of the problems, the variability of the problems, and the style of your instructor. Quizzes written by your
instructor are also a good resource.
You might not necessarily want to do these problems in the order written. Good Luck!
CHAPTER 8
1. Which chemical and associated bond type are incorrectly paired?
a. The bond in O2(g) is nonpolar covalent.
b. The bond in MgCl2(s) is ionic.
c. The bond in C–H bond in CH4(g) is nonpolar covalent.
d. The bond in O–H in water is polar covalent.
e. The bond in Ag(s) is metallic.
2. Which of the following statements about the S–O bond in SO2 is incorrect?
a. The S–O bond is nonpolar and there is a partial negative charge on the S.
b. The S–O bond is nonpolar and there is a partial negative charge on the O.
c. The S–O bond is polar and there is a partial negative charge on the S.
d. The S–O bond is polar and there is a partial negative charge on the O.
e. The S–O bond is ionic and there is a negative charge on the O.
3. Draw the Lewis dot structures for the following molecules.
a. NF3 b. CH3CH3 c. PO3-3 d. Cl2 e. SO2 f. HNO2 g. CH2Cl2
h. N2 i. C2H2 j. COCl2 k. CO3-2 l. NO3- m. K2O n. CaS o. B p. O
q. BF3 r. BeF2 s. OH (not OH-) t. IF4- u. KrF4+2 v. IBr3
4. Which molecule, if any, violates the octet rule?
a. NH4+
b. CF4
c. SF4
d. BrO4-
e. none do
5. Using Lewis dot structures, show how the reactions below occur (not all instructors cover LDS of ionic compounds).
a. 2Li(s) + F2(g) → 2LiF(s)
b. 2Ca(s) + O2(g) → 2CaO(s)
6. Which chemical species has the highest melting point (not all instructors cover melting points of lattices)?
a. BeO
b. NH3
c. MgS
d. KBr
7. Which molecule has the strongest nitrogen-nitrogen bond?
a. N2
b. N3c. N2H2
8. What is the formal charge on the S in SO3?
a. -2
b. -1
c. 0
(watch some TV)
d. N2H4
e. all are equivalent
d. +1
e. +2
9. Which pairs of atoms will form the most ionic compound?
a. nitrogen and oxygen
b. chlorine and fluorine
d. sodium and oxygen
e. phosphorus and oxygen
c. oxygen and oxygen
10. Shown below are four possible Lewis dot structures for SO3-2 without resonance structures drawn. Decide
which structure is best based on formal charges. Explain your answer.
O
O
S
O
S
O
I
O
-2
O
O
S
O
O
II
S
O
III
O
O
IV
11. Draw the 3 Lewis dot resonance structures for thiocyanate, SCN- (C is in the middle). Based on formal
charges, which resonance structure would be the better structure? The electronegativity values (χ) are:
ΕΝS = 2.5, ΕΝC = 2.5, and ΕΝN = 3.0.
12. I. In the molecule shown, CF2O, choose the structure with the correct locations of the δ+ or δ- symbols.
+
Oδ
Cδ
F -F
δ
δ
Oδ
δ
-F
C δ+
F δ
Oδ
Cδ
+
Oδ
+
Cδ
F
F δδ
F
δ+
a.
b.
c.
d.
II. Would the C–O bond be nonpolar covalent, polar covalent, or ionic?
F +
δ
13. Identify each of the following bonds as nonpolar covalent, polar covalent, or ionic.
a. F–I _________________________
b. C–C _________________________
c. Si-Si _________________________
d. Ca–O _________________________
e. H–O _________________________
14. (Not all instructors cover Born-Haber cycles; skip if not covered)
Use the Born-Haber cycle to answer the following questions.
a. Identify the reaction in the Born-Haber cycle labeled E.
b. What is the energy for the heat of vaporization of Mg?
c. What is the reaction in the Born-Haber cycle labeled F?
d. Write the reaction for the lattice energy of MgO(s).
e. Identify the letter for the reaction in the Born-Haber cycle that
corresponds to the lattice energy reaction.
f. Calculate the lattice energy for MgO(s).
15. Calculate the change in enthalpy for the reaction using the bond
energies (kJ/mol) listed below.
C2H4 + H2O → CH3–CH2–OH
Bond
C–C
C–H
C–O
O–H
(time for a nap)
Bond enthalpy
(kJ/mol)
348
413
358
463
Bond
C=C
C≡C
C=O
Bond enthalpy
(kJ/mol)
614
839
799
16. What is the Cl–Cl bond dissociation energy given the bond energies (kJ/mol) listed below.
H2CO(g) + 2Cl2(g) → Cl2CO(g) + 2HCl(g)
ΔHrxn = -208kJ
Bond
C–C
C–H
C–O
Bond enthalpy
(kJ/mol)
348
413
358
Bond
O–H
C–Cl
H–Cl
Bond enthalpy
(kJ/mol)
463
328
431
Bond
C=C
C≡C
C=O
Bond enthalpy
(kJ/mol)
614
839
799
CHAPTER 9-VSEPR
17. A central atom with 2 lone pairs and 3 bonding pairs of e- will have a molecular shape of:
a. linear
b. trigonal pyramid c. trigonal planar
d. T-shape
e. trigonal bipyramid
18. In Lewis dot structures, which electron interactions repel the most?
a. bonding pair–bonding pair
b. bonding pair–lone pair
c. lone pair–lone pair
d. since these are all electrons they are equivalent
19. Given the molecule IBr5 identify the statements below as being true (T) or false (F)?
a. The I–Br bonds are polar covalent. ______
b. There is one lone pair located on the central atom. ______
c. There are six electron domains around the central atom. ______
d. The molecular geometry is trigonal pyramid. ______
e. The molecule is polar. ______
f. The angles within the molecule are 90˚, 120˚, and 180˚. ______
20. (Not all instructors require students to draw molecular shapes and net dipole vectors.) For each molecule below, draw the
Lewis dot structure, draw the electron domain geometry, and draw the molecular geometry. Re-draw the
molecular geometry diagram and draw in vectors representing the bond dipoles, and draw the net vector for the
net dipole for polar molecules otherwise write “no dipole” for nonpolar molecules when no net vector exists.
a. H2O b. NH3 c. XeF2 d. PF421. The molecular geometry for BrF4+ is
a.
b.
c.
d.
e.
Images: Public Domain from Wikipedia.org (http://en.wikipedia.org/wiki/VSEPR_theory)
22. Which of the following molecules are planar (i.e., which have a molecular geometry that is flat)?
I. BClF2
II. NCl3
III. H2Se
IV. SeF4
V. XeF4
a. I
b. I, II, IV
c. I, III
d. I, III, V
e. I, III, IV, V
23. Which molecule will have an angle of ~120˚?
a. NF3
b. SF6
c. CF4
24. Which molecule is polar?
a. PH4+
b. GeCl4-2
(pizza time)
c. SiCl4
d. SeF4
e. none
d. XeF4
e. none
25. Which of the following statements is incorrect? If all are correct select answer “e”.
a. A molecule with two electron domains will be linear in shape.
b. Polar molecules interact with magnetic fields.
c. Electrons in multiple bonds repel more than electrons in single bonds.
d. The reduction in the angle found as the molecular geometry changes from tetrahedral to trigonal pyramid to a
bent structure is because of greater repulsions between the lone pairs of electrons.
e. All of the above statements are correct.
26. How many of these molecules have a tetrahedral electron domain geometry around the central atom?
CCl4
SF4
SiCl4-2
SeI2
KrF2
a. 1
b. 2
c. 3
d. 4
e. 5
27. There are two possible structures, one polar and one nonpolar, for each chemical given. Draw the two
possible electron domain geometry structures, one polar and one nonpolar, for each chemical (i.e., two electron
domain geometry structures for “a” and two for “b”). a. PCl2F3 b. XeF2Cl2
28. Consider the following covalent single bonded pairs of atoms. Which bond will be the shortest?
a. N–F
b. P–Cl
c. N–Cl
d. P–F
e. none of the above
Valence Bond Theory
29. Which of the following statements is incorrect?
a. To reduce lone pair-lone pair electron repulsions in an octahedral electron domain geometry, the first lone
pair of electrons are placed in one of the four equatorial positions rather than the axial positions since the six
locations are not equivalent.
b. The dipole moment, µ, is equal to Qr where Q is the charge magnitude and r is the distance between nuclei.
c. Pi bonds are created with the unhybridized atomic p orbitals overlapping.
d. When forming a bond, if the atomic nuclei distance decreases past the minimum energy, electron-electron
and proton-proton repulsions increase the system’s energy.
e. Sigma bonds have electron density that lies on the internuclear axis.
30. (Not all instructors cover the calculation of dipoles.) If the dipole moment, µ, for CO is 0.122D (D = Debye) and
the C–O bond length, d, is 122.8pm, what will the magnitudes of the partial charges, Q, be on the C and O
atoms? (1 D = 3.34 x 10-30 Cm; 1 x 1012pm = 1m)
31. a. What is the hybridization for the S atom in SO3?
b. What is the hybridization for the S atom in H2S?
32. Answer the questions below about the structure and bonding in this molecule.
First, draw in any lone pairs needed to complete octets.
a. What is the hybridization on the N atom marked “a”?
b. What is the bond order for the bond marked “b”?
c. What is the angle for the C–O–C with the O atom marked “c”?
d. What orbitals overlap to form the bond marked “d”?
e. What is the bond order for the bond marked “e”?
f. What orbitals overlap to form the bond marked “f”?
g. What is the electron domain geometry at the N atom marked “a”?
h. What is the molecular geometry of the O atom marked “c”?
i. How many σ and π bonds are in the molecule?
(time for another nap)
H
33. How many sigma (σ) and pi (π) bonds are present in the molecule shown?
a. 9σ, 3π
b. 10σ, 3π
c. 4σ, 3π
d. 7σ, 5π
C
C
C
C
N
H
H
H
H
e. 7σ, 3π
34. Answer the questions below about the structure and bonding in this molecule.
a. Draw in any lone pairs needed to complete octets.
b. What is the bond order for the bond marked “a”?
c. What is the hybridization on the C atom marked “b”?
d. What is the molecular geometry around the C atom marked “c”?
e. What is the angle on the C-S-H group marked “d”?
f. What is the hybridization on the H atom marked “e”?
g. What is the bond order for the C-O bond marked “f”?
h. What orbitals overlap to form the bonds marked “a”?
i. How many σ and π bonds are in the molecule?
35. Which of the following statements is incorrect?
a. Molecules can have polar bonds and while the molecule can be nonpolar.
b. To reduce lone pair-lone pair repulsions in a trigonal bipyramidal electron domain geometry, the lone pairs
are found in the equatorial positions.
c. The four large lobes of the sp3 hybridized orbitals lie in a tetrahedral arrangement.
d. A covalent bond forms when two orbitals overlap and two electrons with the same spin reside in the overlap
region.
e. The two atoms found above the plane in a trigonal bipyramidal electron domain geometry are axial atoms.
36. State whether each of the following are True or False.
a. The central atom in a trigonal pyramidal molecule has a sp2 hybridization.
b. sp2 hybridized atoms will always have a double bond.
c. The O atom in water is sp hybridized.
d. A pi (π) bond has most of its electron density between the atomic nuclei.
e. A triple bond is comprised of 1 pi (π) and 2 sigma (σ) bonds.
f. It is not possible to rotate through a carbon–carbon double bond without breaking the π bond.
37. For each molecule, draw the hybrid orbital picture and include all labeled hybrid and atomic orbitals (e.g.,
all 3 sp2 hybrid orbitals), all valence electrons, show orbital overlap by drawing a bond between the atoms,
and label the bonds σ or π . (Note: Not all instructors have students draw HO molecule pictures; skip if not covered.)
a.
b.
c.
d.
Molecular Orbital Theory
38. Write the entire molecular electronic configuration including core electrons and determine the bond order
for each molecule. (Note: Some instructors provide the MO diagram; others do not.) a. Li2- b. O239. Using molecular orbital theory, which molecule(s) could not exist?
I. He2+2 II. B2-2
III. H2-2 IV. Be2-2 V. Li2-2
a. II
b. III
c. IV
d. III, V
(go for a walk)
e. II, III, V
40. Which of the following statements is incorrect?
a. As the overlap between orbitals increase as a covalent bond forms, the energy of the interaction decreases; at
this minimum energy the distance between the nuclei is the bond length.
b. Single bonds in molecular orbital theory are comprised of one sigma or pi bond.
c. Molecular orbital theory explained why oxygen is paramagnetic.
d. Hund’s rule applies to electrons in molecular orbitals.
e. Molecular orbital theory explained why electrons follow the Pauli Exclusion Principle while Valence Bond
Theory did not.
41. Using MO theory, place the following in order of increasing bond order (smallest bond order on the left).
I. N2
II. N2+1
III. N2-2
a. I, II, III
b. II, I, III
c. II, III, I
d. III, I, II
e. III, II, I
42. Based on MO theory, which of the following are diamagnetic?
I. N2+2
II. O2-2
III. He2-2
IV. C2+2
a. I, II
b. II, III
c. I, II, III
d. I, II, IV
e. I, II, III, IV
43. For the two molecules, I. N2+2 and II. O2-, answer the following questions. (Note: Some instructors provide the
MO diagram; others do not.)
a. Draw the MO diagram. Include the electron configuration for each atom, label all AOs and MOs, and include
electrons for both the AOs and MOs.
b. What is the molecular orbital electron configuration?
c. What is the bond order? Is this theoretical molecule stable?
d. Is the molecule paramagnetic or diamagnetic?
e. Does the bond length increase or decrease when one electron is removed?
CHAPTER 10
Gas Laws & Stoichiometry
44. If 64.5L acetylene, C2H2, was at 50.0˚C and 1.00atm, what will its volume be at 400.0˚C and 1.00atm?
a. 2.43
b. 79.8
c. 134
d. 156
e. 269
45. At 2.00 x 103˚C and 25.0 torr, the density of a certain element in the gas state is 1.153 x 10-2 g/L. What
element is it?
a. Zn
b. H
c. He
d. Ni
e. Ne
46. A valve between a 10L tank at 1atm and another 20L tank at 10atm both containing the same gas is opened.
Assuming that the temperature remains constant at 300K, what is the final pressure in the tanks after the value is
opened?
a. 4atm
b. 5atm
c. 6atm
d. 7atm
e. 8atm
47. Two different gases that have equal volumes, and the same pressure and temperature will have the same
a. boiling point
b. melting point
c. mass
d. density
e. number of molecules
48. If 8.0L of oxygen gas at 25˚C has the temperature increased to 50. ˚C while the pressure is held constant,
what is the new volume?
a. 4.0L
b. 8.0L
c. 8.7L
d. 16L
e. 25L
(pizza time)
49. What will the new pressure be if a container holding gas has its volume tripled and its temperature cut to 1/4
of its original value?
a. unchanged.
b. 3/4 times the original pressure.
c. 4/3 times the original pressure.
d. 1/12 times the original pressure.
e. 12 times the original pressure.
50. What volume does 1.0 x 102g of SO2 at 750torr and 95˚C occupy?
a. 3100L
b. 12L
c. 0.063L
d. 48L
e. 3400L
51. Two glass bulbs are connected with a valve. One bulb is 2.0L with 7.0 x 102torr He and the second bulb is
3.0L and empty. The valve is opened and He travels between the 2 bulbs; what is the final He pressure in torr?
52. A container holding gas has its volume reduced to half of what it was while at the same time the number of
molecules in the flask is tripled. During this change, the temperature is held constant. What has happened, if
anything, to the pressure in the flask?
a. It is unchanged.
b. It is doubled.
c. It is tripled.
d. It is increased by a factor of one and a half. e. It is increased by a factor of six.
53. During an experiment, a 3.07g sample of gas occupied 2.56L at 25.0˚C and 700. torr. The gas could be:
a. N2
b. CO2
c. Kr
d. O2
e. C2H2
54. How many gas molecules are in a 1.74L gas sample at 0.136atm and 25.0˚C?
a. 5.82 x 1021
b. 0.00967
c. 6.94 x 1022
d. 6.23 x 1025
e. 103
55. a. Draw a graph that shows the relationship between P and V as described by Boyle’s law, PV = constant.
What variables are assumed to be constant in this relationship?
b. Draw a graph showing the velocity distribution of a gas. On the same graph show how the distribution
changes as the temperature changes. Label which graph has the lower and higher temperatures.
56. From the reaction P4S3(s) + 8O2(g) → P4O10(s) + 3SO2(g), how many grams of P4S3 (MW = 220.1g/mol)
are needed to produce 10.0L of SO2(g) at 25.0˚C and 700. torr?
a. 0.125g
b. 27.6g
c. 82.8g
d. 195g
e. 2.10 x 104g
57. 2.00L of butane (C4H10) at 1.00atm and 0.00˚C is combusted with excess oxygen gas to yield water vapor
and carbon dioxide gas. The products are captured at 235˚C and 1.00atm. What volume of carbon dioxide was
captured? (Start by writing a balanced reaction.)
58. A mixture is prepared from 8.0L of ammonia and 9.0L chlorine measured at the same pressure and
temperature; these compounds react according to the following equation:
2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g)
When the reaction is completed, what volume of HCl was produced? Assume the final volumes are also
measured under the same pressure and temperature.
Partial Pressure & Kinetic Molecular Theory
59. What is the partial pressure (in torr) of oxygen in a container that contains 2.0mol of oxygen gas, 3.0mol of
nitrogen gas, and 1.0 mol of carbon dioxide gas with a total pressure is 900torr?
a. 100
b. 200
c. 300
d. 400
e. 600
(another nap)
60. If the gases below were in one container, which gas will have the greatest partial pressure?
a. 1g CO
b. 1g O2
c. 1g Ar
d. 1g N2O
e. all have the same
61. Two glass bulbs are connected with a valve. One bulb is 3.0L and contains 500. torr H2 while the other
bulb is 7.0L and contains 200. torr He. The valve is turned and the H2 and He mix within the 2 bulbs. What is
the partial pressure in torr of He in the bulbs?
62. If a sample of H2 is collected from a 2.00L container over water at 12.0 ˚C and 100.0 mmHg, how many
grams of H2 (MW = 2.016 g/mol) were collected? The vapor pressure of water at 12.0˚C is 10.5 mmHg.
a. 2.03 x 10-2g
b. 2.26 x 10-2g
c. 2.51 x 10-2g
d. 4.12 x 10-2g
e. 15.4g
63. 0.90 grams of water are placed into an empty 1.0-L flask at 45˚C and some of the water vaporizes. The
vapor pressure of water at 45˚C is 61.5 mmHg. How many grams of water remain at the bottom of the flask?
a. 0.90g
b. 0.84g
c. 0.51g
d. 0.39g
e. 0.060g
64. Place the following gases in order of decreasing root-mean square velocities. CO2, N2H4, Kr, C2H6
a. CO2 > Kr > C2H6 > N2H4
b. Kr > CO2 > N2H4 > C2H6
c. N2H4 > CO2 > C2H6 > Kr
d. C2H6 > N2H4 > CO2 > Kr
e. C2H6 > CO2 > N2H4 > Kr
65. What is the temperature of a gas (molar mass = 72.0g/mol) with a root mean square velocity of 525m/s?
a. 1.52 K
b. 796 K
c. 1520 K
d. 7.96 x 104 K
e. 8.06 x 104 K
66. Which system will have the greatest kinetic energy? All systems contain 1 mol of gas at STP.
a. H2
b. He
c. CO
d. not enough info e. all are the same
67. If the average kinetic energy of Ne gas was 25kJ, what is the temperature of the Ne?
68. If a gas has a root mean square velocity of 500. m/s at 50.0˚C, what is the temperature when the root mean
square velocity is 750. m/s?
a. 50.0˚C
b. 144˚C
c. 250.˚C
d. 454˚C
e. 485˚C
69. Five 1-L containers are at the same T and P and contain the following gases: C3H8
SO2
F2
Ar
Answer the next 7 questions using the above information.
A. Which container will have the highest density?
a. C3H8
b. SO2
c. F2
d. Ar
e. all are the same
B. Which container will have the greatest mass?
a. C3H8
b. SO2
c. F2
d. Ar
e. all are the same
C. Which container will have the greatest number of molecules?
a. C3H8
b. SO2
c. F2
d. Ar
e. all are the same
D. Which container will have molecules with the greatest kinetic energy?
a. C3H8
b. SO2
c. F2
d. Ar
e. all are the same
E. Which container will have molecules with the greatest velocity?
a. C3H8
b. SO2
c. F2
d. Ar
e. all are the same
F. Which container will have molecules with the greatest number of atoms?
a. C3H8
b. SO2
c. F2
d. Ar
e. all are the same
G. If all the gases were combined into one flask which gas would have the greatest partial pressure?
a. C3H8
b. SO2
c. F2
d. Ar
e. all are the same
(almost done!)
70. Which molecule has a root-mean-square velocity equal to that of carbon dioxide at the same temperature T?
a. CH4
b. C2H6
c. C3H8
d. C4H10
e. C5H12
(yea done!)
ANSWERS
1. c {the C–H bond in CH4 is polar covalent; only when the atoms are the same will we classify them as nonpolar covalent}
2. d {since the atoms are close to one another and are nonmetals the bond is a covalent one (eliminates choice “e”); since the
atoms are different the bond is polar (eliminates choices “a” and “b”); since ENO > ENS the partial negative charge resides on O
(eliminates choice “c”)}
3. See below for Lewis dot structures.
4. c {draw LDS for each molecule; 10e- around S in SF4}
5. a.
b.
6. a {NH3 is a molecular compound (2 nonmetals) and molecular compounds have much lower melting points than ionic
compounds; as Lattice Energy ↑ ⇒ mp ↑ → KBr has +1/-1 charges while BeO and MgS have +2/-2 charges; higher charges will
yield higher melting points so drop KBr; the two ionic compounds have the same charges:
Be+2O-2 and Mg+2S-2; to differentiate between these the smaller the ions the higher the lattice energy and the higher the melting
point; ions higher in a column on the Periodic Table are smaller, and therefore Be+2 is smaller than Mg+2; likewise, O-2 will be
smaller than S-2; so BeO will have the highest melting point}
7. a {draw LDS of each structure and determine BO; as BO↑ → bond strength↑ → N2 has largest BO and strongest bond
a:
BO = 3; b:
BO = 2; c:
BO = 2; d:
BO = 1}
8. e {Draw Lewis dot structure; has resonance but this does not change FC for the S atom; S: 6 - 4 = 2}
9. d {Most ionic means greatest ΔEN which means further apart from one another on the PT.}
10. Structure IV can be dropped because of the high FC. Structure I can be dropped because it has fewer zeros
than Structures II or III. The difference between Structures II and III is a -1 FC on the O atom and a FC of 0
on the S (Structure II) and the reverse of that for Structure III. Since O is more electronegative than S the O
atom should have the more negative FC. Therefore Structure II is the best structure.
(-1)
(0)
(0)
(0)
O
O
O
O
S
(+1)
S
(0)
S
(-1)
-2
(-2)
S
O
O
O
O
O
O
O
O
(-1)
(-1)
(-1)
(-1)
(-1)
(0)
(0)
(0)
I
II
III
IV
11. Lower formal charges (closer to zero) are better. Two resonance structures have two 0’s and one –1 for formal charges. These
are better structures than the structure with a –2 formal charge. To determine between these 2 resonance structures which is
better, consider the electronegativity of S and N. Since N has a greater EN, it should have the more negative formal charge.
Hence, the resonance structure with 2 double bonds is the best structure in which the –1 formal charge is on the N and not the S.
:.S.
C N
. .:
(0)
(0) (-1)
-
..
: .S. C N :
(-1) (0)
(0)
:S
.. C N
. .:
(+1) (0) (-2)
12. I. b {compare each pair of atoms; the more EN atom in the pair gets a δ- while the less EN atom gets a δ+}
II. polar covalent
13. a. polar covalent b. nonpolar covalent c. nonpolar covalent d. ionic e. polar covalent
14. a. IE2(Mg) {Reaction E: Mg+(g) + 1/2O2(g) → Mg+2(g) + 1/2O2(g); the only change is Mg+(g) → Mg+2(g) which is IE2}
b. 129kJ
{Mg(l) + 1/2O2(g) → Mg(g) + 1/2O2(g); the only change is Mg(l) → Mg(g) which is ΔHvap for Mg}
c. Bond enthalpy or bond dissociation energy for O2
{Reaction F: Mg+2(g) + 1/2O2(g) → Mg+2(g) + O(g); the only
change is 1/2O2(g) → O(g) which is O2 being broken; i.e., bond enthalpy}
MgO(s) → Mg+2(g) + O-2(g)
d.
e. the reverse of Reaction I {Reaction I is: Mg+2(g) + O-2(g) → MgO(s) which is the reverse of the lattice energy reaction}
f. 3929kJ {lattice energy = A + B + C + D + E + F + G + H = 601 + 9 + 129 + 738 + 1451 + 299 + -142 + 844 = 3929kJ}
H
H
C
+ H
C
H
15. -42kJ
O
H
H
H
{Write LDS for reaction:
ΔHrxn = Σ(bonds broken) – Σ(bonds made);
H
H
C
C
H
H
O
H
;
ΔHrxn = [1(DC=C) + 4(DC–H) + 2(DO–H)] – [1(DC–C) + 5(DC–H) + 1(DC–O) + 1(DO–H)] =
[1(614) + 4(413) + 2(463)] – [1(348) + 5(413) + 1(358) + 1(463)] = -42kJ}
O
O
+ 2 Cl
C
16. 242kJ/mol {Write LDS for reaction:
H
Cl
+ 2H
C
H
Cl
Cl
Cl
{ΔHrxn = Σ(bonds broken) – Σ(bonds made) = [2(DC–H) + 1(DC=O) + 2(DCl–Cl)] – [(2(DC–Cl) + 1(DC=O) + 2(DH–Cl)];
-208 = [2(413) + 1(799) + 2(x)] – [(2(328) + 1(799) + 2(431)]; x = 242}
17. d {from VSEPR table}
18. c {repulsions: lone pair–lone pair > bonding pair–lone-pair > bonding pair > bonding pair}
19. a. T {ΔEN ≠ 0 → bonds are polar; since they have similar EN and are both nonmetals → covalent bonds}
b. T
c. T
d. F
e. T
f. F
{from LDS;
}
{from LDS}
{the molecular geometry is square pyramid}
{all molecules with square pyramid molecular geometry are polar}
{from molecular geometry there are only angles of 90˚ and 180˚, no 120˚ angles}
20. a.
b.
F
F
F
Xe
Xe
Xe
-
F
-
F
F
F
Xe
F
F
P
F
P
c.
21. d
22. d
23. d
24.
F
EDG
F
MG
F
MG; vectors cancel;
no dipole; nonpolar
d.
F
F
LDS
EDG
F
P
F
F
MG
-
F
F
P
F
LDS
-
F
F
net dipole
F
MG; polar
{draw LDS; 1 lone pair and 4 bonding pairs around the central atom;
}
{Draw Lewis dot structures and compare to VSEPR table; planar = linear; trigonal planar; bent; T-shaped; and square planar}
{Draw Lewis dot structures; use VSEPR to determine shapes and angles. NF3: trigonal pyramid; SF6: octahedral;
CF4: tetrahedral; SeF4: seesaw – 120˚ angles}
b {GeCl4-2: “seesaw” structure - always polar; “a” and “c” – tetrahedral with same atoms around central atom →
nonpolar; “d” – square planar with same atoms around central atom → nonpolar}
25. e
26. b
{CCl4: tetrahedral; SF4: trigonal bipyramid; SiCl4-2: trigonal bipyramid; SeI2: tetrahedral; KrF2: trigonal bipyramid}
27. a.
b.
28. a {the six locations in an octahedral electron domain are equivalent; in addition, they are generally not referred to as equatorial
and axial}
29. a {this is an atomic size question; smaller atoms → closer together, smaller bond length and stronger bond energy}
30. 3.32 x 10-21C {µ = Q x d; 0.122D(3.34 x 10-30Cm/1D) = Q (122.8pm)(1m/(1 x 1012pm));
4.075 x 10-31Cm = Q(1.228 x 10-10m); Q = 4.075 x 10-31Cm/(1.228 x 10-10m); Q = 3.318 x 10-21C}
a. sp2 {draw the Lewis dot structure; 3 atoms + 0 lone pairs = 3 → sp2; or trigonal planar → sp2}
31.
b. sp3 {draw the Lewis dot structure; 2 atoms + 2 lone pairs = 4 → sp3; or tetrahedral → sp3}
32. Lone pairs drawn:
a. sp3 b. 2 c. 109.5˚ d. sp2-sp2
e. 1.5 {BOresonance = #bonds/#locations = 9/6 = 1.5} f. σ bond: 1s(H)-sp(C) g. tetrahedral h. bent i. 23σ, 7π
33. a {every bond has 1σ bond; a double bond = 1σ + 1π; a triple bond = 1σ + 2π}
34. a. See picture to right.
b. BO = 3 {BO = 1 for single bond; BO = 2 for double bond; BO = 3 for triple bond}
c. sp3 {4 domains ⇒ sp3}
d. trigonal planar {3 domains ⇒ trigonal planar}
e. ~109.5˚ (or 104.5˚) {tetrahedral electron domain around the S atom}
f. 1s {H atoms are not hybridized and remain 1s}
g. BO = 1.5 {resonance:
; BO =
# bonds
; BO = 3/2 = 1.5}
# locations
h. σ = sp(C)–sp(C); π = p(C)–p(C); π = p(C)–p(C)
i. 23s bonds; 6p {single bond = σ bond; double bond = σ bond + π bond; triple bond = σ bond + 2π bonds}
35. d {electrons must have opposite spins}€
36. a. F {sp3} b. F {for example, the B in BF3 is sp2 but does not have a double bond.} c. F {sp3}
d. F {π bond electrons lie above and below the internuclear axis} e. F {1 σ and 2 π bonds} f. T
37. a.
b.
c.
d.
Note: The p orbitals in CO2 can not be drawn this way since the px and py orbitals have to lie on different axes:
38.
- this structure wrong!
2
*
2
2
*
1
a. (σ1s) (σ1s ) (σ2s) (σ2s ) ; BO = 0.5 {BO = (4 – 3)/2 = 1/2}
b. (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3; BO = 1.5 {BO = (10 – 7)/2 = 3/2}
{when BO = 0 ⇒ molecule doesn’t exist; when total #e- is 4, 8, or 20 ⇒ BO = 0; He2+2 = 2e- ⇒ exists; B2-2 = 12e- ⇒
exists; H2-2 = 4e- ⇒ doesn’t exists as BO = 0; Be2-2 = 10e- ⇒ exists; Li2-2 = 8e- ⇒ doesn’t exists as BO = 0}
e {MO Theory explained why oxygen was paramagnetic, why some chemicals were colored, and how electrons could be
excited. Neither MOT nor VBT address the Pauli Exclusion Principle.}
e {#bond and #antibonding e- determined from MO e- config;
39. d
40.
41.
BO N2 = 14e- ⇒ (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4(σ2p)2 = (10 – 4)/2 = 6/2 = 3;
BO N2+1 = 13e- ⇒ (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4(σ2p)1 = (9 – 4)/2 = 5/2 = 2.5;
BO N2-2 = 16e- ⇒ (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)2 = (10 – 6)/2 = 4/2 = 2}
42. c {N2+2 = 12e- ⇒ (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4 ⇒ all e- paired;
O2-2 = 18e- ⇒ (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)4 ⇒ all e- paired;
He2-2 = 6e- ⇒ (σ1s)2(σ1s*)2(σ2s)2 ⇒ all e- paired; C2+2 = 10e- ⇒ (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)2 ⇒ 2e- unpaired;
I, II, and III are diamagnetic}
43. I. a. See below. b. (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(π2p)4
c. BO = 2; stable because BO > 0 {BO = (8 – 4)/2 = 4/2 = 2} d. diamagnetic
e. increase {BO = (7 – 4)/2 = 3/2 = 1.5 → BO decreased from 2.0 to 1.5 → bond length increases}
II. a. See below. b. (σ1s)2(σ1s*)2(σ2s)2(σ2s*)2(σ2p)2(π2p)4(π2p*)3
c. BO = 1.5; stable because BO > 0 {BO = (10 – 7)/2 = 3/2 = 1.5} d. paramagnetic
e. decrease {BO = (10 – 6)/2 = 4/2 = 2.0 → BO increased from 1.5 to 2.0 → bond length decreases}
I. a.
II. a.
44. c {P1V1/T1 = P2V2/T2 and solve for V2; V2 = P1V1T2/P2T1 = (1)(64.5)(273+400)/(1)(273+50) = 134.4L}
45. a {MW = (DRT)/P and solve for MW; MW = [(1.153 x 10-2)(0.0821)(2000+273.15)]/[(25torr)(1atm/760torr)] = 65.38}
46. d {use P1V1 = P2V2 two times and add the partial pressures; for the 20L tank: (10)(20) = (P)(30);
P = 6.677atm; for the 10L tank: (1)(10) = (P)(30); P = 0.333atm; PT = 6.677+0.333 = 7.0atm; another way to do this would be to
calculate moles in each tank, add the moles, and then use the total moles to find the total pressure: mol in 10 L tank: n = PV/RT =
(1)(10)/[(0.0821)(300)] = 0.406 mol; mol in 20 L tank: n = PV/RT = (10)(20)/[(0.0821)(300)] = 8.120 mol;
nT = 0.406 + 8.120 = 8.526 mol; VT = 10L + 20L = 30L; PT = nTRT/VT = [(8.526)(0.0821)(300)]/(30) = 7.00 atm}
47. e {2 gases with same P, V, and T have same number of moles and the same number of molecules}
48. c {V1/T1 = V2/T2 and solve for V2; V2 = V1T2/T1 = [(8)(50+273.15)]/(25+273.15) = 8.67L}
49. d {P1V1/T1 = P2V2/T2 and solve for P2; let V2 = 3V1 and let T2 = 1/4T1; P1V1/T1 = P23V1/(1/4)T1; simplify:
P1 = P23/(1/4); solve for P2 → P2 = P1(1/4)(1/3) = 1/12P1}
50. d {PV = nRT and solve for V; V = nRT/P =
[(100g SO2)(1mol SO2/64g SO2)(0.0821)(95+273.15)]/[(750torr)(1atm/760torr)] = 47.8L}
51. 280 torr {P1V1 = P2V2 and solve for P2; (700)(2) = PHe(5); PHe = 280 torr}
52. e {Use P1V1 = n1RT and divide by P2V2 = n2RT; P1V1/P2V2 = n1RT/n2RT; simplify: P1V1/P2V2 = n1/n2; solve for P2:
P2 = P1V1n2/V2n1; plug in values: V1 = 1 and V2 = 1/2, n1 = 1 and n2 = 3; P2 = P1[(1)(3)]/[(1/2)(1)] = 6P1; P2 has increased
by a factor of 6 times larger than the original P1}
53. d
{MW = gRT/PV and solve for MW; MW = gRT/PV = [(3.07)(0.0821)(25+273.15)]/[(700torr)(1atm/760torr)(2.56)] =
31.85g/mol ≈ 32g/mol → O2; all other choices have different MW}
54. a {PV = nRT and solve for n; n = PV/RT = [(0.136)(1.74)]/(0.0821)(25+273.15)] =
(0.00967mol)(6.022 x 1023molecules/1mol) = 5.823 x 1021 molecules}
55. a. See graph at right. T and n are assumed constant.
b. See graph at right.
56. b {P,V,TA → gB; PV = nRT and solve for nSO2; n = PV/RT = [(700 torr)(1atm/760torr)(10)]/[(0.082057)(25+273.15) =
0.3765mol SO2; (0.3765mol SO2)(1mol P4S3/3mol SO2)(220.1g P4S3/1mol P4S3) = 27.62 g P4S3}
57. 14.9L {P,V,TA → P,V,TB; write the balanced reaction: 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g); PV = nRT and solve
for mol C4H10; nC4H10 = PV/RT = [(1)(2)]/[(0.082057)(0+273.15)] = 0.08923 mol C4H10; find mol CO2 →
(0.08923mol C4H10)(8mol CO2/2mol C4H10) = 0.3569 mol CO2; find volume CO2; PV = nRT and solve for V;
V = nCO2RT/P = [(0.3569)(0.082057)(235+273.15)]/(1) = 14.88L CO2}
58. 18.0L HCl {Avogadro’s Law of V ∝ n since same T and P; limiting reagent problem since 2 reactant quantities are given;
LA → LB for each reactant and then the smaller amount of product is the amount that can be made;
(9L Cl2)(6L HCl/3L Cl2) = 18L HCl; (8L NH3)(6L HCl/2L NH3) = 24L HCl; the smaller amount can be produced → 18L HCl}
59. c {PO2 = χO2PT and solve for PO2; χO2 = (2mol O2)/(2+3+1) = 0.333; PO2 = (0.333)(900) = 300 torr}
60. a {the one with smallest molar mass has more molecules and a greater mole fraction and hence, a greater partial pressure}
61. 140 torr {since the 2 gases can be treated independently, this is a P1V1 = P2V2 problem; (200)(7) = PHe(10);
solve for PHe = 140 torr}
62. a {find PH2 from PT = PH2O + PH2; PH2 = PT - PH2O = 100 - 10.5 = 89.5 mmHg; use PH2V = nH2RT and solve for nH2;
nH2 = PH2V/RT = [(89.5 mmHg)(1atm/760mmHg)(2)]/[(0.0821)(12.0+285.15)] = 0.01007mol H2;
(0.01007mol H2)(2.016g H2/1mol H2) = 0.0203 g H2}
63. b
64.
{find grams of H2O in the gas phase; original mass of water – amount of water in gas phase = amount of water that remains;
use PV = nRT to find nH2O; n = PV/RT = [(61.5mmHg)(1atm/760mmHg)(1)]/[(0.0821)(45+273.15)] = 0.003098mol H2O →
(0.003098mol H2O)(18.0g H2O/1mol H2O) = 0.0558g H2O(g) vaporized; 0.90g H2O(l) – 0.0558g H2O(g) = 0.844g H2O(l)
remains at the bottom of the flask}
d {Smaller molecules travel faster than larger molecules at the same T; determine MW and arrange from smallest to largest}
3RT
3(8.314)T
; 525 =
; 525 = 346.42T ; 275,625 = 346.42T; T = 796.1K}
MW
0.072
66. e {KE α T; all are at the same temperature so same KE}
67. 2000K {KE = (3/2)RT; 25,000J = (3/2)(8.314)(T); T = 2004K ≈ 2000K}
€
u €
T1
T1
750
€
=
68.
d {use rms1
;
; 2.25 = T1/323; T1 = 727K - 273 = 454˚C; or use u rms =
=
u rms2
T2 500
50 + 273
65. b
{ u rms =
3RT
MW
3(8.314)(323)
8,056
; 250,000 =
;
MW
MW
€
€
3(8.314)(T)
€
MW = 0.03223kg/mol; then use the MW = 0.03223kg/mol and urms = 750 to find T: 750 =
;
0.03223
2 times: first find MW with T = 323K and urms = 500: 500 =
562,500 =
3(8.314)(T)
; T = 727K – 273˚C =€454˚C}
0.03223
€
€
€
69. A. b
B. b
C. e
D. e
E. c
F. a
G. e
70.
3.
{under the same T, P, and V conditions, gas with greatest molar mass has greatest D}
{under the same T, P, and V, 2 gases have the same #moles; the one with greater molar mass will have the greater mass}
{under the same T, P, and V, 2 gases will have the same number of moles and hence the same number of molecules}
{all have the same kinetic energy since they are at the same T and kinetic energy is proportional to T}
{since it is the smallest molecule and smaller molecules travel faster than larger ones at the same T}
{since they contain the same number of moles, the molecule with the most atoms, 11 for C3H8, has the most atoms total}
{since they contain same #moles, they’ll have the same partial pressure since partial pressure is determined by #molecules;
Χ for each will be the same}
c {At the same T, a molecule must weigh the same to have the same velocity; CO2 weighs 44g/mol;
C3H8 weighs 44g/mol}