CHEMISTRY 103 – Practice Problems #3
Chapters 7 – 9 (with some Ch 6 review questions)
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 6-Review
1. What neutral atom has the electron configuration in the ground state: 1s22s22p63s23p64s23d6?
2. Which atom or ion in the ground state has the following electron configuration, 1s22s22p63s23p63d3?
a. Cu
b. V
c. Ti+2
d. Mn+4
e. Sc-2
3. Which atom or ion is the most paramagnetic in the ground state?
a. Si
b. Mn+2
c. Nd. Fe+2
e. P
4. Which ion/molecule is isoelectronic with PH3?
a. O
b. N c. CH
e. none of the above
3
3
d. F2
4
CHAPTER 7
5. Place the following atoms in order of decreasing atomic radii: Cl, Se, Ge, Sn
a. Se > Ge > Sn > Cl
b. Sn > Ge > Se > Cl
c. Cl > Se > Ge > Sn
d. Cl > Ge > Se > Sn
e. None of the above are correct.
6. Sort these atoms/ions from smallest to largest in size.
(smallest)
<
<
O-, Se-2, O-2, F-, Ne
<
<
(largest)
7. Which of the following statements is incorrect.
a. The size of atoms increase right to left on the periodic table because Zeff increases.
b. The ionization energy increases as you move up a column because the atom is smaller and it is more difficult
to remove an electron.
c. Metals have low first ionization energies and hence easily become cations.
d. Nonmetals have high first electron affinities and hence easily become anions.
e. Metals are more reactive as you move down the column because the first ionization energy is decreasing.
8. Which of the following reactions corresponds to the third ionization energy reaction for bromine?
a. Br-2(g) + e- → Br-3(g)
b. Br-3(g) → Br(g) + 3e-
d. Br2+2(g) → Br2+3(g) + e-
e. Br+2(g) → Br+3(g) + e-
c. Br(g) → Br+3(g) + 3e-
9. Rank the atoms from lowest to highest second ionization energy? Ca, Rb, K
a. Ca < Rb < K
b. Rb < Ca < K
c. Rb < K < Ca
d. K < Rb < Ca
(take a nap)
e. Ca < K < Rb
10. Place the following atoms, N, O, S, in order of increasing first electron affinity (more exothermic/more
negative values on the right).
a. N < O < S
b. N < S < O
c. S < O < N
d. O < S < N
e. S < N < O
11. Which has the greatest effective nuclear charge?
a. 2p e- in B
b. 2s e- in Li
c. 2p e- in Na
d. 3s e- in Al
e. all have the same Zeff
12. Which atom has the lowest third ionization energy?
a. O
b. F
c. Mg
d. P
e. all have same IE3
13. Which reaction represents the first electron affinity for fluorine?
a. F2(g) + e- → F2(g)b. F(g) → F(g)+ + ed. F(g)- + e- → F(g)-2
c. F2(g) + 2e- → 2F(g)-
e. F(g) + e- → F(g)-
14. Which reaction listed below would be the same as Cl+2(g) + e- → Cl+(g)?
a. The reverse of the first ionization energy reaction for Cl.
b. The first electron affinity reaction for Cl.
c. The second ionization energy reaction for Cl.
d. The second electron affinity reaction for Cl.
e. None of these reactions.
15. Which atom would most likely align with the following ionization energies:
IE1 = 578kJ/mol IE2 = 1817kJ/mol IE3 = 2745kJ/mol IE4 = 11,577kJ/mol
a. Li
b. Mg
c. C
d. Al
e. none of the above
16. Which statement below is incorrect?
a. Metallic character increases as you go right to left on the periodic table.
b. Nonmetal oxides when reacted with water form bases.
c. Metals are ductile and malleable while nonmetals are not.
d. The reactivity of sodium is greater than the reactivity of potassium.
e. The size of anions are greater than neutral atoms because of electron-electron repulsions.
17. Which atom/ion will require the most energy to remove an electron?
a. Na
b. Pc. F+
d. Cl+
e. Si
18. Which atom would require the greatest amount of energy to remove the third electron?
a. Mg
b. N
c. Ca
d. Al
e. Na
19. Determine the energy for each reaction below using the data given on the right.
IE1 for F = 1681kJ/mol
I. F(g) + Cl-(g) → F(g)- + Cl(g)
IE2 for F = 3374kJ/mol
IE3 for F = 6050kJ/mol
EA1 for F = -328kJ/mol
II. O(g)- + Al(g)+2 → O(g)-2 + Al(g)+3
(eat an apple)
IE1 for O = 1314kJ/mol
IE2 for O = 3388kJ/mol
IE3 for O = 5300kJ/mol
EA1 for O = -141kJ/mol
EA2 for O = 844kJ/mol
IE1 for Cl = 1251kJ/mol
IE2 for Cl = 2297kJ/mol
IE3 for Cl = 3826kJ/mol
EA1 for Cl = -349kJ/mol
IE1 for Al = 577kJ/mol
IE2 for Al = 1817kJ/mol
IE3 for Al = 2745kJ/mol
EA1 for Al = -42kJ/mol
CHAPTER 8
20. 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.
21. 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.
22. 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
23. Which molecule, if any, violates the octet rule?
a. NH4+
b. CF4
c. SF4
d. BrO4-
e. none do
24. Using Lewis dot structures, show how the reactions below occur.
a. 2Li(s) + F2(g) → 2LiF(s)
b. 2Ca(s) + O2(g) → 2CaO(s)
25. Place the 4 chemicals in order of lowest to highest melting point? BeO; NH3; MgS; KBr
(lowest mp)
<
<
<
(highest mp)
26. Which molecule has the strongest nitrogen-nitrogen bond?
a. N2
b. N3c. N2H2
d. N2H4
e. all are equivalent
27. What is the formal charge on the S in SO3?
a. -2
b. -1
c. 0
d. +1
e. +2
28. 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
O
S
O
II
-2
O
O
S
O
III
O
O
IV
29. 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.
(watch some TV)
30. I. In the molecule shown, CF2O, choose the structure with the correct locations of the δ+ and δ- symbols.
+
Oδ
Cδ
F
F δ
δ-
Oδ
Cδ
+
Oδ
+
Cδ
F
F δδ
Oδ
C δ+
F
F δ
δ-
F
δ+
a.
b.
c.
d.
II. Would the C–O bond be nonpolar covalent, polar covalent, or ionic?
F +
δ
31. 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).
32. 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
Bond enthalpy
(kJ/mol)
348
413
358
463
Bond
C=C
C≡C
C=O
Bond enthalpy
(kJ/mol)
614
839
799
33. 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
34. Name the molecular geometry of a chemical with a central atom that has 2 lone pairs and 3 bonding pairs of e-?
35. 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
(time for a walk)
36. 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˚. ______
37. 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 representing the net dipole if the molecule is polar otherwise write “no dipole”
if the molecule is nonpolar and there is no net vector.
a. H2O b. NH3 c. XeF2 d. PF438. The molecular geometry for BrF4+ is
a.
b.
c.
d.
e.
Images: Public Domain from Wikipedia.org (http://en.wikipedia.org/wiki/VSEPR_theory)
39. 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
40. Which molecule has an angle of ~120˚?
a. NF3
b. SF6
c. CF4
d. SeF4
e. none
41. Which molecule is polar?
a. PH4+
b. GeCl4-2
d. XeF4
e. none
c. SiCl4
42. 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.
43. 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
44. 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
(time for another nap)
45. 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
46. 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. A polar molecule placed in a microwave will heat up when the microwave is turned on.
c. Pi bonds are created by 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.
47. a. What is the hybridization for the S atom in SO3? b. What is the hybridization for the S atom in H2S?
48. 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?
49. How many sigma (σ) and pi (π) bonds are present in the molecule shown?
H
C
C
a. 9σ, 3π
C
C
N
H
H
H
H
b. 10σ, 3π
c. 4σ, 3π
d. 7σ, 5π
50. Answer the questions below about the structure and bonding in this molecule.
O
First, draw in any lone pairs needed to complete octets.
f
a. What is the bond order for the bond marked “a”?
O
C
b. What is the hybridization on the C atom marked “b”?
c. What is the molecular geometry around the C atom marked “c”?
H
C
d. What is the angle on the C-S-H group marked “d”?
e. What is the hybridization on the H atom marked “e”?
c
f. What is the bond order for the C-O bond marked “f”?
H
g. What orbitals overlap to form the bonds marked “a”?
h. How many σ and π bonds are in the molecule?
(pizza time)
e. 7σ, 3π
e
H
C
C
C
C
C
S
H
H
a
C
C
C
H
H
b
H
d
C
H
51. State whether each of the following are True (T) or False (F).
a. The central atom in a trigonal pyramid 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. ______
52. 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 .
a.
b.
c.
d.
Molecular Orbital Theory
53. Write the entire molecular electronic configuration including core electrons and determine the bond order for
each molecule. a. Li2- b. B254. 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
e. II, III, V
55. For the two molecules, I. N2+2 and II. O2-, answer the same set of questions:
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?
56. 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.
57. 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
(almost done!)
58. 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
(yea done!)
e. I, II, III, IV
ANSWERS
1. a. Fe
2. d {With charged transition metals, remove the s electrons first. Cu: [Ar]4s13d10; V: [Ar]4s23d3; Ti+2: [Ar]3d2;
Mn+4: [Ar]3d3; Sc-2: [Ar]4s23d3}
b {Mn: [Ar]4s23d5; Mn+2: [Ar]3d5 since s electrons are removed 1st}
3.
4. d {The number of electrons in PH3 is 15 + 3(1) = 18; only F2 has that number: 2(9) = 18e-}
5. b {Radii increase as you go to the left and down on the Periodic Table}
6. Ne < F- < O- < O-2 < Se-2 {Ne is farthest to the right and neutral so it is smallest; O is larger than F and by making both
atoms a -1 charge the O- is still larger than F-; as an ion becomes more anionic is becomes larger so -2 anion it will be larger than a
-1 anion and yields: O- < O-2; between O-2 and Se-2 since O < Se when both are neutral when both atoms become a -2 charge this
yields: O-2 < Se-2; combining this information yields: Ne < F- < O- < O-2 < Se-2}
a {The size of atoms increase right to left on the periodic table because Zeff decreases.}
7.
8. e {the third ionization energy refers to the third electron being removed from a single atom in the gas phase}
9. a {Look at positions of the +1 ions, Ca+, Rb+, K+ since the IE2 involves removing the 2nd electron from the
+1 ion. IE2 trend is higher as you go up and to the right on the Periodic Table. K+ is in the Ar position.}
a {N has EA ≈ 0; EA increase (more negative) up a period but 3rd period > 2nd period}
10.
11. c {Zeff = Z – S where S is the number of core electrons in shells below the electron of interest; a. B2p: Zeff = 5 – 2 = +3;
b. Li2s: Zeff = 3 – 2 = +1; c. Na2p: Zeff = 11 – 2 = +9; d. Al2p: Zeff = 13 – 10 = +3}
12. d
13.
14.
{IE increase as you go to the right and up on the Periodic Table; also, since we are considering the 3rd IE, think of the atoms
as +2 ions: O+2, F+2, Mg+2, and P+2; assign these locations on the PT, i.e., O+2 has 6 electrons so it is in the C position and so
on; P+2 is the only atom on the 3rd row and therefore has a lower IE3}
e {EA1: X(g) + e- → X(g)-}
e {correct reaction would have been IE2 reversed (IE2: Cl+(g) → Cl+2(g) + e-; now reversed: Cl+2(g) + e- → Cl+(g)); the
other reactions: “a”: Cl+(g) + e- → Cl(g); “b”: Cl(g) + e- → Cl-(g); “c”: Cl+(g) → Cl+2(g) + e-; “d”: Cl-(g) + e- → Cl-2(g)}
15. d {Since the IE4 is so large as compared to the other IE, the removal of the 4th electron breaks a noble gas
configuration. Only Al breaks a noble gas configuration during IE4: Al+3 → Al+4 + e-}
16. d {metal reactivity increases as you go down a column; hence, Kreactivity > Nareactivity}
17. c {removing an electron from a cation is more difficult than removing an electron from an anion or a neutral atom so the possible
answers are F+ and Cl+; the smaller the cation the harder it is to remove an electron; since F+ is smaller than Cl+ (recall F is
smaller than Cl) it is harder to remove the next electron from F+}
18. a {highest IE3 would be when a noble gas configuration was broken; this occurs with both Mg and Ca; since Mg would break a
Ne core and Ca would break a Ar core, Mg has higher IE3 than Ca}
19. I. 21kJ/mol {F(g) → F(g)-: EA1(F); Cl-(g) → Cl(g) : -EA1(Cl); ΔHrxn = EA1(F) + (-EA1(Cl)) = -328 + (-(-349)) = 21kJ/mol
II. 3589kJ/mol {O(g)- → O(g)-2: EA2(O); Al(g)+2 → Al(g)+3: IE3(Al); ΔHrxn = EA2(O) + IE3(Al) = 844 + 2745 =
3589kJ/mol}
III. 1455kJ/mol {O(g)- → O(g) + e-: -EA1(O); O(g) → O(g)+ + e-: IE1(O); ΔHrxn = -EA1(O) + IE1(O) = -(-141) + 1314 =
1455kJ/mol}
20. c {the C–H bond in CH4 is polar covalent; only when the atoms are the same will we classify them as nonpolar covalent}
21. 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”)}
22. See below for Lewis dot structures.
23. c {draw LDS for each molecule; 10e- around S in SF4}
24. a.
25. NH3 < KBr < MgS < BeO
b.
{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 KBr would be after NH3; the two remaining ionic compounds have the
same charges: Be+2O-2 and Mg+2S-2; to differentiate between these the smaller the ions have the higher 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 and
O-2 is smaller than S-2; so BeO will have the highest melting point}
26. 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}
27. e {Draw Lewis dot structure; has resonance but this does not change FC for the S atom; S: 6 - 4 = 2}
28. 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
29. 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)
30. 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
31. 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
32. -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
33. 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}
34. T-shaped {from VSEPR table}
35. c {repulsions: lone pair–lone pair > bonding pair–lone-pair > bonding pair > bonding pair}
36. 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}
37. a.
b.
F
F
F
Xe
Xe
Xe
-
F
-
F
F
F
Xe
F
F
P
F
P
c.
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
38. d {draw LDS; 1 lone pair and 4 bonding pairs around the central atom;
}
39. d {Draw Lewis dot structures and compare to VSEPR table; planar = linear; trigonal planar; bent; T-shaped; and square planar}
40. d {Draw Lewis dot structures; use VSEPR to determine shapes and angles. NF3: trigonal pyramid; SF6: octahedral;
41.
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}
42. e
43. b
{CCl4: tetrahedral; SF4: trigonal bipyramid; SiCl4-2: trigonal bipyramid; SeI2: tetrahedral; KrF2: trigonal bipyramid}
44. a.
b.
45. a {this is an atomic size question; smaller atoms → closer together, smaller bond length and stronger bond energy}
46. a {the six locations in an octahedral electron domain are equivalent; in addition, they are generally not referred to as equatorial
and axial}
47. a. sp2 {draw the Lewis dot structure; 3 atoms + 0 lone pairs = 3 → sp2; or trigonal planar → sp2
b. sp3 {draw the Lewis dot structure; 2 atoms + 2 lone pairs = 4 → sp3; or tetrahedral → sp3
}
}
48. 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
49. a {every bond has 1σ bond; a double bond = 1σ + 1π; a triple bond = 1σ + 2π}
50. See picture to right.
O
a. BO = 3 {BO = 1 for single bond; BO = 2 for double bond; BO = 3 for triple bond} f
e
H
O
C
b. sp3 {4 domains ⇒ sp3}
a
H
H
C
C
c. trigonal planar {3 domains ⇒ trigonal planar}
H
C
C
C
C
C
C
d. <109.5˚ (or 104.5˚) {tetrahedral electron domain around the S atom}
e. 1s {H atoms are not hybridized and remain 1s}
c
C
C
H
H
S
H
f. BO = 1.5 {resonance:
; BO =
# bonds
; BO = 3/2 = 1.5}
# locations
b
H
d
g. σ = sp(C)–sp(C); π = p(C)–p(C); π = p(C)–p(C)
h. 23s bonds; 6p {single bond = σ bond; double bond = σ bond + π bond; triple bond = σ bond + 2π bonds}
51. 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
52. 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:
53.
54.
- 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(π2p)3; BO = 1.5 {BO = (9 – 4)/2 = 3/2}
d {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}
H
55. 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.
56. 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.}
57. e {#bond and #antibonding e- determined from MO e- config;
58.
22.
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}
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}