Key to Even Numbered Assigned Homework, Zumdahl 9th Edition
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Exam 1
Chpt. 1 -
56.
20 g
68.
V = 1.0 x 102 cm3 , D = 6.2 g/cm3
78.
5.1 x 103 cm
84.
Initially we have a mixture of two substances, Mg and S. After the
reaction we have a pure substance since we have only the compound MgS.
118.
Density of dry sand = 1.45 g/mL
Density of methanol = 0.7913 g/mL
(average) Density of a sand particle = 1.9 g/mL
Chpt. 2 -
40.
Mass ratios of R to Q are the following.
Compound 1
4.67 g R : 1.00 g Q
Compound 2
1.56 g R : 1.00 g Q
The ratio of the mass ratios is 4.67/1.56 = 2.99 (essentially 3). The ratio
of the mass ratios is itself a small whole number ratio (3/1) and therefore
consistent with the law of multiple proportions.
94.
(a)
26 p+, 24 e-,
FeO,
iron (II) oxide
(b)
26 p+, 23 e-,
Fe2O3,
iron (III) oxide
(c)
56 p+, 54 e-,
BaO,
barium oxide
(d)
55 p+, 54 e-,
Cs2O,
cesium oxide
(e)
16 p+, 18 e-,
Al2S3 ,
aluminum sulfide
(f)
15 p+, 18 e-,
AlP,
aluminum phosphide
(g)
35 p+, 36 e-,
AlBr3 ,
aluminum bromide
(h)
7 p+, 10 e-,
AlN,
aluminum nitride
Chpt. 3 -
126.
99.8 g
134.
empirical formula is C3H5O2
molecular formula is C6H10O4
150.
(a)
If more than 40 g of Na is added, Na is present in excess and Cl2 is
the limiting reactant.
(b)
50.8 g of NaCl
(c)
61.7 g of Cl2
(d)
102 g NaCl
(e)
for part (b), 30.9 g of Cl2 remain.
for part (a), 10.0 g of Na remain.
Exam 2
Chpt. 4 -
60.
2.9 g AgCl, [Ca2+] = 0.075 M, [Cl-] = 0.050 M, [NO3-] = 0.10 M
78.
0.01528 M
92.
173 mL
94.
(a) 14.2% MgCl2
(b) 8.95 mL of AgNO3
Chpt. 5 -
70.
2.28 L
78.
C2H2Cl2
116.
(a) 1.224 atm
(b) 1.223 atm
(c) results agree to within 0.08%
(d) In problem 115 the pressure is significantly higher
(approximately 12 atm) and the results differ by 0.91%. The ideal
gas equation works better at lower pressures.
Chpt. 6 -
154.
XCO = 0.291, XCO2 = 0.564, XO2 = 0.145
72.
H = -108.7 kJ
100.
(a)
By the surroundings on the system.
(b)
No work is done
(c)
By the system on the surroundings.
(d)
By the surroundings on the system.
(e)
By the system on the surroundings.
106.
124.
6.66 kJ/OC
Pathway 1
-4.05 x 103 J
Pathway
-2.03 x 103 J
Work is not a state function. This is demonstrated here since the initial
and final states are the same for both pathways but the value of the work is
different for the two pathways.
130.
HO = 285.8 kJ/mol
EO = 282.1 kJ/mol
Exam 3
Chpt. 7 -
40.
m
52.
134.4 nm
76.
78.
80.
82.
86.
1p,
0 electrons
6dx2-y2,
2 electrons
4f,
14 electrons
7py,
2 electrons
2s,
2 electrons
n=3,
18 electrons
(a)
0
(b)
1
(c)
9
(d)
0
(e)
2
Cu: [Ar]4s23d9 (using periodic table), [Ar]4s13d10 (actual)
O:
1s22s22p4
La:
[Xe]6s24f1 expected but actual [Xe]6s25d1
Y:
[Kr]5s24d1
100.
Ba:
[Xe]6s2
Tl:
[Xe]6s24f145d106p1
Bi:
[Xe]6s24f145d106p3
(a)
excited state of boron,
1 unpaired electron,
B ground state: 1s22s22p1,
1 unpaired electron
ground state of neon,
0 unpaired electrons
Ne ground state: 1s22s22p6,
0 unpaired electrons
excited state of fluorine,
3 unpaired electrons
F ground state: 1s22s22p5,
1 unpaired electron
excited state of iron,
6 unpaired electrons
Fe ground state: [Ar]4s23d6,
4 unpaired electrons
(b)
(c)
(d)
114.
Be<B<C<O<N
B and O are exceptions to the general ionization energy trend. The
ionization energy of O is lower because of the extra electron-electron
repulsions present when two electrons are paired in the same orbital. This
makes it slightly easier to remove an electron from O compared to N. B is
an exception because of the smaller penetrating ability of the 2p electron
in B compared to the 2s electrons in Be. The smaller penetrating ability
makes it slightly easier to remove an electron from B compared to Be. The
correct ionization energy ordering, taking into account the two exceptions,
is B < Be < C < O < N.
116.
The general ionization energy trend says that ionization energy increases
going left to right across the periodic table. However, one of the
exceptions to this trend occurs between Groups 2A and 3A. Between
these two groups, Group 3A elements usually have a lower ionization
energy than Group 2A elements. Therefore, Al should have the lowest
first ionization energy value, followed by Mg, with Si having the largest
ionization energy. Looking at the values for the first ionization energy in
the graph, the green plot is Al, the blue plot is Mg, and the red plot is Si.
Mg (the blue plot) is the element with the huge jump between I2 and I3.
Mg has two valence electrons, so the third electron removed is an inner
core electron. Inner core electrons are always much more difficult to
remove than valence electrons since they are closer to the nucleus, on
average, than the valence electrons.
.
124.
O; the electron-electron repulsions will be much more severe for O− + e− 
O2− than for O + e−  O−, resulting in O having the more exothermic
(favorable) electron affinity.
126. (a) The electron affinity (EA) of Mg2+ is ΔH for Mg2+(g) + e → Mg+(g); this is just the
reverse of the second ionization energy (I2) for Mg. EA(Mg2+) = I2(Mg) = 1445
kJ/mol (Table 7.5). Note that when an equation is reversed, the sign on the
equation is also reversed.
(b) I1 of Cl is ΔH for Cl(g)  Cl(g) + e; I1(Cl ) = EA(Cl) = 348.7 kJ/mol (Table 7.7)
(c) Cl+(g) + e  Cl(g)
ΔH =  I1(Cl) = 1255 kJ/mol = EA(Cl+) (Table 7.5)
(d) Mg(g)  Mg(g) + e
ΔH = EA(Mg) = 230 kJ/mol = I1(Mg )
138.
1.57 x 104 J, 7.70 x 1027 photons
164.
n=4
174.
Size also decreases going across a period. Sc and Ti along with Y and Zr are adjacent
elements. There are 14 elements (the lanthanides) between La and Hf, making Hf
considerably smaller.
Chpt. 8 58.
Mg(s)  Mg(g)
ΔH = 150. kJ
(sublimation)
Mg(g)  Mg+(g) + e
ΔH = 735 kJ
Mg+(g)  Mg2+(g) + e
ΔH = 1445 kJ
F2(g)  2 F(g)
ΔH = 154 kJ
2 F(g) + 2 e  2 F(g)
(IE1)
(IE2)
(BE)
ΔH = 2(328) kJ (EA)
Mg2+(g) + 2 F(g)  MgF2(s)
ΔH = 2913 kJ
Mg(s) + F2(g)  MgF2(s)
ΔH of = 1085 kJ/mol
(LE)
60.
Na(g)  Na+(g) + e
F(g) + e  F(g)
Na(g) + F(g)  Na+(g) + F(g)
ΔH = IE = 495 kJ (Table 7.5)
ΔH = EA = 327.8 kJ (Table 7.7)
ΔH = 167 kJ
The described process is endothermic. What we haven’t accounted for is the extremely favorable lattice energy. Here, the lattice energy is a large negative (exothermic) value, making the
overall formation of NaF a favorable exothermic process.
64.
Lattice energy is proportional to the charge of the cation times the charge of the
anion Q1Q2.
Compound
Q1Q2
Lattice Energy
FeCl2
(+2)( 1) = 2
2631 kJ/mol
FeCl3
(+3)( 1) = 3
5359 kJ/mol
Fe2O3
(+3)( 2) = 6
14,744 kJ/mol
82. a. H2CO has 2(1) + 4 + 6 = 12 valence
b. CO2 has 4 + 2(6) = 16 valence electrons.
electrons.
c.
HCN has 1 + 4 + 5 = 10 valence electrons.
84.
a. POCl3 has 5 + 6 + 3(7) = 32 valence electrons.
O
Cl
O
P
Cl
Cl
P
Cl
Cl
Cl
Skeletal
structure
Lewis
structure
Note: This structure uses all 32 e while satisfying the octet rule for all atoms. This is a valid Lewis
structure.
SO42 has 6 + 4(6) + 2 = 32 valence electrons.
2-
O
O
S
O
O
Note: A negatively charged ion will have additional electrons to those that come from the valence shell
of the atoms. The magni-tude of the negative charge indicates the number of extra electrons to add in.
XeO4, 8 + 4(6) = 32 e
O
O
Xe
O
O
PO43, 5 + 4(6) + 3 = 32 e
d. Molecules ions that have the same number of valence electrons and the same number of
atoms will have similar Lewis structures.
88.
SF6, 6 + 6(7) = 48 e
ClF5, 7 + 5(7) = 42 e
XeF4, 8 + 4(7) = 36 e
OCN has 6 + 4 + 5 + 1 = 16 valence electrons.
104.
Formal
charge
O
C
N
O
C
N
O
C
N
0
0
-1
-1
0
0
+1
0
-2
Only the first two resonance structures should be important. The third places a positive formal
charge on the most electronegative atom in the ion and a –2 formal charge on N.
CNO will also have 16 valence electrons.
Formal
charge
C
N
O
C
N
O
C
N
O
-2
+1
0
-1
+1
-1
-3
+1
+1
All the resonance structures for fulminate (CNO) involve greater formal charges than in cyanate
(OCN), making fulminate more reactive (less stable).
106.
N
The nitrogen-nitrogen bond length of 112 pm is between a double (120 pm) and a triple (110
pm) bond. The nitrogen-oxygen bond length of 119 pm is between a single (147 pm) and a
double bond (115 pm). The third resonance structure shown below doesn’t appear to be as
important as the other two since there is no evidence from bond lengths for a nitrogen-oxygen
triple bond or a nitrogen-nitrogen single bond as in the third resonance form. We can
adequately describe the structure of N2O using the resonance forms:
N
O
N
N
O
Assigning formal charges for all three resonance forms:
N
N
O
N
N
O
N
N
O
-1
+1
0
0
+1
-1
-2
+1
+1
For:
, FC = 5 - 4 - 1/2(4) = -1
N
, FC = 5 - 1/2(8) = +1 , Same for
N
, FC = 5 - 6 - 1/2(2) = -2 ;
N
O ,
FC = 6 - 4 - 1/2(4) = 0 ;
O ,
FC = 6 - 2 - 1/2(6) = +1
N
O
N
and
,
FC = 5 - 2 - 1/2(6) = 0
,
FC = 6 - 6 - 1/2(2) = -1
N
We should eliminate N‒N≡O because it has a formal charge of +1 on the most electronegative
element (O). This is consistent with the observation that the N‒N bond is between a double and
triple bond and that the N‒O bond is between a single and double bond
112.
From the Lewis structures (see Exercise 88), XeF4 would have a square planar molecular
structure, and ClF5 would have a square pyramid molecular structure.
134.
X
The general structure of the trihalide ions is:
X
X
Bromine and iodine are large enough and have low-energy, empty d orbitals to accommodate the
expanded octet. Fluorine is small, and its valence shell contains only 2s and 2p orbitals (four orbitals)
and cannot expand its octet. The lowest-energy d orbitals in F are 3d; they are too high in energy
compared with 2s and 2p to be used in bonding.
158.
This molecule has 30 valence electrons. The only C–N bond that can possibly have a doublebond character is the N bound to the C with O attached. Double bonds to the other two C–N
bonds would require carbon in each case to have 10 valence electrons (which carbon never
does). The resonance structures are:
O
H
C
H
N
O
C
H
H
C
H
N
H
H
C
C
H
H
H
H
H
C
H
H
Chpt. 9 32.
No, the CH2 planes are mutually perpendicular to each other. The center C atom is sp
hybridized and is involved in two π bonds. The p orbitals used to form each π bond
must be perpendicular to each other. This forces the two CH2 planes to be
perpendicular.
H
H
C
H
C
C
H
50.
Considering only the 12 valence electrons in O2, the MO models would be:
p*




 p*

 p



p


2s*


2s

O2 ground state
Arrangement of electrons consistent
with the Lewis structure (double bond
and no unpaired electrons).
It takes energy to pair electrons in the same orbital. Thus the structure with no unpaired
electrons is at a higher energy; it is an excited state.
72.
a. Yes, both have four sets of electrons about the P. We would predict a tetrahedral structure
for both. See part d for the Lewis structures.
b. The hybridization is sp3 for P in each structure since both structures exhibit a tetrahedral
arrangement of electron pairs.
c. P has to use one of its d orbitals to form the π bond since the p orbitals are all used to form
the hybrid orbitals.
d. Formal charge = number of valence electrons of an atom − [(number of lone pair electrons)
+ 1/2(number of shared electrons)]. The formal charges calculated for the O and P atoms
are next to the atoms in the following Lewis structures.
O -1
Cl
+1
P
0
O
Cl
Cl
Cl
P
0
Cl
Cl
In both structures, the formal charges of the Cl atoms are all zeros. The structure with the
P=O bond is favored on the basis of formal charge since it has a zero formal charge for all
atoms.
74.
N2 (ground state): (σ2s)2(σ2s*)2(π2p)4(σ2p)2; B.O. = 3; diamagnetic (0 unpaired e)
N2 (1st excited state): (σ2s)2(σ2s*)2(π2p)4(σ2p)1(π2p*)1
B.O. = (7 ‒ 3)/2 = 2; paramagnetic (2 unpaired e)
The first excited state of N2 should have a weaker bond and should be paramagnetic.
76.
dxz
+
pz
x
x
z
z
The two orbitals will overlap side to side, so when the orbitals are in phase, a π bonding
molecular orbital would be expected to form.
90.
One of the resonance structures for benzene is:
H
C
H
H
H
C
C
C
C
C
H
H
To break C6H6(g) into C(g) and H(g) requires the breaking of 6 C‒H bonds, 3 C=C bonds, and 3
C‒C bonds:
C6H6(g)  6 C(g) + 6 H(g)
ΔH = 6DC‒H + 3DC=C + 3DC‒C
ΔH = 6(413 kJ) + 3(614 kJ) + 3(347 kJ) = 5361 kJ
The question asks for H f for C6H6(g), which is ΔH for the reaction:
6 C(s) + 3 H2(g) → C6H6(g)
ΔH = ΔH f , C 6 H 6 ( g )
To calculate ΔH for this reaction, we will use Hess’s law along with the value ΔH f for C(g) and
the bond energy value for H2 ( D H 2 = 432 kJ/mol).
6 C(g) + 6 H(g)  C6H6(g)
6 C(s)  6 C(g)
ΔH2 = 6(717 kJ)
3 H2(g)  6 H(g)
ΔH3 = 3(432 kJ)
6 C(s) + 3 H2(g)  C6H6(g)
kJ/mol
ΔH1 = 5361 kJ
ΔH = ΔH1 + ΔH2 + ΔH3 = 237 kJ; ΔH f , C 6 H 6 ( g ) = 237
The experimental ΔH of for C6H6(g) is more stable (lower in energy) by 154 kJ than the
ΔH of calculated from bond energies (83  237 = 154 kJ). This extra stability is related to
benzene’s ability to exhibit resonance. Two equivalent Lewis structures can be drawn for
benzene. The π bonding system implied by each Lewis structure consists of three localized π
bonds. This is not correct because all C‒C bonds in benzene are equivalent. We say the π
electrons in benzene are delocalized over the entire surface of C6H6 (see Section 9.5 of the text).
The large discrepancy between ΔH of values is due to the delocalized π electrons, whose effect
was not accounted for in the calculated ΔH of value. The extra stability associated with benzene
can be called resonance stabilization. In general, molecules that exhibit resonance are usually
more stable than predicted using bond energies.
92.
The π bonds between S atoms and between C and S atoms are not as strong. The p atomic
orbitals do not overlap with each other as well as the smaller p atomic orbitals of C and O
overlap.