CHAPTER 2
CHEMICAL BONDS
2.1
The coulombic attraction is directly proportional to the charge on each ion
(Equation 1) so the ions with the higher charges will give the greater
3
2
coulombic attraction. The answer is therefore (b) Ga , O .
2.3


The Li ion is smaller than the Rb ion (58 vs 149 pm). Because the
lattice energy is related to the coulombic attraction between the ions, it
will be inversely proportional to the distance between the ions (see
Equation 2). Hence the larger rubidium ion will have the lower lattice
energy for a given anion.
2.5
(a) 5;
2.7
(a) [Ar];
2.9
(a) [Ar]3d ;
2.11
(b) 4;
(c) 7;
(d) 3
10
2
(b) [Ar]3d 4s ;
10
5
(c) [Kr]4d ;
14
10
2
(b) [Xe]4 f 5d 6s ;
10
2
(d) [Ar]3d 4s
10
(c) [Ar]3d ;
14
10
(d) [Xe]4 f 5d
10
2
(a) [Kr]4d 5s ; same; In+ and Sn2+ lose 5p valence electrons.
(b) none
10
(c) [Kr] 4d ; Pd
2
(b) Fe ;
3
(b) Fe ;
2.13
(a) Co ;
2.15
(a) Co ;
2.17
(a) 4s;
2
(c) Mo ;
2
2
(d) Nb
3
(c) Ru ;
3
3
(d) Mo

(b) 3p;
(c) 3p;
(d) 4s
SM-73
2.19
(a) 1;
(b) 2 ;
(c) +1;
(d) +3 (+1 sometimes observed);
(e) +2
2.21
(a) 3;
(b) 6;
(c) 6;
(d) 2
2.23
10
2
(a) [Kr]4d 5s ; no unpaired electrons;
10
(b) [Kr] 4d ; no unpaired
14
4
electrons; (c) [Xe]4 f 5d ; four unpaired electrons;
unpaired electrons;
2.25
(a) 3p;
(b) 5s;
2.27
(a) +7;
(b) 1;
(d) [Kr]; no
(e) [Ar]3d 8 ; two unpaired electrons
(c) 5p;
(d) 4d
(c) [Ne] for +7, [Ar] for 1;
(d) electrons are lost
or added to give noble-gas configuration.
2.29
(a) Mg 3As 2;
2.31
(a) Bi 2O 3;
(b) In2S3 ;
(b) PbO2;
(c) AlH 3;
(d) H 2Te;
(e) BiF3
(c) Tl2 O 3
2.33
Cl
(a)
Cl
C
O
(b)
Cl
Cl
C
Cl
Cl
(c)
O
N
F
F
(d)
F
SM-74
N
F
2.35
H
(a) H
B
Br
(b)
H
O
H
H
(c)
N
H
2.37
O
(a)
H
H
C
(b)
H
H
C
O
H
H
(c)
2.39
H
H
O
N
C
C
H
H
O
H
The structure has a total of 32 electrons; of these, 21 are accounted for by
the chlorines (3 Cl’s  7 valence electrons each); of the 11 electrons
remaining, 6 come from the oxygen. This leaves 5 electrons unaccounted
for; these must come from E. Therefore, E must be a member of the
nitrogen family and since it is a third period element E must be
phosphorous (P)
SM-75
2.41
(
H
(a)
H N H
K
b K
) K
Cl
H
(
3_
P
Na
c
Cl O
)
2.43
H
H
H
C
C
C
C
C
H
H
C
C
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
H
H
H
C
C
C
H
H
C
C
C
Cl
N
O
N
C
C
H
H
H
2.47
(a)
0
+1
N
O
-1
(d) C
-1
C
+
(b)
-2
(e)
0
0
N
N
-1
0
C
N
SM-76
-
C
C
O
-1
+1
(c) C
O
H
C
C
Cl
O
H
C
C
2.45
O
C
C
C
H
C
C
C
H
H
C
C
C
H
H
C
C
C
H
C
C
C
H
C
C
C
H
C
C
C
C
H
H
H
2.49
2.51
1-
H
-1
..
.. 0
0
..
.. +1
S
..
O
..
..
..
O
..
0
lower energy
H
-1
0
..
O
-1
..
S
..
O
..
-1
10
O
..
O
..
..
..
O
..
+2
H
..
O
..
..
..
-1
+1
-1
..
S
..
(b)
..
O
..
1-
..
..
O
..
..
0
0
O
..
-1
..
..
O
-1
O
..
..
S
..
O
..
..
(a)
1-
H
lower energy
2.53
(a) In the first structure, the formal charges at Xe and F are 0, whereas, in
the second structure, Xe is 1, one F is 0, and the other F is +1. The first
structure is favored on the basis of formal charges.
(b) In the first
structure, all of the atoms have formal charges of 0, whereas, in the second
SM-77
structure, one O atom has a formal charge of +1 and the other O has a
formal charge of 1. The first structure is thus preferred.
2.55
(a) The sulfite ion has one Lewis structure that obeys the octet rule:
-2
O
S
O
O
and three with an expanded octet:
-2
-2
O
S
O
O
S
-2
O
O
S
O
O
O
O
The structures with expanded octets have lower formal charges.
(b) There is one Lewis stucture that obeys the octet rule:
-
O
S
H
O
O
The formal charge at sulfur can be reduced to 0 by including one double
bond contribution. This change gives rise to two expanded octet
structures.
-
O
S
H
O
-
O
O
S
H
O
O
Notice that, unlike the sulfite ion, which has three resonance forms, the
presence of the hydrogen ion restricts the electrons to the oxygen atom to
which it is attached. Because H is electropositive, its placement near an
oxygen atom makes it less likely for that oxygen atom to donate a lone
pair to an adjacent atom.
SM-78
(c) The perchlorate ion has one Lewis structure that obeys the octet rule:
..
..
..
O
..
..
Cl
..
..
O
..
O
..
..
..
O
..
The formal charge at Cl can be reduced to 0 by including three doublebond contributions, thereby giving rise to four resonance forms.
..
..
O
..
..
O
..
Cl O
..
..
..
O
..
..
O
..
..
Cl
O
..
..
..
O
..
..
..
O
..
..
..
..
O
..
..
O
..
..
..
Cl
O
..
O
..
..
O
..
..
Cl
O
..
..
..
O
..
..
O
For the nitrite ion, there are two resonance forms, both of which obey the
octet rule:
O
2.57
N
O
O
N
O
Radicals are species with an unpaired electron, therefore only (b) and (c)
are radicals since they have an odd number of electrons while (a) and (d)
have an even number of electrons allowing Lewis structures to be drawn
with all electrons paired.
SM-79
2.59
The Lewis Structures are
(a)
Cl
(b)
O
Cl
O
O
Cl
O
(c)
O
N
O
Cl
Radicals are species with an unpaired electron, therefore (a) and (c) are
radicals.
2.61
(a)
+
Cl
I
Cl
lone pairs
-
Cl
(b)
I has 2 bonding pairs and 2
Cl
I has 4 bonding pairs and 2
lone pairs
I
Cl
Cl
I has 3 bonding pairs and 2
Cl
(c)
lone pairs
Cl
I
Cl
I has 5 bonding pairs and 1
Cl
(d)
Cl
Cl
lone pair
I
Cl
Cl
SM-80
2.63
(a)
(b)
F
F
F
F
S
F
Xe
F
F
F
12 electrons
10 electrons
F
(c)
F
F
Cl
Cl
(d)
Te
As
F
Cl
F
Cl
F
12 electrons
10 electrons
2.65
(a)
(b)
O
F
Xe
(c)
F
F
F
Xe
F
2 lone pairs
O
F
Xe
F
2 lone pairs
F
F
1 lone pair
2.67
I (2.7) < Br (3.0) < Cl (3.2) < F (4.0)
2.69
In (1.8) < Sn (2.0) < Sb (2.1) < Se (2.6)
2.71
(a) The bond in HCl would be more ionic. The electronegativity
difference is greater between H and Cl then between H and I, making the
HCl bond more ionic.
(b) The bonds in CF4 would be more ionic. The electronegativity
difference is greater between C and F than between C and H, making the
C—F bonds more ionic.
SM-81
F
(c) C and S have nearly identical electronegativities, so the C—S bonds
would be expected to be almost completely covalent, whereas the C—O
bonds would be more ionic.
2.73
Rb  < Sr2 < Be 2 ; smaller, more highly charged cations have greater
polarizing power. The ionic radii are 149 pm, 116 pm, 27 pm,
respectively.
2.75
O 2 < N 3 < Cl  < Br  ; the polarizability increases as the ion gets larger
and less electronegative. The ionic radii for these species are 140 pm, 171
pm, 181 pm, 196 pm, respectively.
2.77
(a) CO 3 2 > CO 2 > CO
2
CO 3 will have the longest C—O bond length. In CO there is a triple
bond and in CO 2 the C—O bonds are double bonds. In carbonate, the
bond is an average of three Lewis structures in which the bond is double in
one form and single in two of the forms. We would thus expect the bond
order to be approximately 1.3. Because the bond length is inversely related
to the number of bonds between the atoms, we expect the bond length to
be longest in carbonate.
(b) SO32 > SO2 ~ SO3
Similar arguments can be used for these molecules as in part (a). In SO2
and SO3 , the Lewis structures with the lowest formal charge at S have
double bonds between S and each O. In the sulfite ion, however, there are
three Lewis structures that have a zero formal charge at S. Each has one
S—O double bond and two S—O single bonds. Because these S—O
bonds would have a substantial amount of single bond character, they
would be expected to be longer than those in SO2 or SO3 . This is
consistent with the experimental data that show the S—O bond lengths in
SM-82
SO2 and SO3 to be 143 pm, whereas those in SO32 range from about 145
pm to 152 pm depending on the compound.
(c) CH 3NH2 > CH 2NH > HCN
The C—N bond in HCN is a triple bond, in CH 2NH it is a double bond,
and in CH 3NH2 it is a single bond. The C—N bond in the last molecule
would, therefore, be expected to be the longest.
2.79
(a) The covalent radius of N is 75 pm, so the N—N single bond in
hydrazine would be expected to be ca. 150 pm. The experimental value is
145 pm.
(b) The C—O bonds in carbon dioxide are double bonds. The
covalent radius for doubly bonded carbon is 67 pm and that of O is 60 pm.
Thus we predict the CRO in CO 2 to be ca. 127 pm. The experimental
bond length is 116.3 pm.
(c) The C—O bond is a double bond so it
would be expected to be the same as in (b), 127 ppm. This is the
experimentally found value. The C—N bonds are single bonds and so one
might expect the bond distance to be the sum of the single bond C radius
and the single bond N radius (77 plus 75 pm) which is 152 pm. However,
because the C atom is involved in a multiple bond, its radius is actually
smaller. The sum of that radius (67 pm) and the N single bond radius gives
142 pm, which is close to the experimental value of 133 pm.
(d) The
N—N bond is a double bond so we expect the bond distance to be two
times the double bond covalent radius of N, which is 2  (60 pm) or 120
pm. The experimental value is 123.0 pm.
2.81
(a) 77 pm + 72 pm = 149 pm
(b) 111 pm + 72 pm = 183 pm
(c) 141 pm + 72 pm = 213 pm. Bond distance increases with size going
down Group 14/IV.
SM-83
2.83
+1
-1
+1
N
N
N
N
N
+1
N
N
N
-1
+1
N
N
+1
-1
N
N
+1
N
N
2.85
(a)
-2
O
0
-2
O
-1
-1
O
C
O
0
0
C
O
C
O
C
O
0
-2
O
C
O
O
-2
O
O
C
C
O
O
C
O
(b)
-1
0
Br
O
O
(c)
-1
-1
C
C
-2
2.87
(a)
O
CH3
C
O
CH3
O
SM-84
C
O
N
(b)
H
H
C
H
C
O
H
C
C
CH3
O
CH3
(c)
CH3
2.89
O
H
C
N
CH3
O
H
C
N
P and S are larger atoms that are less able to form multiple bonds to
themselves, unlike the small N and O atoms. All bonds in P4 and S8 are
single bonds, whereas N 2 has a triple bond and O 2 a double bond.
2.91
(a)
SM-85
(b)
N
N
N
N
N
N
+ suitable resonance form
SM-86
2.93
(a) In H2O2, hydrogen peroxide, the oxidation number of oxygen is -1; the
oxygen in water has an oxidation number of -2. Since oxygen is gaining
electrons on going from H2O2 to H2O it is undergoing reduction.
Ascorbic acid (C6H8O6), on the other hand is undergoing oxidation: the
oxidation number of carbon changes from +2/3 to +1 on conversion from
ascorbic acid to C6H8O6 which means it is losing elecctrons.
(b) The Lewis structures for hydrogen peroxide and water, respectively,
are
H
0
0
O
O
0
H
H
O
H
Note that all atom have formal charges of zero, suggesting that there is no
driving force for this conversion based solely on formal charges; however,
since we know that hydrogen peroxide does react to form water in the
lung, and that this conversion is borne by the change in oxidation states,
oxidation numbers are far more useful in determining whether a material
has been oxidized or reduced.
2.95
(a)
MI
d(M–I), pm
(1 − d*/d)/d
Lattice Energy, kJ∙mol-1
Li
274
3.19 × 10-3
759
NaI
294
3.00 × 10-3
700
KI
329
2.72 × 10-3
645
RbI
345
2.61 × 10-3
632
CsI
361
2.51 × 10-3
601
A high correlation (R2 = 0.9731) exists between lattice energy and d(M–I).
A better fit (R2 = 0.9847) is obtained between lattice energy and
(1 − d*/d)/d.
(b) Using the equation L.E. = 218331(1 − d*/d)/d) + 54.887 and the Ag-I
distance (309 pm), the estimated AgI lattice energy is 683 kJ∙mol-1.
SM-87
(c) There is not close agreement between the estimated (683 kJ∙mol-1) and
experimental (886 kJ∙mol-1) lattice energies. A possible explanation is that
the Ag+ ion is more polarizable than the alkali metal cations of similar size
and therefore the bonding in AgI is more covalent.
2.97
(a)
2_
O
H
H
H
H
O
(b) All the atoms have formal charge 0 except the two oxygen atoms,
which are -1. The negative charge is most likely to be concentrated at the
oxygen atoms.
(c) The protons will bond to the oxygen atoms. Oxygen atoms are the
most negative sites in the molecule and act as Lewis bases due to their
lone pairs of electrons. The resulting compound is named hydroquinone.
O H
H
H
H
H
O H
SM-88
2.99
Number of
Z
Configuration
unpaired e
Element
Charge
Energy state
26 [Ar]3d6
4
Fe
2+
ground
52 [Kr] 4d105s25p56s1
2
Te
2-
excited
16 [Ne]3s23p6
0
S
2-
ground
39 [Kr]4d1
1
Y
2+
ground
30 [Ar] 3d84s2
2
Zn
2+
excited
2.101
(a)
H
O
N
C
H
O
H
O
N
C
H
N
C
O
C
N
O
H
C
S
O
C
H
(c)
O
-1
+1
(b)
-1
+1
H
S
O
+1
-1
H
C
H
(d)
N
N
+1
-1
-1
+1
O
N
C
N
SM-89
O
N
2.103
H
R
H
H
H
R'
C
C
C
C
C
H
H
H
R
H
H
H
R'
C
C
C
C
C
R
H
H
H
R'
C
C
C
C
C
H
2.105 (a) I: Tl2O3; II: Tl2O; (b) 3+; 1+; (c) [Xe]4f145d10; [Xe] 4f145d106s2; (d)
Because compound II has a lower melting point, it is probably more
covalent which is consistent with the fact that the 3+ ion is more
polarizing.
2.107 (a)
-1
..
-1
..
..
-1
..
..-1
-1
..-1
Cl O
..
+2
O
..
..
-1
..
O
..
+2
O
..
..
O
..
..
..
Cl O
..
-1
..
..
O
..
..
SM-90
+3
O
..
..
..
-1
..-1
..
..
+2
O
..
..
O
..
O
Cl O
..
..
.. -1
Cl O
..
..
-1
-1
..
..
..
..
-1
..
+2
O
..
..
O
..
..
..
..
Cl O
..
..
..
..
O
..
-1
O
-1
..
O
..
..
..
Cl O
..
O
..
..
O
.. Cl O
..
..
..
-1
..
..
..
-1
..
..
..
O
..
..
..
O
O
O
O
-1
..
..
..
..
-1
+1
..
..
O
.. Cl O
..
..
..
..
..
..
O
..
..
..
..
-1
..
..
..
..
..
..
O
.. Cl O
..-1
O
..
O
O
Cl O
..
O
..
O
..
..
O
.. Cl O
..
O
.. -1
+1
-1
..
..
..
..
O
O
O
..
..
..
+1
-1
..
Cl O
..
..
..
..
Cl O
..
O
..
..
O
..
O
..
..
..
..
..
-1
..
..
..
..
..
..
-1
O
O
..
-1
+1
O
-1
..
..
Cl O
..
..
+1
O
..
-1
..
O
..
-1
..
+1
O
..
..
Cl O
..
..
O
..
O
..
..
..
Cl O
..
..
-1
..
..
..
O
..
..
..
-1
..
..
..
O
..
O
H
The four structures with three double bonds (third row) and the one with
four double bonds are the most plausible Lewis structures according to
formal charge arguments because these five structures minimize the
formal charges. (b) The structure with four double bonds fits these
observations best since its bond lengths would all be 140 pm, or only 4 pm
shorter than the observed length. However, the four structures with three
double bonds also fit because, if the double bonds are delocalized by
resonance, we can estimate the average bond length to be
1
4
(170 pm)  34 (140 pm) 147.5 pm , or just 3.5 pm longer than observed.
(c) 7+; The structure with all single bonds fits this criterion best. (d)
Approaches (a) and (b) are consistent but approach (c) is not. This result
is reasonable because oxidation numbers are assigned by assuming ionic
bonding.
2.109 The alkyne group has the stiffer C—H bond because a large force
constant, k, results in a higher-frequency absorption.
2.111 Look at the Lewis structures for the molecules:
-1
+1
(a) C
O
(d)
-1
+1
O
S
(b) O
O
(e)
(c)
O
-1
+1
N
N
O
-1
+1
O
O
O
(f)
Ar
Of these, a, c, d, and e can all function as greenhouse gases
2.113 Since the halogen all have an odd number of valence electrons (7),
interhalogen compounds of the type XX’n will be radicals and therefore
SM-91
extremely reactive unless the total number of halogens is an even number.
This can only be achieved if n is odd. Look at ICl2 vs. ICl3 as examples:
I
Cl
I
Cl
Cl
Cl
Cl
2.115
(a) Compare the length of a S-S bond to that of a Cl-Cl bond. From
Table 2.5, the Cl-Cl bond length is 199 pm; here the S-S bond length in
S2F2 is reported to be 190 pm. From Table 2.5 it can be determined that,
on average, an X-Y bond is approximately 20 pm longer that a X=Y
bond; this suggests that the S-S bond (which is 9 pm shorter than a Cl-Cl
single bond) has some double bond character.
(b) and (c) The Lewis structures for the two possible S2F2 are:
F
S
S
F
F
+1
-1
S
S
F
isomer 1
favored
F
+1
-1
S
S
F
S
S
isomer 2
F
F
favored
If resonance is occurring then one would expect that the S-S bond length
is indeed between a single and a double bond in length.
2.117
(a) The Lewis structures of NO and NO2 are:
N
O
-1
+1
O
N
(best possible structure)
O
O
+1
-1
N
O
(equivalent resonance structures)
SM-92
From Table 2.4, the average bond dissociation energy of a N=O bond is
630 kJ mol-1, which is right in line with the Lewis structure of NO. The
bond dissociation energy of each NO bond in NO2 is 469 kJ mol-1, which
is about half-way between a N-O double and an N-O single bond,
suggesting that the resonance picture of NO2 is a reasonable one.
(b) An N-O single bond should have a bond length of 149 pm while an
NO double bond should have a bond length of 120 pm (values are
obtained by summing the respective covalent radii values from Figure
2.21). From Table 2.5 the length of a N-O triple bond can be estimated to
be between 105 and 110 pm. Since NO itself has a bond length of 115
pm, this suggests that its actual bond order is somewhere between that of
a double and a triple bond.
0
0
(c)
N
O
O
N
O
0
+1
-1
O
0
(d)
0
0
+1
N
O
+1
O
N
O
O
-1
-1
(e) N2O5(g) + H2O(l)  2 HNO3(aq); Nitric acid is produced
(f)
4.05 g N 2O5 
Molarity =
1 mol N 2O5
2 mol HNO3

= 7.50  10-2 mol HNO3
108.02 g N 2O5
1 mol N 2O5
7.50

10-2 mol HNO3
= 7.50
1.00 L

10-2 mol  L-1 HNO3
(g) The oxidation number of nitrogen for the various nitrogen oxides are
as follows: NO: +2; NO2: +4; N2O3: +3; and N2O5: +5. An oxidizing
agent is a species that wants to gain electrons; based on oxidation
number, N2O5 should be the most potent oxidizing agent of the nitrogen
oxides, as it possesses the most positive oxidation number for N of the
group.
SM-93