Download CHAPTER 9 CHEMICAL BONDING I

CHAPTER 9
CHEMICAL BONDING I
9.1
A Lewis dot symbol consists of the symbol of an element and one dot for each valence electron in an atom of
the element. Lewis dot symbols mainly apply to representative elements.
9.2
Lithium (Group 1A, 1 valence electron, 2s ), Beryllium (Group 2A, 2 valence electrons, 2s ), Boron (Group
2 1
2 2
3A, 3 valence electrons, 2s 2p ), Carbon (Group 4A, 4 valence electrons, 2s 2p ), Nitrogen (Group 5A, 5
2 3
2 4
valence electrons, 2s 2p ), Oxygen (Group 6A, 6 valence electrons, 2s 2p ), Fluorine (Group 7A, 7 valence
2 5
electrons, 2s 2p ).
9.3
Please see Figure 9.1 of the text.
1
9.4
(a) Li
(b) HCl
9.5
(a) HI
(g) Na
(b) HI
+
Pb

+
+
(m)

(h) Na
(c) H2S
2
2
(d) Sr
2+
(e)
3
NH3
(e)
PH3
(f)
(k)
Al
(l) Al

(c) H2S
(i) Mg
(d) H2S
(j) Mg
2+

PH3
3+

(n)
Pb
9.6
An ionic bond is the electrostatic force that holds the cations and anions in an ionic compound.
9.7
An atom with a low ionization energy and another atom with a high electron affinity will tend to combine to
form an ionic compound.
9.8
Metals: Na, K, Rb, Ca, Sr. Nonmetals: N, O, F, Cl, Br. Sodium combined with each of the nonmetals gives:
Na3N (sodium nitride), Na2O (sodium oxide), NaF (sodium fluoride), NaCl (sodium chloride), and NaBr
(sodium bromide).
9.9
Ammonium chloride, NH4Cl.
9.10
Ammonium nitrate, NH4NO3.
9.11
It would take too much energy to remove four or more electrons from an atom or to add four or more
electrons to an atom. The energy released from Coulombic attraction would not make up for the ionization
energy or electron affinity terms.
9.12
Ionic compounds do not exist as molecules. Therefore, the term molecular mass is not appropriate for ionic
compounds, whereas molar mass is the mass of one mole of the empirical formula of the ionic compound.
9.13
A substance can conduct electricity if ions can flow between electrodes. This movement of ions sets up an
electrical current that is equivalent to the flow of electrons along a metal wire. This movement of ions can
occur in (b) and (c).
9.14
Melt the compound to see if the liquid conducts electricity.
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.15
9.16
9.17
9.18
9.19
9.20
We use Coulomb’s law to answer this question: E  k
201
Qcation Qanion
r
(a)
Doubling the radius of the cation would increase the distance, r, between the centers of the ions. A
larger value of r results in a smaller energy, E, of the ionic bond. Is it possible to say how much smaller
E will be?
(b)
Tripling the charge on the cation will result in tripling of the energy, E, of the ionic bond, since the
energy of the bond is directly proportional to the charge on the cation, Qcation.
(c)
Doubling the charge on both the cation and anion will result in quadrupling the energy, E, of the ionic
bond.
(d)
Decreasing the radius of both the cation and the anion to half of their original values is the same as
halving the distance, r, between the centers of the ions. Halving the distance results in doubling the
energy.
(a) RbI, rubidium iodide
(b) Cs2SO4, cesium sulfate
(c) Sr3N2, strontium nitride
(d) Al2S3, aluminum sulfide
Lewis representations for the ionic reactions are as follows.
+

+
2
(a) Na
+
F
Na F
(b) 2K
+
S
2K S
(c) Ba
+
O
Ba O
(d) Al
+
N
Al N
2
2+
3+
3
The Lewis representations for the reactions are:
2
2+
(a)
Sr
+
Se
Sr
(b)
Ca
+
2H
Ca 2H
(c)
3Li
+
N
3Li N
(d)
2Al
+
3S
(a)
I and Cl should form a molecular compound (covalent); both elements are nonmetals. One possibility
would be ICl, iodine chloride.
(b)
Mg and F will form an ionic compound; Mg is a metal while F is a nonmetal. The substance will be
MgF2, magnesium fluoride.
(a) Covalent (BF3, boron trifluoride)
Se
2+
+
3+

3
2
2Al 3 S
(b) ionic (KBr, potassium bromide)
202
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.21
Lattice energy is the energy required to completely separate one mole of a solid ionic compound into
gaseous ions. The larger the lattice energy, the more stable the solid and the more tightly held the ions. It
takes more energy to melt such a solid, and so the solid has a higher melting point than one with a smaller
lattice energy.
9.22
The Born-Haber cycle relates lattice energies of ionic compounds to ionization energies, electron affinities,
and other atomic and molecular properties. As an example, see the procedure for determining the lattice
energy of LiF in Section 9.3 of the text. Lattice energy is based on Coulomb’s law, which states that the
potential energy (E) between two ions is directly proportional to the product of their charges and inversely
proportional to the distance of separation between them. The procedure for calculating the lattice energy of
an ionic compound is based on Hess’s Law.
9.23
(a) MgO
(b) LiF
(c) Mg3N2
Lattice energy is determined by the charge of the ions and the distance between the ions.
9.24
In each case the more stable compound of each pair is listed. (a) LiF, (b) Cs2O, (c) CaBr2.
9.25
(1) Na(s)  Na(g)
(2)
1
2
H1  108 kJ/mol
H 2  121.4 kJ/mol
Cl2(g)  Cl(g)


H 3  495.9 kJ/mol
(3) Na(g)  Na (g)  e


H 4   349 kJ/mol
(4) Cl(g)  e  Cl (g)


(5) Na (g)  Cl (g)  NaCl(s)
Na(s) 
1
2
Cl2(g)  NaCl(s)
H 5  ?

H overall
  411 kJ/mol

H 5  H overall
 H1  H 2  H 3  H 4  (411)  (108)  (121.4)  (495.9)  ( 349)   787 kJ/mol
The lattice energy of NaCl is 787 kJ/mol.
9.26
(1)
Ca(s)  Ca(g)
H1  121 kJ/mol
(2)
Cl2(g)  2Cl(g)
H 2  242.8 kJ/mol
(3)
Ca(g)  Ca (g)  e



2
H 3 '  589.5 kJ/mol

Ca (g)  Ca (g)  e


H 3 "  1145 kJ/mol
H 4  2(349 kJ/mol)   698 kJ/mol
(4)
2[Cl(g)  e  Cl (g)]
(5)
Ca (g)  2Cl (g)  CaCl2(s)
H 5  ?
Ca(s)  Cl2(g)  CaCl2(s)

H overall
  795 kJ/mol
2

CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
203
Thus we write:

H overall
 H1  H 2  H 3 '  H 3 "  H 4  H 5
H 5  (795  121  242.8  589.5  1145  698)kJ/mol   2195 kJ / mol
The lattice energy is represented by the reverse of equation (5); therefore, the lattice energy is
2195 kJ/mol.
9.27
Gilbert Lewis deduced that a chemical bond can involve atoms sharing a pair of electrons. This type of
electron-pair sharing is an example of a covalent bond, a bond in which two electrons are shared by two
atoms. Covalent compounds are compounds that contain only covalent bonds.
9.28
A Lewis dot symbol is for an atom. A Lewis structure is for a molecule.
9.29
HBr, 3 lone pairs, H2S, 2 lone pairs, and CH4, 0 lone pairs.
9.30
In a single bond, two atoms are held together by one electron pair (The bond in F2 is a single bond). In a
double bond, two atoms are held together by two electron pairs (The bond in O2 is a double bond). In a triple
bond, two atoms are held together by three electron pairs (The bond in N2 is a triple bond).
9.31
Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond; whereas,
electron affinity is the negative of the energy change that occurs when an electron is accepted by an atom in
the gaseous state to form an anion. In general, electronegativity increases from left to right across a period in
the periodic table, as the metallic character of the elements decreases. Within each group, electronegativity
decreases with increasing atomic number, and increasing metallic character.
9.32
In a polar covalent bond, or simply a polar bond, the electrons in the bond spend more time in the vicinity of
one atom than the other. Hydrogen fluoride and water contain polar covalent bonds.
9.33
The degree of ionic character in a bond is a function of the difference in electronegativity between the two
bonded atoms. Figure 9.5 lists electronegativity values of the elements. The bonds in order of increasing
ionic character are: NN (zero difference in electronegativity) < SO (difference 1.0)  ClF (difference 1.0)
< KO (difference 2.7) < LiF (difference 3.0).
9.34
Strategy: We can look up electronegativity values in Figure 9.5 of the text. The amount of ionic character
is based on the electronegativity difference between the two atoms. The larger the electronegativity
difference, the greater the ionic character.
Solution: Let EN  electronegativity difference. The bonds arranged in order of increasing ionic
character are:
CH (EN  0.4) < BrH (EN  0.7) < FH (EN  1.9) < LiCl (EN  2.0)
< NaCl (EN  2.1) < KF (EN  3.2)
9.35
We calculate the electronegativity differences for each pair of atoms:
DE: 3.8  3.3  0.5
DG: 3.8  1.3  2.5
EG: 3.3  1.3  2.0
The order of increasing covalent bond character is: DG < EG < DF < DE
DF: 3.8  2.8  1.0
204
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.36
The order of increasing ionic character is:
ClCl (EN  0) < BrCl (EN  0.2) < SiC (EN  0.7) < CsF (EN  3.3).
9.37
9.38
(a)
The two carbon atoms are the same. The bond is covalent.
(b)
The elelctronegativity difference between K and I is 2.5  0.8  1.7. The bond is polar covalent.
(c)
The electronegativity difference between N and B is 3.0  2.0  1.0. The bond is polar covalent.
(d)
The electronegativity difference between Cl and O is 3.5  3.0  0.5. The bond is polar covalent.
(a)
The two silicon atoms are the same. The bond is covalent.
(b)
The electronegativity difference between Cl and Si is 3.0  1.8  1.2. The bond is polar covalent.
(c)
The electronegativity difference between F and Ca is 4.0  1.0  3.0. The bond is ionic.
(d)
The electronegativity difference between N and H is 3.0  2.1  0.9. The bond is polar covalent.
9.39
The octet rule, formulated by Lewis, states that an atom other than hydrogen tends to form bonds until it is
surrounded by eight valence electrons. In other words, a covalent bond forms when there are not enough
electrons for each individual atom to have a complete octet. By sharing electrons in a covalent bond, the
individual atoms can complete their octets. The octet rule works mainly for elements in the second period of
the periodic table. These elements have only 2s and 2p subshells, which can hold a total of eight electrons.
When an atom of one of these elements forms a covalent compound, it can attain the noble gas electron
configuration [Ne] by sharing electrons with other atoms in the same compound.
9.40
An atom’s formal charge is the electrical charge difference between the valence electrons in an isolated atom
and the number of electrons assigned to that atom in a Lewis structure. It does not represent the actual
separation of charges in a molecule.
9.41
The lewis dot structures are:
(a)
I
Cl
(b) H
P
H
(c)
(d) H
P
P
H
S
H
P
P
(e) H
9.42
N
N
H
H
H
(f) H
O
Cl
(g)
O
Br
C
Br
O
O
Strategy: We follow the procedure for drawing Lewis structures outlined in Section 9.6 of the text.
Solution:
(a)
Step 1: It is obvious that the skeletal structure is: O
O
2
4
Step 2: The outer-shell electron configuration of O is 2s 2p . Also, we must add the negative charges to the
number of valence electrons, Thus, there are
(2  6)  2  14 valence electrons
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
205
Step 3: We draw a single covalent bond between each O, and then attempt to complete the octets for the O
atoms.

O
O

Because this structure satisfies the octet rule for both oxygen atoms, step 4 outlined in the text is not required.

Check: As a final check, we verify that there are 14 valence electrons in the Lewis structure of O2 .
Follow the same procedure as part (a) for parts (b), (c), and (d). The appropriate Lewis structures are:
H
(b)

C

C
(c)
N

O
(d)

H
N
H
9.43
(a)
Too many electrons. The correct structure is
H
(b)
(c)
(d)
C
N
Hydrogen atoms do not form double bonds. The correct structure is
H
C
O
Sn
C
H
Too few electrons.
O
Too many electrons. The correct structure is
F
F
B
F
(c)
Fluorine has more than an octet. The correct structure is
H
(f)
O
F
Oxygen does not have an octet. The correct structure is
O
H
C
F
H
206
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
(g)
Too few electrons. The correct structure is
F
N
F
F
9.44

(a)
Neither oxygen atom has a complete octet. The left-most hydrogen atom is forming two bonds (4 e ).
Hydrogen can only be surrounded by at most two electrons.
(b)
The correct structure is:
H O
H C C O H
H
Do the two structures have the same number of electrons? Is the octet rule satisfied for all atoms other than
hydrogen, which should have a duet of electrons?
9.45
Bond length is defined as the distance between the nuclei of two covalently bonded atoms in a molecule. The
term resonance means the use of two or more Lewis structures to represent a particular molecule. A
resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented
accurately by only one Lewis structure.
9.46
No. Resonance structures of a molecule do not adequately represent the actual molecule, which has its own
unique, stable structure.
9.47
The description involving a griffin and a unicorn is more appropriate. Both mule and donkey are real
animals whereas resonance structures are nonexistent.
9.48
In (a) there is a lone pair on the C atom and the negative formal charge is on the less electronegative C atom.
9.49
The resonance structures are:
O
H
-
O
H
C
C
O
O
O
H
C
H
-
O
H
+
-
N
C
O
-
H
-
O
H
-
+
C
N
O
H
+
N
O
-
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.50
207
Strategy: We follow the procedure for drawing Lewis structures outlined in Section 9.6 of the text. After
we complete the Lewis structure, we draw the resonance structures.
Solution: Following the procedure in Section 9.6 of the text, we come up with the following Lewis

structure for ClO3 .
O

O

Cl
O

We can draw two more equivalent Lewis structures with the double bond between Cl and a different oxygen
atom.
The resonance structures with formal charges are as follows:
O
Cl
O
+

O
Cl
O




O
Cl
O
N


H
+
N
N
N
H
+
+
N
N
N
2
The structures of the most important resonance forms are:
H
C
H


N
N

H

C
N
N

Three reasonable resonance structures for OCN are:
O
9.54

N
N
H
9.53

O
The structures of the most important resonance forms are:
H
9.52
O

O
9.51



C
O
N
C
+
N
O
C
N
2
Three reasonable resonance structures with the formal charges indicated are:

O C S
HO C

S
O
N N
9.55
O
O
O

O


N

O
N
C


O

O

SH
+
N N
O


O

9.56
In the third period of the periodic table and beyond, atoms have d orbitals in addition to the s and p orbitals
that can be used in bonding. These orbitals enable an atom to form an expanded octet.
9.57
Fluorine is the most electronegative element. It attracts electrons toward itself most strongly in a chemical
bond. Fluorine would not share the electrons necessary to from bonds to multiple atoms. In fact, fluorine
typically only forms single bonds in covalent compounds. Furthermore, FH7 would surely violate the octet
rule!
208
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.58
A coordinate covalent bond (also referred to as a dative bond), is defined as a covalent bond in which one of
the atoms donates both electrons. Although the properties of a coordinate covalent bond do not differ from
those of a normal covalent bond (because all electrons are alike no matter what their source), the distinction
is useful for keeping track of valence electrons and assigning formal charges.
9.59
The resonance structures are:

B
Cl

Cl
Cl
B
Cl
Cl
B

Cl
9.60

Cl


Cl
Cl
Strategy: We follow the procedure outlined in Section 9.6 of the text for drawing Lewis structures. We
assign formal charges as discussed in Section 9.7 of the text.
Solution: Drawing the structure with single bonds between Be and each of the Cl atoms, the octet rule for
Be is not satisfied. The Lewis structure is:
Cl
Be
Cl
An octet of electrons on Be can only be formed by making two double bonds as shown below:

Cl

Be
Cl

This places a high negative formal charge on Be and positive formal charges on the Cl atoms. This structure
distributes the formal charges counter to the electronegativities of the elements. It is not a plausible Lewis
structure.
9.61
For simplicity, the three, nonbonding pairs of electrons around the fluorine atoms are omitted.
F
F
(a)
F
Xe
F
(b)
F
Xe
F
F
F
F
F
O
O
F
(d)
F
Xe
The octet rule is exceeded in each case.
Xe
(c)
F
F
(e)
F
Xe
O
F
F
F
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.62
2
209
3
The outer electron configuration of antimony is 5s 5p . The Lewis structure is shown below. All five
valence electrons are shared in the five covalent bonds. The octet rule is not obeyed. (The electrons on the
chlorine atoms have been omitted for clarity.)
Cl
Cl
Cl
Sb Cl
Cl
Can Sb have an expanded octet?
9.63
For simplicity, the three, nonbonding pairs of electrons around the fluorine are omitted.
F
Se
F
F
F
F
Se
F
F
F
F
F
The octet rule is not satisfied for Se in both compounds (why not?).
9.64
The reaction can be represented as:
Cl
Cl
Al
Cl
+
Cl

Cl
Cl

Al
Cl
Cl
The new bond formed is called a coordinate covalent bond.
9.65
Bond enthalpy is the enthalpy change required to break a particular bond in 1 mole of gaseous molecules.
For polyatomic molecules, we speak of the average bond enthalpy of a particular bond. For example, we can
measure the energy of the OH bond in 10 different polyatomic molecules and obtain the average OH bond
enthalpy by dividing the sum of the bond enthalpies by 10.
9.66
Bond enthalpies in solids and liquids are affected by neighboring molecules. To break a bond, attractive
forces between atoms must be overcome. This requires the input of energy (endothermic). As a bond forms,
the potential energy of the molecule is lower than the separate atoms. Energy is released (exothermic).
9.67
The enthalpy change for the equation showing ammonia dissociating into a nitrogen atom and three hydrogen
atoms is equal to three times the average bond ethalpy of the NH bond (Why three?).
NH3(g)  N(g)  3H(g)
H  3H(NH)
The equation is the sum of the three equations given in the problem, and by Hess's law (Section 6.6 of the
text) the enthalpy change is just the sum of the enthalpies of the individual steps.
NH3(g)  NH2(g)  H(g)
NH2(g)  NH(g)  H(g)
NH(g)  N(g)  H(g)
NH3(g)  N(g)  3H(g)
H  435 kJ/mol
H  381 kJ/mol
H  360 kJ/mol
H  1176 kJ/mol
210
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
H (N  H) 
9.68
1176 kJ/mol
 392 kJ/mol
3
Strategy: Keep in mind that bond breaking is an energy absorbing (endothermic) process and bond making
is an energy releasing (exothermic) process. Therefore, the overall energy change is the difference between
these two opposing processes, as described in Equation (9.3) of the text.
Solution: There are two oxygen-to-oxygen bonds in ozone. We will represent these bonds as OO.
However, these bonds might not be true oxygen-to-oxygen single bonds. Using Equation (9.3) of the text,
we write:
H  BE(reactants)  BE(products)
H  BE(OO)  2BE(OO)
In the problem, we are given H for the reaction, and we can look up the OO bond enthalpy in Table 9.3
of the text. Solving for the average bond enthalpy in ozone,
2BE(OO)  H  BE(OO)
BE (O  O) 
BE (O=O)  H 
498.7 kJ/mol  107.2 kJ/mol

 303.0 kJ/mol
2
2
Considering the resonance structures for ozone, is it expected that the OO bond enthalpy in ozone is
between the single OO bond enthalpy (142 kJ) and the double OO bond enthalpy (498.7 kJ)?
9.69
When molecular fluorine dissociates, two fluorine atoms are produced. Since the enthalpy of formation of
atomic fluorine is in units of kJ/mol, this number is half the bond enthalpy of the fluorine molecule.
F2(g)  2F(g)
H  156.9 kJ
H   2H f (F)  H f (F2 )
156.9 kJ/mol  2H f (F)  (1)(0)
H f (F) 
9.70
(a)
Bonds Broken
CH
CC
OO
Bonds Formed
CO
OH
156.9 kJ/mol
 78.5 kJ/mol
2
Number Broken
12
2
7
Number Formed
8
12
Bond Enthalpy
(kJ/mol)
414
347
498.7
Bond Enthalpy
(kJ/mol)
799
460
H  total energy input  total energy released
 (4968  694  3491)  (6392  5520)  2759 kJ/mol
Enthalpy Change
(kJ)
4968
694
3491
Enthalpy Change
(kJ)
6392
5520
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
(b)
211
H   4H f (CO2 )  6H f (H 2 O)  [2H f (C2 H6 )  7 H f (O2 )]
H  (4)(393.5 kJ/mol)  (6)(241.8 kJ/mol)  [(2)(84.7 kJ/mol)  (7)(0)]  2855 kJ/mol
The answers for part (a) and (b) are different, because average bond enthalpies are used for part (a).
9.71
9.72
(a)
electron affinity of fluorine
(b)
bond enthalpy of molecular fluorine
(c)
ionization energy of sodium
(d)
standard enthalpy of formation of sodium fluoride
Recall that you can classify bonds as ionic or covalent based on electronegativity difference.
The melting points (C) are shown in parentheses following the formulas.
Ionic:
NaF (993)
MgF2 (1261)
AlF3 (1291)
Covalent:
SiF4 (90.2)
PF5 (83)
SF6 (121)
ClF3 (83)
Is there any correlation between ionic character and melting point?
9.73
By Hess's law, the overall enthalpy (energy) change in a reaction is equal to the sum of the enthalpy (energy)
changes for the individual steps. The reactions shown in the problem are just the sums of the ionization
energy of the alkali metal and the negative of the electron affinity of the halogen. Recall that a positive
electron affinity corresponds to an exothermic reaction.
(a)
Taking data from Tables 8.2 and 8.3 of the text we have


Li(g)  Li (g)  e


I(g)  e  I (g)
H  520 kJ/mol
H  295 kJ/mol


Li(g)  I(g)  Li (g)  I (g)
H  225 kJ/mol
Parts (b) and (c) are solved in an exactly analogous manner.
(b)
(c)


Na(g)  F(g)  Na (g)  F (g)


K(g)  Cl(g)  K (g)  Cl (g)
H  168 kJ/mol
H  70 kJ/mol


9.74
KF is an ionic compound. It is a solid at room temperature made up of K and F ions. It has a high melting
point, and it is a strong electrolyte. Carbon dioxide, CO2, is a covalent compound that exists as discrete
molecules. It is a gas at room temperature. It has a low melting point, is only slightly soluble in water, and is
a nonelectrolyte.
9.75
The three pairs of nonbonding electrons around each fluorine have been omitted for simplicity.
F
F
F
Br
F
F
F
Cl
F
F
F
F
F
I
F
F
F
F
The octet rule is not obeyed in any of the compounds. In order for the octet rule to be obeyed, what would
the value of n in the compound ICln have to be? [Hint: see Problem 9.41 (a)]
212
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.76
The resonance structures are:


N
N
N

N

N
N



N
N
N
Which is the most plausible structure based on a formal charge argument?
9.77
A resonance structure is:
O
+
N
-
C
H
9.78

An example of an aluminum species that satisfies the octet rule is the anion AlCl4 . The Lewis dot
structure is drawn in Problem 9.64.
3
An example of an aluminum species containing an expanded octet is anion AlF6 . (How many pairs
of electrons surround the central atom?)
An aluminum species that has an incomplete octet is the compound AlCl3. The dot structure is given in
Problem 9.64.
(a)
(b)
(c)
9.79
Four resonance structures together with the formal charges are

O

O
+
P F
O
9.80

O
O
P
O

O
F
O
P
O



F
O
O
P

F
O

CF2 would be very unstable because carbon does not have an octet. (How many electrons does it have?)
CH5 does not exist because it violates the octet rule for carbon.

FH2 does not exist because F cannot exceed the octet rule. There are 10 valence electrons, so F would have
to exceed the octet rule.
PI5 appears to be a reasonable species (compared to PF5 in Example 9.10 of the text). However, the iodine
atoms are too large to have five of them "fit" around a single P atom.
9.81
Reasonable resonance structures are:
O
H
(a)
O
S
O
+
O

O

H
O

+
S
O

O
O
H
O
S

+
O
O
O

H
O

2+
S
O
O

CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
O
O
-O
S
O
-
-
+
O
S
O
O
(b)
213
O
O
-O
S
2+
O
-
O-
-
There are five more equivalent resonance structures to the first structure above, and three more
equivalent resonance structures to the second structure above.
O
O
H
O
S
O

H
O
S

O
O
H
O

+
S
O

(c)
O

O
S
O

O

O

S+ O

(d)
There are two more equivalent resonance structures to the first structure.
9.82
(a) false
(b) true
(c) false
(d) false
For question (c), what is an example of a second-period species that violates the octet rule?
9.83
If the central atom were more electronegative, there would be a concentration of negative charges at the
central atom. This would lead to instability. In compounds like H2O and NH3, the more electronegative
atom is the central atom. This is due to the fact that hydrogen cannot be a central atom. With only a 1s
valence orbital, a hydrogen atom can only share two electrons. Therefore, hydrogen will always be a
terminal atom in a Lewis structure.
9.84
The formation of CH4 from its elements is:
C(s)  2H2(g) 
 CH4(g)
The reaction could take place in two steps:
Step 1: C(s)  2H2(g) 
 C(g)  4H(g)

H rxn
 (716  872.8)kJ/mol  1589 kJ/mol
Step 2: C(g)  4H(g) 
 CH4(g)

H rxn
  4  (bond enthalpy of C  H bond)
 4  414 kJ/mol  1656 kJ/mol
Therefore, H f (CH 4 ) would be approximately the sum of the enthalpy changes for the two steps. See
Section 6.6 of the text (Hess’s law).


H f (CH 4 )  H rxn
(1)  H rxn
(2)
H f (CH 4 )  (1589  1656)kJ/mol   67 kJ/mol
The actual value of H f (CH 4 )   74.85 kJ/mol.
214
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.85
(a)
CH
CCl
Bond broken:
Bond made:
H  414 kJ/mol
H  338 kJ/mol

H rxn
 414  338  76 kJ/mol
(b)
CH
HCl
Bond broken:
Bond made:
H  414 kJ/mol
H  431.9 kJ/mol

H rxn
 414  431.9   18 kJ/mol
Based on energy considerations, reaction (b) will occur readily since it is exothermic. Reaction (a) is
endothermic.
9.86
Only N2 has a triple bond. Therefore, it has the shortest bond length.
9.87
The rest of the molecule (in this problem, unidentified) would be attached at the end of the free bond.
(a)
O
O
O
O
C
H
(b)
O
C


C
O

9.88
To be isoelectronic, molecules must have the same number and arrangement of valence electrons. NH4 and
CH4 are isoelectronic (8 valence electrons), as are CO and N2 (10 valence electrons), as are B3N3H6 and
C6H6 (30 valence electrons). Draw Lewis structures to convince yourself that the electron arrangements are
the same in each isoelectronic pair.
9.89
The Lewis structures are:
(a) C
(d) H
9.90
(b) O
H paramagnetic
+
-
N
C diamagnetic
(e) H
H paramagnetic
C
(c) C
O paramagnetic
The reaction can be represented as:
H N
H
+
H O
H
H N H
H
+ O H
C
diamagnetic
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.91
215
The Lewis structures are:
(a)
F
F
C
C
F
F
(b)
H
(c)
H
H
(d)
H
C
C
C
H
O
H
H
H
H
H
C
C
C
H
H
H
C
C
C
C
H
H
H
H
H
(e)
O
C
H
Note, the lone pairs of electrons on F and O have been omitted.

9.92
The central iodine atom in I3 has ten electrons surrounding it: two bonding pairs and three lone pairs. The
central iodine has an expanded octet. Elements in the second period such as fluorine cannot have an

expanded octet as would be required for F3 .
9.93
The bond enthalpy for F2 is:
F2(g)  F(g)  F(g)

H  156.9 kJ

The energy for the process F2(g)  F (g)  F (g) can be found by Hess's law. Thus,
F2(g)  F(g)  F(g)


F(g)  F (g)  e


F(g)  e  F (g)


F2 ( g )  F ( g )  F ( g )
H  156.9 kJ/mol
H  1680 kJ/mol [I1(F), see Table 8.2 of the text]
H  328 kJ/mol [EA1(F), see Table 8.3 of the text]
H  1509 kJ/mol
It is much easier to dissociate F2 into two neutral F atoms than it is to dissociate it into a fluorine cation and
anion.
9.94
The skeletal structure is:
H
H C N C O
H
The number of valence electron is: (1  3)  (2  4)  5  6  22 valence electrons
216
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
We can draw two resonance structures for methyl isocyanate.
H
H
C
H
N
C
O
H
H
9.95

C
N
C
O

H
A reasonable Lewis structure is:
O
Cl
9.96
9.97
O
+
O
-
(a)
This is a very good resonance form; there are no formal charges and each atom satisfies the octet rule.
(b)
This is a second choice after (a) because of the positive formal charge on the oxygen (high
electronegativity).
(c)
This is a poor choice for several reasons. The formal charges are placed counter to the
electronegativities of C and O, the oxygen atom does not have an octet, and there is no bond between
that oxygen and carbon!
(d)
This is a mediocre choice because of the large formal charge and lack of an octet on carbon.
For C4H10 and C5H12 there are a number of structural isomers.
C 2 H6
H
C5H12
H
9.98
N
H
H
C
C
H
H
C4H10
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
The nonbonding electron pairs around Cl and F are omitted for simplicity.
Cl
Cl
F
F H
F C Cl
F C F
H C F
F C C F
Cl
Cl
Cl
F
F
H
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.99
9.100
217
The structures are (the nonbonding electron pairs on fluorine have been omitted for simplicity):
H
H
H
H
H
C
C
C
C
C
H
F
H
(a)
H
H
H
H
H
C
C
C
C
H
H
H
H
H
Using Equation (9.3) of the text,
H  BE(reactants)  BE(products)
H  [(436.4  151.0)  2(298.3)]  9.2 kJ/mol
(b)
Using Equation (6.18) of the text,
H   2H f [HI( g )]  {H f [H 2 ( g )]  H f [I2 ( g )]}
H  (2)(25.9 kJ/mol)  [(0)  (1)(61.0 kJ/mol)]  9.2 kJ/mol
9.101
Note that the nonbonding electron pairs have been deleted from oxygen, nitrogen, sulfur, and chlorine for
simplicity.
(a)
H
H
C
O
H
(b)
H
H
H
C
C
H
H
H
(d)
H
H
H
C
N
H
(f)
H
O
H
H
N
C
N
H
H
H
Cl
C
C
H
H
S
(g)
H
H
O
H
N
C
C
H
Note: in part (c) above, ethyl = C2H5 = H
H
C2H5
H5C2
Pb
C2H5
(e)
H
O
(c)
H
H
C
C
H
H
H
H
C
C
H
H
O
H
Cl
C2H5
218
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.102
The Lewis structures are:


(a) C

(b) N
O

(c) C
O
(d) N
N
N
9.103
O
2
O O
oxide
2
O O
peroxide

superoxide
9.104
True. Each noble gas atom already has completely filled ns and np subshells.
9.105
(a)



The bond enthalpy of F2 is the energy required to break up F2 into an F atom and an F ion.


F2 ( g ) 
 F(g)  F (g)
We can arrange the equations given in the problem so that they add up to the above equation.


F2 ( g ) 
 F2(g)  e
H  290 kJ/mol
F2 ( g ) 
 2F(g)
H  156.9 kJ/mol

F(g)  e


 F (g)

H  333 kJ/mol

F2 ( g ) 
 F(g)  F (g)

The bond enthalpy of F2 is the sum of the enthalpies of reaction.

BE(F2 )  [290  156.9  (333 kJ)]kJ/mol  114 kJ/mol
(b)
9.106

The bond in F2 is weaker (114 kJ/mol) than the bond in F2 (156.9 kJ/mol), because the extra electron
increases repulsion between the F atoms.
The resonance structures are:

C

N


2
C
O
N
3
O
C

N

O
The most likely structure is on the left and the least likely structure is on the right.
9.107
H Ar
F
9.108
There are four CH bonds in CH4, so the average bond enthalpy of a CH bond is:
1656 kJ/mol
 414 kJ/mol
4
The Lewis structure of propane is:
H
H
H
H
C
C
C
H
H
H
H
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
There are eight CH bonds and two CC bonds. We write:
8(CH)  2(CC)  4006 kJ/mol
8(414 kJ/mol)  2(CC)  4006 kJ/mol
2(CC)  694 kJ/mol
So, the average bond enthalpy of a CC bond is:
9.109
F
F
H
C
C
F
Cl
F
Br
H
F
C
C
F
H
C
C
F
Cl
(a)
O
C
F
F
enflurane
H
O
C
Cl
F
H
F
C
C
H
O
Cl F
F
C
H
H
methoxyflurane
isoflurane
9.110
H
Cl F
halothane
F
694
kJ/mol  347 kJ/mol
2
Using Equation (9.3) of the text,
H  BE(reactants)  BE(products)
H  [(941.4  3(436.4)  6(393)]  107 kJ/mol
(b)
Using Equation (6.18) of the text,
H   2H f [NH3 ( g )]  {H f [N 2 ( g )]  3H f [H 2 ( g )]}
H  (2)(46.3 kJ/mol)  [(0)  (3)(0)]  92.6 kJ/mol
9.111
(a)
O H
(b)
The OH bond is quite strong (460 kJ/mol). To complete its octet, the OH radical has a strong
tendency to form a bond with a H atom.
(c)
One CH bond is broken and one OH bond is formed.
H  BE(reactants)  BE(products)
H  414  460  46 kJ/mol
(d)
Energy of one O  H bond 
 
460  103 J
1 mol

 7.64  1019 J/bond
1 mol
6.022  1023 bonds
hc
(6.63  1034 J  s)(3.00  108 m / s)

 2.60  107 m  2.60  102 nm
H
7.64  1019 J
219
220
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
9.112
(a)
H
H
H
C
C
Cl
Cl
H
H
H
C
C
H
Cl
vinyl chloride
ethylene dichloride
In both molecules, the carbon-carbon bonds are covalent (nonpolar). The CH bonds and CCl bonds
are polar (EN  0.4 for CH bond, and EN  0.5 for CCl bond).
Note, the lone pairs of electrons on Cl have been omitted from the structures.
(b)
(c)
H
H
H
H
H
H
C
C
C
C
C
C
H
Cl
H
Cl
H
Cl
In the formation of poly(vinyl chloride) form vinyl chloride, for every CC double bond broken, 2 CC
single bonds are formed. No other bonds are broken or formed. The energy changes for 1 mole of
vinyl chloride reacted are:
total energy input (breaking CC bonds)  620 kJ
total energy released (forming CC bonds)  2  347 kJ  694 kJ
H  620 kJ  694 kJ  74 kJ
The negative sign shows that this is an exothermic reaction. To find the enthalpy change when
3
1.0  10 kg of vinyl chloride react, we proceed as follows:
H   (1.0  106 g C2 H3Cl) 
1 mol C2 H3Cl
74 kJ

  1.2  106 kJ
62.49 g C2 H3Cl 1 mol C2 H3Cl
This process generates a substantial amount of heat. The design would need to incorporate a means of
cooling the reaction vessel.
9.113
The equations for the preparation of sulfuric acid starting with sulfur are:
S(s)  O2(g)  SO2(g)
2SO2(g)  O2(g)  2SO3(g)
SO3(g)  H2O(l)  H2SO4(aq)
The equations for the preparation of sulfuric acid starting with sulfur trioxide are:
SO3(g)  H2SO4(aq)  H2S2O7(aq)
H2S2O7(aq)  H2O(l)  2H2SO4(aq)
Formation of oleum
Generation of sulfuric acid
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
221
There are two resonance structures for oleum. Formal charges other than zero are shown in the structures.
O
O
O
O
S
O
S
O
O
H
O
H

H
9.114
O

2+
O
S
O
O
H
H+
H
H
(a)
H
(b)
This is an application of Hess’s Law.

2H  H  H3
H2  2H

H  849 kJ/mol
H  436.4 kJ/mol
H  H2  H3

H  413 kJ/mol

2+
S
O
O
H

H+
H
H+


H

The energy released in forming H3 from H and H2 is almost as large as the formation of H2 from 2 H
atoms.
9.115
From Table 9.3 of the text, we can find the bond enthalpies of CN and CN. The average can be calculated,
and then the maximum wavelength associated with this enthalpy can be calculated.
The average bond enthalpy for CN and CN is:
(276  615) kJ/mol
 446 kJ/mol
2
We need to convert this to units of J/bond before the maximum wavelength to break the bond can be
calculated. Because there is only 1 CN bond per molecule, there is Avogadro’s number of bonds in 1 mole
of the amide group.
446 kJ
1 mol
1000 J


 7.41  1019 J/bond
23
1 mol
1 kJ
6.022  10 bonds
The maximum wavelength of light needed to break the bond is:
 max 
9.116
9.117
N
+
N
N
+
N
hc
(6.63  1034 J  s)(3.00  108 m/s)

 2.68  107 m  268 nm
E
7.41  1019 J
N


N
+
N
N
+
N
N
N
+
N
AgNO3 – used as an anti-infective agent. Used in eye drops for newborn infants.
BaSO4 – used as an image enhancer in X-rays (barium enema).
CaSO4 – used for making casts for broken bones.
KI – used for thyroid treatment.
Li2CO3 – used to treat manic depression.
Mg(OH)2 – used as an antacid and laxative.
MgSO4 – used to treat preeclampsia (a sharp rise in blood pressure) in pregnant woman.
NaHCO3 – used as an antacid.
Na2CO3 – used as an antacid.

N
+
N
N
222
CHAPTER 9: CHEMICAL BONDING I: THE COVALENT BOND
NaF – used to prevent tooth decay.
TiO2 – used in sunscreens.
ZnO – used in sunscreens.
9.118
There are no lone pairs on adjacent atoms in C2H6, there is one lone pair on each nitrogen atom in N2H4, and
there are two lone pairs on each oxygen atom in H2O2. Draw Lewis structures to determine the number of
lone pairs in each molecule. Looking at Table 9.3 of the text which lists bond enthalpies of diatomic
molecules and average bond enthalpies of polyatomic molecules, we can estimate the bond enthalpies of CC
in C2H6, NN in N2H4, and OO in H2O2 by looking up the average bond enthalpy values. These will not
be exact bond enthalpy values for the given molecules, but the values will be approximate and give us a
measure of what effect lone pairs on adjacent atoms have on the strength of the particular bonds. The values
given in Table 9.3 are:
CC
347 kJ/mol
NN
193 kJ/mol
OO
142 kJ/mol
Comparing these values, it is clear that lone pairs on adjacent atoms weaken the particular bond. The CC
bond in C2H6, with no adjacent lone pairs is the strongest bond, the NN bond in N2H4, with one lone pair
on each nitrogen atom is a weaker bond, and the OO bond in H2O2, with two lone pairs on each oxygen
atom is the weakest bond.