sm Chemistry 9th Edition by Zumdahl
CHAPTER 8
BONDING: GENERAL CONCEPTS
Questions
15. a. This diagram represents a polar covalent bond as in HCl. In a polar covalent bond,
there is an electron rich region (indicated by the red color) and an electron poor region
(indicated by the blue color). In HCl, the more electronegative Cl atom (on the red side
of the diagram) has a slightly greater ability to attract the bonding electrons than does H
(on the blue side of the diagram), which in turn produces a dipole moment.
b. This diagram represents an ionic bond as in NaCl. Here, the electronegativity differences
between the Na and Cl are so great that the valence electron of sodium is transferred to
the chlorine atom. This results in the formation of a cation, an anion, and an ionic bond.
c. This diagram represents a pure covalent bond as in H2. Both atoms attract the bonding
electrons equally, so there is no bond dipole formed. This is illustrated in the
electrostatic potential diagram as the various red and blue colors are equally distributed
about the molecule. The diagram shows no one region that is red nor one region that is
blue (there is no specific partial negative end and no specific partial positive end), so the
molecule is nonpolar.
16. In F2 the bonding is pure covalent, with the bonding electrons shared equally between the two
fluorine atoms. In HF, there is also a shared pair of bonding electrons, but the shared pair is
drawn more closely to the fluorine atom. This is called a polar covalent bond as opposed to
the pure covalent bond in F2.
17. Of the compounds listed, P2O5 is the only compound containing only covalent bonds.
(NH4)2SO4, Ca3(PO4)2, K2O, and KCl are all compounds composed of ions, so they exhibit
ionic bonding. The polyatomic ions in (NH4)2SO4 are NH4
+
and SO4
2
. Covalent bonds exist
between the N and H atoms in NH4
+
and between the S and O atoms in SO4
2
. Therefore,
(NH4)2SO4 contains both ionic and covalent bonds. The same is true for Ca3(PO4)2. The
bonding is ionic between the Ca2+ and PO4
3
ions and covalent between the P and O atoms in
PO4
3
. Therefore, (NH4)2SO4 and Ca3(PO4)2 are the compounds with both ionic and covalent
bonds.
18. Ionic solids are held together by strong electrostatic forces that are omnidirectional.
i. For electrical conductivity, charged species must be free to move. In ionic solids, the
charged ions are held rigidly in place. Once the forces are disrupted (melting or
dissolution), the ions can move about (conduct).
ii. Melting and boiling disrupts the attractions of the ions for each other. Because these
electrostatic forces are strong, it will take a lot of energy (high temperature) to
accomplish this.
CHAPTER 8 BONDING: GENERAL CONCEPTS 251
iii. If we try to bend a piece of material, the ions must slide across each other. For an
ionic solid the following might happen:
Just as the layers begin to slide, there will be very strong repulsions causing the solid
to snap across a fairly clean plane.
iv. Polar molecules are attracted to ions and can break up the lattice.
These properties and their correlation to chemical forces will be discussed in detail in
Chapters 10 and 11.
19. Electronegativity increases left to right across the periodic table and decreases from top to
bottom. Hydrogen has an electronegativity value between B and C in the second row and
identical to P in the third row. Going further down the periodic table, H has an electronegativity value between As and Se (row 4) and identical to Te (row 5). It is important to
know where hydrogen fits into the electronegativity trend, especially for rows 2 and 3. If you
know where H fits into the trend, then you can predict bond dipole directions for nonmetals
bonded to hydrogen.
20. Linear structure (180° bond angle)
Polar; bond dipoles do not cancel. Nonpolar; bond dipoles cancel.
Trigonal planar structure (120° bond angle)
+ 2 other resonance
structures
Polar; bond dipoles do not cancel. Nonpolar; bond dipoles cancel.
Tetrahedral structure (109.5° bond angles)
Polar; bond dipoles do not cancel. Nonpolar; bond dipoles cancel.
strong attraction strong repulsion
252 CHAPTER 8 BONDING: GENERAL CONCEPTS
21. For ions, concentrate on the number of protons and the number of electrons present. The
species whose nucleus holds the electrons most tightly will be smallest. For example, anions
are larger than the neutral atom. The anion has more electrons held by the same number of
protons in the nucleus. These electrons will not be held as tightly, resulting in a bigger size
for the anion as compared to the neutral atom. For isoelectronic ions, the same number of
electrons are held by different numbers of protons in the various ions. The ion with the most
protons holds the electrons tightest and is smallest in size.
22. Two other factors that must be considered are the ionization energy needed to produce more
positively charged ions and the electron affinity needed to produce more negatively charged
ions. The favorable lattice energy more than compensates for the unfavorable ionization
energy of the metal and for the unfavorable electron affinity of the nonmetal as long as
electrons are added to or removed from the valence shell. Once the valence shell is empty, the
ionization energy required to remove the next (inner-core) electron is extremely unfavorable;
the same is true for electron affinity when an electron is added to a higher n shell. These two
quantities are so unfavorable after the valence shell is complete that they overshadow the
favorable lattice energy, and the higher charged ionic compounds do not form.
23. Fossil fuels contain a lot of carbon and hydrogen atoms. Combustion of fossil fuels (reaction
with O2) produces CO2 and H2O. Both these compounds have very strong bonds. Because
stronger product bonds are formed than reactant bonds broken, combustion reactions are very
exothermic.
24. Statements a and c are true. For statement a, XeF2 has 22 valence electrons, and it is
impossible to satisfy the octet rule for all atoms with this number of electrons. The best Lewis
structure is:
For statement c, NO+ has 10 valence electrons, whereas NO
has 12 valence electrons. The
Lewis structures are:
Because a triple bond is stronger than a double bond, NO+
has a stronger bond.
For statement b, SF4 has five electron pairs around the sulfur in the best Lewis structure; it is
an exception to the octet rule. Because OF4 has the same number of valence electrons as SF4,
OF4 would also have to be an exception to the octet rule. However, row 2 elements such as O
never have more than 8 electrons around them, so OF4 does not exist. For statement d, two
resonance structures can be drawn for ozone:
When resonance structures can be drawn, the actual bond lengths and strengths are all equal
to each other. Even though each Lewis structure implies the two OO bonds are different,
this is not the case in real life. In real life, both of the OO bonds are equivalent. When
resonance structures can be drawn, you can think of the bonding as an average of all of the
resonance structures.
CHAPTER 8 BONDING: GENERAL CONCEPTS 253
25. CO2, 4 + 2(6) = 16 valence electrons
The formal charges are shown above the atoms in the three Lewis structures. The best Lewis
structure for CO2 from a formal charge standpoint is the first structure having each oxygen
double bonded to carbon. This structure has a formal charge of zero on all atoms (which is
preferred). The other two resonance structures have nonzero formal charges on the oxygens,
making them less reasonable. For CO2, we usually ignore the last two resonance structures
and think of the first structure as the true Lewis structure for CO2.
26. Only statement c is true. The bond dipoles in CF4 and KrF4 are arranged in a manner that they
all cancel each other out, making them nonpolar molecules (CF4 has a tetrahedral molecular
structure, whereas KrF4 has a square planar molecular structure). In SeF4, the bond dipoles in
this see-saw molecule do not cancel each other out, so SeF4 is polar. For statement a, all the
molecules have either a trigonal planar geometry or a trigonal bipyramid geometry, both of
which have 120° bond angles. However, XeCl2 has three lone pairs and two bonded chlorine
atoms around it. XeCl2 has a linear molecular structure with a 180° bond angle. With three
lone pairs, we no longer have a 120° bond angle in XeCl2. For statement b, SO2 has a Vshaped molecular structure with a bond angle of about 120°. CS2 is linear with a 180° bond
angle, and SCl2 is V-shaped but with an approximate 109.5° bond angle. The three compounds do not have the same bond angle. For statement d, central atoms adopt a geometry to
minimize electron repulsions, not maximize them.
Exercises
Chemical Bonds and Electronegativity
27. Using the periodic table, the general trend for electronegativity is:
(1) Increase as we go from left to right across a period
(2) Decrease as we go down a group
Using these trends, the expected orders are:
a. C < N < O b. Se < S < Cl c. Sn < Ge < Si d. Tl < Ge < S
28. a. Rb < K < Na b. Ga < B < O c. Br < Cl < F d. S < O < F
29. The most polar bond will have the greatest difference in electronegativity between the two
atoms. From positions in the periodic table, we would predict:
a. Ge‒F b. P‒Cl c. S‒F d. Ti‒Cl
30. a. Sn‒H b. Tl‒Br c. Si‒O d. O‒F
254 CHAPTER 8 BONDING: GENERAL CONCEPTS
31. The general trends in electronegativity used in Exercises 27 and 29 are only rules of
thumb. In this exercise, we use experimental values of electronegativities and can begin to
see several exceptions. The order of EN from Figure 8.3 is:
a. C (2.5) < N (3.0) < O (3.5) same as predicted
b. Se (2.4) < S (2.5) < Cl (3.0) same
c. Si = Ge = Sn (1.8) different
d. Tl (1.8) = Ge (1.8) < S (2.5) different
Most polar bonds using actual EN values:
a. Si‒F and Ge‒F have equal polarity (Ge‒F predicted).
b. P‒Cl (same as predicted)
c. S‒F (same as predicted) d. Ti‒Cl (same as predicted)
32. The order of EN from Figure 8.3 is:
a. Rb (0.8) = K (0.8) < Na (0.9), different b. Ga (1.6) < B (2.0) < O (3.5), same
c. Br (2.8) < Cl (3.0) < F (4.0), same d. S (2.5) < O (3.5) < F (4.0), same
Most polar bonds using actual EN values:
a. C‒H most polar (Sn‒H predicted)
b. Al‒Br most polar (Tl‒Br predicted). c. Si‒O (same as predicted).
d. Each bond has the same polarity, but the bond dipoles point in opposite directions.
Oxygen is the positive end in the O‒F bond dipole, and oxygen is the negative end in the
O‒Cl bond dipole (O‒F predicted).
33. Use the electronegativity trend to predict the partial negative end and the partial positive end
of the bond dipole (if there is one). To do this, you need to remember that H has electronegativity between B and C and identical to P. Answers b, d, and e are incorrect. For d (Br2),
the bond between two Br atoms will be a pure covalent bond, where there is equal sharing of
the bonding electrons, and no dipole moment exists. For b and e, the bond polarities are
reversed. In ClI, the more electronegative Cl atom will be the partial negative end of the
bond dipole, with I having the partial positive end. In OP, the more electronegative oxygen
will be the partial negative end of the bond dipole, with P having the partial positive end. In
the following, we used arrows to indicate the bond dipole. The arrow always points to the
partial negative end of a bond dipole (which always is the most electronegative atom in the
bond).
Cl I O P
CHAPTER 8 BONDING: GENERAL CONCEPTS 255
34. See Exercise 33 for a discussion on bond dipoles. We will use arrows to indicate the bond
dipoles. The arrow always points to the partial negative end of the bond dipole, which will
always be to the more electronegative atom. The tail of the arrow indicates the partial positive
end of the bond dipole.
a. b. PH is a pure covalent (nonpolar) bond because
P and H have identical electronegativities.
c. d.
e. The actual electronegativity difference between Se and S is so small that
this bond is probably best characterized as a pure covalent bond having
no bond dipole.
35. Bonding between a metal and a nonmetal is generally ionic. Bonding between two nonmetals
is covalent, and in general, the bonding between two different nonmetals is usually polar
covalent. When two different nonmetals have very similar electronegativities, the bonding is
pure covalent or just covalent.
a. ionic b. covalent c. polar covalent
d. ionic e. polar covalent f. covalent
36. The possible ionic bonds that can form are between the metal Cs and the nonmetals P, O, and
H. These ionic compounds are Cs3P, Cs2O, and CsH. The bonding between the various
nonmetals will be covalent. P4, O2, and H2 are all pure covalent (or just covalent) with equal
sharing of the bonding electrons. PH will also be a covalent bond because P and H have
identical electronegativities. The other possible covalent bonds that can form will all be polar
covalent because the nonmetals involved in the bonds all have intermediate differences in
electronegativities. The possible polar covalent bonds are PO and OH.
Note: The bonding among cesium atoms is called metallic. This type of bonding between
metals will be discussed in Chapter 10.
37. Electronegativity values increase from left to right across the periodic table. The order of
electronegativities for the atoms from smallest to largest electronegativity will be H = P < C <
N < O < F. The most polar bond will be F‒H since it will have the largest difference in
electronegativities, and the least polar bond will be P‒H since it will have the smallest
difference in electronegativities (ΔEN = 0). The order of the bonds in decreasing polarity will
be F‒H > O‒H > N‒H > C‒H > P‒H.
38. Ionic character is proportional to the difference in electronegativity values between the two
elements forming the bond. Using the trend in electronegativity, the order will be:
Br‒Br < N‒O < C‒F < Ca‒O < K‒F
least most
ionic character ionic character
C O
H Cl Br Te
Se S
256 CHAPTER 8 BONDING: GENERAL CONCEPTS
Note that Br‒Br, N‒O, and C‒F bonds are all covalent bonds since the elements are all nonmetals. The Ca‒O and K‒F bonds are ionic, as is generally the case when a metal forms a
bond with a nonmetal.
39. A permanent dipole moment exists in a molecule if the molecule has one specific area with a
partial negative end (a red end in an electrostatic potential diagram) and a different specific
region with a partial positive end (a blue end in an electrostatic potential diagram). If the
blue and red colors are equally distributed in the electrostatic potential diagrams, then no
permanent dipole exists.
a. Has a permanent dipole. b. Has no permanent dipole.
c. Has no permanent dipole. d. Has a permanent dipole.
e. Has no permanent dipole. f. Has no permanent dipole.
40. a. H2O; both H2O and NH3 have permanent dipole moments in part due to the polar OH
and NH bonds. But because oxygen is more electronegative than nitrogen, one would
expect H2O to have a slightly greater dipole moment. This diagram has the more intense
red color on one end and the more intense blue color at the other end indicating a larger
dipole moment.
b. NH3; this diagram is for a polar molecule, but the colors are not as intense as the diagram
in part a. Hence, this diagram is for a molecule which is not as polar as H2O. Since N is
less electronegative than O, NH3 will not be as polar as H2O.
c. CH4; this diagram has no one specific red region and has four blue regions arranged
symmetrically about the molecule. This diagram is for a molecule which has no dipole
moment. This is only true for CH4. The C‒H bonds are at best, slightly polar because
carbon and hydrogen have similar electronegativity values. In addition, the slightly polar
C‒H bond dipoles are arranged about carbon so that they cancel each other out, making
CH4 a nonpolar molecule. See Example 8.2.
Ions and Ionic Compounds
41. Al3+: [He]2s2
2p6
; Ba2+: [Kr]5s2
4d105p6
; Se2
: [Ar]4s2
3d104p6 ; I
−
: [Kr]5s2
4d105p6
42. Te2
: [Kr]5s2
4d105p6
; Cl−
: [Ne]3s2
3p6
; Sr2+: [Ar]4s2
3d104p6
; Li+
: 1s2
43. a. Li+
and N3
are the expected ions. The formula of the compound would be Li3N (lithium
nitride).
b. Ga3+ and O2
; Ga2O3, gallium(III) oxide or gallium oxide
c. Rb+
and Cl
; RbCl, rubidium chloride d. Ba2+ and S2
; BaS, barium sulfide
44. a. Al3+ and Cl
; AlCl3, aluminum chloride b. Na+
and O2
; Na2O, sodium oxide
CHAPTER 8 BONDING: GENERAL CONCEPTS 257
c. Sr2+ and F
; SrF2, strontium fluoride d. Ca2+ and S2
; CaS, calcium sulfide
45. a. Mg2+: 1s2
2s2
2p6
; K
+
: 1s2
2s2
2p6
3s2
3p6
; Al3+: 1s2
2s2
2p6
b. N
3
, O2
, and F
: 1s2
2s2
2p6
; Te2-
: [Kr]5s2
4d105p6
46. a. Sr2+: [Ar]4s2
3d104p6
; Cs+
: [Kr]5s2
4d105p6
; In+
: [Kr]5s2
4d10; Pb2+: [Xe]6s2
4f145d10
b. P
3− and S2−: [Ne]3s2
3p6
; Br−
: [Ar]4s2
3d104p6
47. a. Sc3+: [Ar] b. Te2
: [Xe] c. Ce4+: [Xe] and Ti4+: [Ar] d. Ba2+: [Xe]
All these ions have the noble gas electron configuration shown in brackets.
48. a. Cs2S is composed of Cs+
and S2
. Cs+ has the same electron configuration as Xe, and S2
has the same configuration as Ar.
b. SrF2; Sr2+ has the Kr electron configuration, and F
has the Ne configuration.
c. Ca3N2; Ca2+ has the Ar electron configuration, and N3
has the Ne configuration.
d. AlBr3; Al3+ has the Ne electron configuration, and Br
has the Kr configuration.
49. a. Na+
has 10 electrons. F−
, O2
, and N3
are some possible anions also having 10 electrons.
b. Ca2+ has 18 electrons. Cl−
, S2
, and P3
also have 18 electrons.
c. Al3+ has 10 electrons. F−
, O2
, and N3
also have 10 electrons.
d. Rb+
has 36 electrons. Br−
, Se2
, and As3
also have 36 electrons.
50. a. Ne has 10 electrons. AlN, MgF2, and Na2O are some possible ionic compounds where
each ion has 10 electrons.
b. CaS, K3P, and KCl are some examples where each ion is isoelectronic with Ar; i.e., each
ion has 18 electrons.
c. Each ion in Sr3As2, SrBr2, and Rb2Se is isoelectronic with Kr.
d. Each ion in BaTe and CsI is isoelectronic with Xe.
51. Neon has 10 electrons; there are many possible ions with 10 electrons. Some are N
3
, O2
,
F
, Na+
, Mg2+
, and Al3+. In terms of size, the ion with the most protons will hold the
electrons the tightest and will be the smallest. The largest ion will be the ion with the fewest
protons. The size trend is:
Al3+ < Mg2+
< Na+
< F
< O2
< N3
smallest largest
258 CHAPTER 8 BONDING: GENERAL CONCEPTS
52. All these ions have 18 e
; the smallest ion (Sc3+) has the most protons attracting the 18 e
,
and the largest ion has the fewest protons (S2
). The order in terms of increasing size is Sc3+
< Ca2+ < K+
< Cl
< S
2
. In terms of the atom size indicated in the question:
53. a. Cu > Cu+
> Cu2+ b. Pt2+ > Pd2+ > Ni2+ c. O
2
> O
> O
d. La3+ > Eu3+ > Gd3+ > Yb3+ e. Te2
> I
> Cs+
> Ba2+ > La3+
For answer a, as electrons are removed from an atom, size decreases. Answers b and d follow
the radius trend. For answer c, as electrons are added to an atom, size increases. Answer e
follows the trend for an isoelectronic series; i.e., the smallest ion has the most protons.
54. a. V > V2+ > V3+ > V5+ b. Cs+
> Rb+
> K+
> Na+
c. Te2
> I
> Cs+ > Ba2+
d. P
3
> P2
> P
> P e. Te2
> Se2
> S2
> O2
55. Lattice energy is proportional to Q1Q2/r, where Q is the charge of the ions and r is the
distance between the centers of the ions. The more negative the lattice energy, the more
stable the ionic compound. So greater charged ions as well as smaller sized ions lead to more
negative lattice energy values and more stable ionic compounds.
a. NaCl; Na+
is smaller than K+
. b. LiF; F
is smaller than Cl
.
c. MgO; O2
has a greater charge than OH-
. d. Fe(OH)3; Fe3+ has a greater charge than
Fe2+
.
e. Na2O; O2
has a greater charge than Cl
. f. MgO; both ions are smaller in MgO.
56. a. LiF; Li+
is smaller than Cs+
. b. NaBr; Bris smaller than I
.
c. BaO; O2
has a greater charge than Cl-
. d. CaSO4; Ca2+ has a greater charge than Na+
.
e. K2O; O2
has a greater charge than F-
. f. Li2O; both ions are smaller in Li2O.
57. K(s) K(g) ΔH = 90. kJ (sublimation)
K(g) K
+
(g) + e ΔH = 419 kJ (ionization energy)
1/2 Cl2(g) Cl(g) ΔH = 239/2 kJ (bond energy)
Cl(g) + e Cl
(g) ΔH = 349 kJ (electron affinity)
K
+
(g) + Cl
(g) KCl(s) ΔH = 690. kJ (lattice energy)
K(s) + 1/2 Cl2(g) KCl(s)
o ΔHf
= 411 kJ/mol
CHAPTER 8 BONDING: GENERAL CONCEPTS 259
58. Mg(s) Mg(g) ΔH = 150. kJ (sublimation)
Mg(g) Mg+
(g) + e ΔH = 735 kJ (IE1)
Mg+
(g) Mg2+(g) + e ΔH = 1445 kJ (IE2)
F2(g) 2 F(g) ΔH = 154 kJ (BE)
2 F(g) + 2 e 2 F
(g) ΔH = 2(328) kJ (EA)
Mg2+(g) + 2 F
(g) MgF2(s) ΔH = 2913 kJ (LE)
Mg(s) + F2(g) MgF2(s)
o ΔHf
= 1085 kJ/mol
59. From the data given, it takes less energy to produce Mg+
(g) + O
(g) than to produce
Mg2+(g) + O
2
(g). However, the lattice energy for Mg2+O
2 will be much more exothermic
than that for Mg+O
due to the greater charges in Mg2+O
2
. The favorable lattice energy term
dominates, and Mg2+O
2
forms.
60. Na(g) Na+
(g) + e ΔH = IE = 495 kJ (Table 7.5)
F(g) + e
F
(g) ΔH = EA = 327.8 kJ (Table 7.7)
Na(g) + F(g) Na+
(g) + F
(g) 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.
61. Use Figure 8.11 as a template for this problem.
Li(s) Li(g) ΔHsub = ?
Li(g) Li+
(g) + e
ΔH = 520. kJ
1/2 I2(g) I(g) ΔH = 151/2 kJ
I(g) + e I
(g) ΔH = 295 kJ
Li+
(g) + I
(g) LiI(s) ΔH = 753 kJ
Li(s) + 1/2 I2(g) LiI(s) ΔH = 292 kJ
ΔHsub + 520. + 151/2 295 753 = 292, ΔHsub = 161 kJ
62. Let us look at the complete cycle for Na2S.
2 Na(s) 2 Na(g) 2ΔHsub, Na = 2(109) kJ
2 Na(g) 2 Na+
(g) + 2 e
2IE = 2(495) kJ
S(s) S(g) ΔHsub, S = 277 kJ
S(g) + e S
(g) EA1 = 200. kJ
S
(g) + e S
2
(g) EA2 = ?
2 Na+
(g) + S2
(g) Na2S(s) LE = 2203 kJ
2 Na(s) + S(s) Na2S(s)
o ΔHf
= 365 kJ
o ΔHf
=
2ΔHsub, Na
+ 2IE + ΔHsub, S + EA1 + EA2 + LE, 365 = 918 + EA2, EA2 = 553 kJ
260 CHAPTER 8 BONDING: GENERAL CONCEPTS
For each salt:
o ΔHf
= 2ΔHsub, M + 2IE + 277 200. + LE + EA2
K2S: 381 = 2(90.) + 2(419) + 277 200. 2052 + EA2, EA2 = 576 kJ
Rb2S: 361 = 2(82) + 2(409) + 277 200. 1949 + EA2, EA2 = 529 kJ
Cs2S: 360. = 2(78) + 2(382) + 277 200. 1850. + EA2, EA2 = 493 kJ
We get values from 493 to 576 kJ.
The mean value is:
4
553 576 529 493
= 538 kJ. We can represent the results as EA2 =
540 ±50 kJ.
63. Ca2+ has a greater charge than Na+
, and Se2
is smaller than Te2
. The effect of charge on the
lattice energy is greater than the effect of size. We expect the trend from most exothermic
lattice energy to least exothermic to be:
CaSe > CaTe > Na2Se > Na2Te
(2862) (2721) (2130) (2095) This is what we observe.
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
Bond Energies
65. a.
Bonds broken: Bonds formed:
1 H‒H (432 kJ/mol) 2 H‒Cl (427 kJ/mol)
1 Cl‒Cl (239 kJ/mol)
ΔH = ΣDbroken ΣDformed, ΔH = 432 kJ + 239 kJ 2(427) kJ = 183 kJ
b.
Bonds broken: Bonds formed:
1 N≡N (941 kJ/mol) 6 N‒H (391 kJ/mol)
3 H‒H (432 kJ/mol)
ΔH = 941 kJ + 3(432) kJ 6(391) kJ = 109 kJ
+ 3 H H 2 H N H
H
N N
CHAPTER 8 BONDING: GENERAL CONCEPTS 261
66. Sometimes some of the bonds remain the same between reactants and products. To save
time, only break and form bonds that are involved in the reaction.
a.
Bonds broken: Bonds formed:
1 C≡N (891 kJ/mol) 1 CN (305 kJ/mol)
2 HH (432 kJ/mol) 2 CH (413 kJ/mol)
2 NH (391 kJ/mol)
ΔH = 891 kJ + 2(432 kJ) [305 kJ + 2(413 kJ) + 2(391 kJ)] = 158 kJ
b.
Bonds broken: Bonds formed:
1 NN (160. kJ/mol) 4 HF (565 kJ/mol)
4 NH (391 kJ/mol) 1 N ≡ N (941 kJ/mol)
2 FF (154 kJ/mol)
ΔH = 160. kJ + 4(391 kJ) + 2(154 kJ) [4(565 kJ) + 941 kJ] = 1169 kJ
67.
Bonds broken: 1 C‒N (305 kJ/mol) Bonds formed: 1 C‒C (347 kJ/mol)
ΔH = ΣDbroken ΣDformed, ΔH = 305 347 = 42 kJ
Note: Sometimes some of the bonds remain the same between reactants and products.
To save time, only break and form bonds that are involved in the reaction.
68.
N N
H
H
H
H
+ 2 F F 4 H F + N N
H C C N
H
H
H C N C
H
H
H C C O H
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