Download Chapter 2 Chemical Bonds Ionic Bonds

Chapter 2 Chemical Bonds
Ionic Bonds (Sections 2.1–2.4)
Key Concepts
chemical bond, ionic bond, covalent bond, valence electrons, coulombic attraction, ionic
model, ionic solid, crystalline solid, lattice energy, Madelung constant, Born-Meyer
equation, refractory material, octet of electrons, duplet of electrons, noble-gas
configuration, Lewis symbols, variable valence
2.1 The Formation of Ionic Bonds
Example 2.1a
Calculate the energy change for the reaction K(g) + Cl(g) → K+(g) + Cl−(g).
Solution
Break the reaction into two parts: ionization and electron gain. Use the ionization energy
and electron affinity data from Chapter 1 in the text.
K(g) → K+(g) + e−(g)
Ionization:
I 1 = 418 kJ⋅mol–1
Electron gain:
Cl(g) + e−(g) → Cl−(g)
− E ea = −349 kJ⋅mol–1
Overall:
K(g) + Cl(g) → K+(g) + Cl−(g)
∆E = 1 mol × (I 1 − E ea) = 69 kJ
The energy of the separated ions is higher than that of the separated atoms. Bringing the ions
together to form solid KCl releases a larger amount of energy, leading to the stability of KCl.
Example 2.1b
Given that I 2 for Ca(g) is 1150 kJ⋅mol−1, calculate the energy change for the reaction:
Ca(g) + 2 Br(g) → Ca2+(g) + 2 Br−(g)
Answer
∆E = 1 mol × (I 1 + I 2) − (2 mol × E ea) = [590 + 1150 − (2 × 325)] kJ = 1090 kJ
2.2 Interactions Between Ions
Example 2.2a
Calculate the energy change for the reactions: (A) K+(g) + Cl−(g) → KCl(g)
and (B) KCl(g) → K(g) + Cl(g), given that r12 = 0.267 nm for KCl(g).
Note: The bonding is considered to be totally ionic and only the coulomb
energy between ions is considered. The repulsive energy and other sources of
attractive energy are ignored. In (B), use the results of (A) and Example 2.1a.
Solution
(A) The energy between two ions at infinite separation is 0. The energy change ∆E for one
molecule is then EP,12(product) − EP,12(reactants).
∆E = EP,12 ( r12 = 0.267 nm) − EP,12 ( r12 = ∞ )
=
2.307( +1)( −1)
× 10−19 J⋅nm − 0 = −8.64 × 10−19 J
(0.267 nm)
or NA × ∆E = −520 kJ for 1 mol of KCl molecules.
Note: Compare K+(g) + Cl−(g) → KCl(s), where NA × ∆E = −717 kJ
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Chapter 2
(B) Break the reaction into two parts:
+
KCl(g) → K+(g) + Cl−(g)
−(−520 kJ)
K (g) + Cl−(g) → Κ(g) + Cl(g)
−(69 kJ)
_____________________________________
(A)
Example 2.1a
KCl(g) → K(g) + Cl(g)
451 kJ
The experimentally derived value for this reaction is 429 kJ.
Example 2.2b
Answer
Calculate the energy change for the reactions (A) Ca2+(g) + 2 Br−(g) → CaBr2(g)
and (B) CaBr2(g) → Ca(g) + 2 Br(g), given that r12 = 0.296 nm (sum of the ionic
radii of the cation and anion) for CaBr2(g), in which the bonding is considered to
be totally ionic and the molecule is assumed to be linear.
Consider the coulomb interaction for each cation-anion pair and single anion-anion
pair separately. As before, the energy at infinite separation is zero.
Beware of the cation charge and the anion-anion distance. In (B), use the
results of (A) and Example 2.1b.
(A) NA × ∆E = −1642 kJ for one mol of CaBr2 molecules
(B) −(−1642 kJ) −1090 kJ = 552 kJ
Note: The ionic radii are obtained from structural data for ionic solids. In Example 2.2a,
the actual bond distance for a gas-phase KCl molecule is used.
Example 2.2c
Calculate the ratio of the lattice energy of KCl to that of RbCl. Use the ionic radii
of the cations and anions in Chapter 1 of the text to estimate d. Notice that both
KCl and RbCl have the NaCl crystal structure (neglect repulsion energies).
Solution
The Madelung constants and ion charges are the same for KCl and RbCl. Therefore, the
ratio of lattice energies is simply
EKCl
d
149 pm + 181 pm
= RbCl =
= 1.03
ERbCl
d KCl 138 pm + 181 pm
Example 2.2d
Answer
(A) Calculate the ratio of the lattice energy of CsCl to that of CsI.
Both solids have the CsCl crystal structure.
(B) In which solid are the ions more strongly bound?
(A) (170 + 220)/(170 + 181) = (390/351) = 1.11.
(B) From part (A), CsCl has the larger potential energy, EP, and therefore the ions are
bound together more strongly in this salt.
2.3 The Electron Configurations of Ions
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Example 2.3a
Determine the electron configuration of Fe(II) and Fe (III).
Answer
Iron (II), Fe 2+, [Ar] 3d 6 and iron (III), Fe 3+, [Ar]3d 5
Example 2.3b
Determine the electron configuration of the carbide (methanide) and hydride anions.
Answer
Carbide (methanide), C 4− , [He] 2s2 2p6 or [Ne]
Hydride, H −, 1s2 or [He]
Example 2.3c
Determine the formula of calcium bromide.
Answer
Ca, [Ar] 4s2, loses two 4s-electrons to become Ca2+, [Ar].
Br, [Ar] 3d 10 4s2 4p5, gains an electron to become Br−, [Ar]3d 10 4s2 4p6 or [Kr].
{octet}
{duplet}
Chemical Bonds
Electrical neutrality: combining two Br− ions for each Ca2+ gives calcium bromide,
CaBr2, an ionic solid.
2.4 Lewis Symbols (Atoms and Ions)
Example 2.4a
Write the Lewis symbols for the neutral atoms in the first two periods of the Periodic
Table.
Answer
H
He
Li
Be
B
C
N
Note: The ground-state structures for B and C are
Example 2.4b
+
Example 2.4c
Ne
and C .
2+
Be
N
O
F
Write the Lewis symbols for the reaction between calcium atoms and oxygen atoms
to form calcium oxide.
2−
2+
Ca
Answer
B
F
Write the Lewis symbols for the lithium cation, beryllium cation, nitride anion,
oxide anion, and the fluoride anion.
−
3−
2−
Li
Answer
O
+ O
Ca
O
Covalent Bonds (Sections 2.5–2.9)
Key Concepts
covalent bond, Lewis structures, octet rule, valence, lone pairs, single bond,
double bond, triple bond, multiple bonds, bond order, resonance, resonance hybrid,
delocalized electrons, Kekulé structure, formal charge, plausibility of a structure
2.5 The Nature of the Covalent Bond
Example 2.5
List several nonmetallic elements that form covalent bonds. Also, include the molecular
formula.
Answer
Nonmetallic elements such as H2, N2, O2, F2, Cl2, Br2, I2, P4, and S8
2.6 Lewis Structures
Example 2.6
Write Lewis structures for the hydrogen and nitrogen molecules.
Answer
H–H, (single bond) duplet on each atom (valence of hydrogen = 1) and :N≡N:,
(triple bond and lone pair) octet on each atom (valence of nitrogen = 3)
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Chapter 2
2.7 Lewis Structures for Polyatomic Species
Example 2.7a
Determine the Lewis structure of ethyne (acetylene), C2H2.
Solution
Count the number of valence electrons [2(4) + 2(1) = 10 valence electrons or 5 electron
pairs]. Draw a symmetrical arrangement of atoms with C atoms as central atoms because
H atoms are terminal. There are a minimum of three single bonds [H–C–C–H ]
accounting for three pairs of electrons. Four electron pairs are required to complete the
octets on carbon atoms, but there are only two pairs left. Make two additional bonds.
Because H atoms make only one bond, the two additional bonds are located between the
carbon atoms.
Answer
The final Lewis structure is H–C≡C–H with one triple bond between the carbon atoms
and no lone pair (nonbonding pair) electrons.
Example 2.7b
Determine the Lewis structure of hydrogen cyanide, HCN.
Solution
Count the number of valence electrons [1 + 4 + 5 = 10 valence electrons or 5 electron
pairs]. Draw an arrangement of atoms with the C atom as the central atom because C has
the lowest I 1 and the H atom is terminal. There are a minimum of two single bonds
[H–C–N ] accounting for two pairs of electrons. Five electron pairs are required to
complete the octets on the carbon and nitrogen atoms, but there are only three pairs left.
One of the three pairs remaining is assigned as a lone pair and the other two pairs make
additional bonds. Because H atoms make only one bond, the two additional bonds are
located between the carbon and nitrogen atoms. The lone pair is assigned to the nitrogen
atom to complete its octet.
Answer
The final Lewis structure is H–C≡N: with one triple bond between the carbon and
nitrogen atoms and a lone pair (nonbonding pair) of electrons on the nitrogen atom.
Example 2.7c
Determine the Lewis structure of the ammonium ion, NH4+.
Answer
Valence electrons = 5 + 4(1) − 1 (electron removed) = 8 or four pairs.
H
H N H
H
NH4+
has a Lewis structure similar to that of methane, CH4 .
2.8 Resonance
Example 2.8a
Determine two Lewis structures for nitrous oxide, N2O.
Solution
N2O, nitrous oxide, has 2(5) + 6 = 16 valence electrons or eight pairs.
Answer
N N O
N N O
Blending of two structures, both of which obey the octet rule
Note: A triple bond to an O atom is found in species such as BO−, CO, NO+, and O22+.
Note: The central N atom is the least electronegative atom (Section 2.13).
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Example 2.8b
Determine Lewis structures for benzene, C6H6 .
Solution
C6H6, benzene, has 6(4) + 6(1) = 30 valence electrons or 15 pairs.
Chemical Bonds
or
Answer
Kekulé structures
Resonance hybrid
Note: Carbon atoms are at the corners of the hexagon (the six C–H bonds radiating from each
corner are not shown). The last structure depicts six valence electrons delocalized around
the ring.
2.9 Formal Charge
Example 2.9a
Calculate the formal charges on each atom in the first Lewis structure of N2O given above,
namely
N N O . Also, calculate the formal charges in the second structure, which has a
triple bond between the N atoms, and in one with a triple bond between N and O.
Solution
Consider the first structure.
Left N atom:
formal charge = 5 − [4 + (0.5)4] = 5 − 6 = −1
Central N atom:
formal charge = 5 − [0 + (0.5)8] = 5 − 4 = +1
Right O atom:
formal charge = 6 − [4 + (0.5)4] = 6 − 6 = 0
The sum of the formal charges equals the charge on the species.
Answer
Finally,
N N O . For the second structure, N N O .
−1
+1
0
0
For the structure with a triple bond between N and O,
+1
−1
N N O . In this case, the formal
−2
+1
+1
charges are not as close to 0 as in the previous two structures. The last structure does not
contribute appreciably to the resonance hybrid and is normally not included.
Example 2.9b
Write the Lewis structures for hydrogen cyanide, HCN, and hydrogen isocyanide, HNC.
Calculate the formal charges and predict the isomer that is favored.
Answer
H–C≡N: and H–N≡C:
0 0
Example 2.9c
Answer
0
Lower formal charges on HCN, which is favored.
0 +1 −1
Write the Lewis structure with two double bonds for nitrous oxide, NNO, and its isomer,
NON. Calculate the formal charges and predict the isomer that is favored.
N N O
−1
+1
0
and N O N
−1
+2
−1
Lower formal charges on NNO, which is energetically more favorable. Note that there are
usually only one or two bonds to an oxygen atom except for species such as BO−, CO, NO+,
O22+, and O3 (central O atom).
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Chapter 2
Exceptions to the Octet Rule (Sections 2.10–2.12)
Key Concepts
radicals, antioxidant, biradicals, expanded valence shell, variable covalence,
coordinate covalent bond
2.10 Radicals and Biradicals
Example 2.10
Write the Lewis structure(s) for the ozonide ion, O3−.
Solution
First determine the number of valence electrons. Locate them in the molecular ion
to satisfy the octet rule, if possible. Because this is a radical, one atom will be
unsatisfied. If more than one structure is possible, show the blending of Lewis
structures (resonance hybrid).
Answer
Valence electrons = 6 + 6 + 6 + 1 (charge on anion) = 19. The all-single-bond
structure is O–O–O with 15 electrons unaccounted for. A total of 16 electrons are
needed to satisfy the octet rule, but only 15 remain. Three choices for locating the
odd electron are
O O O
O O O
O O O
.
2.11 Expanded Valence Shells
Example 2.11
Write the major Lewis structures for sulfuric acid, H2SO4.
Solution
Count the number of valence electrons and first produce a structure satisfying the
octet rule. Then consider octet expansion, which is allowed for sulfur atoms.
Answer
The number of valence electrons = 2(1) + 6 + 4(6) = 32. A minimum structure requires 6
single bonds or 12 electrons. The remaining 20 electrons may occupy lone pair sites, thus
satisfying the octet/duplet rule. Recall that oxoacids have H atoms bonded to O atoms, and S
is a central atom. When the structure is drawn with only single bonds, the formal charge on S
is +2. This charge can be reduced to +1 by forming an additional bond from a lone pair on one
terminal O atom. It can be further reduced to 0 by forming an additional bond from a lone pair
on the other terminal O atom. Three representative structures are
O
O
O
H O S O H
H O S O H
H O S O H
O
O
O
Note: In the structure on the right, all atoms have a formal charge of 0 and the S atom is surrounded
by 12 electrons in an expanded octet. Because it has the lowest formal charges, we expect the
structure on the right to be the most favored one energetically and the one to make the greatest
contribution to the resonance hybrid.
There is one additional Lewis structure with a single double bond not shown.
Other examples similar to this one include SO42−, SO2, and S2O (S central atom).
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Chemical Bonds
2.12 The Unusual Structures of Some Group 13/III Compounds
Example 2.12a Write Lewis structures for the reaction of BF3 with NH3.
F
Answer
F
H
B
N H
F
H
F
F
H
B
N H
F
H
Both electrons in the B-N bond are derived from the N lone pair.
Example 2.12b Write the Lewis structure for the dimer Al2Cl6 molecule (vapor at high temperatures) that is
formed from the reaction of two molecules of AlCl3.
Solution
Each coordinate covalent bond is formed by the donation of a lone pair of electrons on Cl to
complete the octet on Al. The resulting Al–Cl–Al bridging bonds (bold) are equivalent.
Cl
Cl
Al
Answer
Cl
Cl
Al
Cl
Cl
Ionic versus Covalent Bonds (Sections 2.13–2.14)
Key Concepts
partial charges, polar covalent bond, electric dipole, electric dipole moment, debye,
electronegativity, ionic character, covalent character, polarizability, polarizable,
polarizing power
2.13 Correcting the Covalent Model: Electronegativity
Example 2.13a In the HF molecule, the measured dipole moment is 1.82 D and the distance between
partial charges (bond distance) is 91.7 pm. Estimate the partial charges on the HF
molecule by rearranging the above relationship to solve for δ.
Answer
δ=
1.82 D
µ
=
(4.80 D) × (distance in pm/100 pm) (4.80 D) × (91.7 pm /100 pm)
δ = 0.413
This suggests a negative charge of 0.413 electrons on the F atom and a corresponding
positive charge on the H atom.
Example 2.13b Estimate the dipole moment of the HF molecule by using the electronegativity values of
H and F. Compare with the measured value of 1.82 D.
Answer
µ ≈ (χF − χH) D = (4.0 − 2.2) D = 1.8 D, very close to the measured value.
The HF bond has considerable ionic character. It is a polar covalent bond.
7
Chapter 2
Note: The approximate relationship does not apply to ionic bonds.
Note: Rough rules of thumb:
(χA − χB) ≥ 2
0.5 ≤
bond is essentially ionic
(χA − χB) ≤ 1.5 bond is polar covalent
(χA − χB) ≤ 0.5 bond is essentially covalent
2.14 Correcting the Ionic Model: Polarizability
Example 2.14
Give examples of covalent bonding enhanced by polarizing power and polarizability.
Answer
Bonds between highly polarizing cations and highly polarizable anions have significant
covalent character. The Be2+ cation is highly polarizing and the Be–Cl bond is significantly
covalent (even though there is an electronegativity difference of 1.6). In the series AgCl to
AgI, the bonds become more covalent as the polarizability (and size) of the anion increases
(Cl− < Br− < I− ).
The Strengths and Lengths of Covalent Bonds
(Sections 2.15–2.17)
Key Concepts
dissociation energy, bond strength, bond length, covalent radius
2.15 Bond Strengths
Example 2.15
Describe the property of dissociation energy and its relationship to bond strength.
Answer
Dissociation energy is the energy required to separate bonded atoms. The greater the value of
the dissociation energy, the stronger the bond becomes. Dissociation energy is defined exactly
for diatomic molecules. For polyatomic molecules, the dissociation energy also depends on the
other bonds in the molecule. However, for many molecules, this dependence is slight.
2.16 The Variation of Bond Strength
Example 2.16
Provide some examples of the factors influencing bond strength.
Answer
For polyatomic molecules, bond strength is defined as the average dissociation energy
for one type of bond found in different molecules.
Factors influencing bond strength are resonance [C=C > C…C (benzene) > C–C], bond
multiplicity (C≡C > C=C > C–C), lone pairs on neighboring atoms (F–F < H–H), and atomic
radii (HF > HCl > HBr > HI). (The smaller the radius, the stronger the bond.)
8
Chemical Bonds
2.17 Bond Lengths
Example 2.17a Use the covalent radii displayed in Fig. 2.21 in the text to estimate the various bond lengths in
the acetic acid molecule, CH3COOH.
Solution
First write a Lewis structure for acetic acid. Add together the covalent radii of the atoms in
each bond to estimate its length.
Answer
Lewis structure:
H O
H C C O H
H
Bond type
C–H
C–C
C–O
C=O
O–H
Bond length estimate
Actual
77 pm + 37 pm = 114 pm ≈109 pm
77 pm + 77 pm = 154 pm ≈150 pm
77 pm + 74 pm = 151 pm 134 pm
67 pm + 60 pm = 127 pm 120 pm
74 pm + 37 pm = 111 pm
97 pm
Example 2.17b Use the covalent radii displayed in Fig. 2.21 in the text to estimate the various bond lengths in
the hydroxyl amine molecule, NH2OH.
Answer
N–H, 112 pm
N–O, 149 pm
O–H, 111 pm
Note: The H atom covalent radius of 37 pm obtained from H2 overestimates bond lengths in
other molecules. A better value is 30 pm for –H bonds except H–H.
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