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.
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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.)
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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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