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Basic Concepts
of Chemical
Bonding
Visualizing Concepts
8.1 Analyze/Plan. Count the number of electrons in the Lewis symbol. This corresponds to the ‘A’-group number of the family. Solve. (a) Group 14 or 4A (b) Group 2 or 2A (c) Group 15 or 5A (These are the appropriate groups in the s and p blocks, where Lewis symbols are most useful.) 8.2 (a) No. Oppositely charged ions combine (via Coulombic attraction) to form ionic compounds. A 1 and A 2 have like charges and repel each other. (b) A 1 Z 1 has the largest lattice energy. Lattice energy increases as ionic charge increases and decreases as inter-ionic distance increases. Since the magnitude of all charges is 1, this factor doesn’t vary among possible ion combinations. A 1 and Z 1 have the smallest radii, which leads to the smallest inter-ionic distance and the largest lattice energy. (c) A 2 Z 2 has the smallest lattice energy, because it has the largest inter- ion distance. 8.3 Analyze/Plan. Count the valence electrons in the orbital diagram, take ion charge into account, and find the element with this orbital electron count on the periodic chart. Write the complete electron configuration for the ion. Solve. (a) This ion has seven 3d electrons. Transition metals, or d-block elements, have valence electrons in d-orbitals. Transition metal ions first lose electrons from the 4s orbital, then from 3d if required by the charge. This 2+ ion has lost two electrons from 4s, none from 3d. The transition metal with seven 3d-electrons is cobalt, Co.
(b) The electron configuration of Co is [Ar]4s 2 3 d^7. (The configuration of Co^2 +^ is [Ar]3d^7 ). 8.4 Analyze/Plan. This question is a ”reverse” Lewis structure. Count the valence electrons shown in the Lewis structure. For each atom, assume zero formal charge and determine the number of valence electrons an unbound atom has. Name the element. Solve. A: 1 shared e–^ pair = 1 valence electron + 3 unshared pairs = 7 valence electrons, F E: 2 shared pairs = 2 valence electrons + 2 unshared pairs = 6 valence electrons, O D: 4 shared pairs = 4 valence electrons, C Q: 3 shared pairs = 3 valence electrons + 1 unshared pair = 5 valence electrons, N X: 1 shared pair = 1 valence electron, no unshared pairs, H Z: same as X, H Check. Count the valence electrons in the Lewis structure. Does the number correspond to the molecular formula CH 2 ONF? 12 e–^ pair in the Lewis structure. CH 2 ONF = 4 + 2 + 6 + 5 + 7 = 24 e–^ , 12 e–^ pair. The molecular formula we derived matches the Lewis structure. 8.5 Analyze/Plan. Since there are no unshared pairs in the molecule, we use single bonds to H to complete the octet of each C atom. For the same pair of bonded atoms, the greater the bond order, the shorter and stronger the bond. Solve. (a) Moving from left to right along the molecule, the first C needs two H atoms, the second needs one, the third needs none, and the fourth needs one. The complete molecule is: (b) In order of increasing bond length: 3 < 1 < 2 (c) In order of increasing bond enthalpy (strength): 2 < 1 < 3 8.6 (a) The central atom Xe, is a member of Group 8A, currently known as noble gases. Prior to 1960, this group was known as inert gases, because there were no known compounds of group 8A elements. Noble gas elements have 8 valence electrons, so they satisfy the octet rule without forming chemical bonds. Since forming compounds wasn’t necessary for these elements, it was assumed that no compounds of group 8A elements existed. (b) There are a total of three resonance structures for the given Lewis structure of XeO 3 , each with the single bond in a different Xe–O bonding domain. (c) The given Lewis structure does not satisfy the octet rule, because the
electrons. These valence electrons are available for chemical bonding, while the core electrons do not participate in bonding.
Ionic Bonding
8.15 (a) AlF 3 (b) K 2 S (c) Y 2 O 3 (d) Mg 3 N (^2) 8.16 (a) BaO (b) RbI (c) Li 2 S (d) MgBr (^2) 8.17 (a) Sr^2 +: [Kr], noble-gas configuration (b) Ti^2 +: [Ar]3d^2 (c) Se^2 – : [Ar]4s 2 3 d^1 04 p^6 = [Kr], noble-gas configuration (d) Ni^2 +: [Ar]3d^8 (e) Br–^ : [Ar]4s 2 3 d^1 04 p^6 = [Kr], noble-gas configuration (f) Mn 3 +: [Ar]3d^4 8.18 (a) Zn 2 +: [Ar]3d^1 (b) Te^2 – : [Kr]5s^2 4 d^1 05 p^6 = [Xe], noble-gas configuration (c) Se^3 +: [Ar]4s^2 3 d^1 04 p^1 (This is a very unlikely ion. A more stable and commonly found ion would be Sc^3 +.) Sc^3 +: [Ar], noble-gas configuration (d) Ru 2 +: [Kr]4d^6 (e) Tl+^ : [Xe]6s 2 4 f^1 45 d^1 (f) Au +^ [Xe]4f^1 45 d^1 8.19 (a) Lattice energy is the energy required to totally separate one mole of solid ionic compound into its gaseous ions. (b) The magnitude of the lattice energy depends on the magnitudes of the charges of the two ions, their radii, and the arrangement of ions in the lattice. The main factor is the charges, because the radii of ions do not vary over a wide range. 8.20 (a) NaF, 910 kJ/mol; MgO, 3795 kJ/mol The two factors that affect lattice energies are charge and ionic radii. K Si^ Mg 2 + or Mg^2 +^ P 3 (a) (^) (b) (c)^ (d) or^ P^3
The Na–F and Mg–O separations are similar (Na+^ is larger than Mg^2 +, but F –^ is smaller than O^2 – ). The charges on Mg^2 +^ and O^2 –^ are twice those of Na+^ and F–^ , so according to Equation 8.4, the lattice energy of MgO is approximately four times that of NaF. (b) MgCl 2 , 2326 kJ/mol; SrCl 2 , 2127 kJ/mol The two factors that affect lattice energies are charge and ionic radii. The ionic charges are the same in the two compounds. The ionic radius of Mg^2 +^ is smaller than that of Sr 2 +^ so the Mg–Cl distance is slightly smaller than the Sr–Cl distance. Since lattice energy is inversely proportional to the ion separation, the lattice energy of MgCl 2 is slightly larger than that of SrCl 2. 8.21 KF, 808 kJ/mol; CaO, 3414 kJ/mol; ScN, 7547 kJ/mol The sizes of the ions vary as follows: Sc^3 +^ < Ca 2 +^ < K+^ and F –^ < O^2 –^ < N^3 –. Therefore, the inter-ionic distances are similar. According to Coulomb’s law for compounds with similar ionic separations, the lattice energies should be related as the product of the charges of the ions. The lattice energies above are approximately related as (1)(1): (2)(2): (3)(3) or 1:4:9. Slight variations are due to the small differences in ionic separations. 8.22 (a) According to Equation 8.4, electrostatic attraction increases with increasing charges of the ions and decreases with increasing radius of the ions. Thus, lattice energy (i) increases as the charges of the ions increase and (ii) decreases as the sizes of the ions increase. (b) RbBr < NaBr < LiCl < MgO. This order is confirmed by the lattice energies given in Table 8.2. MgO has the highest lattice energy because the ions have 2+ and 2– charges. The other compounds have cations with 1+ charges and anions with 1– charges. They are placed in order of decreasing ionic separation. Rb+^ and Br –^ have the largest radii; Na+^ is smaller than Rb+^ , Li+^ is smaller than Na +^ , and Cl–^ is smaller than Br –^. 8.23 Since the ionic charges are the same in the two compounds, the K–Br and Cs–Cl separations must be approximately equal. Since the radii are related as Cs+^ > K +^ and Br–^ > Cl–^ , the difference between Cs +^ and K+^ must be approximately equal to the difference between Br –^ and Cl–^. This is somewhat surprising, since K +^ and Cs+^ are two rows apart and Cl–^ and Br– are only one row apart. 8.24 (a) In MgO, the magnitude of the charges on both ions is 2; in MgCl 2 , the magnitudes of the charges are 2 and 1. Also, the Cl–^ ion is larger than the O^2 –^ ion, so the charge separation is greater in MgCl 2. Thus, the lattice energy of MgO is greater, because the product of the ionic charges is greater and the ion separation is smaller. (b) The ions have 1+ and 1– charges in all three compounds. In NaCl the cationic and anionic radii are smaller than in the other two
8.30 K and Ar. K is an active metal with one valence electron. It is most likely to achieve an octet by losing this single electron and to participate in ionic bonding. Ar has a stable octet of valence electrons; it is not likely to form chemical bonds of any type. 8.31 Analyze/Plan. Follow the logic in Sample Exercise 8.3. Solve. Check. Each pair of shared electrons in SiCl 4 is shown as a line; each atom is surrounded by an octet of electrons.
8.33 (a) (b) A double bond is required because there are not enough electrons to satisfy the octet rule with single bonds and unshared pairs. (c) The greater the number of shared electron pairs between two atoms, the shorter the distance between the atoms. If O 2 has a double bond, the O–O distance will be shorter than the O–O single bond distance.
The C–S bonds in CS 2 are double bonds, so the C–S distances will be shorter than a C–S single bond distance. 8.35 (a) Electronegativity is the ability of an atom in a molecule (a bonded atom) to attract electrons to itself. (b) The range of electronegativities on the Pauling scale is 0.7–4.0. (c) Fluorine, F, is the most electronegative element. (d) Cesium, Cs, is the least electronegative element that is not radioactive. 8.36 (a) The electronegativity of the elements increases going from left to right across a row of the periodic chart. (b) Electronegativity decreases going down a family of the periodic chart. (c) Generally, the trends in electronegativity are the same as those in ionization energy and opposite those in electron affinity. That is, the more positive the ionization energy and the more negative the electron affinity (ignoring a few exceptions), the greater the electronegativity of an element. 8.37 Plan. Electronegativity increases going up and to the right in the periodic
table. Solve. (a) Br (b) C (c) P (d) Be Check. The electronegativity values in Figure 8.6 confirm these selections. 8.38 Electronegativity increases going up and to the right in the periodic table. (a) O (b) Al (c) Cl (d) F 8.39 The bonds in (a), (c) and (d) are polar because the atoms involved differ in electronegativity. The more electronegative element in each polar bond is: (a) F (c) O (d) I 8.40 The more different the electronegativity values of the two elements, the more polar the bond. (a) O–F < C–F < Be–F. This order is clear from the periodic trend. (b) S–Br < C–P < O–Cl. Refer to the electronegativity values in Figure 8. to confirm the order of bond polarity. The 3 pairs of elements all have the same positional relationship on the periodic chart. The more electronegative element is one row above and one column to the left of the less electronegative element. This leads us to conclude that EN is similar for the 3 bonds, which is confirmed by values in Figure 8.6. The most polar bond, O–Cl, involves the most electronegative element, O. Generally, the largest electronegativity differences tend to be between row 2 and row 3 elements. The 2 bonds in this exercise involving elements in row 2 and row 3 do have slightly greater EN than the S–Br bond, between elements in rows 3 and 4. (c) C–S < N–O < B–F. You might predict that N–O is least polar since the elements are adjacent on the table. However, the big decrease going from the second row to the third means that the electronegativity of S is not only less than that of O, but essentially the same as that of C. C– S is the least polar. 8.41 Analyze/Plan. Q is the charge at either end of the dipole. Q = /r. From Table 8.3, the values for HF are = 1.82 D and r = 0.92 Å. Change Å to m and use the definition of the Debye and the charge of an electron to calculate the charge in units of e. Solve.
- 41 e
- 60 10 C 1 e 1 D
3. 34 10 C
1 10 m
1 Å
0.92 Å
1. 82 D
r
Q 19
30 10 m
Check. The calculated charge on H and F is 0.41 e. This can be thought of as the amount of charge “transferred” from H to F. This value is consistent with our idea that HF is a polar covalent molecule; the bonding electron pair is unequally shared, but not totally transferred from H to F. 8.42 (a) The more electronegative element, Br, will have a stronger attraction for the shared electrons and adopt a partial negative charge.
atom. (d) 32 valence e–^ , 16 e–^ pairs (Choose the Lewis structure that obeys the octet rule, Section 8.7.) (e) Follow Sample Exercise 8.8. 20 valence e–^ , 10 e–^ pairs (f) 14 valence e–^ , 7 e–^ pairs Check. In each molecule, bonding e–^ pairs are shown as lines, and each atom is surrounded by an octet of electrons (duet for H). 8.46 (a) 12 e–^ , 6 e–^ pairs (b) 14 valence e–^ , 7 e– pairs (c) 50 valence e–^ , 25 e–^ pairs (d) 26 valence e–^ , 13 e–^ pairs (Choose the Lewis structure that obeys the octet rule, Section 8.7) (e) 26 valence e–^13 e–^ pairs (f) 10 e–^ , 5 e–^ pairs (Choose the Lewis structure that obeys they octet rule, Section 8.7.) 8.47 (a) Formal charge is the charge on each atom in a molecule, assuming all atoms have the same electronegativity. (b) Formal charges are not actual charges. They assume perfect covalency, one extreme for the possible electron distribution in a molecule. (c) The other extreme is represented by oxidation numbers, which
assume that the more electronegative element holds all electrons in a bond. The true electron distribution is some composite of the two extremes. 8.48 (a) 26 e–^ , 13 e–^ pairs The octet rule is satisfied for all atoms in the structure. (b) F is more electronegative than P. Assuming F atoms hold all shared electrons, the oxidation number of each F is –1. The oxidation number of P is +3. (c) Assuming perfect sharing, the formal charges on all F and P atoms are
(d) The oxidation number on P is +3; the formal charge is 0. These represent extremes in the possible electron distribution, not the best picture. By virtue of their greater electronegativity, the F atoms carry a partial negative charge, and the P atom a partial positive charge. 8.49 Analyze/Plan. Draw the correct Lewis structure: count valence electrons in each atom, total valence electrons and electron pairs in the molecule or ion; connect bonded atoms with a line, place the remaining e–^ pairs as needed, in nonbonded pairs or multiple bonds, so that each atom is surrounded by an octet (or duet for H). Calculate formal charges: assign electrons to individual atoms [nonbonding e–^ + 1/2 (bonding e–^ )]; formal charge = valence electrons – assigned electrons. Assign oxidation numbers, assuming that the more electronegative element holds all electrons in a bond. Solve. Formal charges are shown near the atoms, oxidation numbers (ox. #) are listed below the structures. (a) 10 e–^ , 5 e–^ pairs (b) 32 valence e–^ , 16 e– pairs ox. #: N, +3; O, – ox #: P, +5; Cl, –1; O, – 2
(c) Since each N–O bond has partial double bond character, the N–O bond length in NO 2 –^ should be shorter than in species with formal N–O single bonds. 8.52 (a) 16 e–^ , 8 e–^ pairs (b) More than one correct Lewis structure can be drawn, so resonance structures are needed to accurately describe the structure. (c) NO 2 +^ has 16 valence electrons. Consider other triatomic molecules involving second-row nonmetallic elements. O 3 2 +^ or C 3 4 –^ are not “common” (or stable). CO 2 is common and matches the description (as does N 3 –^ , azide ion). 8.53 Plan/Solve. The Lewis structures are as follows: 5 e–^ pairs 8 e–^ pairs 12 e–^ pairs The more pairs of electrons shared by two atoms, the shorter the bond between the atoms. The average number of electron pairs shared by C and O in the three species is 3 for CO, 2 for CO 2 , and 1.33 for CO 3 2 –. This is also the order of increasing bond length: CO < CO 2 < CO 3 2 –. 8.54 The Lewis structures are as follows: 5 e–^ pairs 9 e–^ pairs 12 e–^ pairs The average number of electron pairs in the N–O bond is 3.0 for NO+^ , 1. for NO 2 –^ , and 1.33 for NO 3 –^. The more electron pairs shared between two atoms, the shorter the bond. Thus the N–O bond lengths vary in the order NO+^ < NO 2 –^ < NO 3 –^.
8.55 (a) Two equally valid Lewis structures can be drawn for benzene. Each structure consists of alternating single and double C–C bonds; a particular bond is single in one structure and double in the other. The concept of resonance dictates that the true description of bonding is some hybrid or blend of the two Lewis structures. The most obvious blend of these two resonance structures is a molecule with six equivalent C–C bonds, each with some but not total double-bond character. If the molecule has six equivalent C–C bonds, the lengths of these bonds should be equal. (b) The resonance model described in (a) has six equivalent C–C bonds, each with some double bond character. That is, more than one pair but less than two pairs of electrons is involved in each C–C bond. This model predicts a uniform C–C bond length that is shorter than a single bond but longer than a double bond. 8.56 (a) (b) The resonance model of this molecule has bonds that are neither single nor double, but somewhere in between. This results in bond lengths that are intermediate between C–C single and C^ C^ double bond lengths. (c)
Exceptions to the Octet Rule
8.57 (a) The octet rule states that atoms will gain, lose, or share electrons until they are surrounded by eight valence electrons. (b) The octet rule applies to the individual ions in an ionic compound. That is, the cation has lost electrons to achieve an octet and the anion has gained electrons to achieve an octet. For example, in MgCl 2 , Mg loses 2 e–^ to become Mg^2 +^ with the electron configuration of Ne. Each Cl atom gains one electron to form Cl–^ with the electron configuration of Ar.
(e) 40 e–^ , 20 e–^ pairs Does not obey octet rule; 10 e–^ around central Sb 8.62 (a) 16 e–^ , 8 e–^ pairs Three resonance structures, all obey the octet rule. The left structure minimizes formal charge. (b) 26 e–^ , 13 e–^ pairs Other resonance structures with one, two, or three double bonds can be drawn. While a structure with three double bonds minimizes formal charges, all structures with double bonds violate the octet rule. Theoretical calculations show that the single best Lewis structure is the one that doesn’t violate the octet rule. Such a structure is shown above. (c) 6 e–^ , 3 e–^ pairs (d) 32 e–^ , 16 e–^ pairs 6 electrons around B. (e) 22 e–^ , 11 e–^ pair Obeys octet rule. Violates the octet rule; 10 e–^ around central Xe. 8.63 (a) 16 e–^ , 8 e–^ pairs This structure violates the octet rule; Be has only 4 e–^ around it. (b) (c) The formal charges on each of the atoms in the four resonance structures are:
Formal charges are minimized on the structure that violates the octet rule; this form is probably most important. Note that this is a different conclusion than for molecules that have resonance structures with expanded octets that minimize formal charge. 8.64 (a) 19 e–^ , 9.5 e–^ pairs, odd electron molecule (b) None of the structures satisfies the octet rule. In each structure, one atom has only 7 e–^ around it. If a molecule has an odd number of electrons in the valence shell, no Lewis structure can satisfy the octet rule. (c) Formal charge arguments predict that the two resonance structures with the odd electron on O are most important. This contradicts electronegativity arguments, which would predict that the less electronegative atom, Cl, would be more likely to have fewer than 8 e– around it.
Bond Enthalpies
8.65 Analyze. Given: structural formulas. Find: enthalpy of reaction. Plan. Count the number and kinds of bonds that are broken and formed by the reaction. Use bond enthalpies from Table 8.4 and Equation 8.12 to calculate the overall enthalpy of reaction, H. Solve. (a) H = 2D(O–H) + D(O–O) + 4D(C–H) + D(C=C) –2D(O–H) – 2D(O–C) – 4D(C–H) – D(C–C) H = D(O–O) + D(C=C) – 2D(O–C) – D(C–C) = 146 + 614 – 2(358) – 348 = –304 kJ (b) H = 5D(C–H) + D(C N) + D(C=C) – 5D(C–H) – D(C N) – 2D(C–C) = D(C=C) – 2D(C–C) = 614 – 2(348) = –82 kJ (c) H = 6D(N–Cl) – 3D(Cl–Cl) – D(N N) = 6(200) – 3(242) – 941 = –467 kJ 8.66 (a) H = 3D(C–Br) + D(C–H) + D(Cl–Cl) – 3D(C–Br) – (C–Cl) – D(H–Cl) = D(C–H) + D(Cl–Cl) – D(C–Cl) – D(H–Cl)
(b) H = 3D(Si–H) + D(Si–Cl) + 3D(Cl–Cl) – 4D(Si–Cl) – 3D(H–Cl) = 3D(Si–H) + 3D(Cl–Cl) – 3D(Si–Cl) – 3D(H–Cl) = 3(323) + 3(242) – 3(464) – 3(431) = –990 kJ (c) Plan. Use bond enthalpies to calculate H for the reaction with S 8 (g) as a product. Then, H [ H H S(g )] H forS(g )
H
8HS(g) 8H(g) S (s) S(g) S (s) 8 HS(g) 8H(g) S (g) 8 º rxn f 8 º f 2 2 8 8 8 2 2 8
H = 16D(S–H) – 8(H–H) – 8(S–S)
= 16(339) – 8(436) – 8(266) = –192 kJ H (^) rxn HHfS 8 (g) 192 kJ 102. 3 kJ 294. 3 294 kJ 8.69 Plan. Draw structural formulas so bonds can be visualized. Then use Table 8.4 and Equation 8.12. Solve. (a) H = D(N N) + 3D(H–H) – 6(N–H) = 941 kJ + 3(436 kJ) – 6(391 kJ) = –97 kJ/2 mol NH 3 ; exothermic (b) Plan. Use Eq. 5.31 to calculate Hrxn from H f values. H (^) rxn nHf(products)nHf(reactants).Hf NH 3 (g) 46. 19 kJ. Solve. rxn 3 rxn f 3 f 2 f 2 H 2 ( 46. 19 ) 3 ( 0 ) 0 92. 38 kJ/2mol NH H 2 H NH(g) 3 H H (g) H N (g)
The H calculated from bond enthalpies is slightly more exothermic (more negative) than that obtained using Hf values. 8.70 (a) H = 4D(C–H) + D(C=C) + D(H–H) – 6D(C–H) – D(C–C)
= D(C=C) + D(H–H) – 2D(C–H) – D(C–C)
H = 614 + 436 – 2(413) – 348 = –124 kJ (b) H ^ HfC 2 H 6 (g)Hf C 2 H 4 (g)Hf^ H 2 (g) = –84.68 – 52.30 – 0 = –136.98 kJ The values of H for the reaction differ because the bond enthalpies used in part (a) are average values that can differ from one compound to another. For example, the exact enthalpy of a C–H bond in C 2 H 4 is probably not equal to the enthalpy of a C–H bond in C 2 H 6. Thus, reaction enthalpies calculated from average bond enthalpies are estimates. On the other hand, standard enthalpies of formation are measured quantities and should lead to accurate reaction enthalpies. The advantage of average bond enthalpies is that they can be used for reactions where no measured enthalpies of formation are available. 8.71 The average Ti–Cl bond enthalpy is just the average of the four values listed. 430 kJ/mol. 8.72 (a) (i) H = 2D(F–F) – 4D(C–F) = 2(155) – 4(485) = –1630 kJ (ii) H = D(CO) + 3D(F–F) – 4D(C–F) – 2D(D–F) = 1072 + 3(155) – 4(485) – 2(190) = –783 kJ (iii) H = 2D(C=O) + 4D(F–F) – 4D(C–F) – 4D(O–F) = 2(799) + 4(155) – 4(485) – 4(190) = –482 kJ Reaction (i) is most exothermic. (b) The more oxygen atoms bound to carbon, the less exothermic the reaction in this series.
Additional Exercises
8.73 Six nonradioactive elements in the periodic table have Lewis symbols with
