133

Chapter 10: Molecular Structure and Bonding Theories 10.1 See Section 10.1. The main premise of the VSEPR model is that the electron pairs within the valence shell of an atom repel each other and determine the molecular geometry of the molecule or ion of interest. 10.3 See Section 10.1. Formaldehyde, H2CO, is an example of a trigonal planar molecule in which the carbon forms four bonds, one of which is part of a double bond to the oxygen atom. Phosgene, Cl2CO, is another example of a trigonal planar molecule in which carbon forms four bonds. CO2 is an example of a molecule with a central C atom that makes four bonds but has a linear-bonded atom lone-pair arrangement. 10.5 See Section 10.1 and Figure 10.6

IFF

F

I

F

FF

I

F

F

F

The VSEPR model states that the order of importance of repulsions within the valence shell of an atom is lp-lp > lp-bp > bp-bp. In addition, repulsions diminish as the angle between the electron pairs increases from 90º to 120º to 180º. Hence, lone pairs always occupy equatorial rather than axial positions in trigonal bipyramidal arrangements because there are only two interactions at 90º for equatorial positions in trigonal bipyramidal positions compared to three interactions at 90º for axial positions. The structure on the right is favored and is observed experimentally. 10.7 See Section 10.2 and Examples 10.4, 10.5. Any molecule with a totally symmetrical arrangement of atoms and lone pairs of electrons is nonpolar. Hence, any of the electron-pair arrangements shown in Figure 10.1 having like atoms attached to a central atom that is different has polar bonds but is nonpolar overall. For example, BeCl2(g) as is nonpolar because the bond dipoles cancel each other. This always occurs when the molecule is totally symmetrical.

Be ClCl

10.9 See Section 10.3. Valence bond theory describes bonds as being formed by atoms sharing valence electrons in overlapping valence orbitals. These overlaps are caused by the attraction between the nuclear charge of one of the bonded atoms and the electron cloud of the other atom and vice versa.

134

10.11 See Section 10.3 and Example 10.6.

Cl [Ne]

↑↓3s

↑↓

↑↓3p

↑

F [He]

↑↓2s

↑↓

↑↓2p

↑

The partially filled 3p orbital of Cl overlaps with the partially filled 2p orbital of F to form the bond in ClF. 10.13 See Section 10.1, 10.3, Figure 10.15, 10.16, 10.17 and Example 10.7.

BCl3 B

Cl

Cl Cl

SN (steric number for B = 3, trigonal planar electron-pair arrangement, sp2

hybrids for B.

BCl 4−

Cl

BCl

Cl

Cl

¯SN for B = 4, tetrahedral electron-pair arrangement, sp3 hybrids for B.

10.15 See Section 10.3 and Table 10.1. H Cl Hybrid orbitals are predicted by looking at Lewis structures, observing the steric number,

predicting the electron-pair arrangement and selecting the corresponding hybrid orbitals for the central atom. However, we can only determine the positions of attached atoms and are unable to determine the positions of lone pairs. Hence, any prediction of use of hybrid atomic orbitals by Cl in HCl would be a matter of pure conjecture that could not be verified by experiment. 10.17 See Section 10.4 and Figures 10.22, 10.24. A sigma (σ) bond is a bond in which the shared pair of electrons is symmetric about the axis joining the two nuclei of the bonded atoms. A pi (π) bond is a bond that places electron density above and below the line joining the bonded atoms and can be formed by the sideways overlap of p orbitals.

pz pz

σ

py py

π

135

10.19 See Section 10.5 and Figures 10.35, 10.36.

H H

1s 1s

σ1s

σ*1s

H2 10.21 See Section 10.6 and Figure 10.48. According to molecular orbital theory, there is a weak interaction between the 2s orbital of Li and the 2p orbital of F pointing toward Li resulting in a sigma bonding molecular orbital. This orbital is mainly centered around the fluorine atom, and the valence electron of Li is nearly completely transferred to the fluorine atom. According to the ionic bonding description, the valence electron of lithium is completely transferred from Li to F giving Li+ and F−. Hence, the two descriptions differ slightly in terms of electron transfer from Li to F. 10.23 See Section 10.1 and Figure 10.1. (a) SN = 3, trigonal planar (b) SN = 4, tetrahedral (c) SN = 4, tetrahedral (d) SN = 5, trigonal bipyramidal Remember that “bonded-atom lone-pair arrangement” takes into account both the outer atoms bound to the central atom and any lone pairs present on the central atom. 10.25 See Section 10.1, Figures 10.1, 10.6, and Examples 10.1, 10.2.

(a) CF4 (b) CS2 (c) AsF5

F

CF

F

F

C SS

As

F

F

FF

F

SN = 4 tetrahedral SN = 2, linear SN = 5, trigonal bipyramidal (d) CF2O (e) NH 4

+

C

O

F F

H

NH

H

H

+

SN = 3, trigonal planar SN = 4, tetrahedral

136

10.27 See Section 10.1, Figures 10.1, 10.6, and Examples 10.1, 10.2.

(a) SeO2 (b) N2O (c) H3O+

Se

OO

N N O

OH H

H

+

SN = 3, trigonal planar SN = 2, linear SN = 4, tetrahedral shape: bent shape: linear shape: trigonal pyramidal (d) IF5 (e) SCl4

I

FF F

F F

S

Cl

Cl

Cl

Cl

SN = 6, octahedral SN = 5, trigonal bipyramidal shape: square pyramidal shape: see - saw 10.29 See Section 10.1, Figures 10.1, 10.6, and Examples 10.1, 10.2. (a) BCl3 NCl3 (b) OF2 SF6

B

Cl

Cl Cl

N

ClCl

Cl O

F F S

FF F

F FF

SN = 3 SN = 4 SN = 4 SN = 6 trigonal planar tetrahedral tetrahedral octahedral electron pair electron pair electron pair electron pair arrangement arrangement arrangement arrangement 120º bond angles 109º bond angles 109º bond angles 90º & 180º

bond angles NCl3 has smaller bond angles than BCl3. SF6 has smaller bond angles than OF2. 10.31 See Section 10.1, Figures 10.1, 10.6, and Examples 10.1, 10.2. (a) Cl2NH NH 4

+ (b) SF2 IF 4−

NCl

ClH

H

NH

H

H

+

S

F F

IF F

F F

SN = 4 SN =4 SN = 4 SN = 6 tetrahedral electron tetrahedral electron tetrahedral electron octahedral electron pair arrangement pair arrangement pair arrangement pair arrangement

137

109º bond angles 109º bond angles 109º bond angles 90º & 180º bond angles

Cl2NH has slightly smaller bond angles due to the lone pair on the N. IF−4 has smaller bond angles

than SF2. 10.33 See Section 10.1, Figures 10.1, 10.6, and Examples 10.2, 10.3. (a) H3CCCH (b) Br2CCH2 (c) H3CNH2

C C HC

H

HH

109 Þ 180 Þ180 Þ

C C

Br

Br

H

H

120 Þ 120 Þ

C N

H

HH H

H

109 Þ

109 Þ

10.35 See Section 10.1, Figures 10.1, 10.6, and Examples 10.2, 10.3. (a) SO2 (b) ClO 3

− (c) SCN−

S

120 Þ

O O

ClO

O

O

109 Þ

S C N

180 Þ

expanded valence shell expanded valence shell 10.37 See Section 10.1, Figures 10.1, 10.6, and Examples 10.2, 10.3.

(a)(CH3)2CO (b)CH3CN

CH

H H

CC

H

HH

O

120 Þ109 Þ 109 Þ

C NC

H

HH

109 Þ180 Þ

138

10.39 See Section 10.1, Figures 10.1, 10.6, and Examples 10.2, 10.3.

(a)CH3ONCl2

C

H

H H

N Cl

ClO

109 Þ

109 Þ109 Þ

b)HC2PH2

C P

HH

CH

109 Þ180 Þ180 Þ 10.41 See Section 10.1, Figures 10.1, 10.6, and Examples 10.2, 10.3. (a) C2H4O (b) C3H6O (c) XeF2

CC

H

H H

H

O

120 Þ109 Þ

OC

H

H H

C C

H

HH

120 Þ

109 Þ 109 Þ

Xe FF

180 Þ

10.43 See Section 10.1, Figures 10.1, 10.6, and Examples 10.2, 10.3. (a) CH3SH (b) N2H2

SC

H

H H

H109 Þ

109 Þ

N N

H

H

120 Þ

10.45 See Section 10.2 and Examples 10.4, 10.5. (a) CF4 (b) CS2 (c) AsF5 (d) F2CO

F

CF F

F

C SS

F AsF

F

F

F

C O

F

F symmetrical symmetrical symmetrical unsymmetrical nonpolar nonpolar nonpolar polar 10.47 See Section 10.2 and Examplse 10.4, 10.5. (a) SeO2 (b) N2O (c) SCl4

Se

OO

N N O

S

Cl

Cl

Cl

Cl

139

expanded valence shell expanded valence shell unsymmetrical unsymmetrical unsymmetrical polar polar polar 10.49 See Section 10.2 and Example 10.4. (a) HCN (b) I2 (c) NO

C NH I⎯ I N O unsymmetrical symmetrical unsymmetrical polar nonpolar polar 10.51 See Section 10.2 and Example 10.4. (a) NF3 (b) CBr4 (c) BeI2

NF

FF

Br

CBr

Br

Br Be II

unsymmetrical symmetrical symmetrical polar nonpolar nonpolar 10.53 See Section 10.2 and Example 10.4. (a) C CF F (b) C CH F

C CF F C CH F symmetrical unsymmetrical nonpolar polar bond dipoles cancel bond dipoles do not cancel 10.55 See Section 10.3. (a) 120º, sp2 or sp3d (b) 90º, sp3d or sp3d2 (c) 180º, sp, sp3d or sp3d2

hybrids hybrids hybrids

10.57 See Section 10.3 and Examples 10.7, 10.8. (a) CF4 (b) SbCl 6

− (c) AsF5 (d) SiH4

F

CF

FF

Sb

Cl

Cl

Cl Cl

Cl Cl

F AsF

F

F

F

H

SiH

H

H

SN = 4 SN = 6 SN = 5 SN = 4 tetrahedral octahedral trigonal bipyramidal tetrahedral sp3 for C sp3d2 for Sb sp3d for As sp3 for Si

140

(e) NH 4+

H

NH

H

H

+

SN = 4 tetrahedral sp3 for N 10.59 See Section 10.3 and Examples 10.7, 10.8. (a) N2O (b) SnCl2(g) (c) I3

− (d) SeO2

N N O

SnCl Cl

I I I

Se

OO

SN = 2 SN = 3 SN = 5 SN = 3 linear trigonal planar trigonal bipyramidal trigonal

planar sp for central N sp2 for Sn sp3d for central I sp2 for Se 10.61 See Section 10.3 and Examples 10.7, 10.8. (a) CO3

2- (b) CH2F2 (c) H2CO

CO O

O2-

H

CH

FF

C

H H

O

SN = 3 SN = 4 SN = 3 trigonal planar tetrahedral trigonal planar sp2 for C sp3 for C sp2 for C 10.63 See Section 10.3 and Examples 10.7, 10.8. (a) H3O+ (b) H3COH (c) Cl2O

OH H

H

+

H

CH

HO

H

OCl Cl

SN = 4 SN = 4 SN = 4 tetrahedral tetrahedral tetrahedral sp3 for O sp3 for O sp3 for O

141

10.65 See Section 10.3 and Examples 10.7, 10.8. (a) OF2 SN = 4, tetrahedral

OF F

sp3 for O and [He] ↑↓

2s

↑↓ 2p

↑↓

↑ for F.

A sp3 orbital O containing one electron overlaps with a 2p orbital from F containing one electron to form an O-F bond in OF2. The lone pairs of electrons are in sp3 orbitals of O. (b) NH3 SN = 4, tetrahedral

NH

HH

sp3 for N and

↑1s

for H.

A sp3 orbital from N containing one electron overlaps with a 1s orbital of H containing one electron to form a N-H bond in NH3. The lone pair of electrons is in a sp3 orbital of N. (c) BCl3 SN = 3, trigonal planar

B

Cl

Cl Cl

sp2 for B and [Ne] ↑↓3s

↑↓

↑↓3p

↑

for Cl.

A sp2 orbital from B containing one electron overlaps with a 3p orbital of Cl containing one electron to form a B-Cl bond in BCl3. 10.67 See Section 10.3 and Examples 10.7, 10.8. SeF4 SN = 5, trigonal bipyramidal

Se

F

F

F

F

sp3d for Se and [He] ↑↓2s

↑↓ ↑↓2p

↑ for F.

A sp3d orbital from Se containing one electron overlaps with a 2p orbital of F containing one electron to form a Se-F bond in SeF4. The lone pair of electrons is in a sp3d orbital of Se. 10.69 See Section 10.3 and Examples 10.7, 10.8. (a) For C: SN = 4, tetrahedral, sp3.

C

H

H H

N Cl

ClO

For O: SN = 4, tetrahedral, sp3.

For N: SN = 4, tetrahedral, sp3.

(b) For both C: SN = 2, linear, sp.

C P

HH

CH For P: SN = 4, tetrahedral, sp3.

142

10.71 See Section 10.4, Figures 10.22-31.

(a)

pz pz

σ (b)

py py

π (c)

spz pz

σ

10.73 See Section 10.3, 10.4, and Examples 10.7, 10.8, 10.9. H3CCN

C1

H

HH

C2 N

Bond Orbital Overlaps Bond Type

C-H sp3-1s σ C1-C2 sp3-sp σ C2αN sp-pz σ py-py π px-px π SN C1 = 4, tetrahedral, sp3 SN C2 = 2, linear, sp There are a total of five σ bonds and two π bonds inH3CCN. 10.75 See Section 10.3, 10.4, and Examples 10.7, 10.8, 10.9.

C

H

H

N

H

C N

σ

σ

π

Bond Orbital Overlaps Bond Type C-H sp2-1s σ N-H sp2-1s σ C=N sp2-sp2 σ px-px π SN C = 3, trigonal planar, SN N = 3, trigonal planar, sp2 Bond overlaps are similar to those shown for C2H4 in Figure 10.25.

143

10.77 See Section 10.3, 10.4, and Examples 10.7, 10.8, 10.9.

(a) C

CC

CC

C

H

H

H H

H

H

HH H

H The single bonded carbon atoms are tetrahedral and sp3.

The double bonded carbon atoms are trigonal planar and sp2.

(b) C

Cl

Cl

O The C is trigonal planar and sp2.

(c) C1N

HH

C2

O

H H

O

H

The N is tetrahedral and sp3.

The C1 is tetrahedral and sp3. The C2 is trigonal planar and sp2. The O of C2-O-H is tetrahedral and sp3. 10.79 See Section 10.3, 10.4, and Examples 10.7, 10.8, 10.9.

NO O

N

O O

sp2 for N sp2 for N

10.81 See Section 10.3, 10.4, and Examples 10.7, 10.8, 10.9. (a)

NC

H

HH

HH

C is tetrahedral and sp3. N is tetrahedral and sp3.

(b)

C2C1

H

HH

C3 H C1 is tetrahedral and sp3. C2 and C3 are linear and sp.

144

10.83 See Section 10.3, 10.4, and Examples 10.7, 10.8, 10.9.

C1 C2

H

H

H

C3 N

C1 is trigonal planar and sp2. C2 is trigonal planar and sp2.

C3 is linear and sp. 10.85 See Section 10.5, Figure 10.37, and Example 10.10.

He+

1s 1s

σ1s

σ*1s

He22+ He+

For He22+ the electron configuration is (σ1s)2 there are no unpaired electrons, the bond order is

12

[2 − 0] = 1, and it is predicted to be stable.

10.87 See Section 10.5, Figures 10.37, 10.40 Table 10.2, and Example 10.10.

Li

2s 2s

σ2s

σ*2s

Li2 Li For Li2 the electron configuration is (σ2s)2 there are no unpaired electrons, the bond order is

12

[2 − 0] = 1, and it is predicted to be stable.

10.89 See Section 10.5, Figure 10.40, and Examples 10.11, 10.12. (a) C2

+ has (4 + 4 − 1) = 7 valence electrons: (σ2s)2 (σ*2s)2 (π2p)3.

The bond order is

12

[5 − 2] = 1.5, and there is one unpaired electron in a π2p orbital.

(b) N2− has (5 + 5 + 1) = 11 valence electrons: (σ2s)2 (σ*2s)2 (π2p)4 (σ2p)2 (π*2p)1.

The bond order is

12

[8 − 3] = 2.5, and there is one unpaired electron in a π*2p orbital.

(c) Be2− has (2 + 2 + 1) = 5 valence electrons: (σ2s)2 (σ*2s)2 (π2p)1.

The bond order is

12

[3 − 2] = 0.5, and there is one unpaired electron in a π2p orbital.

145

10.91 See Section 10.5, Figure 10.40, and Examples 10.11, 10.12. N2 has (5 + 5) + 10 valence electrons: (σ2s)2(σ*

2s)2(π2p)4(σ2p)2.

The bond order is σ

12

[8−2] = 3.0.

N2− has (5+5+1) = 11 valence electrons: (σ2s)2(σ*

2s)2(π2p)4(σ2p)2(π*2p)1.

The bond order is

12

[8−3] = 2.5.

N2 has a higher bond order than N2− because the additional electron in N2

− occupies an antibonding orbital. 10.93 See Section 10.5, Figure 10.40, and Examples 10.11, 10.12. (a) B2 has (3+3) = 6 valence electrons: (σ2s)2(σ*

2p)2(π2p)2.

The bond order is

12

[4−2] = 1.0.

B−2 has (3+3+1) = 7 valence electrons: (σ2s)2(σ*

2p)2(π2p)3.

The bond order is

12

[5−2] = 1.5.

B2− has a higher bond order and stronger bond than B2 because the additional electron in B2

− occupies a bonding orbital. (b) C2

− has (4+4+1) = 9 valence electrons: (σ2s)2(σ*2s) 2(π2p)4(σ2p)1.

The bond order is

12

[7−2] = 2.5.

C2+ has (4+4−1) = 7 valence electrons: (σ2s)2(σ*

2s)2(π2p)3.

The bond order is

12

[5−2] = 1.5.

C2− has a higher bond order and stronger bond than C2

+ . Forming C2− from C2 involves adding an

electron to a bonding orbital, whereas forming C2+ from C2 involves removing an electron from a

bonding orbital. (c) O2

2+ has (6+6−2) = 10 valence electrons: (σ2s)2(σ*2s)2(π2p)4(σ2p)2.

The bond order is

12

[8−2] = 3.0.

O2 has (6+6) = 12 valence electrons: (σ2s2)(σ*

2s)(σ2p)2(σ*2p)2.

The bond order is

12

[8−4] = 2.0.

O22+ has a higher bond order and stronger bond than O2 because two electrons are removed from

antibonding orbitals of O2 to form O22+ .

146

10.95 See Section 10.5 and Exercises 10.89, 10.91, 10.93.

(a) O2− and F2

+ have 13 valence electrons. A bond order of 12

[8−5] = 1.5 indicates these species

should be stable.

(b) C22− , N2, and O2

2+ have 10 valence electrons. A bond order of 12

[8 − 2] = 3.0 indicates these

species should be stable.

(c) Li22− , Be2, and B2

2+ have 4 valence electrons. A bond order of 12

[2 − 2] = 0 indicates these species

would not be stable. 10.97 See Section 10.5 and Figure 10.40. (a) N2 and CO have 10 valence electrons and (σ2s)2(σ*

2s)2(π2p)4(σ2p)2. (b) B2 and BeC have 6 valence electrons and (σ2s)2(σ*

2s)2(π2p)2. 10.99 See Section 10.5 and Figure 10.47 Number of Number of Species valence electrons Electron Configuration Bond Order unpaired electrons

(a) CN 9 (σ2s)2 (σ*2s)2 (π2p)4 (σ2p)1

12

[7 − 2] = 2.5 1 in σ2p

(b) CO− 11 (σ2s)2 (σ*2s)2 (π2p)4 (σ2p)2 (π*

2p)1 12

[8 − 3] = 2.5 1 in π*2p

(c) BeB− 6 (σ2s)2 (σ*2s)2 (π2p)2

12

[4 − 2] = 1.0 2 in π2p

(d) BC+ 6 (σ2s)2 (σ*2p)2 (π2p)2

12

[4 − 2] = 1.0 2 in π2p

10.101 See Section 10.5 and Figure 10.47.

2s

2s

σ2s

σ*2s

O F-OF-

2p

2p

π2p

σ2p

π∗2p

σ∗2p

The bond order for OF− is

12

[8 − 6]=1.

147

10.103 See Section 10.6 and Figure 10.49. NO2

− is isoelectronic with O3. The delocalized π molecular orbital is formed by the 2p orbitals of N and O that are perpendicular to the plane of the atoms. It looks like the delocalized π molecular orbital shown for O3 in Figure 10.49.

N

O O

10.105 See Section 10.1, 10.3, 10.4, Figures 10.15, 10.17, and Example 10.7. N2O5

NO

NO

O

O

O Each N is trigonal planar with 120º bond angles and sp2.

The central O is tetrahedral with 109º bond angles and sp3.

10.107 See Section 10.1, Figures 10.1, 10.6, and Examples 10.1, 10.2, 10.3. ClF 2

− ClF 2+

Cl

F

F

F

ClF

+

<F − Cl − F = 180º <F − Cl − F = 109º 10.109 See Section 10.3, 10.4, and Examples 10.8, 10.9.

148

NC

CN

CC

OC

H

H HO

C H

C

HH

C

C

C

C

C

H

H

H

HH

H

O

HC

C

H

H

OH O

H

H

(a) There are thirty-nine sigma bonds and six pi bonds shown in the Lewis structure. The pi bonding in the six-membered carbon ring is actually delocalized pi bonding. (b) The SN for each carbon atom forming a double bond with oxygen is three, the electron-pair arrangement is trigonal planar and the hybridization for each of these carbon atoms is sp2. (c) The SN for each N atom is four, the electron-pair arrangement is tetrahedral and the hybridization for each N atom is sp3. 10.111 See Section 10.1, 10.2, 10.3, 10.4, and Examples 10.7, 10.8, 10.9.

C

Cl

Cl

O

The C has a SN of 3, a trigonal planar electron-pair arrangement and uses sp2 hybrids.

The molecule is unsymmetrical and therefore polar.

10.113 See Sections 10.1, 10.3, 10.4, and Examples 10.1, 10.9.

2N

CC1

N

C

H

H H

C

C

C4

O5

H

H H

H N3

O

H H Bond ∠ 1: SN = 3, trigonal planar and therefore 120º. Bond ∠ 2: SN = 4, tetrahedral and therefore 109º. Bond ∠ 3: SN = 4, tetrahedral and therefore 109º. Bond ∠ 4: SN = 3, trigonal planar and therefore 120º. Bond ∠ 5: SN = 4, tetrahedral and therefore 109º.

149

10.115 See Section 10.5 and Figure 10.47.

2s

2s

σ2s

σ*2s

N O-NO-

2p

2p

π2p

σ2p

π∗2p

σ∗2p

2s

2s

σ2s

σ*2s

N+ ONO+

2p

2p

π2p

σ2p

π∗2p

σ∗2p

The bond order for NO− =

12

[8 − 4] = 2. The bond order for NO+ = 12

[8 − 2] = 3.

Hence, NO+ has a higher bond order than NO−.

150

10.117 See Section 10.1, 10.2, 10.3, 10.4. (a) NH2NO2 Total valence electrons = [2 × 5(N) + 2 × 1(H) + 2 × 6(O)] = 24.

N N

O

OH

H

Fourteen electrons remain after assigning five single bonds, and sixteen

unshared electrons are needed to give each atom a noble gas configuration (2 for each N and 6 for each O). Hence, two electrons (16-14) must be used to form one additional bond. Formal charge considerations indicate the additional bond is formed by N and O. SN of 4 for N on the left indicates bond angles of approximately 109º, and SN of 3 for N on right indicates bond angles of 120º. The N on the left uses sp3 hybrids, and the N on the right uses sp2 hybrids. The molecule is unsymmetrical and polar.

N N

O

OH

H

(b) HN3 Total valence electrons = [1 × 1(H) + 3 × 5(N)] = 16

H N N N

H N N N

Ten electrons remain after assigning three single bonds, and fourteen

unshared electrons are needed to give each atom a noble gas configuration (4 for left N, 4 for center N and 6 for right N). Hence, four electrons (14-10) must be used to form two additional bonds. SN of 3 for N on left indicates bond angles of approximately 120º and SN of 2 for N in the center indicates bond angles of 180º. The N on the left uses sp2 hybrids and the N in the center uses sp hybrids. The molecule is unsymmetrical and polar.

Note: H N N N

is plausible with sp3 for left N and sp for central N.

10.119 See Sections 10.1, 10.2, 10.3, 10.4. (a) OCN− Total valence electrons =[1 × 6(O) + 1 × 4(C) + 1 × 5(N) + 1(charge)] = 16. O C N Twelve electrons remain after assigning two single bonds, and sixteen unshared electrons are needed to give each atom a noble gas configuration (6 for O, 4 for C, and 6 for N). Hence, four electrons (16-12) must be used to form two additional bonds. This gives the following resonance possibilties with lowest formal charges:

A B C

O C N O C N

O C N

C hybrids: sp sp sp Structure B is likely to be the most important because it has the lowest formal charges and places the negative formal charge on O rather than N, on the more electronegative atom. SN of 2 for indicates bond angle of 180º. (b) NO3

− Total valence electrons =[1 × 5(N) + 3 × 6(O) + 1(charge)] = 24. O

NO O

Eighteen electrons remain after assigning three single bonds, and twenty unshared

151

Electrons are needed to give each atom a noble gas configuration. Hence, two electrons (20-18) must be used to form one additional bond. This gives the following resonance possibilities:

A B C

O

NO O

O

N

O

O

ON

O

O

N hybrids: sp2 sp2 sp2 These structures are equivalent and equally important. SN of 3 for N indicates bond angles of 120º. 10.121 See Section 10.1, 10.2, 10.3, 10.4.

CC C

CC

CC

CC

CC

C

C

C

HH

H

H

H

H

H

H

C

C

H

H

C

H

O

H H

H

H H H

HHH H

H

H

H H H HH

H

H HH

Vitamin-A contains ten sp2 hybridized carbon atoms. These are the ten carbon atoms that are involved in double bonds. Vitamin-A also contains ten sp3 hybridized carbon atoms. These are the four carbon atoms that are part of the ring and not part of double bonds, the six that are part of CH3 groups and the one that is bonded to oxygen. 10.123 See Section 3.3, 10.1, 10.4, and Example 3.12. Assume the sample has a mass of 100.00 g and therefore contains 54.53 g C, 9.15 g C, 9.15 g H, and 36.32 g O.

? mol C = 54.53 g C ×

1 mol C12.01 g C

= 4.54 mol C relative mol C = 4.54 mol C

2.27 = 2.00 mol C

? mol H = 9.15 g H ×

1 mol H1.008 g H

= 9.08 mol H relative mol H = 9.08 mol H

2.27 = 4.00 mol H

? mol O = 36.32 g O ×

1 mol O16.0 g O

= 2.27 mol O relative mol O = 2.27 mol O

2.27 = 1.00 mol O

152

The simplest formula is C2H4), and the simplest formula molar mass is 44.0 g/mol.

n =

molar mass compoundmolar mass C2H4O

= 44.0 g / mol44.0 g / mol

= 1, so the molecular formula is also C2H4O.

The two possible Lewis structures for C2H4O are:

A B

CC

H

HH H

O109 Þ

120 Þ

C C

H

H

H

O H

120 Þ 120 Þ

109 Þ

Hybrids: C, sp3; C, sp2 C, sp2; C, sp2; O, sp3

10.125 See Section 10.1, 10.2, 10.3, 10.4. S2F2 Total valence electron = [2 × 7(F) + 2 × 6(s)] = 26 F S S F Twenty electrons remaining after assigning three single bonds and twenty unshared electrons are needed to give each atom a noble gas electron configuration, no unshared electrons are leftover.

F S S F

SN = 4 for each sulfur atom, therefore hybridization for both S atoms is sp3.

S S

F

F Twenty electrons remaining after assigning three single bond and eighteen unshared

electrons are needed to give the fluorine atoms and the sulfur atom bonded to the other sulfur atom a noble gas electron configuration , two unshared electrons are assigned to the sulfur atom bonded to fluorine atoms.

S S

F

F

An alternate representation closer to reality is:

S S

FF

This will indicate sp3 hybridization in S atom bonded to fluorine atoms, the other S atom is using atomic orbitals, p orbitals to form the covalent bond. Since sulfur can violate the octet rule, a double bond between the two sulfur atoms can be formed to eliminate the positive and negative formal charges on both S atoms.

SS

FF

This still leaves the central sulfur atom with sp3 hybridization; the double bond is

formed by overlap of a p orbital on the terminal sulfur and a d orbital on the central sulfur atom.

153

10.127 See Sections 3.2, 3.3, 5.4, 10.1, 10.3, 10.4. To determine the value of x, the mol H per mol C are determined and then the mol H per 2 mol C.

? mol C in C2H2 = 1.30 g C2H2 ×

1 mol C2H2

26.0 g C2H2

× 2 mol C

1 mol C2H2

= 0.100 mol C

? mol H in C2H2 = 1.30 g C2H2 ×

1 mol C2H2

26.0 g C2H2

× 2 mol H

1 mol C2H2

= 0.100 mol H

? mol H in 1.22 L H2 = 2 × mol H2. Known Quantities: P = 1.01 atm V = 1.22 L T = 27 + 273 = 300 K

Solving PV = nRT for n gives n = PVRT

. n = (1.01 atm) (1.22 L)

0.0821 L g atmmol g K

⎛⎝⎜

⎞⎠⎟

(300 K) = 0.500 mol H2

in 1.22 L H2

= 0.0500 mol H2 ×

2 mol H1 mol H2

= 0.100 mol H

? mol Hmol C

= 0.100 mol H + 0.100 mol H0.100 mol C

= 2.00 mol H1 mol C

Hence, there are 4.00 mol H per 2.00 mol C, and the compound is C2H4. The Lewis structure for C2H4 is:

C C

H

H

H

H The carbon-carbon double bond is formed by a sp2-sp2 σ overlap and a 2p-2p π overlap. See Figures 10.23, 10.24, 10.25 for illustration of these overlaps.