Chapter 10: Molecular Structure and Bonding Theories

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1 hapter 10: Molecular Structure and Bonding Theories 10.1 See Section 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 See Section ormaldehyde,, 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, l, is another example of a trigonal planar molecule in which carbon forms four bonds. is an example of a molecule with a central atom that makes four bonds but has a linear-bonded atom lone-pair arrangement See Section 10.1 and igure 10.6 I I 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 10º to 180º. ence, 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 See Section 10. and Examples 10.4, Any molecule with a totally symmetrical arrangement of atoms and lone pairs of electrons is nonpolar. ence, any of the electron-pair arrangements shown in igure 10.1 having like atoms attached to a central atom that is different has polar bonds but is nonpolar overall. or example, Bel (g) as is nonpolar because the bond dipoles cancel each other. This always occurs when the molecule is totally symmetrical. I l Be l 10.9 See Section 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. 133

2 10.11 See Section 10.3 and Example l [e] 3s 3p [e] s p The partially filled 3p orbital of l overlaps with the partially filled p orbital of to form the bond in l See Section 10.1, 10.3, igure 10.15, 10.16, and Example l Bl 3 B S (steric number for B = 3, trigonal planar electron-pair arrangement, sp l l hybrids for B. l Bl 4 l B l l S for B = 4, tetrahedral electron-pair arrangement, sp 3 hybrids for B See Section 10.3 and Table l ybrid 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. owever, we can only determine the positions of attached atoms and are unable to determine the positions of lone pairs. ence, any prediction of use of hybrid atomic orbitals by l in l would be a matter of pure conjecture that could not be verified by experiment See Section 10.4 and igures 10., 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. p z p z σ p y p y π 134

3 10.19 See Section 10.5 and igures 10.35, σ* 1s 1s 1s σ 1s 10.1 See Section 10.6 and igure According to molecular orbital theory, there is a weak interaction between the s orbital of Li and the p orbital of 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 giving Li + and. ence, the two descriptions differ slightly in terms of electron transfer from Li to See Section 10.1 and igure (a) S = 3, trigonal planar (b) S = 4, tetrahedral (c) S = 4, tetrahedral (d) S = 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 See Section 10.1, igures 10.1, 10.6, and Examples 10.1, 10.. (a) 4 (b) S (c) As 5 S S S = 4 tetrahedral S =, linear S = 5, trigonal bipyramidal (d) (e) As S = 3, trigonal planar S = 4, tetrahedral 135

4 10.7 See Section 10.1, igures 10.1, 10.6, and Examples 10.1, 10.. (a) Se (b) (c) 3 + Se S = 3, trigonal planar S =, linear S = 4, tetrahedral shape: bent shape: linear shape: trigonal pyramidal (d) I 5 (e) Sl 4 I S = 6, octahedral shape: square pyramidal S l l l l S = 5, trigonal bipyramidal shape: see - saw 10.9 See Section 10.1, igures 10.1, 10.6, and Examples 10.1, (a) Bl 3 l 3 (b) S 6 l l B l S l l l S = 3 S = 4 S = 4 S = 6 trigonal planar tetrahedral tetrahedral octahedral electron pair electron pair electron pair electron pair arrangement arrangement arrangement arrangement 10º bond angles 109º bond angles 109º bond angles 90º & 180º bond angles l 3 has smaller bond angles than Bl 3. S 6 has smaller bond angles than See Section 10.1, igures 10.1, 10.6, and Examples 10.1, 10.. (a) l + 4 (b) S I 4 + l l S I S = 4 S =4 S = 4 S = 6 tetrahedral electron tetrahedral electron tetrahedral electron octahedral electron pair arrangement pair arrangement pair arrangement pair arrangement 136

5 109º bond angles 109º bond angles 109º bond angles 90º & 180º bond angles l has slightly smaller bond angles due to the lone pair on the. I 4 has smaller bond angles than S See Section 10.1, igures 10.1, 10.6, and Examples 10., (a) 3 (b) Br (c) 3 Br Br 180 Þ180 Þ 10 Þ 10 Þ See Section 10.1, igures 10.1, 10.6, and Examples 10., (a) S (b) l 3 (c) S S 10 Þ expanded valence shell l expanded valence shell S 180 Þ See Section 10.1, igures 10.1, 10.6, and Examples 10., (a)( 3 ) 10 Þ (b) Þ 137

6 10.39 See Section 10.1, igures 10.1, 10.6, and Examples 10., (a) 3 l l l b) P P 180 Þ 180 Þ See Section 10.1, igures 10.1, 10.6, and Examples 10., (a) 4 (b) 3 6 (c) Xe Xe 10 Þ 180 Þ 10 Þ See Section 10.1, igures 10.1, 10.6, and Examples 10., (a) 3 S (b) S 10 Þ See Section 10. and Examples 10.4, (a) 4 (b) S (c) As 5 (d) As S S symmetrical symmetrical symmetrical unsymmetrical nonpolar nonpolar nonpolar polar See Section 10. and Examplse 10.4, (a) Se (b) (c) Sl 4 l Se l S l l 138

7 expanded valence shell expanded valence shell unsymmetrical unsymmetrical unsymmetrical polar polar polar See Section 10. and Example (a) (b) I (c) I I unsymmetrical symmetrical unsymmetrical polar nonpolar polar See Section 10. and Example (a) 3 (b) Br 4 (c) BeI Br Br Br Br unsymmetrical symmetrical symmetrical polar nonpolar nonpolar I Be I See Section 10. and Example (a) (b) symmetrical unsymmetrical nonpolar polar bond dipoles cancel bond dipoles do not cancel See Section (a) 10º, sp or sp 3 d (b) 90º, sp 3 d or sp 3 d (c) 180º, sp, sp 3 d or sp 3 d hybrids hybrids hybrids See Section 10.3 and Examples 10.7, (a) 4 (b) Sbl 6 (c) As 5 (d) Si 4 l l l Sb l l l S = 4 S = 6 S = 5 S = 4 tetrahedral octahedral trigonal bipyramidal tetrahedral sp 3 for sp 3 d for Sb sp 3 d for As sp 3 for Si As Si 139

8 (e) S = 4 tetrahedral sp 3 for See Section 10.3 and Examples 10.7, (a) (b) Snl (g) (c) I 3 (d) Se l Sn I I I Se l S = S = 3 S = 5 S = 3 linear trigonal planar trigonal bipyramidal trigonal planar sp for central sp for Sn sp 3 d for central I sp for Se See Section 10.3 and Examples 10.7, (a) (b) (c) S = 3 S = 4 S = 3 trigonal planar tetrahedral trigonal planar sp for sp 3 for sp for See Section 10.3 and Examples 10.7, (a) 3 + (b) 3 (c) l + l l S = 4 S = 4 S = 4 tetrahedral tetrahedral tetrahedral sp 3 for sp 3 for sp 3 for 140

9 10.65 See Section 10.3 and Examples 10.7, (a) S = 4, tetrahedral sp 3 for and [e] for. s p A sp 3 orbital containing one electron overlaps with a p orbital from containing one electron to form an - bond in. The lone pairs of electrons are in sp 3 orbitals of. (b) 3 S = 4, tetrahedral sp 3 for and for. 1s A sp 3 orbital from containing one electron overlaps with a 1s orbital of containing one electron to form a - bond in 3. The lone pair of electrons is in a sp 3 orbital of. (c) Bl 3 S = 3, trigonal planar l l B l sp for B and [e] 3s 3p for l. A sp orbital from B containing one electron overlaps with a 3p orbital of l containing one electron to form a B-l bond in Bl See Section 10.3 and Examples 10.7, Se 4 Se S = 5, trigonal bipyramidal sp 3 d for Se and [e] s p for. A sp 3 d orbital from Se containing one electron overlaps with a p orbital of containing one electron to form a Se- bond in Se 4. The lone pair of electrons is in a sp 3 d orbital of Se See Section 10.3 and Examples 10.7, (a) or : S = 4, tetrahedral, sp 3. l or : S = 4, tetrahedral, sp 3. l or : S = 4, tetrahedral, sp 3. (b) or both : S =, linear, sp. P or P: S = 4, tetrahedral, sp

10 10.71 See Section 10.4, igures (a) p z p z σ (b) p y p y π (c) sp z p z σ See Section 10.3, 10.4, and Examples 10.7, 10.8, Bond rbital verlaps Bond Type - sp 3-1s σ 1 - sp 3 -sp σ α sp-p z σ p y -p y π p x -p x π S 1 = 4, tetrahedral, sp 3 S =, linear, sp There are a total of five σ bonds and two π bonds in See Section 10.3, 10.4, and Examples 10.7, 10.8, π σ Bond rbital verlaps Bond Type - sp -1s σ - sp -1s σ = sp -sp σ p x -p x π S = 3, trigonal planar, S = 3, trigonal planar, sp Bond overlaps are similar to those shown for 4 in igure σ 14

11 10.77 See Section 10.3, 10.4, and Examples 10.7, 10.8, (a) (b) l The single bonded carbon atoms are tetrahedral and sp 3. The double bonded carbon atoms are trigonal planar and sp. The is trigonal planar and sp. l (c) 1 The is tetrahedral and sp 3. The 1 is tetrahedral and sp 3. The is trigonal planar and sp. The of -- is tetrahedral and sp See Section 10.3, 10.4, and Examples 10.7, 10.8, sp for sp for See Section 10.3, 10.4, and Examples 10.7, 10.8, (a) is tetrahedral and sp 3. is tetrahedral and sp 3. (b) is tetrahedral and sp 3. and 3 are linear and sp. 143

12 10.83 See Section 10.3, 10.4, and Examples 10.7, 10.8, is linear and sp See Section 10.5, igure 10.37, and Example is trigonal planar and sp. is trigonal planar and sp. σ* 1s 1s 1s σ 1s e + + e e + or e + the electron configuration is (σ 1s ) there are no unpaired electrons, the bond order is 1 [ 0] = 1, and it is predicted to be stable See Section 10.5, igures 10.37, Table 10., and Example σ* s s s σ s Li Li Li or Li the electron configuration is (σ s ) there are no unpaired electrons, the bond order is 1 [ 0] = 1, and it is predicted to be stable See Section 10.5, igure 10.40, and Examples 10.11, (a) + has ( ) = 7 valence electrons: (σ s ) (σ* s ) (π p ) 3. The bond order is 1 [5 ] = 1.5, and there is one unpaired electron in a π p orbital. (b) has ( ) = 11 valence electrons: (σ s ) (σ* s ) (π p ) 4 (σ p ) (π* p ) 1. The bond order is 1 [8 3] =.5, and there is one unpaired electron in a π* p orbital. (c) Be has ( + + 1) = 5 valence electrons: (σ s ) (σ* s ) (π p ) 1. The bond order is 1 [3 ] = 0.5, and there is one unpaired electron in a π p orbital. 144

13 10.91 See Section 10.5, igure 10.40, and Examples 10.11, has (5 + 5) + 10 valence electrons: (σ s ) (σ * s) (π p ) 4 (σ p ). The bond order is σ 1 [8 ] = 3.0. has (5+5+1) = 11 valence electrons: (σ s ) (σ * s) (π p ) 4 (σ p ) (π * p) 1. The bond order is 1 [8 3] =.5. has a higher bond order than because the additional electron in orbital. occupies an antibonding See Section 10.5, igure 10.40, and Examples 10.11, (a) B has (3+3) = 6 valence electrons: (σ s ) (σ * p) (π p ). The bond order is 1 [4 ] = 1.0. B has (3+3+1) = 7 valence electrons: (σ s ) (σ * p) (π p ) 3. The bond order is 1 [5 ] = 1.5. B has a higher bond order and stronger bond than B because the additional electron in B occupies a bonding orbital. (b) has (4+4+1) = 9 valence electrons: (σ s ) (σ * s) (π p ) 4 (σ p ) 1. The bond order is 1 [7 ] =.5. + has (4+4 1) = 7 valence electrons: (σ s ) (σ * s) (π p ) 3. The bond order is 1 [5 ] = 1.5. has a higher bond order and stronger bond than +. orming from involves adding an electron to a bonding orbital, whereas forming + from involves removing an electron from a bonding orbital. (c) + has (6+6 ) = 10 valence electrons: (σ s ) (σ * s) (π p ) 4 (σ p ). The bond order is 1 [8 ] = 3.0. has (6+6) = 1 valence electrons: (σ s )(σ * s)(σ p ) (σ * p). The bond order is 1 [8 4] =.0. + has a higher bond order and stronger bond than because two electrons are removed from antibonding orbitals of to form

14 10.95 See Section 10.5 and Exercises 10.89, 10.91, (a) and + have 13 valence electrons. A bond order of 1 [8 5] = 1.5 indicates these species should be stable. (b),, and + have 10 valence electrons. A bond order of 1 species should be stable. [8 ] = 3.0 indicates these (c) Li, Be, and B + have 4 valence electrons. A bond order of 1 [ ] = 0 indicates these species would not be stable See Section 10.5 and igure (a) and have 10 valence electrons and (σ s ) (σ * s) (π p ) 4 (σ p ). (b) B and Be have 6 valence electrons and (σ s ) (σ * s) (π p ) See Section 10.5 and igure umber of umber of Species valence electrons Electron onfiguration Bond rder unpaired electrons (a) 9 (σ s ) (σ * s) (π p ) 4 (σ p ) 1 1 [7 ] =.5 1 in σ p (b) 11 (σ s ) (σ * s) (π p ) 4 (σ p ) (π * p) 1 1 [8 3] =.5 1 in π* p (c) BeB 6 (σ s ) (σ * s) (π p ) 1 [4 ] = 1.0 in π p (d) B + 6 (σ s ) (σ * p) (π p ) 1 [4 ] = 1.0 in π p See Section 10.5 and igure σ p p π p p σ p π p The bond order for is 1 [8 6]=1. s σ* s s σ s

15 See Section 10.6 and igure is isoelectronic with 3. The delocalized π molecular orbital is formed by the p orbitals of and that are perpendicular to the plane of the atoms. It looks like the delocalized π molecular orbital shown for 3 in igure See Section 10.1, 10.3, 10.4, igures 10.15, 10.17, and Example Each is trigonal planar with 10º bond angles and sp. The central is tetrahedral with 109º bond angles and sp See Section 10.1, igures 10.1, 10.6, and Examples 10.1, 10., l l + + l l < l = 180º < l = 109º See Section 10.3, 10.4, and Examples 10.8,

16 (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 S 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 sp. (c) The S for each atom is four, the electron-pair arrangement is tetrahedral and the hybridization for each atom is sp See Section 10.1, 10., 10.3, 10.4, and Examples 10.7, 10.8, l The has a S of 3, a trigonal planar electron-pair arrangement and uses sp hybrids. l The molecule is unsymmetrical and therefore polar See Sections 10.1, 10.3, 10.4, and Examples 10.1, Bond 1: S = 3, trigonal planar and therefore 10º. Bond : S = 4, tetrahedral and therefore 109º. Bond 3: S = 4, tetrahedral and therefore 109º. Bond 4: S = 3, trigonal planar and therefore 10º. Bond 5: S = 4, tetrahedral and therefore 109º

17 See Section 10.5 and igure σ p p π p p σ p π p s σ* s s σ s - - σ p p π p p σ p π p s σ* s s σ s + + The bond order for = 1 [8 4] =. The bond order for + = 1 [8 ] = 3. ence, + has a higher bond order than. 149

18 See Section 10.1, 10., 10.3, (a) Total valence electrons = [ 5() + 1() + 6()] = 4. ourteen electrons remain after assigning five single bonds, and sixteen unshared electrons are needed to give each atom a noble gas configuration ( for each and 6 for each ). ence, two electrons (16-14) must be used to form one additional bond. ormal charge considerations indicate the additional bond is formed by and. S of 4 for on the left indicates bond angles of approximately 109º, and S of 3 for on right indicates bond angles of 10º. The on the left uses sp 3 hybrids, and the on the right uses sp hybrids. The molecule is unsymmetrical and polar. (b) 3 Total valence electrons = [1 1() + 3 5()] = 16 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, 4 for center and 6 for right ). ence, four electrons (14-10) must be used to form two additional bonds. S of 3 for on left indicates bond angles of approximately 10º and S of for in the center indicates bond angles of 180º. The on the left uses sp hybrids and the in the center uses sp hybrids. The molecule is unsymmetrical and polar. ote: is plausible with sp 3 for left and sp for central See Sections 10.1, 10., 10.3, (a) Total valence electrons =[1 6() + 1 4() + 1 5() + 1(charge)] = 16. Twelve electrons remain after assigning two single bonds, and sixteen unshared electrons are needed to give each atom a noble gas configuration (6 for, 4 for, and 6 for ). ence, four electrons (16-1) must be used to form two additional bonds. This gives the following resonance possibilties with lowest formal charges: A B 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 rather than, on the more electronegative atom. S of for indicates bond angle of 180º. (b) 3 Total valence electrons =[1 5() + 3 6() + 1(charge)] = 4. Eighteen electrons remain after assigning three single bonds, and twenty unshared 150

19 Electrons are needed to give each atom a noble gas configuration. ence, two electrons (0-18) must be used to form one additional bond. This gives the following resonance possibilities: A B hybrids: sp sp sp These structures are equivalent and equally important. S of 3 for indicates bond angles of 10º See Section 10.1, 10., 10.3, Vitamin-A contains ten sp hybridized carbon atoms. These are the ten carbon atoms that are involved in double bonds. Vitamin-A also contains ten sp 3 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 3 groups and the one that is bonded to oxygen See Section 3.3, 10.1, 10.4, and Example 3.1. Assume the sample has a mass of g and therefore contains g, 9.15 g, 9.15 g, and 36.3 g. 1 mol 4.54 mol? mol = g = 4.54 mol relative mol = =.00 mol 1.01 g.7? mol = 9.15 g 1 mol g? mol = 36.3 g 1 mol 16.0 g 9.08 mol = 9.08 mol relative mol =.7.7 mol =.7 mol relative mol =.7 = 4.00 mol = 1.00 mol 151

20 The simplest formula is 4 ), and the simplest formula molar mass is 44.0 g/mol. molar mass compound 44.0 g / mol n = = = 1, so the molecular formula is also molar mass g / mol 4. The two possible Lewis structures for 4 are: A B 10 Þ 10 Þ 10 Þ ybrids:, sp 3 ;, sp, sp ;, sp ;, sp See Section 10.1, 10., 10.3, S Total valence electron = [ 7() + 6(s)] = 6 S S 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. S S S = 4 for each sulfur atom, therefore hybridization for both S atoms is sp 3. S S 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 An alternate representation closer to reality is: S S This will indicate sp 3 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. S S This still leaves the central sulfur atom with sp 3 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. 15

21 10.17 See Sections 3., 3.3, 5.4, 10.1, 10.3, To determine the value of x, the mol per mol are determined and then the mol per mol.? mol in = 1.30 g 1 mol mol = mol 6.0 g 1 mol? mol in = 1.30 g 1 mol 6.0 g? mol in 1. L = mol. Known Quantities: P = 1.01 atm V = 1. L Solving PV = nrt for n gives n = PV in 1. L mol 1 mol = mol T = = 300 K RT. n = (1.01 atm) (1. L) L g atm = mol mol g K (300 K) = mol mol = mol 1 mol? mol mol mol.00 mol = = mol mol 1 mol ence, there are 4.00 mol per.00 mol, and the compound is 4. The Lewis structure for 4 is: The carbon-carbon double bond is formed by a sp -sp σ overlap and a p-p π overlap. See igures 10.3, 10.4, 10.5 for illustration of these overlaps. 153

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