Chem 121 Problem Set V Lewis Structures, VSEPR and Polarity

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1 hemistry 121 Problem set V olutions - 1 hem 121 Problem et V Lewis tructures, VEPR and Polarity AWER 1. pecies Elecronegativity difference in bond Bond Polarity Mp 3 E = = 0 for - very weakly polar covalent < -40 Al 3 E = = 1.5 for -Al strongly polar covalent 190 at 2.5 atm 3 E = = 1.0 for - polar covalent -73 K E = = 1.7 for -K ionic (barely) E = = 0.4 for - weakly polar covalent E = = 1.5 for - strongly polar covalent -150 Why does 4 have a similar mp to 4 when the bond polarities are very different? 2. 2 a) valence electrons = 10 b) first structure gives bp 2 lp 12 = 14, = -4 or two extra bonds c) second structure gives bp 6 lp 4 = 10 d) ormal charge() = = 0 a) valence electrons = = 10 b) first structure gives bp 2 lp 12 = 14, = -4 or two extra bonds c) second structure gives bp 6 lp 4 = 10 d) () = = 0, () = = 1 2 a) valence electrons = 12 b) first structure gives bp 2 lp 12 = 14, = -2 or one extra bond c) second structure gives bp 4 lp 8 = 12 d) () = = a) valence electrons = = 26 b) first structure gives bp 6 lp 20 = 26, = 0 ok c) second structure has () = = -1 () = = a) valence electrons = 14 b) first structure gives bp 2 lp 12 = 14, = 0 ok c) () = = 0 Xe 3 a) valence electrons = 8 18 = 26 b) first structure gives bp 6 lp 20 = 26, = 0 ok

2 hemistry 121 Problem set V olutions - 2 c) second structure has (Xe) = = 3 () = = -1 Xe 3 Xe Xe 3 a) valence electrons = 5 21 = 26 b) first structure gives bp 6 lp 20 = = 0 ok c) first structure has () = = 0 () = = 0 Xe 4 a) valence electrons = 8 28 = 36 b) first structure gives bp 8 lp 24 = = 4 need 4 more electrons on central atom c) second structure has (Xe) = = 0 () = = 0 Xe Xe B 3 a) valence electrons = 3 21 = 24 b) first structure gives bp 6 lp 20 = = -2 need to lose two electrons but do not use multiple bonds in this case as boron is electron deficient and only 6 electrons on boron c) second structure has (B) = = 0 () = = 0 B B Mn 4 - a) valence electrons = = 32 b) first structure gives bp 8 lp 24 = = 0 c) first structure has (Mn) = = 3 () = = -1 d) resonance Mn 3 Mn Mn Mn Mn Mn Mn 2 a) valence electrons = 6 12 = 18 b) first structure has bp 4 lp 16 = = -2 need to lose 2 electrons as double bonds c) second structure has bp 6 lp 12 = 18 d) second structure has () = = 1 (-) = = -1 (=) = = 0

3 hemistry 121 Problem set V olutions - 3 e) resonance (the 3d obitals on are close in energy to the 3p orbitals and can take more than 8 electrons) 3 a) valence electrons = 18 b) first structure has bp 4 lp 16 = = -2 need to lose 2 electrons as double bonds c) second structure has bp 6 lp 12 = 18 d) second structure has (-=) = = 1 (-) = = -1 (=) = = 0 e) resonance (no two double bond hybrid as the lowest energy empty orbitals are too high in energy and so oxygen can only take 8 electrons, no 2d orbitals. 3. Xe 4 a) valence electrons = 8 28 = 36 b) first structure has bp 8 lp 28 = 36 (problem 12) c) VEPR (central atom) bp 4 lp 2 = 6 VEPR shape is octahedral (second structure) d) Molecular shape is square planar e) third structure: E (-Xe) = 1.38 and all 4 E (-Xe) vectors cancel; E (lp-xe) = 1.1 and the two lp E vectors cancel f) nonpolar Xe Xe 6 a) valence electrons = 6 42 = 48 b) first structure has bp 12 lp 36 = 48 c) VEPR (central atom) bp 6 lp 0 = 6 VEPR shape is octahedral (second structure) d) Molecular shape is octahedral e) third structure: E (-) = 1.4 and all 6 E (-) vectors cancel f) nonpolar 2 a) valence electrons = 6 14 = 20 b) first structure has bp 4 lp 16 = 20 c) VEPR (central atom) bp 2 lp 2 = 4 VEPR shape is tetrahedral (second structure) d) Molecular shape is bent

4 hemistry 121 Problem set V olutions - 4 e) third structure: E (-) = 0.54 and resultant E (-) vector between - bonds; E (lp-) = 0.22 and resultant E (lp-) vector between lp- bonds; both resultant vectors come close to cancelling ( = 0.28). f) weakly polar, E = 0.3 (the dipole moment is 0.3 D) 2 2 a) valence electrons = = 20 b) first structure has bp 8 lp 12 = 20 c) VEPR (central atom) bp 4 lp 0 = 4 VEPR shape is tetrahedral (second structure) d) Molecular shape is tetrahedral e) third structure: E (-) = 0.61 and resultant E (-) vector between - bonds; E (-) = 0.35 and resultant E (-) vector between - bonds; resultant vectors are additive ( = 0.96). f) polar, E = 0.96 (the dipole moment is 1.6 D) 3 a) valence electrons = 5 21 = 26 b) first structure has bp 6 lp 20 = 26 c) VEPR (central atom) bp 3 lp 1 = 4 VEPR shape is tetrahedral (second structure) d) Molecular shape is trigonal pyramidal e) third structure: E (-) = 0.12 and resultant small E (-) vector between - bonds; E (lp-) =0.66 and E (lp-) vector overcomes the resultant from the E (-) vectors ( = 0.54). f) polar, E = 0.54 (the dipole moment is 0.39 D) 3 a) valence electrons = = 26 b) first structure has bp 8 lp 18 = 26 c) VEPR (central atom) bp 4 lp 0 = 4 VEPR shape is tetrahedral (second structure) d) Molecular shape is tetrahedral e) third structure: E (-) = 1.43 and resultant E (-) vector between - bonds; E (-) = 0.34 and E (-) vector adds to the resultant from the E (-) vectors ( = 1.78). f) strongly polar, E = 1.78 (the dipole moment is 1.65 D) 4 a) valence electrons = 4 28 = 32 b) first structure has bp 8 lp 24 = 32 c) VEPR (central atom) bp 4 lp 0 = 4 VEPR shape is tetrahedral (second structure) d) Molecular shape is tetrahedral e) third structure: E (-) = 1.43 and all E (-) vectors cancel.

5 hemistry 121 Problem set V olutions - 5 f) nonpolar 4. (a) (b) (c) (c) is the least stable as it has the most formal charge (including two negatives on one nitrogen) (b) is the most stable as the negative charge is on oxygen (a) is less stable than (b) with the negative charge on nitrogen rder of contribution to resonance hybrid is (b) > (a) >> (c) The resonance hybrid will have the largest contribution from (b) and a smaller contribution from (a) so the - bond will have an electron density between a triple and a double bond and thus its length will be between 110 and 120 ppm, in fact it is 112 ppm. The - bond will have an electron density between a single and a double bond and so its length will be between 115 and 147 ppm, in fact it is 119 ppm. 5a) (i) (ii) (iii) Lewis structure (iii) is not legitimate as carbon is forming 5 bonds and has 10 valence electrons. arbon only has orbitals for 8 valence electrons. (i) is the most stable as the negative charge is on nitrogen (ii) is less stable than (i) with the negative charge on carbon rder of contribution to resonance hybrid is (i) > (ii) b) ybridization sp 2 sp sp 2 sp 3 sp sp c) (i) 120º, 180º; (ii) 109.5º, 180º 6. pecies bond length 113 pm 120 pm 143 pm

6 hemistry 121 Problem set V olutions - 6 ( ) 143 = = 135pm 3 rder with respect to bond length (longest to shortest): 3 > 3 2 > 2 > 7. E () = = 0.9 The electronegativity vector is toward oxygen, i.e. oxygen has a higher electron density than carbon, but the formal charges indicate that in, oxygen is deficient in electrons and carbon has an excess, which would lead to a lower separation of charge and a smaller dipole moment than expected from electronegativity values. 8. yanate, (i) (ii) (iii) The most stable Lewis structure is (ii) with the negative charge on oxygen, though (i) with the negative charge on nitrogen will contribute to the resonance hybrid, making cyanate a stable anion. ulminate, (i) (ii) (iii) arbon forms three bonds, under special conditions it will form three bonds and carry a single negative charge, but such species are very reactive. Lewis structure (ii) is the only significant contributor to the resonance hybrid, structures (i) and (iii) with two and three negative charges on carbon will contribute insignificantly to the resonance hybrid. Thus the cyanate ion with two Lewis structures contributing to the resonance hybrid (negative charge on oxygen and nitrogen) would be expected to be reasonably stable, while the fulminate ion with only one significant structure contributing to the resonance hybrid and with the negative charge on carbon would be expected to be quite unstable. 9. a) P 4

7 hemistry 121 Problem set V olutions - 7 P thus 4 has the most Lewis structures contributing to the resonance hybrid and the negative charge is delocalized to the greatest extent, making it the weakest base (and the most stable anion). 2 P 4 has the least contributors to the resonance hybrid and the negative charge is delocalized to the smallest extent, making it the least stable and thus the most basic. o the order of increasing basicity is: 4 < 4 < 2 P 4 b) 2 P 3 thus is the least stable with the negative charge localized on one oxygen, making it the most basic; 3 is the most stable with the negative charge delocalized on three oxygens, making it the least basic. o the order of increasing basicity is: 3 < 2 < 10. E(-) = 1.40 and E(lp-) = net vector = 0.32 net vector = 0.32 net vector = 0.32 net vector = 0.32 permanent dipole permanent dipole permanent dipole permanent dipole 5 5 6

8 hemistry 121 Problem set V olutions - 8 vectors cancel net vector = 0.32 vectors cancel no permanent dipole permanent dipole no permanent dipole 11a) To obtain a square planar arrangement, we need an octahedral VEPR shape (6 electron pairs) with two lone pairs on the central atom. Thus () = = 1 thus the formula is 4 b) To obtain a pyramidal arrangement, we need an tetrahedral VEPR shape (4 electron pairs) with one lone pair on the central atom. Thus () = = 2 thus the formula is 3 2 c) To obtain a T-shaped arrangement, we need an trigonal bipyramid VEPR shape (5 electron pairs) with two lone pairs on the central atom. Thus () = = 0 thus the formula is 3 d) To obtain a linear arrangement, we need an trigonal bipyramid VEPR shape (5 electron pairs) with three lone pairs on the central atom. Thus () = = 1 thus the formula is 2 e) To obtain a square pyramid arrangement, we need an octahedral VEPR shape (6 electron pairs) with one lone pair on the central atom. Thus () = = 0 thus the formula is 5 f) To obtain a octahedral arrangement, we need an octahedral VEPR shape (6 electron pairs). Thus () = = 1 thus the formula is 6

9 hemistry 121 Problem set V olutions ince b and have different electronegativities, a nonpolar species will have all the electronegativity vectors cancel. E(-b) = 1.93 and E(lp-b) = 1.61 o, working through the possible species with an even number of valence electrons and zero or a single charge we have: b, the single vector cannot cancel so it is polar. b 2 b 2 b 3 b 4 b b b b b b b net vector = 0.28 net vector = 0.28 net vector = 0.28 vectors cancel permanent dipole permanent dipole permanent dipole no permanent dipole b 4 b 5 b 6 b 6 b b b net vector = 0.28 vectors cancel can not exist vectors cancel permanent dipole no permanent dipole no permanent dipole 13. The only VEPR shape where all adjacent -X- angles are 90º is the octahedron, so we want X 6. The Lewis structure must have no lone pairs on the central atom.

10 hemistry 121 Problem set V olutions - 10 X which gives 6(6) 6(2) = 48 valence electrons thus: (valence electrons on X) 6(7) = 48 and so: (valence electrons on X) = 6 X can be, e, Te but not. The only VEPR shape where all adjacent -X- angles are slightly less than 90º is the octahedron with one lone pair on the central atom, so we want X 5. which gives 5(6) 2 5(2) = 42 valence electrons thus: (valence electrons on X) 5(7) = 42 and so: (valence electrons on X) = 7 X can be, Br, but not. 14. The VEPR shape where X 4 has a permanent dipole can not be a tetrahedron (4 bp) or an octahedron with two lone pairs (4 bp and 2 lp) as in each case all the electronegativity vectors will cancel. t can not be a trigonal planar as this shape has only 3 regions of electron density and we have 4 bp. The VEPR shape can only be a trigonal bipyramid. X which gives 4(6) 2 4(2) = 34 valence electrons thus: (valence electrons on X) 4(7) = 34 and so: (valence electrons on X) = 6 X can be, e, Te but not (as oxygen can have a maximum of 8 electrons on the central atom). X

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