Homework #7. Chapter 14. Covalent Bonding Orbitals
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1 omework #7 hapter 14 ovalent Bonding rbitals 7. Both M theory and LE model use quantum mechanics to describe bonding. In the LE model, wavefunctions on one atom are mixed to form hybridized orbitals. In M theory wavefunctions on all of the atoms in the structure are mixed to form new molecular orbits. When mixing orbitals in either M or LE the same number of orbitals that you put in is the number that you get out. LE model breaks down for some structures, for instants it is not able to predict the magnetic properties of or the existence of B 6. In addition, resonance structures are not explained by LE model. M theory is able to fix all of these problems. 1. ingle bonds have their electron density concentrated between the two atoms (on the internuclear axis). Therefore, an atom can rotate freely on the internuclear axis. Double and triple bonds have their electron density located above and to the side of the internuclear axis (see below). Therefore, in order for an atom to rotate on the internuclear axis (while the other atom is stationary) the double and/or triple bonds would need to be broken. 16. LE (Local Electron) Model ybridizations () Total Valence e - (1) Wanted e - () wanted valence 0 1 # bonds = = = 4 # e = valence (#bonds) = 1 (4) = 4 is a trigonal planer molecule. The central carbon atom is sp hybridized and the oxygen atom is sp hybridized. Two of the sp hydride orbitals, on the carbon, are used in - single bonds (σ bond). The - single bonds are formed from the overlap of the s orbital on the hydrogen and the sp hybridized orbital on the carbon. There is a double bond between the - (1 σ bond and 1 π bond). The σ bond is formed from the overlap of the sp hybridized orbital, 1
2 on the carbon, and the sp hybridized orbital on the oxygen. The π bond is formed from the overlap of the unhybridized p orbitals on the carbon and the oxygen atoms. The loan pair electrons on the oxygen atom are located in sp hybridized orbitals. () () Total Valence e - (1) (4) 10 Wanted e - () (8) 0 wanted valence 0 10 # bonds = = = 5 # e = valence (#bonds) = 10 (5) = 0 Wanted Valence 0 10 # bonds 5 # e Valence # bonds is a linear molecule. The central carbon atoms are sp hybridized. n each carbon atom, one of the sp hybridized orbitals overlaps with a s orbital on the hydrogen atom to form a - single bonds (σ bonds). There is a triple bond between the carbons (1 σ bond and π bonds). The σ bond is formed from the overlap of the sp hybridized orbitals on each carbon atom. The two π bonds are formed from the overlap of the 4 unhybridized p orbitals ( on each carbon) on the carbons. 3. a) b) 4() Total Valence e - 4 4(7) 3 Wanted e - 8 4(8) 40 wanted valence 40 3 # bonds = = = 4 # e = valence (#bonds) = 3 (4) = 4 Molecular tructure =Tetrahedral Bond Angles = ybridization=sp 3 N 3() Total Valence e - 5 3(7) 6 Wanted e - 8 3(8) 3 wanted valence 3 6 # bonds = = = 3 # e = valence (#bonds) = 6 (3) = 0 Polar/Non Polar = Non Polar Molecular tructure =Trigonal Pyramidal Bond Angles = <109.5 N ybridization=sp 3 N Polar/Non Polar = Polar
3 c) () Total Valence e - 6 (7) 0 Wanted e - 8 (8) 4 d) wanted valence 4 0 # bonds = = = # e = valence (#bonds) = 0 () = 16 Molecular tructure = Bent Bond Angles = <109.5 ybridization=sp 3 B 3() Total Valence e - 3 3(7) 4 Boron is known to not obey the octet rule. It is known to form only 3 bonds allowing the formal charge on B to be 0. Polar/Non Polar = Polar Molecular tructure = Trigonal Planer Bond Angles = 10. B ybridization=sp Polar/Non Polar = Non Polar e) Be is a covalent compound because the difference in electronegativity between the Be and is only 0.6. f) g) Be () Total Valence e - (1) 4 Beryllium is known to not obey the octet rule. It is known to form only bonds allowing the formal charge on Be to be 0. Molecular tructure = Linear Bond Angles = 180. Be ybridization=sp Te 4() Total Valence e - 6 4(7) 34 Te must expand its octet to accommodate all of the electrons/atoms Polar/Non Polar = Non Polar Molecular tructure = ee-aw Bond Angles = <10, <90, and <90 Te ybridization=sp 3 d As 5() Total Valence e - 5 5(7) 40 As must expand its octet to accommodate all of the atoms Polar/Non Polar = Polar Molecular tructure = Trigonal Bipyramidal Bond Angles = 10 and 90 As ybridization=sp 3 d B Be As Te Polar/Non Polar = Non Polar 3
4 h) j) k) i) Kr () Total Valence e - 8 (7) Kr must expand its octet to accommodate all of the electrons Molecular tructure = Linear Bond Angles = 180 Kr ybridization=sp 3 d Kr 4() Total Valence e - 8 4(7) 36 Kr must expand its octet to accommodate all of the electrons Molecular tructure = quare Planer Kr ybridization=sp 3 d e 6() Total Valence e - 6 6(7) 48 e must expand its octet to accommodate all of the atoms i) Polar/Non Polar = Non Polar Bond Angles = 90 and 180 Polar/Non Polar = Non Polar Molecular tructure = ctahedral Bond Angles = 90 e ybridization=sp 3 d I 5() Total Valence e - 7 5(7) 4 I must expand its octet to accommodate all of the electrons/atoms Polar/Non Polar = Non Polar Molecular tructure = quare Pyramidal Bond Angles = <90 and <180 I ybridization=sp 3 d I 3() Total Valence e - 7 3(7) 8 I must expand its octet to accommodate all of the electrons/atoms Polar/Non Polar = Polar Molecular tructure = T-haped Bond Angles = <90 and <180 I ybridization=sp 3 d I Kr e I Kr Polar/Non Polar = Polar 4
5 4. a) () Total Valence e - 6 (6) 18 Wanted e - 8 (8) 4 b) wanted valence 4 18 # bonds = = = 3 # e = valence (#bonds) = 18 (3) = 1 Molecular tructure = Bent Bond Angles = <10 ybridization = sp 3() Total Valence e - 6 3(6) 4 Wanted e - 8 3(8) 3 wanted valence 3 4 # bonds = = = 4 # e = valence (#bonds) = 4 (4) = 16 c) d) Molecular tructure = Trigonal Planer Bond Angles = 10 ybridization = sp () 3() e - Total Valence e - (6) 3(6) + 3 Wanted e - (8) 3(8) 40 wanted valence 40 3 # bonds = = = 4 # e = valence (#bonds) = 3 (4) = 4 Molecular tructure = Tetrahedral Bond Angles = ybridization = sp 3 () 8() e - Total Valence e - (6) 8(6) + 6 Wanted e - (8) 8(8) 80 wanted valence 80 6 # bonds = = = 9 # e = valence (#bonds) = 6 (9) = 44 Molecular tructure = Tetrahedral (around ) Bent (around central ) Bond Angles = (around ) and <109.5 (around central ) ybridization = sp
6 e) f) wanted valence 3 6 # bonds = = = 3 # e = valence (#bonds) = 6 (3) = 0 Molecular tructure = Trigonal Pyramidal Bond Angles = <109.5 ybridization = sp 3 wanted valence 0 1 # bonds = = = 4 wanted valence 40 3 # bonds = = = 4 # e = valence (#bonds) = 3 (4) = 4 g) h) 3() e - Total Valence e - 6 3(6) + 6 Wanted e - 8 3(8) 3 4() e - Total Valence e - 6 4(6) + 3 Wanted e - 8 4(8) 40 Molecular tructure = Tetrahedral Bond Angles = ybridization = sp 3 () Total Valence e - 6 (7) 0 Wanted e - 8 (8) 4 wanted valence 4 0 # bonds = = = # e = valence (#bonds) = 4 () = 0 - wanted valence # bonds = = = 4 Molecular tructure = Bent Bond Angles = <109.5 ybridization = sp 3 4() Total Valence e - 6 4(7) 34 ulfur must expand its octet to accommodate all of the electrons/atoms Molecular tructure = ee-aw Bond Angles = <90 and <10 ybridization = sp 3 d
7 i) j) 6() Total Valence e - 6 6(7) 48 ulfur must expand its octet to accommodate all of the electrons/atoms Molecular tructure = ctahedral Bond Angles = 90 ybridization = sp 3 d () 4() Total Valence e - (6) 4(7) 40 ulfur must expand its octet to accommodate all of the electrons/atoms Molecular tructure = ee-aw and Bent Bond Angles = <90, <10, and <109.5 ybridization = sp 3 d and sp () () Total Valence e - 4(1) (4) 1 Wanted e - 4() (8) 4 wanted valence 4 1 # bonds = = = 6 # e = valence (#bonds) = 1 (6) = 0 The hybridization around both carbons is sp hybridized, making the shape trigonal planer around each carbon. The σ bond of the double bond is formed from the overlap of the p orbitals (blue) that are in the plane of the page. The π bond of the double bond is formed form the overlap of a sp hybridized (green) orbital on each. Therefore, the remaining sp orbital are all in plane with each other. 6. In order for the central carbon atom to form two double bonds, one of the double bonds must be formed with p orbitals into/out of the page and the other double bond must be formed with p orbitals in the plane of the page. Therefore, the hydrogens on either end will be oriented 90 from each other. 7
8 7. 6() () 4() Total Valence e - 6(1) (6) 4(4) 34 Wanted e - 6() (8) 4(8) 60 wanted valence # bonds = = = 13 # e = valence (#bonds) = 34 (13) = 8 bond angles all = 10 carbon hybridization (list in order of appearance in structure) = sp 3, sp, sp, sp 3 σ bonds = 11 π bonds= In this structure all of the carbon and oxygen atoms are in the same plane. The two central carbon atoms have a trigonal planer geometry which places all carbon atoms in same plane. 8() () 4() Total Valence e - 8(1) (6) 4(4) 36 Wanted e - 8() (8) 4(8) wanted valence # bonds = = = 14 # e = valence (#bonds) = 36 (14) = 8 bond angles = and 10 hybridization (list in order of appearance in structure) = sp 3, sp 3, sp, sp 3 σ bonds = 13 π bonds = () N 3() Total Valence e - 3(1) 5 3(4) 0 Wanted e - 3() 8 3(8) 38 wanted valence 38 0 # bonds = = = 9 # e = valence (#bonds) = 0 (9) = carbon hybridization (listed in the order that they appear in the structure) = sp, sp, sp Bond angles: a= 10., b=10., c=180. σ bonds = 6 π bonds = 3 All atoms are in the same plane. 8() 5() Total Valence e - 8(1) (6) 5(4) 40 Wanted e - 8() (8) 5(8) 7 wanted valence 7 40 # bonds = = = 16 # e = valence (#bonds) = 40 (16) = 8 carbon hybridization (listed in the order that they appear in the structure) = sp, sp, sp 3, sp, sp 3 Bond angles: d = 10., e = 10., f = <109.5 N 8
9 σ bonds = 14 π bonds = Atoms with circles around them have to be in the same plane. Atoms with squares around them have to be in the same plane. Depending on the rotation, the atoms with the circles and the atoms with the squares can be in the same plane a) a) 6 carbon atoms are sp 3 hybridized (carbons cycles in solid circle) b) 4 carbon atoms are sp hybridized (carbons cycles in dashed line) c) The boxed nitrogen atom is sp hybridized d) 33 σ bonds e) 5 π bonds f) 180 g) <109.5 h) sp 3 x3 b) piperine 6 carbon atoms sp 3 hybridized (circled with solid line) 11 carbons atoms sp hybridized (circles with dashed line) 0 carbon atoms sp hybridized 9
10 capsaicin 9 carbon atoms sp 3 hybridized (circled with solid line) 9 carbons atoms sp hybridized (circles with dashed line) 0 carbon atoms sp hybridized c) piperine The nitrogen is sp 3 hybridized. capsaicin The nitrogen is sp 3 hybridized. d) a. 10 b. 10 c. 10 d. 10 e. <109.5 f g. 10 h i. 10 j k. 10 l a) The bonding orbital is seen above. This is the bonding orbital because there is electron density between the nucleuses of the atoms (black dots). The antibonding orbital is seen above. This is the antibonding orbital because there is no electron density between the nucleuses of the atoms (black dots). b) The bonding molecular orbital is at a lower energy than the antibonding orbital because the electrons in the area between the nuclei are attracted to two nuclei instead of one. 35. a) b) This is a σ p molecular orbital. The phases of the wavefunctions were the same causing an increase in electron density between the nuclei. It is a sigma molecular orbital because there is cylindrical symmetry. It is a bonding molecular orbital because there is electron density between the nuclei. + - c) This is a π p molecular orbital. It is a pie molecular orbital because there is not cylindrical symmetry. It is a bonding molecular orbital because there is electron density above and below the nuclei This is a σ p molecular orbital. It is a sigma molecular orbital because there is cylindrical symmetry. It is antibonding because there is no electron density between the nuclei. 10
11 d) This is a π p molecular orbital. It is a pie molecular orbital because there is no cylindrical symmetry. It is antibonding because there is no electron density above and below the nuclei bond order = # bonding e #antibonding e If the bonding order is greater than zero then the species is predicted to be stable. + a) = ( σ 1s) 1 bond order = 1 0 stable = 1 = ( σ 1s) bond order = 0 = 1 stable - = ( σ 1s) ( σ 1s* ) 1 bond order = 1 = 1 - = ( σ 1s) ( σ 1s*) stable bond order = = 0 not stable b) N - = ( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( π p) 4 ( σ p) ( π p*) bond order = 10 6 = stable - = ( σ 1s) ( σ 1s*) ( σ s) ( σ s*) (σ p) (π p) 4 ( π p*) 4 bond order = 10 8 = 1 stable - = ( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( σ p) ( π p) 4 ( π p*) 4 ( σ p*) bond order = = 0 not stable c) Be = ( σ 1s) ( σ 1s*) ( σ s) ( σ s*) bond order = 4 4 = 0 not stable B = ( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( π p) bond order = 6 4 = 1 stable Ne = ( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( σ p) ( π p) 4 ( π p*) 4 ( σ p*) bond order = bond order = # bonding e #antibonding e N =( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( π p) 4 ( σ p) = 0 not stable bond order = 10 4 = 3 Therefore, one way to get a bond order of.5 is to add 1 more antibonding e - 11
12 N - =( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( π p) 4 ( σ p) ( π p * ) 1 bond order = 10 5 =.5 r another way to get a bond order of.5 is to remove 1 bonding e - N + =( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( π p) 4 ( σ p) 1 bond order = 9 4 = = ( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( σ p) ( π p) 4 ( π p*) 3 bond order = 10 7 = unpaired e - = ( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( σ p) ( π p) 4 ( π p*) 4 bond order = 10 8 = 1 No unpaired e - - = ( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( σ p) ( π p) 4 ( π p*) 4 ( σ p*) bond order = 10 9 = unpaired e - Increasing - bond length - + < < 48. a) N + =( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( π p) 4 bond order = 8 4 = Diamagnetic b) N=( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( π p) 4 ( σ p) 1 bond order = 9 4 =.5 Paramagnetic ) N - =( σ 1s) ( σ 1s*) ( σ s) ( σ s*) ( π p) 4 ( σ p) bond order = 10 4 = 3 Diamagnetic Increasing bond length N - < N < N + Increasing Bond Energy N + < N < N - 1
13 55. Delocalized π bonding: Is when electrons are spread out over several atoms instead of being concentrated between/around atoms. This phenomenon happens anytime that you have resonance structures. It also explains why you only see 1 bond length in structures that have two types of bonds. 57. In 6 6 the delocalized bonding electrons are spread out over the entire ring causing the carbon carbon bond length to be in-between a single and a double bond length. All carbon-carbon bond lengths are the same. In, the electrons are spread over both of the sulfur oxygen bonds causing the sulfur-oxygen bond length to be in-between a single and a double bond length. All sulfur-oxygen bond lengths are the same The LE model says that the central carbon atom is sp hybridized and that the sp hybrid orbitals on the carbon atoms overlap with either the sp ( double bonded atom) or sp 3 (single bonded atoms) to form 3 sigma bonds. The carbon atom has one unhybridized p orbital that overlaps with a p orbital on the double bonded oxygen atom to form 1 π bond. This localized π bond moves from one position to another. In the molecular orbital model all of the oxygen atoms have p orbitals that form a delocalized π bonding network with the π electrons spread out over the molecule. 13
14 91. All of the carbon atoms are sp 3 hybridized except for the two with * by them, they are sp hybridized. All of the carbon atoms are not in the same plane; sp 3 hybridized structures have bond angles of and only atoms, of the four, can be in the same plane. 14
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