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Homework #
Chapter 14
Covalent Bonding Orbitals
- Single 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.
- LE (Local Electron) Model Hybridizations 2(H) C O Total Valence e-^ 2(1) 4 6 12 Wanted e-^ 2(2) 8 8 20
e^ ^ Valence bonds
bonds Wanted Valence
H 2 CO is a trigonal planer molecule. The central carbon atom is sp^2 hybridized and the oxygen atom is sp^2 hybridized. Two of the sp^2 hydride orbitals, on the carbon, are used in C-H single bonds (σ bond). The C-H single bond is formed from the overlap of the s orbital on the hydrogen and the sp^2 hybridized orbital on the carbon. There is a double bond between the C-O (1 σ bond and 1 π bond). The σ bond is formed from the overlap of the sp^2 hybridized orbital, on the carbon, and the sp^2 hybridized orbital on the oxygen. The π bond is formed from the overlap of the 2 unhybridized p orbitals on the carbon and the oxygen atoms. The loan pair electrons on the oxygen atom are located in sp^2 hybridized orbitals. 2(H) 2(C) Total Valence e-^ 2(1) 2(4) 10 Wanted e-^ 2(2) 2(8) 20
e^ ^ Valence bonds
bonds Wanted Valence
H 2 C 2 is a linear molecule. The central carbon atoms are sp hybridized. On each carbon atom, one of the sp hybridized orbitals overlaps with a s orbital on the hydrogen atom to form a H-C single bonds (σ bonds). There is a triple bond between the carbons (1 σ bond and 2π 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 (2 on each carbon) on the carbons.
- a)
e^ ^ Valence bonds
bonds Wanted Valence
Molecular Structure =Tetrahedral Bond Angles = 109.5° C Hybridization=sp^3 Polar/Non Polar = Non Polar b)
e^ ^ Valence bonds
Wanted Valence
bonds
Molecular Structure =Trigonal Pyramidal Bond Angles = <109.5° N Hybridization=sp^3 Polar/Non Polar = Polar
c)
e^ ^ Valence bonds
bonds Wanted Valence
Molecular Structure = Bent Bond Angles = <109.5°
C 4(F) Total Valence e-^4 4(7) 32 Wanted e-^8 4(8) 40
N 3(F) Total Valence e-^5 3(7) 26 Wanted e-^8 3(8) 32
O 2(F) Total Valence e-^6 2(7) 20 Wanted e-^8 2(8) 24
Kr Hybridization=sp^3 d Polar/Non Polar = Non Polar i)
Kr must expand its octet to accommodate all of the electrons Molecular Structure = Square Planer Bond Angles = 90° and 180˚ Kr Hybridization=sp^3 d^2 Polar/Non Polar = Non Polar j)
Se must expand its octet to accommodate all of the atoms Molecular Structure = Octahedral Bond Angles = 90° Se Hybridization=sp^3 d^2 Polar/Non Polar = Non Polar k)
I must expand its octet to accommodate all of the electrons/atoms Molecular Structure = Square Pyramidal Bond Angles = <90°and <180° I Hybridization=sp^3 d^2 Polar/Non Polar = Polar i)
I must expand its octet to accommodate all of the electrons/atoms Molecular Structure = T-Shaped Bond Angles = <90°and <180° I Hybridization=sp^3 d Polar/Non Polar = Polar
- a)
# 2 ^ # ^182 3 12
e^ ^ Valence bonds
Wanted Valence
bonds
Molecular Structure = Bent Bond Angles = <120° S Hybridization = sp^2
Kr 4(F) Total Valence e-^8 4(7) 36
Se 6(F) Total Valence e-^6 6(7) 48
I 5(F) Total Valence e-^7 5(7) 42
I 3(F) Total Valence e-^7 3(7) 28
S 2(O) Total Valence e-^6 2(6) 18 Wanted e-^8 2(8) 24
b)
e^ ^ Valence bonds
bonds Wanted Valence
Molecular Structure = Trigonal Planer Bond Angles = 120° S Hybridization = sp^2
c)
e^ ^ Valence bonds
Wanted Valence
bonds
Molecular Structure = Tetrahedral Bond Angles = 109.5° S Hybridization = sp^3
d)
e^ ^ Valence bonds
Wanted Valence
bonds
Molecular Structure = Tetrahedral (around S) Bent (around central O) Bond Angles = 109.5° (around S) and <109.5°(around central O) S Hybridization = sp^3
e)
e^ ^ Valence bonds
Wanted Valence
bonds
Molecular Structure = Trigonal Pyramidal Bond Angles = <109.5° S Hybridization = sp^3
S 3(O) Total Valence e-^6 3(6) 24 Wanted e-^8 3(8) 32
2(S) 3(O) e-^ Total Valence e-^ 2(6) 3(6) +2 32 Wanted e-^ 2(8) 3(8) 40
2(S) 8(O) e-^ Total Valence e-^ 2(6) 8(6) +2 62 Wanted e-^ 2(8) 8(8) 80
S 3(O) e-^ Total Valence e-^6 3(6) +2 26 Wanted e-^8 3(8) 32
e^ ^ Valence bonds
bonds Wanted Valence
The hybridization around both carbons is sp^2 hybridized, making the shape trigonal planer around each carbon. The double bonds between the carbons are formed from the out-of- page/in-page p orbitals. Therefore, the remaining sp^2 orbital are all in plane with each other.
- 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.
e^ ^ Valence bonds
bonds Wanted Valence
CCO bond angles = all 120° carbon hybridization (list in order of appearance in structure) = sp^3 , sp^2 , sp^2 , sp^3 σ bonds = 11 π bonds= 2 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.
4(H) 2(C) Total Valence e-^ 4(1) 2(4) 12 Wanted e-^ 4(2) 2(8) 24
6(H) 2(O) 4(C) Total Valence e-^ 6(1) 2(6) 4(4) 34 Wanted e-^ 6(2) 2(8) 4(8) 60
# 2 ^ # ^362 14 8
e^ ^ Valence bonds
Wanted Valence
bonds
CCO bond angles = 109.5° and 120° C hybridization (list in order of appearance in structure) = sp^3 , sp^3 , sp^2 , sp^3 σ bonds = 13 π bonds = 1
e^ ^ Valence bonds
bonds Wanted Valence
carbon hybridization (listed in the order that they appear in the structure) = sp^2 , sp^2 , sp Bond angles: a= 120.°, b=120.°, c=180.° σ bonds = 6 π bonds = 3 All atoms are in the same plane.
e^ ^ Valence bonds
bonds Wanted Valence
carbon hybridization (listed in the order that they appear in the structure) = sp^2 , sp^2 , sp^3 , sp^2 , sp^3 Bond angles: d = 120.°, e = 120.°, f = <109.5° σ bonds = 14 π bonds = 2 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.
8(H) 2(O) 4(C) Total Valence e-^ 8(1) 2(6) 4(4) 36 Wanted e-^ 8(2) 2(8) 4(8) 64
3(H) N 3(C) Total Valence e-^ 3(1) 5 3(4) 20 Wanted e-^ 3(2) 8 3(8) 38
8(H) O 5(C) Total Valence e-^ 8(1) 2(6) 5(4) 40 Wanted e-^ 8(2) 2(8) 5(8) 72
109.5˚ 109.5˚
120.˚ 120.˚
capsaicin 9 carbon atoms sp^3 hybridized (circled with solid line) 9 carbons atoms sp^2 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. 120 ° b. 120 ° c. 120 ° d. 120 ° e. <109.5° f. 109.5°
g. 120 ° h. 109.5° i. 120 ° j. 109.5° k. 120 ° l. 109.5°
- 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 that antibonding orbital because the electrons in the area between the nuclei are attracted to two nuclei instead of one.
- a)
This is a 𝜎𝑝 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. b)
This is a 𝜋𝑝molecular 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. c)
-^ +^ -
This is a 𝜎𝑝∗^ 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. d)
This is a 𝜋𝑝∗^ 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) H 2 +^ = ( σ1s)^1 1 0 1 2 2
bond order stable
H 2 = ( σ1s)^2 (^2 0 ) 2
bond order stable
H 2 -^ = ( σ1s)^2 ( σ1s*)^1 2 1 1 2 2
bond order stable
H 2 2-^ = ( σ1s)^2 ( σ1s*)^2 (^2 2 ) 2
bond order not stable
b) N 2 2-^ = ( σ1s)^2 ( σ1s)^2 ( σ2s)^2 ( σ2s)^2 ( π2p)^4 ( σ2p)^2 ( π2p*)^2 (^10 6 ) 2
bond order stable
O 2 2-^ = ( σ1s)^2 ( σ1s)^2 ( σ2s)^2 ( σ2s)^2 (σ (^) 2p)^2 (π (^) 2p)^4 ( π2p*)^4 (^10 8 ) 2
bond order stable
F 2 2-^ = ( σ1s)^2 ( σ1s)^2 ( σ2s)^2 ( σ2s)^2 ( σ2p)^2 ( π2p)^4 ( π2p)^4 ( σ2p)^2 (^10 10 ) 2
bond order not stable
c) Be 2 = ( σ1s)^2 ( σ1s)^2 ( σ2s)^2 ( σ2s)^2 (^4 4 ) 2
bond order not stable
B 2 = ( σ1s)^2 ( σ1s)^2 ( σ2s)^2 ( σ2s)^2 ( π2p)^2 (^6 4 ) 2
bond order stable
Ne 2 = ( σ1s)^2 ( σ1s)^2 ( σ2s)^2 ( σ2s)^2 ( σ2p)^2 ( π2p)^4 ( π2p)^4 ( σ2p)^2
In C 6 H 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 SO 2 , 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^2 hybridized and that the sp^2 hybrid orbitals on the carbon atoms overlap with either the sp^2 ( double bonded O atom) or sp^3 (single bonded O 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.
All of the carbon atoms are sp^3 hybridized except for the two with “*” by them, they are sp^2 hybridized. All of the carbon atoms are not in the same plane; sp^3 hybridized structures have bond angles of 109.5 and only 2 atoms, of the four, can be in the same plane.
