Chemistry 121 Problem set VI solutions - 1
Molecular Orbital Theory, Valence Bond Theory and
Chem 121 Problem Set VI
Hybridization
1.
H2Te
a) Te is in Group VI, so Lewis structure is analogous to H2O (first structure)
b) VSEPR
2 bp + 2 lp = 4
shape is tetrahedral
c) Molecular shape is bent
d) Hybridization is sp3 (VSEPR 4 pairs on central atom so need 4 orbitals)
e) Polar. ∆EN (H-Te) ≈ 0. Resultant vector comes from the two lone pair ∆EN vectors (second structure)
H
H
Te
H
H
CBr4
a) Lewis structure is first structure
b) VSEPR
4 bp + 0 lp = 4
shape is tetrahedral
c) Molecular shape is tetrahedral
d) Hybridization is sp3 (VSEPR 4 pairs on central atom so need 4 orbitals)
e) Nonpolar. ∆EN (Br-C) = 0.3. All 4 ∆EN (C-Br) vectors cancel. Second structure shows the two opposing
resultants for two ∆EN (C-Br) vectors.
Br
Br
C
Br
Br
Br
Br
Br
Br
AsCl5
a) Lewis structure is first structure
b) VSEPR
5 bp + 0 lp = 5
shape is trigonal bipyramid
c) Molecular shape is trigonal bipyramid
d) Hybridization is sp3d (VSEPR 5 pairs on central atom so need 5 orbitals)
e) Nonpolar. ∆EN (Cl-As) = 0.9. All 5 ∆EN (Cl-As) vectors cancel. Second structure shows the two vertical
vectors will cancel and the three vectors in the triangular plane will cancel.
Cl
Cl
Cl
As
Cl
Cl
AsCl3
a) Lewis structure is first structure and has an extra lone pair on the central atom
b) VSEPR
3 bp + 1 lp = 4
shape is tetrahedral
c) Molecular shape is trigonal pyramid (second structure).
d) Hybridization is sp3 (VSEPR 4 pairs on central atom so need 4 orbitals)
e) Weakly polar. ∆EN (Cl-As) = 0.9. The 3 ∆EN (Cl-As) vectors will give a resultant pointing down. The
lone pair gives a vector pointing up and these two vectors will come close to cancelling.
As
Cl
Cl
Cl
As
Cl
Cl
Cl
CO2
a) Lewis structure is first structure
b) VSEPR
2 bp + 0 lp = 2 (multiple bonds count as one bp)
shape is linear
c) Molecular shape is linear
d) Hybridization is sp (VSEPR 2 pairs on central atom so need 2 orbitals)
Chemistry 121 Problem set VI solutions - 2
e) Nonpolar. ∆EN (O-C) = 1. The 2 opposing ∆EN (O-C) vectors cancel.
O
C
O
IF5
a) Lewis structure is first structure and has an extra lone pair on the central atom
b) VSEPR
5 bp + 1 lp = 6
shape is octahedral
c) Molecular shape is square pyramid (second structure).
d) Hybridization is sp3d2 (VSEPR 6 pairs on central atom so need 6 orbitals)
e) Weakly polar. ∆EN (F-I) = 1.5. The 4 ∆EN (F-I) vectors in the square plane will cancel. The ∆EN (F-I)
vector will come close to cancelling with the lone pair vector.
F
F
F
F
F
I
F
F
F
I
F
F
SF4
a) Lewis structure is first structure and has an extra lone pair on the central atom
b) VSEPR
4 bp + 1 lp = 5
shape is trigonal bipyramidal
c) Molecular shape is seesaw (second structure).
d) Hybridization is sp3d (VSEPR 5 pairs on central atom so need 5 orbitals)
e) Weakly polar. ∆EN (F-S) = 1.5. The 2 ∆EN (F-S) vectors in the vertical plane will cancel. The resultant
of the two ∆EN(F-S) vectors will come close to cancelling with the lone pair vector in the horizontal
triangular plane.
F
F
F
S
F
F
F
F
F
ICl3
a) Lewis structure is first structure and has two extra lone pairs on the central atom
b) VSEPR
3 bp + 2 lp = 5
shape is trigonal bipyramidal
c) Molecular shape is T shaped (second structure).
d) Hybridization is sp3d (VSEPR 5 pairs on central atom so need 5 orbitals)
e) Polar. ∆EN (Cl-I) = 0.5. The 2 ∆EN (F-S) vectors in the vertical plane will cancel. The resultant of the
two lone pair vectors will be much greater than the ∆EN (F-S) vector in the horizontal triangular plane.
Cl
Cl
Cl
I
Cl
Cl
Cl
XeF2
a) Lewis structure is first structure and has three extra lone pairs on the central atom
b) VSEPR
2 bp + 3 lp = 5
shape is trigonal bipyramidal
c) Molecular shape is linear, lone pairs go into the horizintal trigonal plane. (second structure).
d) Hybridization is sp3d (VSEPR 5 pairs on central atom so need 5 orbitals)
e) Nonpolar. ∆EN (F-Xe) = 1.4. The 2 ∆EN (F-Xe) vectors in the vertical plane will cancel. The 3 lone pair
vectors in the horizontal triangular plane will cancel.
Chemistry 121 Problem set VI solutions - 3
F
F
Xe
F
F
NO2a) Lewis structure is first structure and and has a formal charge on one oxygen to give a net charge of -1.
Nitrogen forms one double bond and there are two resonance hybrids.
b) VSEPR
2 bp + 1 lp = 3
(multiple bonds count as one bp)
shape is trigonal.
c) Molecular shape is bent. (second structure).
d) Hybridization is sp2 (VSEPR 3 pairs on central atom so need 3 orbitals)
e) Polar and the resultant of the two NO vectors will not cancel with the lone pair vector. Molecule has a
net negative charge.
N
O
2.
3.
O
O
O
CA
VSEPR 4 bp + 0 lp = 4 pair tetrahedral
Hybridization
sp3 (VSEPR 4 pairs on central atom so need 4 orbitals)
CB
VSEPR 3 bp + 0 lp = 3 pair trigonal planar (multiple bonds count as one bp)
Hybridization
sp2 (VSEPR 3 pairs on central atom so need 3 orbitals)
a) CH3CH3
H
H
H
C
C
H
H
Carbon
H
VSEPR: 4 bp + 0 lp = 4 pairs; tetrahedral and hybridization is sp3
2
sp3
2s
Carbon
The six carbon-hydrogen bond orbitals in ethane result from the over lap of sp3 orbitals on the carbon
atoms and 1s orbitals on the hydrogen atoms. The carbon-carbon bond orbital results from the overlap of
two sp3 orbitals, one from each carbon atom. There are (6 x 1) + (2 x 4) = 14 valence electrons in ethane.
Chemistry 121 Problem set VI solutions - 4
Each of the seven bond orbitals is occupied by two valence electrons of opposite spins, accounting for
the seven bonds in ethane.
b) CH3OH
H
H
C
O
H
H
Carbon VSEPR: 4 bp + 0 lp = 4 pairs; tetrahedral and hybridization is sp3 (energy diagram as above)
Oxygen VSEPR: 2 bp + 2 lp = 4 pairs; tetrahedral and hybridization is sp3
2
sp3
2s
Oxygen
The three carbon-hydrogen bond orbitals and the oxygen-hydrogen bond orbital in methanol result from the
overlap of sp3 orbitals and 1s orbitals on the hydrogen atoms. The carbon-oxygen bond orbital results from
the overlap of a carbon sp3 orbital and an oxygen sp3 orbital. Ten of the valence electrons (4 x 1) + 4 + 6 =
14 in methanol occupy five bond orbitals and the other four valence electrons occupy the two remaining sp3
orbitals as lone pairs on the oxygen atom.
c) Ethene
H
H
C
C
H
Carbon
H
VSEPR: 3 bp + 0 lp = 3 pairs; trigonal planar and hybridization is sp2
2
2p
sp2
2s
Carbon
Chemistry 121 Problem set VI solutions - 5
All six atoms of the ethene molecule lie in one plane. The first diagram shows the two sp2 hybridized
carbon atoms, each with three sp2 orbitals and one remaining 2p orbital. Each orbital contains one electron
as shown in the energy diagram. The second diagram shows the σ framework. The four carbon-hydrogen
bond orbitals result from the overlap of carbon sp2 orbitals and hydrogen 1s orbitals. The two carbon atoms
are joined by the overlap of an sp2 orbital from each. The resulting bond orbital is cylindrically symmetric
around the carbon-carbon axis and is therefore a σ bond orbital., as are the carbon-hydrogen bond orbitals.
Finally the last diagram shows the overlap of the two 2p orbitals to form a π bond, the second bond in the
double bond. There are (2 x 4) + 4 x 1) = 12 valence electrons in ethene, 10 of these occupy the five σ
bonds and the last two occupy the π bond.
d) Ethyne
H
C
Carbon
C
H
VSEPR: 2 bp + 0 lp = 2 pairs; linear and hybridization is sp
2
2p
sp
2s
Carbon
The first diagram shows the σ bond framework for ethyne. The carbon-carbon σ bond orbital results from
overlapping two sp orbitals, one from each carbon atom. Each of the two carbon-hydrogen σ bond orbitals
results from overlapping a carbon sp orbital and a hydrogen 1s orbital. When carbon is sp hybridized, there
are two 2p orbitals, each with one electron, unused (energy diagram). The second diagram shows the 2p
orbitals from both carbons (one set along the y-axis and the other set along the z-axis). The 2py orbitals
from each carbon overlap to form a π bond orbital, as do the 2pz orbitals to form a second π bond orbital.
The two π bonds in ethyne (acetylene) constitute a barrel-shaped distribution of electron density in the
bond region.
e) Carbon monoxide
Chemistry 121 Problem set VI solutions - 6
C
O
Carbon
VSEPR: 1 bp + 1 lp = 2 pairs; linear and hybridization is sp
2
2p
sp
2s
Carbon
Oxygen VSEPR: 1 bp + 1 lp = 2 pairs; linear and hybridization is sp
2
2p
sp
2s
Oxygen
The hybridization for carbon is sp and one of the sp orbitals contains a lp and the other a bp. There are two
remaining 2p orbitals each occupied by one electron. Thus carbon has five valence electrons; note the
negative formal charge on carbon in the Lewis structure. The hybridization for oxygen is sp and one of the
sp orbitals contains a lp and the other a bp. There are two remaining 2p orbitals each occupied by one
electron. Thus oxygen has five valence electrons; note the positive formal charge on carbon in the Lewis
structure. The carbon-oxygen σ bond orbital results from overlapping the two sp orbitals, each contributing
one electron. The two π bonds (a triple bond results from a σ bond and two π bonds) result from the
overlap of the two 2py orbitals on carbon and oxygen and the two 2pz orbitals as in ethyne. Carbon
monoxide has (4 + 6) = 10 valence electrons, two occupy the σ orbital, four occupy the two π orbitals and
the other four occupy the two remaining sp orbitals on carbon and oxygen as lone pairs.
Chemistry 121 Problem set VI solutions - 7
4.
*
*
2p x
*2p
*2p
*
2pz
y
2p
2p
*2p
2pz
2p
2p
*
2py
2pz
*
2s
2p
2p
2p x
2py
2pz
*
2s
2s
2s
2s
2pz
*
2s
2s
2s
N2+
2pz
y
2p x
2s
2s
2p x
*
y
2p x
2py
*
2p x
2s
2s
N2
N2-
Valence electrons 9
(9 − 4)
= 2.5
Bond Order
2
10
(10 − 4)
=3
2
11
(10 − 5)
= 2.5
2
Magnetism
diamagnetic
paramagnetic
Species
paramagnetic
Bond strength
N2
>
N2+
N2−
=
5a)
*
*2p
2pz
2py
2pz
2p x
*
Total electrons
Bond Order
*
2pz
y
2py
2pz
2px
*2p
2p x
*
2pz
y
2py
2pz
2p x
*2p
*
2pz
y
2py
2pz
2p x
*
2s
2s
2s
*
2s
2s
2s
2s
2s
*
Species
*2p
*
2p x
2px
*
y
*
*
2p x
*
*
1s
1s
1s
*
1s
1s
1s
1s
1s
*
O2+
O2
O2-
O22-
15
16
17
18
(10 − 5)
= 2.5
2
(10 − 6)
=2
2
(10 − 7)
= 1.5
2
(10 − 8)
=1
2
(i) weakest bond is O22-
Chemistry 121 Problem set VI solutions - 8
(ii) shortest bond is O2+
5b)
σ∗2p
σ∗2p
σ∗2p
π∗2p π∗2p
π∗2p π∗2p
π∗2p π∗2p
σ2p
σ2p
σ2p
π2p π2p
π2p π2p
π2p π2p
σ∗2s
σ∗2s
σ∗2s
σ2s
σ2s
σ2s
σ∗1s
σ∗1s
σ∗1s
σ1s
σ1s
σ1s
Species
NO
CN-
B2
Total electrons
15
14
10
Unpaired electrons
1
0
2
Increasing paramagnetism
CN-
<
NO
<
B2