Week 6
Recitation Problems
1. In H2 in the bonding orbital why is the electron density between the atoms greater
than the sum of the electron densities on the two atoms? Why is the density in the
antibonding orbital lower that the sum of the to AO's?
2. In O2, F2 and Ne2 the order of the bonds is different that for N2. Draw the MO
diagram for O2 and N2 and explain why the order of the bonds changes.
3. Draw MO diagram, predict their stability and magnetic properties:
• H2,
H2+,
H22+,
H2-,
H22-,
• Li2
Li22+,
Li22-,
Li2+,
+
2+
• He2,
He2 ,
He2 ,
He2-,
• F2 ,
F2 - ,
F2+,
+
• O2,
O2 ,
O22+,
O2-,
O22-,
• N2,
N2+,
N22+,
N2-,
N22-,
• Sn2,
Si2
Ge2,
C2
4. Place in order of increasing bond length
• B2,
N2,
C2
• F2 ,
O2,
N2
• F2 ,
Cl2
Br2
• Na2
K2
Li2
• N2
P2
As2
Sb2
5. Draw the bonding and antibonding orbitals for Li2 as a function of the distance
between the two nuclei.
6. Draw Lewis, MO diagrams and tell if paramagnetic
• NF,
NF-,
NF+
• SO,
NO,
SeO
• PO,
SbO
ClF
• BrF
BrCl
HF
• HCl
HBr
CN
CN7. Draw the first 3 excited states of
• Li,
H2,
H2+,
• H2-,
F2 ,
O2,
• N2,
Sn2,
Si2
8. Draw Lewis dot, VSEPR and describe the hybridization for the following
• CH4
CO2
OF2
BeH2
• BF3
BH4ClO3+
ClO2+
• NCCl
N2
N3
9. Hybridization: Fill in the table:
# hybride of orbitals angle between
hybridization
left over AO's
orbitals
2
180
sp
px py
3
4
8. Show the hybridization of all atoms other than H and describe the shape of the
molecule.
a. H4C
H3CCH3
H2CCH2
b. HCCH
NH3
CH3CHCH2
c. CH3CHO
H3CCOOH H3CCOCH3
d. H2CCHCHCH2
CS2 PF3
9. π Bonds: Straight chain hydrocarbons: Draw MO diagrams and show the AO
orbital contribution, predict the relative bond lengths and shapes for:
a. C6H8
H12C10
H7C5
10. π Bonds: Cyclic Hydrocarbons: Draw MO diagram, show AO contributions to
MOs and predict if they are aromatic:
a. C4H4
C4H42+
C4H42•
+
b. C5H5
C5H5
C5H5c. C8H8
C8H82+
C8H82d. C10H10
C10H102+
C10H102-
Problems RS 7
1. In H2 in the bonding orbital why is the electron density between the atoms greater
than the sum of the electron densities of the two atoms? When is it the same?
Why is the density in the antibonding orbital lower that the sum of the to AO's?
Probablity of finding an electron at a particular position is given by Ψ2 since for a
bonding and antibonding orbital on H2 if we label the nuclei as Ha and Hb and the
wavefunction for the H atoms as ψa and ψb we have the bonding and antibonding orbrtials
as " = # a + # b ; and " * = # a $ # b . The the electron probability is given by
2
!
!
2
" 2 = (# a + # b ) = # a 2 + # b 2 + 2# a# b ; and " *2 = (# a $ # b ) = # a 2 + # b 2 $ 2# a# b the cross
terms ψaψb are due to the interfearance between the wavefunctions and are the reason that
the density can be more that just the sum of the densities of the two separate orbitals. At
the center point between the two nuclei the value of are due to the interfearance between
the wavefunctions and are the reason that the density can be more that just the sum of the
densities of the two separate orbitals. At the center point between the two nuclei the
value of ψa=ψb so " 2 = 4# a = 4# b ; and " *2 = 0
2. In O2, F2 and Ne2 the order of the bonds is different that for N2. Draw the MO
diagram for O2 and N2 and explain why the order of the bonds changes.
! for Homonuclear molecules
MO diagram
N
O
N
O
!"
!"
#"
#"
2p
2p
!
2p
2p
#
#
!
!"
2s
2s
!"
!
2s
2s
a. bond order for N (see
left side picture
b. Bond order for O see
right side picture
c. Reason is energy
distance between 2s and
2p orbitals and the
interaction between
them
d. Predict bond order,
number of unpaired
electrons,
!
3. Draw MO diagram, predict their stability and magnetic properties:
Molecule MO diagram
Bond Order
# of unpaired
electrons
H2
above left
1
0
+
H2
above left
½
1
H22+
above left
0
0
H2
above left
½
1
2H2
above left
0
0
Li2
above left
1
0
2+
Li2
above left
0
0
Li22above left
0
0
Li2+
He2
He2+
He22+
He2F2
F2 F2 +
O2
O2+
O22+
O2O22N2
N2+
N22+
N2N22Sn2
Si2
Ge2
C2
above left
above left
above left
above left
above left
Above right
Above right
Above right
Above right
Above right
Above right
Above right
Above right
Above left
Above left
Above left
Above left
Above left
Above left
½
0
½
1
?
1
½
1½
2
2½
3
1½
1
3
2½
2
2½
2
2
2
2
2
1
0
1
0
1
0
1
1
Para 2
Para 1
0
1
0
0
Par 1
0
Para 1
0
Para 2 ePara 2 ePara 2 ePara 2 e-
4. Place in order of increasing bond length
• N2<C2<<B2
• N<O2<F2,
• F2<Cl2<Br2
• Li2< Na2<K2
• N2<P2<As2<Sb2
5. Draw the bonding and antibonding orbitals for Li2 as a function of the distance
between the two nuclei.
!"2s
!2s
!"1s
!1s
equilibrium distance in molecule
distance between nuclei
6. Draw Lewis, MO diagrams and tell if paramagnetic
First need Iz of the various atoms:
Atom
1s
2s
2p
3s
3p
H
110
Li
44
C
157
86
4s
4p
N
O
F
P
S
Cl
206
261
374
106
128
151
151
167
204
NF-,
NF,
•
(5 e- ) N
F(7
82
94
111
NF+
For NF need Iz of N 206, and 106 for 2s and 2p
and F is 374 and 151 for 2s and 2p. So the F2s is
too low energy to bond with anything. Thus use
the R 2p to bond with the N 2s and 2p. For NF get:
There are 2 unpaired electrons and the BO is 2. For
NF- you add one electron and get 1 unparied
electron and a BO of 1½ and for NF+ you loss an
electron from NF and have one uparied electron
and a BO of 2½ .
e- )
2p
2p
2s
For SO the S 3s and 3p are
at 167 and 94 while the O
2s and 2p ae at 261 and
128. Thus the O 2p
interacts with the S3s and
2p. It is similar to the
2s
diagram at left but the N
orbitals would be the S 3s and 2p and the F orbitals
would be for O the 2s and 2p. The O 2p would be
lower than they are shown. See diagram below:
NO,
SeO
PO,
SbO
ClF
BrF
BrCl
• HCl
HBr
7. Draw the first 3 excited states of
• Li2,
H2,
2nd Excited
State Li2
First Excited
state Li2
!"
H2+,
3rd Excited
State Li2
!"
#"
2p
HF
CN
!"
#"
2p
2p
!
#"
2p
2p
!
2p
!
#
#
#
!"
!"
!"
2s
2s
!
2s
2s
!
2s
2s
!
CN-
(6 e- ) S
O(6 e-)
2p
2p
2s
2s
3rd Excited
state O2
2nd Excited
state O2
First Excited
state O2
!"
!"
#"
!"
#"
First Excited
State N2
#"
Second Excited
State N2
!"
2p
2p
2p
2p
2p
2p
#
!
#
!
#
!
!"
!"
!"
#"
2p
2p
2p
!
!
•
H2-,
2s
2s
!
2s
2s
2s
F2 ,
2p
!
2p
!
#
#
!"
2s
!
#"
2p
!
!"
2s
!"
#"
#
2s
Third Excited
State N2
!"
2s
!"
2s
!
2s
2s
!
O2,
• N2,
Sn2,
Si2
8. Draw Lewis dot, VSEPR and describe the hybridization for the following
• CH4
CO2
OF2
BeH2
• BF3
BH4ClO3+
ClO2+
• NCCl
N2
N3
9. Hybridization: Fill in the table:
# hybride of orbitals angle between
hybridization
left over AO's
orbitals
2
180
sp
px py
3
4
11. Show the hybridization of all atoms other than H and describe the shape of the
molecule.
a. H4C. sp3
H3CCH3 sp3 sp3
H2CCH2 sp2 sp2
3
3
b. HCCH, sp sp NH3 sp
CH3CHCH2, sp3 sp2, sp2
c. CH3CHO sp3 sp2
H3CCOOH sp3 sp2
H3CCOCH3 sp3 sp2 sp3
d. H2CCHCHCH2
CS2 PF3
12. π Bonds: Straight chain hydrocarbons: Draw MO diagrams and show the AO
orbital contribution, predict the relative bond lengths and shapes for:
a. C3H3•
C3H3+
C3H3- C6H8 H12C10 H7C5
13. π Bonds: Cyclic Hydrocarbons: Draw MO diagram, show AO contributions to
MOs and predict if they are aromatic:
a. C4H4
C4H42+
C4H42•
+
b. C5H5
C5H5
C5H5c. C8H8
C8H82+
C8H822+
d. C10H10
C10H10
C10H102-