What is the greenhouse effect, and is it affecting our climate?
The greenhouse effect is unquestionably real, and is, in fact, essential for life on Earth. It
is the result of heat absorption by certain gases in the atmosphere (called greenhouse
gases bec
The four orbitals of 1,3-butadiene are of the following symmetries under the
preserved plane (see the orbitals in the Figure above): 1 = e, 2 = o, 3 =e, 4 = o. The
and * and and * orbitals of cyclobutane which evolve from the four active orbitals of
the
on the photochemically induced reaction of 1,3-butadiene to produce cyclobutene. In
contrast, the thermal reaction considered first above has a symmetry-imposed barrier
because the orbital occupancy is forced to rearrange (by the occupancy of two electron
Chapter 12, to symmetry-caused energy barriers on the H2CO => H 2 + CO reaction
potential energy surface.
II. Orbital Correlation Diagrams
Connecting the energy-ordered orbitals of reactants to those of products according
to symmetry elements that are pre
H2CO are energy-ordered and labeled according to C2v symmetry in the Figure shown
below as are the orbitals of the product H2 + CO.
b2
CH *
a1, b2
CO *
CO *
a1
b1
HH *
O lone pairs
b2
a1
CO bond
CO bond
b1
a1
b2
a1
a1
b1, b2
a1
a1
b1, b2
CO *
CO *
C lone
spherical symmetry, can mix to form the g bonding orbital, the u antibonding, as well as
the g and u nonbonding lone-pair orbitals. The fact that 2s and 2p have different l-values
no longer uncouples these orbitals as it did for the isolated atoms, becaus
Chapter 6
Along "Reaction Paths", Orbitals Can be Connected One-to-One According to Their
Symmetries and Energies. This is the Origin of the Woodward-Hoffmann Rules
I. Reduction in Symmetry
As fragments are brought together to form a larger molecule, the
and (iii) the potential energy part of the Hamiltonian is spherically symmetric (and
commutes with L2 and Lz), the energies of atomic orbitals depend upon the l quantum
number and are independent of the m quantum number. This is the source of the 2l+1- fo
For homonuclear diatomic molecules and other linear molecules which have a center
of symmetry, the inversion operation (in which an electron's coordinates are inverted
through the center of symmetry of the molecule) is also a symmetry operation. Each
resu
orbitals one chooses to utilize in conceptualizing a molecule's orbital interactions,
symmetry ultimately returns to force one to form proper symmetry adapted combinations
which, in turn, renders the various points of view equivalent. In the above example
The bonding molecular orbital pair (with m = +1 and -1) is of u symmetry whereas the
corresponding antibonding orbital is of g symmetry. Examples of such molecular orbital
symmetries are shown above.
The use of hybrid orbitals can be illustrated in the li
Hamiltonian contains (h2/2me r2) 2/2 , whereas the potential energy part is independent
of , the energies of the molecular orbitals depend on the square of the m quantum
number. Thus, pairs of orbitals with m= 1 are energetically degenerate; pairs with m=
Chapter 7
The Most Elementary Molecular Orbital Models Contain Symmetry, Nodal Pattern, and
Approximate Energy Information
I. The LCAO-MO Expansion and the Orbital-Level Schrdinger Equation
In the simplest picture of chemical bonding, the valence molecula
1. The diagonal values < |- h2 /2me 2 + V| >, which are usually denoted ,
are taken to be equal to the energy of an electron in the atomic orbital and, as such, are
evaluated in terms of atomic ionization energies (IP's) and electron affinities (EA's):
<
Si 3s
15
Radial Function R(r)
12
9
6
Z=14
Z=4
3
0
-3
0
1
2
3
r (bohr)
Si 3p
5
Radial Function R(r)
4
Z=14
3
2
1
Z=4
0
-1
-2
0
1
2
3
4
5
r (bohr)
Exercises
1. Two Slater type orbitals, i and j, centered on the same point results in the following
overlap in
2
2 ni+ 1
Sij = a i
(2ni)!
0 0
1
2
0
-ir
Y
1
a
2 r(ni-1)e 0
li,m i(,).
0
-jr
nj+1 1 1 (n -1) a0
2j 2
Y
2
j e
a
lj,m j(,).
(2nj)! r
0
r2sindrdd.
For these s orbitals l = m = 0 and Y0,0 (,) =
1
. Performing the integrations over and
4
yields 4 w
N 2px xu
b3u
b1
a'
N 2py yu
b2u
b2
a'
N 2pz u
b1u
a1
a'
2.
v. The Nitrogen molecule is in the yz plane for all point groups except the Cs in
which case it is placed in the xy plane.
Dh
D2h
C2v
Cs
N 1s + N 1s g
ag
a1
a'
N 1s - N 1s u
b1u
b2
a'
N 2s + N 2s
x
y
z
z
x
y
x
x
y
y
z
z
2.
i.In ammonia the only "core" orbital is the N 1s and this becomes an a1 orbital in
C3v symmetry. The N 2s orbitals and 3 H 1s orbitals become 2 a1 and an e set of orbitals.
The remaining N 2p orbitals also become 1 a1 and a set
C-C sp2-sp2 = -5.0 eV
C-H sp2-s = -4.0 eV
a. Determine the C=C (2p ) 1-electron molecular orbital energies and
wavefunctions. Calculate the * transition energy for ethylene within this model.
b. Determine the C-C (sp2) 1-electron molecular orbital energie
energy levels and 1-electron wavefunctions. Draw the orbital correlation diagram for
formation of the N2 molecule. Indicate the symmetry of each atomic and molecular orbital.
Designate each of the molecular orbitals as bonding, non-bonding, or antibonding
Section 2 Exercises, Problems, and Solutions
Review Exercises:
1. Draw qualitative shapes of the (1) s, (3) p and (5) d "tangent sphere" atomic orbitals
(note that these orbitals represent only the angular portion and do not contain the radial
portion of
Atom
# Elec.
es=ep
H
Li
Be
B
C
N
O
F
Na
Mg
Al
Si
P
S
Cl
1
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1.3
0.650
0.975
1.300
1.625
1.950
2.275
2.425
0.733
0.950
1.167
1.383
1.600
1.817
2.033
ed
s(eV)
p(eV)
d(eV)
1.383
1.400
1.500
2.033
-13.6
-5.4
-10.0
-15.2
-21.4
-26.
and antibonding orbitals depends on the overlap between the pair of atomic orbitals. Also,
the energy of the antibonding orbital lies higher above the arithmetic mean Eave = EA + EB
of the energies of the constituent atomic orbitals (EA and EB) than the b
The EA of the 2p orbital is obtained from the
C(1s22s22px2py) => C -(1s22s22px2py2pz)
energy gap, which means that EA2pz = EAC . Some common IP's of valence 2p orbitals in
eV are as follows: C (11.16), N (14.12), N+ (28.71), O (17.70), O+ (31.42), F+ (37.
Expanding the mo i in the LCAO-MO manner, substituting this expansion into the above
Schrdinger equation, multiplying on the left by * , and integrating over the coordinates
of the electron generates the following orbital-level eigenvalue problem:
< |- h