Transcript PPT 4

Molecular Orbitals
-- The VSEPR and valence-bond theories don’t
explain the excited states of molecules, which
come into play when molecules absorb and
emit light.
-- This is one thing that the molecular orbital (MO)
theory attempts to explain.
Molecules respond to the
many wavelengths of light.
The wavelengths that are
absorbed and then re-emitted
determine an object’s color,
while the wavelengths that are
NOT re-emitted raise the
temperature of the object.
molecular orbitals: wave functions that describe the
locations of electrons in molecules
-- these are analogous to atomic orbitals in atoms
(e.g., 2s, 2p, 3s, 3d, etc.), but MOs are possible
locations of electrons in molecules (not atoms)
-- MOs, like atomic orbitals,
can hold a maximum of
two e– with opposite spins
-- but MOs are for
entire molecules
MO theory is more powerful than valence-bond theory;
its main drawback is that it isn’t easy to visualize.
Hydrogen (H2)
The overlap of two atomic
orbitals produces two MOs.
(antibonding MO)
s*1s
1s
+
1s
H atomic orbitals
s1s
(bonding MO)
H2 molecular orbitals
-- The lower-energy bonding molecular orbital
concentrates e– density between nuclei.
-- For the higher-energy antibonding molecular orbital,
the e– density is concentrated outside the nuclei.
-- Both of these are s molecular orbitals.
Energy-level diagram (molecular orbital diagram)
s*1s
1s
1s
H atom s1s
H atom
H2 m’cule
(antibonding MO)
s*1s
1s
+
1s
H atomic orbitals
s1s
(bonding MO)
H2 molecular orbitals
Consider the energy-level diagram for the
hypothetical He2 molecule…
s*1s
1s
1s
He atom s1s
He atom
He2 m’cule
2 bonding e–, 2 antibonding e–
No energy benefit to bonding.
He2 molecule won’t form.
bond order = ½ (# of bonding e– – # of antibonding e–)
-- the higher the bond order,
the greater the bond stability
0 = no bond
1 = single bond
2 = double bond
3 = triple bond
007
7 = James Bond
-- MO theory allows for fractional bond orders as well.
-- a bond order of...
What is the bond order of H2+?
1 e–
total
1 bonding e–,
zero antibonding e–
BO = ½ (1–0) = ½
Second-Row Diatomic Molecules
3
4
Li Be
6.941
9.012
B
5
10.811
C
6
12.011
7
8
N
O
14.007
15.999
F
9
18.998
10
Ne
20.180
1. # of MOs = # of combined atomic orbitals
2. Atomic orbitals combine most effectively
with other atomic orbitals of similar energy.
3. As atomic orbital overlap increases, bonding
MO is lowered in energy, and the antibonding
MO is raised in energy.
4. Both the Pauli exclusion principle and
Hund’s rule apply to MOs.
Use MO theory to
predict whether Li2
and/or Be2 could
possibly form.
s*2s
2s
2s
s2s
s*1s
1s
1s
Li
BO = ½ (4–2) = 1
s1s
Li
Li2
“YEP.”
s*2s
2s
2s
s2s
Bonding and antibonding e– cancel each
other out in core energy levels, so any
bonding is due to e– in bonding orbitals of
outermost shell.
Be
Be
BO = ½ (4–4) = 0 Be2
“NOPE.”