Chemistry 445'

1

• Chemistry 445.
• Lecture 4.
• Molecular Orbital Theory
• of diatomic molecules

2
The non-existent He2 molecule(bond order 0)
BO (2-2)/2 0
The MO diagram for the He2 molecule is similar to
that for the H2 molecule, but we see that the
energy drop of the pair of electrons in the s1s
orbital is negated because the other pair in the
s1s rises in energy by an equal amount. There is
thus no net stabilization, and so the He2
molecule does not exist.
3
The He2 molecule/ion exists, bond order ½
unpaired electron so is paramagnetic
BO (2-1)/2 ½
The logic of the MO diagram suggests that if we
remove an electron from the He2 molecule, we
would obtain a stable He2 cation, which is
true in the gas-phase. This illustrates the power
of the MO approach, since the Lewis dot diagram
does not predict this.
4
The Li2 molecule. Bond order 1
BO (2-0)/2 1
Bond energy for Li2 110 kJ.mol-1, Compared to
436 kJ.mol-1 for H2.
Note. In drawing up an MO diagram, only the
valence shells are considered, so for the
diatomic molecules from Li2 to F2, the
overlaps of the pairs of 1s orbitals are
ingnored. This is valid because these are filled,
and they make no net contribution to the bonding.
5
The non-existent Be2 molecule,bond order 0
BO (2-2)/2 0
Here again, MO theory predicts that Be2 does not
exist, which Lewis dot diagrams do not predict.
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s and p bonding

• In the molecules we have considered so far, only
  s overlaps have
• been of importance. In the formal definition, a
  sbond is one which lies along a rotational
  symmetry axis, (the rest of the molecule is
  ignored). A pbond does not lie along a
  rotational axis (symmetry will be discussed
  later). In practical terms, a sbond lies along
  the bond connecting the two atoms, whereas a
  pbond does not.

z
z

p(pz)
pz orbitals
anti-bonding p MO
z
z
p(pz)

bonding p MO
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O2 molecule, bond order 2
molecules with unpaired electrons are paramagnetic
The ability to predict the number of
unpaired electrons in molecules is where
MO excels, and Lewis-dot fails.
BO (6-2)/2 2
(disregardiing overlap of 2s orbitals)
O O2 O
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F2 molecule, bond order 1
F2 has no unpaired electrons, and so is
diamagnetic
BO (6-4)/2 1
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Variation of the energies of the 2s and 2p
orbitals in crossing the periodic table from Li
to F. (HS Fig. 1.22)
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Energy levels of first-row homonuclear diatomic
molecules (HS Fig 1.23)
crossover point
Molecules Li2, Be2, B2,C2 and N2 have p(2p) lower
in energy than s(2p)
Molecules O2, and F2 have p(2p) higher in energy
than s(2p)
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Be2 molecule, bond order 0(BO 0, means does
not exist)
BO (2-2)/2 0
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B2 molecule, bond order 1
BO (2-0)/2 1
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N2 molecule, bond order 3
diamagnetic
BO (6-0)/2 3
N N2 N
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C2 molecule, bond order 2
diamagnetic
BO (4-0)/2 2
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Orbital parity gerade (g) and ungerade (u)

• Symmetry of orbitals and molecules is of great
  importance, and we should be able to determine
  whether orbitals are gerade (g) or ungerade (u)
  (from German for even or odd). This is because in
  the spectra of inorganic compounds whether
  absorption of a photon to produce an electronic
  transition is determined by whether the two
  orbitals involved are g or u. According to the
  Laporte selection rules,
• transitions from g?u and u?g are allowed, but
  g?g and u?u are forbidden. An orbital is g if it
  has a center of inversion, and u if it does not.
  So looking at atomic orbitals, we see that s and
  d are g, while p orbitals are u (see next page
  for definition of center of sym.)

s-orbital p-orbital
d-orbital gerade (g)
ungerade (u) gerade
(g)
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Orbital parity gerade (g) and ungerade (u)

• Symmetry of orbitals and molecules is of great
  importance, and we should be able to determine
  whether orbitals are gerade (g) or ungerade (u)
  (from German for even or odd). This is because in
  the spectra of inorganic compounds whether
  absorption of a photon to produce an electronic
  transition is determined by whether the two
  orbitals involved are g or u. According to the
  Laporte selection rules,
• transitions from g?u and u?g are allowed, but
  g?g and u?u are forbidden. An orbital is g if it
  has a center of inversion, and u if it does not.
  So looking at atomic orbitals, we see that s and
  d are g, while p orbitals are u (see next page
  for definition of center of sym.)

a
not a center of inversion a ? b
a
center of inversion a b
a
b
b
b
s-orbital p-orbital
d-orbital gerade (g)
ungerade (u) gerade
(g)
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Parity (g or u) of molecular orbitals
a
a
not a center of inversion a ? b (sign of
wave- function is opposite)
center of Inversion a b
b
b
s(1s)u p(2p)g
a
a
not a center of inversion a ? b
b
b
center of inversion a b
s(1s)g p(2p)u
The test for whether an MO is g or u is to find
the possible center of inversion of the MO. If
two lines drawn out at 180o to each other from
the center, and of equal distances, strike
identical points (a and b), then the orbital is g.
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Energy levels of the N2 molecule
see if you can decide which are g or are u, and
bonding or anti-bonding
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Energy levels of the N2 molecule(calculated
using semi-empirical MO theory)
s2pu
p2pg
s2pg
s2su
p2pu
s2sg
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To summarize

• A bonding molecular orbital has overlap of the
  two atomic orbitals, and has no nodal plane. An
  anti-bonding orbital has a nodal plane between
  the two atoms forming the bond.
• g orbitals have even parity, and have a center of
  inversion. u orbitals have odd parity and have no
  center of inversion
• In drawing up an MO diagram, you should fully
  label all atomic orbitals and MOs (indicate
  atomic orbital MO is derived from ( 1s, 2p,
  etc.), g or u, s or p, bonding or non-bonding ()
  ), indicate number of unpaired electrons,
  diamagnetic or paramagnetic, and bond order.