chemistry 445 lecture 4 molecular orbital theory of diatomic molecules n.
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Chemistry 445. Lecture 4. Molecular Orbital Theory of diatomic molecules. The non-existent He 2 molecule (bond order = 0). BO = (2-2)/2 = 0. The MO diagram for the He 2 molecule is similar to that for the H 2 molecule,

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chemistry 445 lecture 4 molecular orbital theory of diatomic molecules
Chemistry 445.

Lecture 4.

Molecular Orbital Theory

of diatomic molecules

the non existent he 2 molecule bond order 0
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 σ1s orbital is

negated because the other pair in the σ*1s rises in energy by an equal amount.

There is thus no net stabilization, and so the He2 molecule does not exist.

the he 2 molecule ion exists bond order
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.

the li 2 molecule bond order 1
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.

the non existent be 2 molecule bond order 0
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.

and bonding
σ and π bonding

In the molecules we have considered so far, only σ overlaps have

been of importance. In the formal definition, a σ–bond is one which lies along a rotational symmetry axis, (the rest of the molecule is ignored). A π–bond does not lie along a rotational axis (symmetry will be discussed later). In practical terms, a σ–bond lies along the bond connecting the two atoms, whereas a π–bond does not.

z

z

+

π*(pz)

pz

orbitals

anti-bonding π* MO

z

z

π(pz)

+

bonding π MO

o 2 molecule bond order 2
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

f 2 molecule bond order 1
F2 molecule, bond order = 1

F2 has no unpaired

electrons, and so

is diamagnetic

BO = (6-4)/2 = 1

slide9
Variation of the energies of the 2s and 2p orbitals in crossing the periodic table from Li to F. (H&S Fig. 1.22)
energy levels of first row homonuclear diatomic molecules h s fig 1 23
Energy levels of first-row homonuclear diatomic molecules (H&S Fig 1.23)

crossover point

Molecules Li2, Be2, B2,C2 and N2

have π(2p) lower in energy than σ(2p)

Molecules O2, and F2 have π(2p) higher in energy than σ(2p)

n 2 molecule bond order 3
N2 molecule, bond order = 3

diamagnetic

BO = (6-0)/2 = 3

N N2 N

c 2 molecule bond order 2
C2 molecule, bond order = 2

diamagnetic

BO = (4-0)/2 = 2

singlet oxygen 1 o 2
Singlet oxygen (1O2)

BO = (6-2)/2 = 2

Singlet Oxygen

is an excited

state of the

ground state

triplet 3O2

molecule. It is much more reactive, and will readily attack organic

molecules.

O O2 O

The O2 molecule in its excited singlet state which is 25 kcal/mol in energy above the ground triplet state. Irradiation with IR light causes excitation to the singlet state, which can persist for hours because the spin-selection rule (see later) inhibits transitions that involve a change of spin state.

orbital parity gerade g and ungerade u
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 can occur 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)

orbital parity gerade g and ungerade u1
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.)

not a

center of inversion

a ≠ b

a

a

center of

inversion

a = b

a

b

b

b

s-orbital p-orbital d-orbital

gerade (g) ungerade (u) gerade (g)

parity g or u of molecular orbitals
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

σ*(1s)u π*(2p)g

a

a

not a center of inversion

a ≠ b

b

b

center of

inversion

a = b

σ(1s)g π(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.

energy levels of the n 2 molecule
Energy levels of the N2 molecule

see if

you can

decide

which are g

or are u, and

bonding or

anti-bonding

energy levels of the n 2 molecule calculated using semi empirical mo theory
Energy levels of the N2 molecule(calculated using semi-empirical MO theory)

σ*2pu

π*2pg

σ2pg

σ*2su

π*2pu

σ2sg

labeling molecular orbitals as g or u
Labeling molecular orbitalsas g or u:

The following little table will help you to label molecular orbitals as g or u. For σ-overlap, the bonding orbitals are g, while the antibonding orbitals are u, while for π–overlap the opposite is true:

bonding MO anti-bonding MO

σ-bonding g u

π-bonding u g

to summarize
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 MO’s (indicate atomic orbital MO is derived from ( 1s, 2p, etc.), g or u, σorπ, bonding or non-bonding (*) ), indicate number of unpaired electrons, diamagnetic or paramagnetic, and bond order.