Construct the valence molecular orbital diagram for four H atoms

Construct the valence molecular orbital diagram for four H atoms arranged in a square as shown below.
Then construct a correlation diagram by stretching r2. Indicate if the MOs go up or down in energy and
explain why. Discuss when a square or a rectangular structure will be more stable.
z
r1
r2
x
y
Forming a MO diagram
1.
2.
3.
4.
5.
6.
7.
8.
9.
determine the molecular shape and identify the point group of the molecule
define the axial system find all of the symmetry operations on the molecule
identify the chemical fragments, and put them along the bottom of the diagram
determine the energy levels and symmetry labels of the fragment orbitals
combine fragment orbitals of the same symmetry, estimate the splitting energy and draw in the MO energy levels and
MOs (in pencil!)
determine the number of electrons in each fragment and hence the central MO region
identify if any MO mixing occurs, determine the mixed orbitals and redraw the MO diagram with shifted energy levels
and the mixed MOs
use the MO diagram check-list!
analyse the MO diagram
shape has been given, but start with the high symmetry structure point group is D4h
axial system has been defined, symmetry operations
C"2(y)
!v(1)
!d(yz)
C'2
!v(2)
y
!d(xz)
i
x
C"2(x)
z
!h(xy)
C4(z), C2, S4
C'2
this is pretty easy to construct from two H2 molecules interacting side on
how do you know the middle two are degenerate?
1. they have the same phase pattern
2. one has the phase pattern of x-axis, the other of the y-axis, in D4h these are degenerate and eu
b2g
eu
H1sAO
a1g
D4 h
C4(z)
r1
y
y
r2
x
H
H
H
H
H
H
x
H
H
start with the high symmetry geometry and then distort looking at what happens to the overlap of the
orbitals.
like dxy b2g
y
antibonding interactions
decrease: lower energy
b1g
x
bonding interactions
decrease: rise in energy
b2u
like x or y axis
eu
antibonding interactions
decrease: lower energy
b3u
totally
symmetric
a1g
bonding interactions
decrease: rise in energy
a1g
r1=r2
r2>>r1
distance r2
D2h
D4 h
C4(z)
r1
C2(z)
r2
x
y
y
x
r1>>r2
rectangular structure is likely to be more stable if the stabilisation of the b3u orbital exceeds the
destabilisation of the a1g orbital. H4- (anion) however could be expected to be more stable as a square
structure due to destabilisation of the b2u orbital.
Extension.
Determine the MO diagram if a TM is placed in the center of the square formed by the four H atoms.
The dAO fragment orbitals will be nearly degenerate with the antibonding FO from the H4 fragment.
Will a TM with 4d electrons be stable?
b2g
larger contribution from
dz2 AO because it is
higher in energy
a1g
FO contributions are equal
because the fragments
orbitals are degenerate,
splitting is large for the
same reason
dz2 -> a1g
dyz -> eg
b2g
eg, b1g
dxz -> eg
y
dxy -> b2g
eu
x
dx2-y2 -> b1g
eu
b2g
a1g
D4 h
C4(z)
r1
larger contribution from
H4 AOs because they are
lower in energy
a1g
r2
x
looking down
dz2 AO
y
Yes a square planar species with 4 electrons in the dAOs will be stable. In addition, since the LUMO is
the non-bonding dAOs more electrons can be added without substantially destabilising the molecule.

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