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314 9MOLECULAR STRUCTURE
other pairs of neighbours. One of the e2u orbitals, the one on the le� in Fig. 9E.4
onpage 377, is antibonding between the same twopairs of neighbouring carbon
atoms as above, and nonbonding between all other pairs of neighbours.�ere-
fore promotion of an electron between these two orbitals decreases the bond
strength between these two pairs of neighbours, and leaves the bond strength
unchanged between all other pairs.
However, the situation ismore complicated due to the degeneracy of theHOMO
and LUMO. Promotion of an electron between all possible pairs of HOMO and
LUMO have to be considered together. Due to the six-fold rotational symme-
try of benzene, the bond strength between each pair of atoms is expected to
change the same way.�e HOMO of benzene is overall bonding, whereas the
LUMO is antibonding, therefore promotion of an electron from the HOMO
to the LUMO will, in fact, decrease the bond strength between all pairs of
neighbouring carbon atoms in the molecule.
P9E.8 To estimate if the given species are stablewith respect to dissociation into smaller
entities, the standard internal energy change for each dissociation reaction is
calculated and its sign examined.�is change in internal energy is given by the
change in the total electron binding energies (Etot).
�e energies of the molecular orbitals are calculated using the expression given
in Problem P9E.2. �e occupied orbitals are identi�ed and with this infor-
mation it is then possible to calculate the total electron binding energy of the
hydrogen ring compounds, as in the table below (VE is the number of valence
electrons)
species VE energies of occupied MOs Etot
H4 4 α + 2β, α (degenerate) 4α + 4β
H5+ 4 α + 2β, α + 0.618β (degenerate) 4α + 5.24β
H5 – 6 α + 2β, α + 0.618β (degenerate) 6α + 6.47β
H6 6 α + 2β, α + β (degenerate) 6α + 8β
H7+ 6 α + 2β, α + 1.25β (degenerate) 6α + 9β
H2 2 α + β 2α + 2β
H3+ 2 α + 2β 2α + 4β
H– 2 α 2α
�e table includes some other species of interest in these calculations. Note that
in ProblemP9E.7 it is shown thatH3+ is lower in energy thanH++H2, therefore
the positively charged hydrogen ring compounds are more likely to dissociate
into H2 and H3+, than to H2 and H+. �e negatively charged hydrogen ring
compounds are likely to give H2 and H− as dissociation products.
�e stability of each species is examined in turn.
(i) �e dissociation of H4 is likely to give two H2 molecules, H4 ÐÐ→ 2H2.
�erefore ∆rU−○ = Etot(products)− Etot(reactants) = 2(2α + 2β)− (4α +