E=α ± 1.62β, α± 0.62β......... (1)
The orbitals and their energy is below the fig. The greatest number of internulear nodes, the higher energy of the orbitals... There is four electron to accommodate the ground state configuration is 1π² 2π². The front orbitals of butadiene are the 2π orbitals (HOMO) Which is largely bonding and the 3π orbitals (LUMO) which is largely antibonding. The Largely bounding means orbitals has both bonding and antibonding interactions between various neighbours but bonding effect is dominate. The largely antibonding indicates but that antibonding effects dominate.
The ver important point emerges when we can calculate the total π- electron binding energy, The Eπ sum of the energy's of each π electron and compare it with that we find in ethene. The total energy of ethene is, Eπ= 2(α+β) =2α + 2β
The Huckel molecular orbital energy level of butadiene and top view of the corresponding π orbitals. The four p electrons occupy the two lower π orbitals, note that the orbital are delocalized.
Eπ=2(α+1.6β) +2(α+0.62β) =4α+4.48β
Butadiene molecular energy is lies lower by 0.48β(110kj mol-¹) the sum of the two individual π electrons bonds. The stabilization of the conjugated system compared with a set of localized π bonds is called the delocalization energy of the molecules.
Closely related quantity is the π bond formation energy, Ebf, is the energy released when π bond is formed, but contribution of α is the same molecule as in the atoms, The we find the π bond formation energy from the π electron the binding energy by writing,
Ebf=Eπ-Ncα,
Where Nc is the number of carbon atoms in the molecule and π- bond formation energy in butadiene, for instance is 4.48β.
Delocalization energy :_
Using the Huckel approximation and find the energies of the π bond orbitals of cyclobutadiene, and estimate the delocalization energy.
Fixing the secular determinant using the same basis for butadiene, but note that atoms A and D are also neighbours, then solve for root of the secular equation and assess the total π bond energy. The delocalization energy is, subtract from the total π bond energy, the energy of two π-bonds
The hamiltonian matrix is,
(α β 0 β
β α β 0
H= 0 β α β
β 0 β α)
(0 1 01
H=(α1+β(1 0 1 0→ Diagonalize
0 1 0 1
1 0 1 0)
( 2 0 0 0
0 0 0 0
0 0 0 0
0 0 0 - 2)
Energies gives of the diagonalization orbitals as
E= α+2 β, α, α, α - 2β
Four electrons must be a accommodated. Two occupy the lowest orbitals of energy α+ 2β, and two occupy the doubly degenerate orbitals of energy α
The total energy is 4α+ 4, β.
Then two isolated π bonds would have energy 4α+ 4β,
But in this case, the delocalization energy is zero.
Super explanation
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