=> Write down the Hiickel secular determinant for the acyclic three C-atoms in terms of x, where x = (α – E)/ β.

=> Solve the secular determinant for x and then rewrite the solution in terms of (α – E)/ β and so find the three allowed quantized energies (eigenvalues).

=> Evaluate the normalized wavefunction (eigenvector) for each energy (eigenvalue).

=> Which of the wavefunctions is the HOMO, LUMO or SOMO (singly occupied MO) as appropriate, for each of the allyl cation, radical and anion molecules?

**Solution**

Since α and β are both have negative values so with increasing energy the eigenvalues are the following

**E _{1 }= α + √2 β**

**E _{2} = α**

**E _{3} = α – √2 β**

To find the three eigenvectors (wavefunctions) corresponding to these three eigenvalues we must separately solve the Bickel matrix for ally’ for each energy For E_{1 }= α + √2 β then (α – E_{1}) = – √2 β.

Subtracting the third equation from the first gives c_{1} = c_{3}. From the first equation c_{2} = √2 c_{1}, and from the third equation c_{2} = √2 c_{3} which when we normalize the coefficients using c_{1}^{2} + c_{2}^{2}+ c_{3}^{2} = 1 gives

From the first and third simultaneous equations c_{2} = 0, and from second equation, c_{1} = – c_{3} If the coefficients are normalized, then c_{1}^{2} + c_{2}^{2} + c_{3}^{2} = 1 then,

Substituting into the Hiickel matrix for ally) with E_{3} = α – √2 β then (α – E_{3}) = √2 β

Subtracting the third from the first simultaneous equations c_{1} = c_{3} from the first equation c_{2} = √2 c_{1}; from the third equation c_{2} = – √2 c_{3} thus when the coefficients are normalised, then c_{1}^{2} + c_{2}^{2} + c_{3}^{2} = 1 we have,

Below Figure shows the MO energy diagrams for the three allyl species.

In the allyl cation CH_{2}=CH—CH_{2}^{+} with 2 π-electrons, then ψ_{1} is the HOMO and ψ_{2} is the LUMO. In the allyI radical CH_{2}=CH—CH_{2}^{+} with 3 π-electrons, then ψ_{1} has 2 π-electrons and ψ_{2} is the SUMO with 1 π-electron and ψ_{3} is the LUMO. In the allyl anion CH_{2}=CH—CH_{2}^{–} with 4 π-electrons, then ψ_{1} has 2 π -electrons and ψ_{2} is the HOMO with 2 π-electrons and ψ_{3} is the LUMO.