Benzene Hückel Molecular Orbitals

In Hückel molecular orbital theory, the pi (π) molecular orbitals of benzene (a cyclic system with 6 carbon atoms) are linear combinations of the atomic p orbitals (\(\chi_i\)) of the carbon atoms.

The energy levels for the molecular orbitals in benzene are given by:

• \(\alpha + 2\beta\) (lowest energy, bonding, non-degenerate)
• \(\alpha + \beta\) (bonding, degenerate pair)
• \(\alpha - \beta\) (antibonding, degenerate pair)
• \(\alpha - 2\beta\) (highest energy, antibonding, non-degenerate)

The molecular orbitals (\(\psi\)) corresponding to these energy levels can be expressed as linear combinations of the atomic orbitals (\(\chi_i\)), \(\psi_k = \sum_{i=1}^{6} c_{ki} \chi_i\), where \(c_{ki}\) are the coefficients for the \(k\)-th molecular orbital on the \(i\)-th atom.

The given molecular orbital is:

\(\psi_{given} = \frac{1}{2}(\chi_2 + \chi_3 - \chi_5 - \chi_6)\)

The coefficients for this MO are (0, 1/2, 1/2, 0, -1/2, -1/2) for \(\chi_1\) through \(\chi_6\).

Degenerate molecular orbitals are orbitals that have the same energy. In benzene, there are two sets of degenerate orbitals at energies \(\alpha + \beta\) and \(\alpha - \beta\). Orbitals within a degenerate set are orthogonal to each other.

The standard Hückel molecular orbitals for benzene are (normalized):

• \(\psi_1\) (\(\alpha + 2\beta\)): \(\frac{1}{\sqrt{6}}\)(\(\chi_1 + \chi_2 + \chi_3 + \chi_4 + \chi_5 + \chi_6\))
• \(\psi_2\) (\(\alpha + \beta\)): \(\frac{1}{\sqrt{12}}\)(2\(\chi_1\) + \(\chi_2\) − \(\chi_3\) − 2\(\chi_4\) − \(\chi_5\) + \(\chi_6\))
• \(\psi_3\) (\(\alpha + \beta\)): \(\frac{1}{2}\)(\(\chi_2\) + \(\chi_3\) − \(\chi_5\) − \(\chi_6\))
• \(\psi_4\) (\(\alpha - \beta\)): \(\frac{1}{2}\)(\(\chi_2\) − \(\chi_3\) + \(\chi_5\) − \(\chi_6\))
• \(\psi_5\) (\(\alpha - \beta\)): \(\frac{1}{\sqrt{12}}\)(2\(\chi_1\) − \(\chi_2\) − \(\chi_3\) + 2\(\chi_4\) − \(\chi_5\) − \(\chi_6\))
• \(\psi_6\) (\(\alpha - 2\beta\)): \(\frac{1}{\sqrt{6}}\)(\(\chi_1\) − \(\chi_2\) + \(\chi_3\) − \(\chi_4\) + \(\chi_5\) − \(\chi_6\))

Comparing the given molecular orbital \(\frac{1}{2}(\chi_2 + \chi_3 - \chi_5 - \chi_6)\) with the standard forms, we see it matches \(\psi_3\). The orbital \(\psi_3\) is part of the degenerate pair at the \(\alpha + \beta\) energy level. The other orbital in this degenerate pair is \(\psi_2\).

Now let's look at the options provided and compare them to \(\psi_2\):

• Option 1: \(\frac{1}{\sqrt{12}}\)(2\(\chi_1\) + \(\chi_2\) − \(\chi_3\) − 2\(\chi_4\) − \(\chi_5\) + \(\chi_6\)). This matches \(\psi_2\).
• Option 2: \(\frac{1}{2}\)(\(\chi_2\) − \(\chi_3\) + \(\chi_5\) − \(\chi_6\)). This matches \(\psi_4\) (at energy \(\alpha - \beta\)).
• Option 3: \(\frac{1}{\sqrt{12}}\)(2\(\chi_1\) − \(\chi_2\) − \(\chi_3\) + 2\(\chi_4\) − \(\chi_5\) − \(\chi_6\)). This matches \(\psi_5\) (at energy \(\alpha - \beta\)).
• Option 4: \(\frac{1}{\sqrt{6}}\)(\(\chi_1\) − \(\chi_2\) + \(\chi_3\) − \(\chi_4\) + \(\chi_5\) − \(\chi_6\)). This matches \(\psi_6\) (at energy \(\alpha - 2\beta\)).

The molecular orbital that is degenerate with \(\frac{1}{2}(\chi_2 + \chi_3 - \chi_5 - \chi_6)\) (which is \(\psi_3\)) is \(\psi_2\). Comparing this with the options, we find that Option 1 is \(\psi_2\).

We can also verify orthogonality. The given MO \(\psi_{given}\) has coefficients (0, 1/2, 1/2, 0, -1/2, -1/2). Option 1 MO \(\psi_1\) has coefficients \(\frac{1}{\sqrt{12}}\)(2, 1, -1, -2, -1, 1).

The dot product of the coefficient vectors should be zero for orthogonal orbitals:

\(\sum c_{given, i} c_{1, i} = (0)(\frac{2}{\sqrt{12}}) + (\frac{1}{2})(\frac{1}{\sqrt{12}}) + (\frac{1}{2})(\frac{-1}{\sqrt{12}}) + (0)(\frac{-2}{\sqrt{12}}) + (\frac{-1}{2})(\frac{-1}{\sqrt{12}}) + (\frac{-1}{2})(\frac{1}{\sqrt{12}})\)

\(= 0 + \frac{1}{2\sqrt{12}} - \frac{1}{2\sqrt{12}} + 0 + \frac{1}{2\sqrt{12}} - \frac{1}{2\sqrt{12}} = 0\)

Since their dot product is zero, they are orthogonal. As they are known to be part of the same energy level (\(\alpha + \beta\)), they are the degenerate pair.

Therefore, the molecular orbital degenerate with the given orbital is the one in Option 1.