Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2B / Section B / 36 / a

The drawn

x, y

axes pass through opposite sites, rather than along the square edges. In the D4h molecular point group class order

(E, 2 C_{4}, C_{2}, 2 C_{2}^{'}, 2 C_{2}^{''}, i, 2 S_{4}, σ_{h}, 2 σ_{v}, 2 σ_{d})

, an s orbital contributes

+ 1

whenever its centre is fixed and zero otherwise. Thus

χ_{Γ} = (4, 0, 0, 2, 0, 0, 0, 4, 2, 0) .

(1)

For example, a transverse twofold rotation about the printed

x

axis fixes sites 1 and 3, while its diagonal counterpart exchanges all four sites. Reflection in the square plane fixes every site. Taking character inner products, with class sizes

(1, 2, 1, 2, 2, 1, 2, 1, 2, 2)

, gives

Γ = A_{1 g} \oplus B_{1 g} \oplus E_{u} .

(2)

These have dimensions

1 + 1 + 2 = 4

. The

B_{1 g}

label is tied to the printed diagonal-axis convention; rotating the transverse coordinate labels by

4 5^{\circ}

would interchange

B_{1 g}

and

B_{2 g}

Solution

 0  0

Treat the site atomic orbitals

s_{1}, \dots, s_{4}

as orthonormal, as in the Hückel method. A normalized set of symmetry-adapted linear combinations is

Φ_{A_{1 g}} Φ_{B_{1 g}} Φ_{E_{u}, x} Φ_{E_{u}, y} = \frac{1}{2} (s_{1} + s_{2} + s_{3} + s_{4}), = \frac{1}{2} (s_{1} - s_{2} + s_{3} - s_{4}), = \frac{1}{2} (s_{3} - s_{1}), = \frac{1}{2} (s_{2} - s_{4}) .

(1)

The first combination is unchanged by every operation; the alternating combination has the transformation signs of

x^{2} - y^{2}

for the printed axes. The last two transform together like

(x, y)

. The squared coefficients sum to one in each case, and their pairwise inner products vanish. Any orthogonal rotation of the two

E_{u}

combinations is an equally valid basis for that degenerate subspace.

Figure 1.
Original coefficient and phase maps for the four square-array symmetry orbitals
. **Details**.

Solution

 0  0

Use a common Coulomb integral

α

on each site, a resonance integral

β

on each adjacent pair, zero interaction across a diagonal, and the identity overlap matrix. In site order

(1, 2, 3, 4)

the Hückel method gives

H = α β 0 β β α β 0 0 β α β β 0 β α .

(1)

Applying it to the four symmetry-adapted linear combinations gives

E_{A_{1 g}} = α + 2 β, E_{E_{u}} = α (twice), E_{B_{1 g}} = α - 2 β .

(2)

For the conventional bonding sign

β < 0

, the all-in-phase combination is lowest and the alternating combination is highest. The

E_{u}

combinations have canceling nearest-neighbour interactions and are nonbonding within this model.

Solution

 0  0

The molecular orbital diagram has one level at

α + 2 β

labelled

A_{1 g}

, two degenerate levels at

α

labelled

E_{u}

, and one level at

α - 2 β

labelled

B_{1 g}

, in increasing energy for

β < 0

. Each spatial molecular orbital admits two opposite-spin electrons. The diagram below also shows how one representative distortion lifts the middle degeneracy; the distorted scale is illustrative, not a fitted molecular spectrum.