 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2a/section-b/23/b/ii/solution

For the collinear centrosymmetric arrangement, the longitudinal coordinates transform in the D infinity h molecular point group. Inversion sends

q_{i} \mapsto - q_{5 - i}

, because displacement is a polar vector. Therefore

Γ_{z} = 2 Σ_{g}^{+} + 2 Σ_{u}^{+} .

Coordinate	Displacement vector	Symmetry	$ω^{2}$ at $γ = 0$	$t$	$(1, 1, 1, 1) /2$	$Σ_{u}^{+}$	$0$	$r$	$(- 1, - 1, 1, 1) /2$	$Σ_{g}^{+}$	$0$	$s$	$(- 1, 1, - 1, 1) /2$	$Σ_{g}^{+}$	$2 k / m$	$a$	$(- 1, 1, 1, - 1) /2$	$Σ_{u}^{+}$	$2 k / m$

Here

t

translates the complete system,

r

separates the molecular centres,

s

stretches both molecules together, and

a

lengthens one while shortening the other. Reversing every arrow changes the coordinate sign, not its physical mode.

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