 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2025/paper-1/a4/b/ii/solution

For operations ordered as

(E, C_{2} (z), σ_{x z}, σ_{yz})

, the characters of the twelve Cartesian displacements are

χ_{3 N} = (12, - 2, 4, 2) .

Only atoms left fixed contribute: all four for

E

and

σ_{x z}

, but only C and O for

C_{2}

and

σ_{yz}

. Multiplying their counts by the Cartesian traces

3, - 1, 1, 1

gives these numbers. Subtract

χ_{trans} = (3, - 1, 1, 1), χ_{rot} = (3, - 1, - 1, - 1)

to obtain

χ_{vib} = (6, 0, 4, 2)

. The character table reduction gives

Γ_{vib} = 3 A_{1} + 2 B_{1} + B_{2} .

6 = 3 N - 6

, as required. The

B_{2}

mode is the out-of-plane wag with this axis convention.

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