 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-1/a3/b/ii/solution

In the printed geometry, one B–F bond lies along the in-plane axis and contributes zero. The other two fluorines have perpendicular distance

R sin 6 0^{\circ} = 3 R /2

; boron is again on the axis. Therefore

I_{b} = 2 m_{F} (\frac{3 R}{2})^{2} = \frac{3}{2} m_{F} R^{2} = \frac{I _{c}}{2} .

The other in-plane principal moment is the same by threefold symmetry. Hence Boron trifluoride is an oblate symmetric top with

I_{c} > I_{b}

and

C = B /2

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