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 $R\sin 60^{\circ }=\sqrt{3}R/2$; boron is again on the axis. ThereforeThe 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 $\tilde{C}=\tilde{B}/2$.