 Solution

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

2 B = 0.686 cm^{- 1}, B = 0.343 cm^{- 1} = 34.3 m^{- 1} .

Using

I_{b} = h / (8 π^{2} c B)

with the wavenumber in inverse metres,

I_{b} = \frac{6.62607015 \times 1 0 ^{- 34}}{8 π ^{2} ( 2.99792458 \times 1 0 ^{8} ) ( 34.3 )} = 8.16115 \times 1 0^{- 46} kg m^{2} .

m_{F} = 19 u

and

I_{b} = 3 m_{F} R^{2} /2

R = \frac{2 I _{b}}{3 ( 19 u )} = 1.31319 \times 1 0^{- 10} m, R ≃ 131 pm .

Using

4 B

for the observed spacing would incorrectly treat BF

_{3}

as a linear diatomic rotor and overestimate the bond length by

2

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