Q19 / 200 · Manipal (MET) 2014 · Chemistry
Order the species $\mathrm{CF}^{+}, \mathrm{CF}$ and $\mathrm{CF}^{-}$according to increasing $\mathrm{C}-\mathrm{F}$ bond length
- Correct answerA$\mathrm{CF}^{+} \lt \mathrm{CF} \lt \mathrm{CF}^{-}$
- B$\mathrm{CF}^{+} \lt \mathrm{CF}^{-} \lt \mathrm{CF}$
- C$\mathrm{CF}^{-} \lt \mathrm{CF} \lt \mathrm{CF}^{+}$
- D$\mathrm{CF} \lt \mathrm{CF}^{-} \lt \mathrm{CF}^{+}$
Solution
The molecular orbital electronic configuration of $\mathrm{CF}(15 \bar{e})$ is $\sigma 1 s^2, \quad{ }^* 1 s^2 \sigma 2 s^2 \quad{ }^* 2 s^2 \sigma 2 p_z^2$ $\pi 2 p_x^2 \approx \pi 2 p_y^2, \pi 2 p_x^1$
Bond order $(\mathrm{BO})=\frac{10-5}{2}=2.5$
In $\mathrm{CF}^{+}$, there is one electron less in antibonding MO therefore, $\mathrm{BO}=\frac{10-4}{2}=3$
In $\mathrm{CF}^{-}$, there is one electron more in antibonding MO therefore, $\mathrm{BO}=\frac{10-6}{2}=2$
Bond length $\propto \frac{1}{\text { Bond order }}$
Hence, the order of bond length in given series is $\mathrm{CF}^{+} \lt \mathrm{CF} \lt \mathrm{CF}^{-}$.
Hence, the correct answer is option: 'A'.