Question

Order the species $CF ^{+}, CF$ and $CF ^{-}$according to increasing $C - F$ bond length

• A. $CF ^{+}< CF < CF ^{-}$
• B. $C F^{+} < C F^{-} < C F$
• C. $CF ^{-}< CF < CF ^{+}$
• D. $CF < CF ^{-}< CF ^{+}$

Answer 1

Answer: (A)

The molecular orbital electronic configuration of $CF (15 \bar{e})$ is
$\sigma 1 s^{2}, \overset{*}{\sigma}1 s^{2} \sigma 2s^2\overset{*}{\sigma} \sigma 2 p_{z}^{2}$
$\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \overset{*}{\pi} 2 p_{x}^{1}$
Bond order $( BO )=\frac{10-5}{2}=2.5$
In $CF ^{+}$, there is one electron less in antibonding MO therefore,
$BO =\frac{10-4}{2}=3$
In $CF ^{-}$, there is one electron more in antibonding $MO$ therefore,
$BO =\frac{10-6}{2}=2$
Bond length $\propto \frac{1}{\text { Bond order }}$
Hence, the order of bond length in given series is $CF ^{+}< CF < CF ^{-}$.