Question

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

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

Answer 1

Answer: (A)

The molecular orbital electronic configuration of $CF(15\bar{e})$ is
$\sigma 1s^2,\stackrel{\ast }{\sigma }1s^2\sigma 2s^2\stackrel{\ast }{\sigma }\sigma 2p_z^2$
$\pi 2p_x^2\approx \pi 2p_y^2,\stackrel{\ast }{\pi }2p_x^1$
Bond order $(BO)=\frac{10-5}{2}=2.5$
In $C{F}^{+}$, there is one electron less in antibonding MO therefore,
$BO=\frac{10-4}{2}=3$
In $C{F}^{-}$, 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 $C{F}^{+}<CF<C{F}^{-}$.