Chemistry - Gauche effect and bond length

Solution 1:

Actually, the more the polarization of a bond, the stronger it is, since both the atoms are drawn close together due to electrostatic attraction, decreasing the bond length and increasing the bond stability (think about ionic bonds. Your reasoning would dictate them to be very weak, while we know how strong they are!). Actually, I would predict $\ce{CH3-CF3}$ to have a really small bond length (the bond is probably the strongest, would someone confirm this please?), since the fluorinated carbon has quite a positive charge, while the other one is relatively negatively charged (not negative in the absolute, just relative to the other carbon).

All the remaining options have equal number of fluorines on either carbon, differing only in their number. These three are tricky. Consider $\ce{CF3-CF3}$. Both of its carbons are severely deficient in electrons. Each badly needs electrons, and since it can't argue with fluorine (fluorine is an unreasonable bully), so both carbons try to drag the shared pair of electrons towards itself. This ultimately results in them both coming closer to each other. As we reduce the number of fluorines on each carbon, their desparation is reduced, thus increasing the bond length. Thus the answer is (B).

Edit: If @pentavalentcarbon's calculations of bond lengths are correct, then my reasoning was too simplistic, and indeed it seems so now. So if we dig deeper, $\ce{CH2F-CH2F}$ could have the second smallest bond length and $\ce{CF3-CF3}$. Why? Well, it is because of hyperconjugation, similar to the gauche effect (read it up to get better understanding of the gauche effect, although I explain it in short here also).

So, what happens is that in the case of $\ce{CH2F-CH2F}$, the bonding C-H molecular orbital (MO) overlaps with the antibonding C-F MO, transferring some of its electron density to the C-F MO (see diagram). This interaction is stabilizing (since delocalization of electrons stabilizes the molecule), and also introduces slight double bond character in the C-C sigma bond, thus serving to shorten it.

Now in $\ce{CH2F-CH2F}$, there will be two such interactions, but then there are also two such interactions in $\ce{CHF2-CHF2}$! There is a difference though. In the latter case, there are two fluorines on each carbon, which heavily draw electron density away from the carbon through the sigma bonds. This decreases the sigma bond density in the C-H bond, which being poor in electrons, gives relatively low stabilization when overlapping with the antibonding C-F MO. So $\ce{CHF2-CHF2}$ has a slightly higher bond length than $\ce{CH2F-CH2F}$.

Using this same reasoning, we can see why $\ce{CH3-CF3}$ has the smallest bond length (three favourable hyperconjugational interactions!) and $\ce{CF3-CF3}$ has the longest bond length (no hyperconjugation).

Solution 2:

Many of these cases hinge on some weird effect. Evidently the gauche form of 1,2-Difluoroethane is more stable than the anti form which is sort of weird.

This wikipedia article on the Gauche Effect list CC bond of gauche 1,2-Difluoroethane as 150 pm and anti form as 151.4 pm.

This reference says 151.2 pm for C-C bond in 1,1,2,2-tetrafluoroethane. (High-Resolution Microwave and Infrared Molecular-Beam Studies of the Conformers of 1,1,2,2-Tetrafluoroethane, Stephen C. Stone et al., Journal of Molecular Spectroscopy 192, 75–85 (1998).)

This reference says the C-C bond length is shorter for hexafluoroethane than then the corresponding hydrocarbon. (Fluorocarbon and Related Chemistry By R. E. Banks, M. G. Barlow, ISBN-13: 978-0851865140, page 10.)

This answer on this website says ethane C-C bond is 153.51 pm.

So to summarize: $$ \small \begin{array}{lcc} \hline \text{Compound} & \text{Bond Length (pm)}\\ \hline \text{gauche 1,2-difluoroethane} & 150 \\ \text{anti 1,2-difluoroethane} & 151.4 \\ \text{1,1,2,2-tetrafluoroethane} & 151.2 \\ \text{Hexafluoroethane} & \text{<ethane C-C} \\ \text{Ethane} & 153.51 \\ \hline \end{array} $$