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I've been asked to compare the relative acidity of $\ce{CF4}$ and $\ce{SiF4}$. As both atoms have noble gas configurations, neither can act as a lewis acid and so I assume the question is inferring that they are in solution.

My guess is that under aqueous conditions they could react via some sort of hydrolysis to give $\ce{H+}$ and $\ce{F-}$ alongside an alcohol in the case of carbon or $\ce{Si(OH)_{(4-x)}F_{x}}$ with silicon.

As the energy match between the $\ce{Si}$ and $\ce{F}$ HOMO and LUMO is worse this means that this is likely to be a weaker bond and so $\ce{SiF4}$ is more acidic? Is there an argument that could be made about the relative stability of the hydrolysis products?

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  • $\begingroup$ SiF4 can act as a lewis acid and they both can't hydrolyse. $\endgroup$ – Mithoron May 8 '15 at 14:05
  • $\begingroup$ So CF4 is not acidic at all and SiF4 would be slightly acidic? $\endgroup$ – Goods May 8 '15 at 14:16
  • $\begingroup$ Sorry, SiF4 does hydrolyse to form hexafluorosilicic acid - nzic.org.nz/ChemProcesses/production/1C.pdf $\endgroup$ – Mithoron May 8 '15 at 15:32
  • $\begingroup$ SiF4 can hydrolyse because of the presence of vacant d-orbitals. $\endgroup$ – Papul May 9 '15 at 20:04
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    $\begingroup$ $\ce{SiF4}$ does not hydrolyse due to the presence or absence of hypothetic d-orbitals. Rather, the $\ce{Si-F}$ bond can be cleaved more easily by nucleophilic oxygens. $\endgroup$ – Jan Jun 9 '15 at 21:06
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Both molecules can theoretically act as a Lewis acid. This would involve the approach of a Lewis base (i.e. a lone pair-carrying group), its interaction with the central atom and the formation of a four-electron-three-centre bond. These bonds with a bond order of $0.5$ can be understood by invoking the following two resonance structures:

$$\ce{\overset{-}{X}\bond{...}AF3-F <-> X-AF3\bond{...}\overset{-}{F}}$$

Where $\ce{A}$ is carbon or silicon in the context of this question. The structure that this arrangement would require is a trigonal bipyramid; the central atom is $\mathrm{sp^2}$ hybridised with its unhybridised p orbital forming the 4e3c bond to the electronegative groups.

You may immediately recognise this as the transition state of an $\mathrm{S_N2}$ reaction which is possible both with carbon and silicon centres.

These four-electron-three-centre bonds are typically more stable if the central atom has a lower and the terminal atoms have a higher electronegativity. If you compare interhalogen compounds, only iodine is able to form such bonds with chlorine ($\ce{(ICl3)2}$) while all other halogens can only form these bonds with fluorine (e.g. $\ce{ClF3}$). This already should allow us to conclude that $\ce{SiF4}$ is the stronger Lewis acid. But the discussion is not finished here.

We also need to discuss steric effects. Carbon tetrahalides are very crowded molecules — even with fluorine even though it is the smallest halogen. Giving up the tetrahedral structure of $\ce{CF4}$ to arrive at trigonal bipyramidal $\ce{CF4X-}$ would require the three fluorines that do not take part in the 4e3c bond to move closer together, increasing steric crowing even more. Indeed, the required interaction would be the Lewis base’s lone pair with $\sigma_{\ce{C-F}}^*$ — an orbital which is already well hidden by the three other fluorines. Thus, steric effects further reduce $\ce{CF4}$’s ability to be a Lewis acid to the point where we shouldn’t even be saying it be a weak Lewis acid: we should just call it Lewis inactive.

Silicon is larger than carbon and better able to stabilise 4e3c bonds due to its lower electronegativity. Thus, 4e3c bonds are sufficiently accessable and stable for silicon in $\ce{SiF4}$ to act as a Lewis acid. Depending on the nature of $\ce{X-}$, the pentacoordinated $\ce{SiF4X-}$ species may either break down under liberation of either $\ce{F-}$ ($\mathrm{S_NSi}$ substitution) or $\ce{X-}$ (reverse reaction) or — especially if $\ce{X} = \ce{F}$ — it may add another to give $\ce{[SiF4X2]^2-}$ like the well-known hexafluoridosilicate anion $\ce{[SiF6]^2-}$.

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