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When comparing the acidic strength of the halogen acids, we see that the acidic strength decreases as $\ce{HI>HBr>HCl>HF}$. It is said that the reason for this is the bond strength of $\ce{HI}$ is the least, while that of $\ce{HF}$ is the greatest. But, the polarity of the bond in $\ce{HF}$ is the greatest, how then is the strongest bond?

I understand that fluorine has smaller valence atomic orbital (2p) than, for example, iodine (5p), and so has more effective overlap with the s-orbital of hydrogen. But how do we reconcile this fact with the fact that the fluorine hydrogen bond is more polar too? The electrons remain more with fluorine in $\ce{HF}$ than with iodine in $\ce{HI}$, so intuitively it seems that the hydrogen fluorine bond may be more easily broken by a base than the hydrogen iodine bond. So what is going on?

In short, how can a bond be stable when it is highly polar?


Edit

My question is regarding bond strength, and not the acidic order. For example, in organic compounds, the leaving group ability of the halide ions decreases as$^1$: $$\ce{I^{-}>Br^{-}>Cl^{-}>>F^-}$$ This means that the fluorine-carbon bond is the hardest to break. But fluorine has very high electronegativity, and so the bond must be highly polar. But inspite of this, fluorine has the highest bond strength!

So is this just a misconception I have? If it isn't a misconception, then why does the highly polar bond have the maximum bond strength?

Thank you!

$^1$: https://chem.ucr.edu/documents/curricularmaterials/neumantextbook/Chapter7.pdf

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marked as duplicate by M.A.R., Jon Custer, Todd Minehardt, ringo, Curt F. Aug 5 '16 at 0:49

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  • $\begingroup$ Why wouldn't it, really? When you break a highly polar bond, you're left with two ions which are attracted to each other with great force. $\endgroup$ – Ivan Neretin Aug 4 '16 at 6:59
  • $\begingroup$ @IvanNeretin Hmm, then when comparing acidities of compunds across a period, we say that because of increasing polarity of the bond with hydrogen (moving from right to left), the compound becomes more acidic? ($\ce{HF>H2O>NH3>CH4}$) $\endgroup$ – FreezingFire Aug 4 '16 at 7:17
  • $\begingroup$ Sounds true to me. $\endgroup$ – Ivan Neretin Aug 4 '16 at 7:32
  • $\begingroup$ @IvanNeretin I agree that what you say makes sense, but we use exactly the opposite explanation while comparing acidities in a period! What is the correct explanation for that, then? $\endgroup$ – FreezingFire Aug 4 '16 at 9:10
  • $\begingroup$ @Mithoron I believe my question is different (I have checked your suggested duplicate). Please check my edit and guide me if I can put it in any better words! $\endgroup$ – FreezingFire Aug 4 '16 at 14:16
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The important thing here to keep in mind is that the bond strength is not a spherical chicken in vacuum, meaning that we need to keep the entire process in mind when analyzing all the factors that can influence it. Although the fluorine has a higher electronegativity, it is significantly smaller than Cl, Br, or I (same goes to their corresponding ions). And without going too deep into intricacies, $\ce{F-}$ is $way$ less stable than $\ce{I-}$ and thus, using the OChem language, a much better nucleophile. Due to it's inability to stabilize the charge, it holds onto the hydrogen with all it's might (pardon me for anthropomorphizing the atoms).

SO, when you're thinking about breaking $\ce{H-F}$ bond, you gotta take into consideration what happens with the products. In the example with organic chemistry and leaving groups that you have pointed out, fluorine is the worst leaving group for the precisely the same reason--it cannot stabilize the negative charge well, so it doesn't "want" to leave. However, if you try to do a, say, $S_N2$ reaction in a protic solution (not the best way to go around, but doable), then $\ce{F-}$ will be a much better leaving group than $\ce{I-}$, b/c $\ce{F-}$ will instantly "catch" a hydrogen from the solution and stop being nucleophilic, which according to Le'Chatelie'r principle shifts the equilibrium to where $\ce{F-}$ is a leaving group.

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  • $\begingroup$ Thank you for your answer! You mentioned that $\ce{F}^-$ ion is less stable. Is it because of the repulsion between its lone pairs, or due to high charge density (making it attractive to electrophiles)? Or is it something else entirely? $\endgroup$ – FreezingFire Aug 5 '16 at 9:32
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    $\begingroup$ @FreezingFire: if explaining it "on fingers" without going into the depths of quantum theory: yes, and yes. Having a high density of a charge (either positive or negative) is never a good thing, which make the corresponding species unstable. $\endgroup$ – ChemistryHelpCenter Aug 6 '16 at 2:38

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