« Stoichometry with gases |
Since fluorine is the most electronegative atom, why isn't HF a strong acid, too? The HCl bond should be less polarized than the HF bond.
Dear 5110 student,
You are right, F is more electronegative than Cl, and if the only factor were electronegativity, HF would be stronger than HCl. But it isn't, HF is a weak acid. There is a competing factor.
The size of the halogens (or any group) changes more dramatically than their electronegativities as you go up and down the group. The F- ion is tiny compared to the Cl- ion, so negative charge is concentrated on this tiny ball, hating itself, so to speak. Big, fluffy iodide with a negative charge is the most stable, because it can distribute the -1 charge over a larger surface. So the charge looks more like zero everywhere.
Your electronegativity rule is a good one and does work as you go across a row, rather than down a column (H2O vs. HF for example) because size changes less than electronegativity across a row in the periodic table. For my own comprehensive review on acid base chemistry, see Things Everyone Should Know About Acid Base Chemistry
Let me know if that clears things up!
Lauren probably won't see this unfortunately but here's the explanation. Since iodine is so large it actually is a weaker bond with the hydrogen -- there is less orbital overlap. Fluorine, being so small, is able to have a lot more orbital overlap with hydrogen, so the bond is a lot stronger. Also, increased electronegativity in a covalent bond actually weakens the bond -- the electronegative atom will tend to pull the electron density towards itself and away from the covalent "sharing" bond. This would indicate that HF is a strong acid (weak H-F bond), and it's why you're confused. However, the stability of an F anion "overpowers" this and it prefers its protonated form, HF.
Nice explanation, thank you, Andrew!
November 16, 2014 at 01:29 PM
You have got some good thinking going on there! I would be happy to have you in one of my classes.
Your only blindspot is that you are thinking only of the bond breaking. You also have to think of the stability of the products, after the bond breaks.
It helps to think of the separation of H from F in terms of Before and After. There are two competing processes; the separation of H from F, and the stability of the products.
You seem a little stuck in the before picture, which asks, how easy is it to separate H from F? You were right the first time when you guessed it would be easy, since F is so electronegative compared to Cl, Br, or I.
The after picture is just as important, and a competing factor. This asks, how stable are the products, H+ and F-?
F- is very unstable. It does not like to be on its own. This is because charge of one flavor does not like to be forced into a small area. It repels itself! F is a very small atom.
If you put the same amount of charge on a Cl atom, it has more room to spread out and be comfortable, so to speak. So a Cl- ion is more likely to persist and be stable than a F- ion.
For every reaction, judging whether it can happen involves looking at the energies (stabilities) of both the reactants AND the products.
Holly Phaneuf Erskine |
April 25, 2014 at 02:16 PM
but oh wait is HF not a strong acid because the such high electronegativity of fluorine and thus such strong bond makes the first ionization energy so high. so yes its a small atom and the bond is really strong but the bond is so strong that it wont readily dissociate like strong acids need to. As opposed to iodine which is bigger in size but still has that polarity with hydrogen but the ionization energy is lower so the hydrogen is more easily removed
April 25, 2014 at 12:56 PM
ok so if Fluorine is smaller than iodine then shouldn't it have a stronger bond with the hydrogen because the positives and negatives are closer. So then the fluorine should be able to attract the shared electrons better. If this is the case then the electron density on the hydrogen would be heavily decreased and then the hydrogen would be able to be easily removed thus making a strong acid. Please let me know where I am not understanding.
April 25, 2014 at 12:41 PM
Thanks Dr H! It all makes sense now :) xx
5036 Student |
November 06, 2008 at 05:23 AM
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