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Electronegativity and acid strength


encipher

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Hi,

 

Does electronegativity play any significant role in the strength of the acids in a given group? (particularly Group 7A binary acids) I know the general trend across periods... But I'm curious as to wether or not EN has a significant effect down a group. Because HF < HCl < .....

 

Thanks

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Electronegativity increases the farther right and up you go.

 

With the least being Francium and the most Flourine, there are howeever exceptions.

 

 

I'm aware of that, my question however is regarding its effect on the strength of an acid going down a group.

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Why would electronegativity play a role in acid strength? There are two conclusions we might draw. First, a greater electronegativity of the atom or atoms attached to the H-O in the oxyacid apparently results in a weaker H-O bond, which is thus more readily ionized. We know that an electronegative atom polarizes bonds by drawing the electrons in the molecule towards it. In this case, the Cl in HOCl and the Br in HOBr must polarize the H-O bond, weakening it and facilitating the ionization of the hydrogen. In comparing HOCl to HOClO, the added oxygen atom must increase the polarization of the H-O bond, thus weakening the bond further and increasing the extent of ionization.

 

A second conclusion has to do with the ion created by the acid ionization. The negative ion produced has a surplus electron, and the relative energy of this ion will depend on how readily that extra electron is attracted to the atoms of ion. The more electronegative those atoms are, the stronger is the attraction. Therefore, the OCl- ion can more readily accommodate the negative charge than can the OBr- ion. And the OClO- ion can more readily accommodate the negative charge than can the OCl- ion.

 

We conclude that the presence of strongly electronegative atoms in an acid increases the polarization of the H-O bond, thus facilitating ionization of the acid, and increases the attraction of the extra electron to the negative ion, thus stabilizing the negative ion. Both of these factors increase the acid strength. Chemists commonly use both of these conclusions in understanding and predicting relative acid strength.

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Riogho, Thanks for that explanation, however I am interested not in Oxyacids but in Binary acids.. I am aware of the reasoning behind the strength an oxyacid in relation to electronegativity.

 

I want to know the effect (if at all significant) on the acidic strength of a given GROUP, specifically, 7A. i.e. HF, HCl, HBr, HI

 

Thanks

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I believe it`s to do with electrostatic forces, the I- is a larger atom than the F-.

so the bond will be stronger in HF than HI as the inter nucleus distance is smaller.

 

so in the case of HI, the H+ will be easier to replace than the H+ in HF.

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  • 2 months later...

In the case of a binary acid, the radius of the non-hydrogen atom effects the acid strength. HF is a weak acid, but as you read down the group the (binary) acids get stronger and stronger (HI>HBr>HCl>HF). This trend cannot be explained by electronegativity, because F is more electronegative than Cl, Br, or I. If the bond between the atom and the hydrogen is long, then it is a weak bond, and the hydrogen dissociates more easily. An atom with a larger radius has to have "longer" bonds, because the electrons take up more space. Think of the atoms as being like...balloons? The larger the balloon (electron cloud) is, the longer the bond between it and an other atom has to be. A "strong" acid is one that readily looses it's hydrogen and becomes it's conjugate base. The weaker bonds make it easier for this to happen. Does this make sense? I'm about halfway though AP Chemistry, so while I have some knowledge, I'm sure other people on this board know a lot more than I do.

 

However, when you have larger acids, electrnegativity does play a role. This is because electronegative atoms draw electrons away from the bond between the central atom and H. This weakens the bond, and weaker bonds create stronger acids. This leads into what Riogho just said, so I'll stop there.

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