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Postby JustinHorriat_4f » Sun Dec 01, 2019 11:50 pm

Can someone explain to me what Professor Lavelle meant when he said "Resulting anion must be stable, and that oxyacids more readily lose H+ if resulting anion is stabilized by e- withdrawing atoms which decline and stabilize negative charge."

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Re: Lecture

Postby jeffreygong1I » Mon Dec 02, 2019 2:41 am

If a molecule has more electronegative atoms like Cl or F attached to it, they pull away the electron density from the H-central atom bond and therefore make that bond weaker and the acid stronger.

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Re: Lecture

Postby CalvinTNguyen2D » Mon Dec 02, 2019 8:16 am

The more stable an acid’s structure is, the stronger it usually is because of how it will run to full completion rather than the reaction being partially completed. The more electronegative atoms attached or the more resonance structures, the more stable these acids are.

Shail Avasthi 3C
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Re: Lecture

Postby Shail Avasthi 3C » Mon Dec 02, 2019 10:08 am

Compare the acids HClO and HBrO. Their resulting anions will be ClO- and BrO- respectively. ClO- will be more stable than BrO for the following reason: Chlorine and Bromine are both considered "electron-withdrawing atoms" because of their electronegativity, and they will have the effect of delocalizing the negative charge on the oxygen. The more delocalized the charge throughout the anion, the more stable it is. Therefore, since Cl is more electronegative than Br, it will delocalize the negative charge more than Br will.

Oxoacids will be more acidic in proportion with how stable their resulting anion is. Therefore, HClO is more acidic than HBrO. This applies for all oxoacids.

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