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Why is HBrO a weaker acid than HBrO3? A. The HO bond in HBrO is less polar than the HO bond in HBrO3 B. The HO bond in HBrO is more polar than the HO bond in HBrO3 C. The HBr bond in HBro is weaker than the HBr in HBrO3 D. The HBr bond in HBrO is stronger than the HBr bond in HBrO3 E. The HO bond in HBrO is weaker than the HO bond in HBrO3 C and D are invalid because there is no HBr bond. None of them really fit right if I had to pick on I'd pick one I'd say the HO bond is less polar because the e- cloud is being pulled toward the Cl. Anyone know?

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Its B. The OH bond is weaker in HBrO3 because the BrO3 ion is resonance stabilized. So Bro3 is more stable than BrO-. So there is a stronger tendency to pull the e away from the H to from the BrO3- ion. So OH bond breaks easily. There is no resonance stabilization in BrO- ion.
How would that make the bond more polar? shouldnt that make it less polar if it doesnt pull the e away as much?
Resonance stabilization of the bromate anion is a good point, but the increased acidity is usually attributed primarily to inductive effects (meaning having three four electronegative atoms attached to the O polarizes the O-H bond much more than only having one). I really have no idea whether anyone has determined the relative contributions of induction and resonance stabilization of the anion. I think in either case you would argue the OH bond is more polar, not weaker. I found one reference that gives the bond lengths as essentially identical (97 pm in HBrO and 98 pm in HOBrO2), which suggests the bond strengths are relatively similar. But a much more polar OH bond (along with resonance stabilization of the product anion) would leave the H much more vulnerable to nucleophilic attack by a water molecule to for the proton-transfer product. So I concur in B.

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Hmm I've got 2 people saying B and 2 people saying A, i do not see how the OH bond in HBrO is more polar than in HBrO3
2 people saying A are obviously not on this site.
The electron could is more pulled away with the more oxygens you add making the bond more polar because there is now a far bigger dipole moment making the bond more polar. Would you not agree?

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