How to Draw the Lewis Structure for Bromic Acid (HBrO3)
Bromic acid isn't something you keep on a shelf. It decomposes readily and is almost always handled in aqueous solution. The HBrO3 molecule itself has a structure that students routinely get wrong because the bonding doesn't follow the simple octet rule they learned in intro chemistry. Here's how it actually works. Start by counting valence electrons. Bromine brings 7. Each of the three oxygens brings 6, so that's 18. Hydrogen contributes 1. Total: 26 valence electrons to distribute. The central atom is bromine. That's the first decision point and where a lot of people go wrong. Hydrogen never acts as a central atom in oxyacids. It bonds to an oxygen, not the other way around. Bromine is the hub, and the three oxygens connect to it. One of those oxygens carries the hydrogen.
Draw single bonds between Br and each oxygen first. That uses 6 electrons. Place the hydrogen on one oxygen with a single bond. That uses 2 more. You've now spent 8 electrons. Subtract from 26 and you have 18 left to work with. Fill the octets on the three oxygens. Each oxygen needs 6 more electrons to complete its octet through single bonds. Three oxygens times 6 equals 18. That uses all remaining electrons. But now bromine only has 3 bonds and 0 lone pairs — it has just 6 electrons around it. That's a problem. Bromine is in period 4, which means it has access to d-orbitals and can expand its octet. That's the key move here. Take a lone pair from each of the two non-hydrogenated oxygens and convert them into double bonds with bromine. Now bromine has 2 double bonds, 1 single bond to the OH group, and 1 lone pair. Count electrons around bromine: 4 from the two double bonds, 2 from the single bond, 2 from the lone pair. That's 10 electrons. Expanded octet. Formal charges all work out to zero on every atom, which is exactly what you want to see.
Geometry around bromine is tetrahedral electron geometry, trigonal pyramidal molecular geometry because of that lone pair. Bond angles should be slightly less than 109.5 degrees, probably closer to 104 or 105 given the double bond repulsion. The Br-OH bond is the longest at roughly 1.65 angstroms, while the two Br=O bonds are shorter at about 1.55 angstroms. These numbers come from computational studies since isolating pure HBrO3 is impractical. I ran into a specific issue once when cross-referencing textbook diagrams. Multiple sources drew HBrO3 with only single bonds and a formal charge of +1 on bromine and -1 on each terminal oxygen. That structure is technically a valid resonance contributor if you strictly enforce the octet rule, but it's not the dominant form. The expanded octet structure with zero formal charges is significantly more stable and is what you'd actually calculate using standard quantum chemistry methods. The workaround I use now is simple: after drawing any Lewis structure, immediately compute formal charges. If you get non-zero charges on highly electronegative atoms like oxygen, reconsider whether expanding the octet on the central atom would improve things. For period 3 and below central atoms, this almost always resolves it. One thing nobody emphasizes enough: HBrO3 is a stronger acid than HBrO2, which makes sense structurally. More terminal oxygens means more electron withdrawal from the O-H bond, making the proton easier to release. The pKa is around 0.7. That puts it in the same ballpark as phosphoric acid's first dissociation. It's strong but not violently so, and that's directly reflected in the Lewis structure — the two double-bonded oxygens are doing the heavy lifting on stabilization through inductive effects.
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Another common pitfall involves resonance. Some sources show the two Br=O bonds as interchangeable through resonance, suggesting the double bond character is delocalized. In practice, the structure is better described as having two distinct Br=O bonds and one Br-OH single bond. The resonance formalism exists but the asymmetry is real and measurable. Don't draw resonance structures that swap the OH oxygen with the double-bonded oxygens — that implies the hydrogen could migrate, which it doesn't in any meaningful way. If you're working with this for computational chemistry or spectrophotometry interpretation, be aware that HBrO3 in water exists in equilibrium with BrO3- and H+. The anion BrO3- has D3h symmetry with three equivalent Br-O bonds, all the same length, around 1.81 angstroms. The loss of the proton and the geometric relaxation is significant. So if your data shows three equivalent oxygen environments, you're looking at bromate ion, not the acid. That distinction matters when you're trying to assign peaks in an IR spectrum or interpret NMR data.
