1. Count Valence Electrons
* Bromine (Br) has 7 valence electrons.
* Oxygen (O) has 6 valence electrons.
* The negative charge adds one extra electron.
* Total valence electrons: 7 + (4 * 6) + 1 = 32
2. Determine the Central Atom
* Bromine (Br) is the least electronegative element and will be the central atom.
3. Connect Atoms with Single Bonds
* Connect the four oxygen atoms to the bromine atom with single bonds. This uses up 8 valence electrons (4 bonds * 2 electrons per bond).
4. Distribute Remaining Electrons as Lone Pairs
* We have 24 valence electrons remaining (32 total - 8 used).
* Distribute these as lone pairs around the oxygen atoms, giving each oxygen atom three lone pairs (6 electrons). This uses up all 24 remaining electrons.
5. Check for Octet Rule Fulfillment
* All oxygen atoms have a full octet (2 electrons from the single bond + 6 from lone pairs).
* Bromine also has a full octet (8 electrons from the four bonds).
6. Formal Charge Calculation
* Formal Charge = (Valence Electrons) - (Non-bonding Electrons) - (1/2 Bonding Electrons)
* Formal Charge on Bromine (Br):
* Valence electrons = 7
* Non-bonding electrons = 0
* Bonding electrons = 8
* Formal Charge = 7 - 0 - (8/2) = +1
* Formal Charge on One of the Oxygen Atoms (O):
* Valence electrons = 6
* Non-bonding electrons = 6
* Bonding electrons = 2
* Formal Charge = 6 - 6 - (2/2) = -1
Lewis Structure:
```
O
|
O - Br - O
|
O
```
Formal Charges:
* Bromine (Br): +1
* Oxygen (O): -1
Important Note: The formal charges are a tool to help us understand the distribution of electrons. The perbromate ion is a resonance hybrid, meaning the actual structure is an average of multiple possible structures. The negative charge is delocalized across all four oxygen atoms.
