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QuestionChemistry

Draw the best Lewis structure for BrO^4− and determine the formal charge on bromine.
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Step 1:
I'll solve this problem step by step, following the specified LaTeX formatting guidelines:

Step 2:
: Determine the total number of valence electrons

- Total valence electrons: $$7 + 24 + 1 = 32$$ electrons
- Bromine (Br): 7 valence electrons - Oxygen (O): 6 valence electrons × 4 = 24 valence electrons - Negative charge: + 1 electron

Step 3:
: Arrange atoms in the Lewis structure

- Place bromine in the center - Surround bromine with four oxygen atoms - Connect each oxygen to bromine with a single bond

Step 4:
: Distribute remaining electrons

- Initial single bonds use $$8$$ electrons
- Remaining electrons: 32 - 8 = 24 electrons - Add three additional electron pairs to each oxygen atom as lone pairs

Step 5:
: Calculate formal charge on bromine

\text{Formal Charge}_{\text{Br}} = 7 - 0 - \frac{1}{2}(8) = 7 - 0 - 4 = +3
- Bromine valence electrons: 7 Formal charge calculation:

Step 6:
: Verify the Lewis structure

- Bromine has a formal charge of $$+3
- Four oxygen atoms each have three lone pairs - Bromine has four single bonds to oxygen atoms - Total electron count matches initial calculation

Final Answer

The Lewis structure for \text{BrO}_{4}^{-} has bromine with a formal charge of + 3, with four single bonds to oxygen atoms, and each oxygen having three lone pairs.