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QuestionChemistry

Draw the structure of bromous acid, [tex]HBrO^2[/tex]. Optimize formal charges.
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Answer

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Step 1:
I'll solve this problem step by step, following the LaTeX formatting guidelines precisely:

Step 2:
: Determine the number of valence electrons

Total valence electrons: $$1 + 7 + (2 \times 6) = 20$$ valence electrons
- Hydrogen (H): 1 valence electron - Bromine (Br): 7 valence electrons - Oxygen (O): 6 valence electrons

Step 3:
: Arrange atoms and create initial skeleton structure

- Bromine will be the central atom - Hydrogen will bond to one oxygen - Remaining oxygen will be bonded to bromine

Step 4:
: Connect atoms and distribute electrons

- Draw single bonds between Br-O atoms - Draw single bond between H-O - Remaining electrons will be placed as lone pairs

Step 5:
: Calculate formal charges

\text{Formal Charge}_{\text{H}} = 1 - 0 - \frac{2}{2} = 0
Where: - V = valence electrons of free atom - N = non-bonding electrons - B = number of bonding electrons For Bromine: - V = 7 - Non-bonding electrons = 4 - Bonding electrons = 4 For Oxygen atoms: - V = 6 - Non-bonding electrons = 4 - Bonding electrons = 4 For Hydrogen: - V = 1 - Non-bonding electrons = 0 - Bonding electrons = 2

Step 6:
: Draw Lewis structure

The optimized structure will have: - Bromine as central atom - Two single bonds to oxygen atoms - One oxygen bonded to hydrogen - Lone pairs on oxygen atoms

Final Answer

The Lewis structure of bromous acid HBrO_{2} has: - Bromine as central atom - Symmetric arrangement with zero formal charges - Hydrogen attached to one oxygen - Double bond between bromine and one oxygen - Single bond between bromine and other oxygen - Lone pairs on oxygen atoms to satisfy octet rule