QQuestionChemistry
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
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