QQuestionChemistry
QuestionChemistry
"Draw the Lewis structure of BH^3 and then determine its electron domain and molecular geometries.
A. trigonal / planar
B. trigonal / tetrahedral
C. trigonal planar / tetrahedral
D. trigonal / bent (120°)"
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Answer
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Step 1:Let's solve this step by step:
Step 2:: Determine the total number of valence electrons
- Total valence electrons: $$3 + (1 \times 3) = 6$$ electrons
- Boron (B): 3 valence electrons - Hydrogen (H): 1 valence electron × 3 hydrogens
Step 3:: Draw the Lewis structure
- Boron is the central atom - Connect three hydrogen atoms to boron using single bonds - Distribute remaining electrons to achieve octet (if possible) \begin{array}{c} H-B-H \ H \end{array}
Step 4:: Electron Domain Geometry Analysis
- Total electron domains: 3 - Electron domain geometry: trigonal planar - No lone pairs on central boron atom - All electron domains are bonding domains
Step 5:: Molecular Geometry Determination
- Since all domains are bonding and arranged trigonally - Molecular geometry is also trigonal planar - Bond angles are approximately 120°
Step 6:: Verify Geometry Characteristics
- Trigonal planar geometry - 120° bond angles - Flat molecular shape
Final Answer
The key characteristics are: - 3 electron domains - Trigonal planar electron domain geometry - Trigonal planar molecular geometry - 120° bond angles
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