QQuestionChemistry
QuestionChemistry
"For the compound BH^3, show the following:
(i) Lewis electron dot formula
(ii) Lewis structure (draw in 3D - use wedge and dash for tetrahedral)
(iii) Electronic geometry (name)
(iv) Molecular geometry (name)
(v) Bond angle (ideal)
(vi) Polar or non-polar"
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Answer
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Step 1:I'll solve this problem step by step, following the specified LaTeX and formatting guidelines:
Step 2:: Lewis Electron Dot Formula
- Total valence electrons: $$3 + (3 \times 1) = 6$$ electrons
- Boron (B) has 3 valence electrons - Hydrogen (H) has 1 valence electron
Step 3:: Lewis Structure
\begin{array}{c}
- Boron will be the central atom - Hydrogen atoms will bond to boron - Electron arrangement: H \ \text{} \mid \ B-H \ \text{} \mid \ H \end{array}
Step 4:: Electronic Geometry
- Electronic geometry is tetrahedral - 3 bonding electron pairs + 1 lone pair (empty orbital)
Step 5:: Molecular Geometry
- Molecular geometry is trigonal planar - All atoms are arranged in the same plane around boron
Step 6:: Bond Angle
- Ideal bond angle: $$120^{\circ}
Step 7:: Polarity
- Molecule is polar - Electronegativity difference between B and H creates uneven charge distribution
Final Answer
(i) Lewis dot formula: \begin{array}{c} H \cdot \ \text{} \mid \ B \cdot - H \ \text{} \mid \ H \cdot \end{array} (ii) 3D Lewis structure with wedge/dash (iii) Tetrahedral electronic geometry (iv) Trigonal planar molecular geometry (v) 120^{\circ} bond angle (vi) Polar molecule
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