Answer
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Step 1:I'll solve this problem step by step, following the specified LaTeX and formatting guidelines:
Step 2:: Determine the Lewis Structure
- First, we need to draw the Lewis structure for PF$$_{3}
- Phosphorus (P) is the central atom - Count the total valence electrons: * P: 5 valence electrons * F: 7 valence electrons × 3 = 21 valence electrons * Total: 26 valence electrons
Step 3:: Arrange Electron Pairs
- Use the Valence Shell Electron Pair Repulsion (VSEPR) theory - Count the number of electron domains around the central phosphorus atom - Three single bonds to fluorine atoms - One lone pair on phosphorus - Total electron domains: 4
Step 4:: Determine Electron Geometry
- Represented by $$AX_{3}E$$ molecular geometry notation
- With 4 total electron domains, the electron geometry is tetrahedral - The electron geometry is determined by the arrangement that minimizes electron pair repulsion
Step 5:: Confirm Electron Geometry
- 4 electron domains arranged in a tetrahedral configuration - Bond angles will be close to 109.5°
Final Answer
The electron geometry of PF_{3} is tetrahedral.
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