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"Determine the electron geometry (eg), molecular geometry (mg) around the Oxygen atom of CH^3OH and the polarity of the whole CH^3OH molecule. a. eg = tetrahedral, mg = linear, nonpolar b. eg = trigonal bipyramidal, mg = trigonal planar, nonpolar c. eg = linear, mg = linear, polar d. eg = octahedral, mg = square planar, nonpolar e. eg = tetrahedral, mg = bent, polar
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Answer

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Step 1:
Let's solve this step by step:

Step 2:
: Analyze the Lewis Structure of CH₃OH

- Total valence electrons: $$4 + 6 + (4 \times 1) = 14$$ electrons
- Carbon (C) has 4 valence electrons - Oxygen (O) has 6 valence electrons - Hydrogen (H) has 1 valence electron

Step 3:
: Determine Electron Geometry Around Oxygen

- Oxygen has two lone pairs and two bonding pairs - This results in a tetrahedral electron geometry - Total electron domains = 4 (2 bonding, 2 non-bonding)

Step 4:
: Determine Molecular Geometry

- With 2 lone pairs and 2 bonding pairs, the molecular geometry is bent - The lone pairs cause a deviation from a linear shape - Bond angle will be less than 109.5° due to lone pair repulsion

Step 5:
: Polarity Analysis

- Oxygen is more electronegative than carbon and hydrogen - This creates a permanent dipole moment - The bent geometry allows unequal distribution of charge - This makes the molecule polar

Step 6:
: Evaluate Given Options

- Option e matches our analysis perfectly: - Electron geometry = tetrahedral - Molecular geometry = bent - Polarity = polar

Final Answer

\text{CH}_3\text{OH has a tetrahedral electron geometry, bent molecular geometry, and is polar}