a) The number of valence electrons for the compound CH3OH can be determined by adding up the valence electrons from each atom in the molecule. Carbon (C) has 4 valence electrons, each of the three hydrogen (H) atoms has 1 valence electron, and oxygen (O) has 6 valence electrons. Therefore, the total number of valence electrons is:
- 1 Carbon: 4
- 3 Hydrogens: 3 (1 x 3)
- 1 Oxygen: 6
Total: 4 + 3 + 6 = 13 valence electrons.
b) The Lewis structure for CH3OH can be drawn by placing the carbon atom in the center, with three hydrogen atoms bonded to it and one hydroxyl (OH) group connected to the carbon. Each bond is formed using two valence electrons. The remaining lone pairs of electrons are placed around the oxygen atom. The resulting Lewis structure looks like this:
H
|
H - C - O - H
|
H
c) In terms of VSEPR (Valence Shell Electron Pair Repulsion) theory, the shape of the CH3OH molecule can be analyzed. The molecule has a tetrahedral electron geometry due to four regions of electron density around the carbon atom (three C-H bonds and one C-O bond). The presence of the lone pair on the oxygen atom also affects the shape around it. The molecular geometry of the CH3OH, considering only the atoms and not the lone pairs, is staggered and generally described as tetrahedral, leading to a bond angle of approximately 109.5 degrees around the carbon atom.