Q. \( \mathrm{CH_3OH} \) molecular geometry.

We want the molecular geometry of methanol, CH₃OH.

Step 1: Identify the central atom and surrounding atoms

The molecule has two types of atoms to consider: carbon and oxygen. The most important geometry questions for methanol are usually about the bond angles around oxygen, because oxygen is the center of the polar “OH group” direction.

The central atom for the geometry of the “functional group” is typically the oxygen atom, which is bonded to:

• the carbon atom (C–O bond)
• the hydrogen atom (O–H bond)
• and it has two lone pairs on oxygen

So the oxygen has a total electron-domain count of:

• 2 bonding pairs (C–O and O–H)
• 2 lone pairs

Total electron domains around O:

\[
2 + 2 = 4
\]

Step 2: Use VSEPR to determine the electron-domain geometry

With 4 electron domains around oxygen, VSEPR predicts a tetrahedral electron geometry.

Step 3: Convert to molecular geometry (shape)

Oxygen has 4 electron domains, but only 2 of them are bonding domains. The 2 remaining domains are lone pairs, which do not appear as part of the molecular shape.

For a molecule of the form \(AX_2E_2\), the molecular shape is:

• bent (because there are two bond directions and a lone-pair “compression”)

Therefore, around the oxygen, the H–O–C part of methanol is a bent (V-shaped) arrangement.

Step 4: Describe the geometry in words for methanol as a whole

So methanol has these key geometric features:

• About the oxygen atom: \(AX_2E_2\) gives a bent H–O–C molecular geometry.
• About the carbon atom: carbon is tetrahedral (\(sp^3\)) with four single bonds, giving an overall tetrahedral arrangement of the C–H, C–H, C–H, and C–O bonds.

Step 5: Final answer (what you should report)

Methanol (\(CH_3OH\)) molecular geometry:

• At oxygen: bent (V-shaped), VSEPR type \(AX_2E_2\).
• At carbon: tetrahedral, \(sp^3\).