Question

What is the typical bond angle between hydrogen and oxygen atoms in a water molecule?

• A. $90^{\circ}$
• B. $104.5^{\circ}$
• C. $120^{\circ}$
• D. $180^{\circ}$

Hint:

Think of water as a distorted tetrahedron. The two bulky non-bonding lone pairs compress the ideal tetrahedral bond angle of \(109.5^{\circ}\) down to \(104.5^{\circ}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Water (\(\text{H}_2\text{O}\)) has a bent molecular geometry due to the hybridization of the central oxygen atom and the repulsion between valence electron pairs.

Step 2: Detailed Explanation:
The central oxygen atom in a water molecule has six valence electrons.
It forms two single covalent bonds with two hydrogen atoms, leaving two non-bonding lone pairs of electrons.
The oxygen atom is \(sp^3\) hybridized, which would ideally form a regular tetrahedral geometry with a bond angle of \(109.5^{\circ}\).
However, according to Valence Shell Electron Pair Repulsion (VSEPR) theory, lone pair-lone pair repulsions are stronger than bonding pair-bonding pair repulsions.
These two lone pairs exert a downward force on the \(\text{O-H}\) bonding pairs, compressing the \(\text{H-O-H}\) bond angle down to approximately \(104.5^{\circ}\).

Step 3: Final Answer:
Therefore, the typical bond angle between hydrogen and oxygen atoms in a water molecule is \(104.5^{\circ}\).