What is the bond angle in a water molecule (H2O) according to VSEPR theory?

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Question: What is the bond angle in a water molecule (H2O) according to VSEPR theory?

Options:

Correct Answer: 104.5 degrees

Solution:

The bond angle in H2O is approximately 104.5 degrees due to the two lone pairs on oxygen.

What is the bond angle in a water molecule (H2O) according to VSEPR theory?

What is the bond angle in a water molecule (H2O) according to VSEPR theory?

Step 1: Understand that a water molecule (H2O) consists of one oxygen atom and two hydrogen atoms.
Step 2: Recognize that the oxygen atom has two lone pairs of electrons that are not involved in bonding.
Step 3: Apply VSEPR theory, which stands for Valence Shell Electron Pair Repulsion theory, to determine the shape of the molecule.
Step 4: Note that the two hydrogen atoms and the two lone pairs create a bent shape for the water molecule.
Step 5: Understand that the presence of lone pairs pushes the hydrogen atoms closer together, affecting the bond angle.
Step 6: Conclude that the bond angle in a water molecule is approximately 104.5 degrees.

VSEPR Theory – Valence Shell Electron Pair Repulsion theory explains the geometry of molecules based on the repulsion between electron pairs.
Molecular Geometry – The shape of the water molecule is bent due to the presence of two lone pairs on the oxygen atom.
Bond Angles – The specific angle formed between the hydrogen atoms in the water molecule, which is influenced by lone pair repulsion.