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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?
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Practice Questions
1 question
Q1
What is the bond angle in a water molecule (H2O) according to VSEPR theory?
120 degrees
109.5 degrees
104.5 degrees
90 degrees
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The bond angle in H2O is approximately 104.5 degrees due to the two lone pairs on oxygen.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the bond angle in a water molecule (H2O) according to VSEPR theory?
Solution:
The bond angle in H2O is approximately 104.5 degrees due to the two lone pairs on oxygen.
Steps: 6
Show Steps
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.
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