In the actual molecule of water (H_2O), which has a bent geometry, what are the approximate values of the bond angles a (H–O–H) and b (O–H–lone pair)?

• A. a = 180°, b = 120°
• B. a = 120°, b = 90°
• C. a = 104.5°, b = 109.5°
• D. a = 109.5°, b = 109.5°

Answer: (C)

Verified step by step guidance

1. Step 1: Understand the molecular geometry of water (H\_2O). Water has a bent shape due to the two hydrogen atoms bonded to oxygen and two lone pairs of electrons on oxygen.
2. Step 2: Recall that the ideal bond angle for a tetrahedral electron pair geometry (which includes bonding pairs and lone pairs) is approximately 109.5°.
3. Step 3: Recognize that lone pairs occupy more space than bonding pairs, causing the bond angle between the hydrogen atoms (H–O–H) to be less than the ideal tetrahedral angle.
4. Step 4: Identify that the bond angle a (H–O–H) is therefore slightly less than 109.5°, typically around 104.5°, due to lone pair repulsion pushing the hydrogen atoms closer together.
5. Step 5: Note that the angle b (O–H–lone pair) is approximately the ideal tetrahedral angle of 109.5°, as it involves one bonding pair and one lone pair.