What is the approximate bond angle in water (H2O)?

• A. 120 degrees
• B. 109.5 degrees
• C. 107 degrees
• D. 104.5 degrees

Answer: (D)

The main idea here is how lone pairs influence bond angles using VSEPR. Oxygen in water has two bonding pairs (to hydrogen) and two lone pairs, giving four electron domains around the atom. These domains arrange themselves roughly in a tetrahedral pattern to minimize repulsion, which would put the H–O–H angle at about 109.5 degrees if all four domains were equivalent. But lone pairs repel more strongly than bonding pairs, so the lone-pair–bonding-pair repulsion pushes the O–H bonds closer together, reducing the angle to around 104.5 degrees. That’s why water is bent rather than having the ideal tetrahedral angle. The other values don’t fit because 120 degrees would come from a trigonal planar arrangement, and 109.5 degrees is the unadjusted tetrahedral angle, while 107 degrees is typical for ammonia with only one lone pair—water, with two lone pairs, ends up slightly smaller.