Based on the bond angles in CH_4, NH_3, and H_2O, which of the following correctly ranks the magnitude of electron pair repulsions from greatest to least?

• A. CH_4 > H_2O > NH_3
• B. CH_4 > NH_3 > H_2O
• C. H_2O > NH_3 > CH_4
• D. NH_3 > H_2O > CH_4

Answer: (C)

Verified step by step guidance

1. Recall that the magnitude of electron pair repulsions is related to the bond angles in molecules: smaller bond angles indicate stronger repulsions between electron pairs because the electron clouds are pushed closer together.
2. Identify the molecular geometries and approximate bond angles for each molecule: CH_4 (tetrahedral, about 109.5°), NH_3 (trigonal pyramidal, about 107°), and H_2O (bent, about 104.5°).
3. Understand that lone pairs exert greater repulsion than bonding pairs because lone pairs occupy more space around the central atom, pushing bonding pairs closer and reducing bond angles.
4. Compare the bond angles: since H_2O has the smallest bond angle, it experiences the greatest electron pair repulsion; NH_3 has a slightly larger bond angle, so its repulsions are less than H_2O but greater than CH_4; CH_4 has no lone pairs and the largest bond angle, indicating the least repulsion.
5. Rank the electron pair repulsions from greatest to least based on the bond angles and lone pair effects: H_2O > NH_3 > CH_4.