Based on the bond angles in CH_4, NH_3, and H_2O, which molecule experiences the greatest magnitude of electron pair repulsions around the central atom?

• A. NH_3
• B. CH_4
• C. All experience equal repulsions
• D. H_2O

Answer: (D)

Verified step by step guidance

1. Identify the molecular geometries of CH_4, NH_3, and H_2O based on their Lewis structures and electron domains around the central atom. CH_4 is tetrahedral, NH_3 is trigonal pyramidal, and H_2O is bent.
2. Recall that the bond angles in these molecules are influenced by the repulsions between electron pairs (bonding and lone pairs) around the central atom. Lone pairs repel more strongly than bonding pairs because lone pairs occupy more space.
3. Note the number of lone pairs on the central atom in each molecule: CH_4 has 0 lone pairs, NH_3 has 1 lone pair, and H_2O has 2 lone pairs.
4. Understand that the presence of lone pairs increases electron pair repulsion, which decreases bond angles compared to the ideal tetrahedral angle of 109.5°. Therefore, H_2O, with 2 lone pairs, experiences the greatest repulsion, followed by NH_3, then CH_4.
5. Conclude that the molecule with the greatest magnitude of electron pair repulsions around the central atom is H_2O due to its two lone pairs causing stronger repulsions and smaller bond angles.