VSEPR theory predicts bond angles from electron pair repulsions Steps:

• Identify central atom and VSEPR type (AX_nE_m) for each: CH4 (C, AX4), CH3OCH3 (O, AX2E2), NH3 (N, AX3E1), NH4+ (N, AX4).
• Recall ideal angles: tetrahedral electron geometry gives 109.5° for AX4 or AX3E1 (actual smaller due to lone pair), AX2E2 bent ~104.5°-111°.
• Measure actual bond angles: CH4 109.5°, CH3OCH3 (C-O-C) 111°, NH3 (H-N-H) 107°, NH4+ 109.5°.
• Compare: smallest is 107° in NH3 due to lone pair repulsion compressing bonds.

Why C is correct:

• NH3 has AX3E1 geometry; one lone pair on N repels bonding pairs more than in AX4, reducing H-N-H angle to 107° per VSEPR.

Why the others are wrong:

• A: CH4 is AX4 tetrahedral with no lone pairs, so 109.5° angle.
• B: CH3OCH3 is AX2E2 bent at O with two lone pairs, but C-O-C angle is 111° (larger than NH3 due to less repulsion from CH3 groups).
• D: NH4+ is AX4 tetrahedral like CH4, with 109.5° angle and no lone pair compression.

Final answer: C