See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Analysis of Molecular Geometry **Step 1: Determine electron configuration and geometry for each species** **(A) $CH_3^+$ (methyl cation)** - C: 3 valence electrons (lost 1) = 3 bonding pairs - Geometry: **Trigonal planar** (sp²) **(B) $BF_3$ (boron trifluoride)** - B: 3 valence electrons, 3 B-F bonds - Geometry: **Trigonal planar** (sp²) **(C) $CH_3^-$ (methyl anion)** - C: 4 + 1 = 5 valence electrons - 3 bonding pairs + 1 lone pair - Geometry: **Pyramidal** (sp³) - The lone pair occupies the apex position **(D) $\cdot CH_3$ (methyl radical)** - C: 3 valence electrons + 1 unpaired electron = 3 bonding pairs - Geometry: **Trigonal planar** (sp²) **Conclusion:** Only $CH_3^-$ has the tetrahedral electron geometry with one lone pair, resulting in a **pyramidal molecular shape** with the 3 hydrogen atoms forming the base and the lone pair pointing upward. **Answer: (C) $CH_3^-$**