Concept: Hybridization and molecular geometry are determined by the number of sigma bonds and lone pairs around the central atom. Changes in the number of bonds or lone pairs can alter both hybridization and shape.

Why (A) is correct:

• In CH₃ (methyl radical): The central carbon atom has 3 sigma bonds and 1 unpaired electron. It is sp² hybridized, and its shape is trigonal planar.
• In C₂H₆ (ethane): Each carbon atom has 4 sigma bonds. It is sp³ hybridized, and its shape is tetrahedral.

Thus, in the conversion from CH₃ to C₂H₆, both the hybridization (sp² to sp³) and the shape (trigonal planar to tetrahedral) of the central carbon atom change.

Option Analysis:

• A) CH₃ → C₂H₆: Hybridization changes from sp² to sp³, and shape changes from trigonal planar to tetrahedral. This option is correct.
• B) NH₃ → NH₄⁺: In NH₃, N is sp³ hybridized (3 bond pairs, 1 lone pair), shape is pyramidal. In NH₄⁺, N is sp³ hybridized (4 bond pairs, 0 lone pairs), shape is tetrahedral. Hybridization remains sp³, but the shape changes from pyramidal to tetrahedral.
• C) H₂O → H₃O⁺: In H₂O, O is sp³ hybridized (2 bond pairs, 2 lone pairs), shape is bent. In H₃O⁺, O is sp³ hybridized (3 bond pairs, 1 lone pair), shape is pyramidal. Hybridization remains sp³, but the shape changes from bent to pyramidal.
• D) All of these: Since only option A shows a change in both hybridization and shape, this option is incorrect.

Correct Answer: (A)