Final answer:
The smaller H–N–H angle in NH3 compared to H–C–H in CH4 is due to the presence of a lone pair in NH3, which takes up more space resulting in greater repulsion and a reduced bond angle. NH4+ has an identical bond angle to CH4 as it lacks a lone pair, making the structure truly tetrahedral.
Step-by-step explanation:
The H–N–H angle in NH3 (ammonia) is smaller than the H–C–H bond angle in CH4 (methane) due to the presence of a lone pair of electrons on the nitrogen atom in ammonia.
Lone pairs occupy more space than bonding electron pairs, which results in increased electron pair repulsion. This shifts the bonded hydrogen atoms closer together, reducing the bond angle from the ideal tetrahedral angle of 109.5° to approximately 107°.
In contrast, the methane molecule has no lone pairs and exhibits a perfect tetrahedral geometry with bond angles of 109.5°.
For the NH4+ ion (ammonium), the central nitrogen atom does not have a lone pair; instead, it forms four equivalent bonds with hydrogen atoms, restoring the ideal tetrahedral bond angles seen in methane (109.5°).
This similarity in bond angle is due to the complete absence of lone pair repulsion in the ammonium ion, similar to methane.
The correct answer to the question posed would therefore be (a) Presence of lone pairs, which are responsible for the difference in the bond angles due to their additional repulsion.