Question

The set of molecules with different geometry and same type of hybridization is

• A. \(CH_4,\;PCl_5,\;SF_6\)
• B. \(H_2O,\;BeF_2,\;PCl_3\)
• C. \(CH_4,\;NH_3,\;H_2O\)
• D. \(CO_2,\;SO_2,\;SO_3\) \bigskip

Hint:

For \(sp^3\) hybridization: \[ CH_4 \rightarrow \text{Tetrahedral} \] \[ NH_3 \rightarrow \text{Trigonal pyramidal} \] \[ H_2O \rightarrow \text{Bent} \] Same hybridization does not necessarily imply same geometry.

Answer 1

The Correct Option is C

Concept: Hybridization depends on the steric number (bond pairs + lone pairs) around the central atom. However, molecular geometry depends on both bond pairs and lone pairs. Thus molecules can have the same hybridization but different geometries due to different numbers of lone pairs.

Step 1: Examine option (C).
For methane: \[ CH_4 \] Central atom carbon has four bond pairs. \[ sp^3 \] Geometry: \[ \text{Tetrahedral} \] For ammonia: \[ NH_3 \] Central atom nitrogen has \[ 3 \text{ bond pairs } +1 \text{ lone pair} \] Hybridization: \[ sp^3 \] Geometry: \[ \text{Trigonal pyramidal} \] For water: \[ H_2O \] Central atom oxygen has \[ 2 \text{ bond pairs }+2 \text{ lone pairs} \] Hybridization: \[ sp^3 \] Geometry: \[ \text{Bent or V-shaped} \]

Step 2: Compare hybridization and geometry.
All three molecules possess \[ sp^3 \] hybridization. But their geometries are \[ \text{Tetrahedral} \] \[ \text{Trigonal pyramidal} \] \[ \text{Bent} \] which are different.

Step 3: Conclude the answer.
Hence the required set is \[ \boxed{CH_4,\ NH_3,\ H_2O} \] Therefore, \[ \boxed{\text{Option (C)}} \]