Question:
Identify the pair of molecules/ions in which hybridisation of central atom is same, while their geometries are different.
Identify the pair of molecules/ions in which hybridisation of central atom is same, while their geometries are different.
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Same hybridisation does not guarantee same geometry; lone pairs can change molecular shape significantly.
Updated On: Jun 20, 2026
- BF\(_3\), SO\(_2\)
- XeF\(_2\), BeCl\(_2\)
- XeF\(_4\), SF\(_4\)
- ClF\(_3\), NH\(_4^+\)
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The Correct Option is A
Solution and Explanation
Step 1: Understand hybridisation vs geometry.
Hybridisation is determined by steric number (bond pairs + lone pairs), while geometry depends on arrangement of both bonding and lone pairs. Two species can have same hybridisation but different geometry due to presence of lone pairs.
Step 2: Analyze BF\(_3\).
In BF\(_3\), boron has 3 bond pairs and no lone pair. Steric number = 3, so hybridisation = \( sp^2 \). Geometry is trigonal planar due to absence of lone pairs.
Step 3: Analyze SO\(_2\).
In SO\(_2\), sulfur has 2 bond pairs and 1 lone pair. Steric number = 3, so hybridisation is also \( sp^2 \). However, due to lone pair presence, geometry becomes bent (angular).
Step 4: Compare geometry difference.
Both BF\(_3\) and SO\(_2\) have \( sp^2 \) hybridisation, but: - BF\(_3\): trigonal planar - SO\(_2\): bent shape Thus, hybridisation same but geometry different.
Step 5: Check other options briefly.
Other options either have different hybridisation or do not satisfy the condition simultaneously. Hence they are rejected.
Step 6: Final verification.
Only BF\(_3\) and SO\(_2\) satisfy same hybridisation but different molecular geometry condition correctly.
Final Answer: \[ \boxed{\text{BF}_3,\ \text{SO}_2} \]
Hybridisation is determined by steric number (bond pairs + lone pairs), while geometry depends on arrangement of both bonding and lone pairs. Two species can have same hybridisation but different geometry due to presence of lone pairs.
Step 2: Analyze BF\(_3\).
In BF\(_3\), boron has 3 bond pairs and no lone pair. Steric number = 3, so hybridisation = \( sp^2 \). Geometry is trigonal planar due to absence of lone pairs.
Step 3: Analyze SO\(_2\).
In SO\(_2\), sulfur has 2 bond pairs and 1 lone pair. Steric number = 3, so hybridisation is also \( sp^2 \). However, due to lone pair presence, geometry becomes bent (angular).
Step 4: Compare geometry difference.
Both BF\(_3\) and SO\(_2\) have \( sp^2 \) hybridisation, but: - BF\(_3\): trigonal planar - SO\(_2\): bent shape Thus, hybridisation same but geometry different.
Step 5: Check other options briefly.
Other options either have different hybridisation or do not satisfy the condition simultaneously. Hence they are rejected.
Step 6: Final verification.
Only BF\(_3\) and SO\(_2\) satisfy same hybridisation but different molecular geometry condition correctly.
Final Answer: \[ \boxed{\text{BF}_3,\ \text{SO}_2} \]
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