Question:
In \( S_8 \) molecules, the type of hybridisation exhibited by sulphur is:
In \( S_8 \) molecules, the type of hybridisation exhibited by sulphur is:
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In molecules like \( S_8 \), where atoms form multiple bonds and have lone pairs, \( sp^3 \) hybridization is common.
Updated On: Apr 22, 2026
- \( sp^2 \)
- \( sp^3 \)
- \( sp \)
- \( sp^3d \)
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The Correct Option is B
Solution and Explanation
Step 1: Understand the bonding in \( S_8 \).
In \( S_8 \) molecules, each sulfur atom forms two single bonds with neighboring sulfur atoms in a cyclic structure. The sulfur atoms in \( S_8 \) are surrounded by four electron pairs, suggesting that the sulfur atoms undergo \( sp^3 \) hybridization.
Step 2: Analyze the hybridization.
- \( sp^2 \): This hybridization occurs when an atom forms three sigma bonds and one lone pair. This is not applicable to sulfur in \( S_8 \).
- \( sp^3 \): This is the correct hybridization for sulfur in \( S_8 \), as each sulfur atom forms four sigma bonds (two with other sulfur atoms and two lone pairs).
- \( sp \): This is incorrect as sulfur does not form two bonds in \( S_8 \).
- \( sp^3d \): This hybridization involves the use of d-orbitals, but sulfur in \( S_8 \) does not require such orbitals for bonding.
Step 3: Conclusion.
The correct answer is (2), \( sp^3 \), because each sulfur atom in \( S_8 \) undergoes \( sp^3 \) hybridization.
In \( S_8 \) molecules, each sulfur atom forms two single bonds with neighboring sulfur atoms in a cyclic structure. The sulfur atoms in \( S_8 \) are surrounded by four electron pairs, suggesting that the sulfur atoms undergo \( sp^3 \) hybridization.
Step 2: Analyze the hybridization.
- \( sp^2 \): This hybridization occurs when an atom forms three sigma bonds and one lone pair. This is not applicable to sulfur in \( S_8 \).
- \( sp^3 \): This is the correct hybridization for sulfur in \( S_8 \), as each sulfur atom forms four sigma bonds (two with other sulfur atoms and two lone pairs).
- \( sp \): This is incorrect as sulfur does not form two bonds in \( S_8 \).
- \( sp^3d \): This hybridization involves the use of d-orbitals, but sulfur in \( S_8 \) does not require such orbitals for bonding.
Step 3: Conclusion.
The correct answer is (2), \( sp^3 \), because each sulfur atom in \( S_8 \) undergoes \( sp^3 \) hybridization.
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