Question:
Among CH$_4$, H$_2$O, NH$_3$ and PH$_3$, the molecule having the smallest percent s character for the covalent bond (X–H) is
Among CH$_4$, H$_2$O, NH$_3$ and PH$_3$, the molecule having the smallest percent s character for the covalent bond (X–H) is
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Down the group, hybridization decreases and p-character increases, lowering s-character.
Updated On: Dec 14, 2025
- CH$_4$
- H$_2$O
- NH$_3$
- PH$_3$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding s character in hybrid orbitals.
Higher hybridization (sp, sp$^2$, sp$^3$) means higher s character. s character decreases as: O–H>C–H>N–H>P–H.
Step 2: Reasoning.
PH$_3$ has almost pure p orbitals involved in bonding, with minimal hybridization. Hence, it has the lowest s character among the given molecules.
Step 3: Conclusion.
PH$_3$ has the smallest percent s character in the X–H bond.
Higher hybridization (sp, sp$^2$, sp$^3$) means higher s character. s character decreases as: O–H>C–H>N–H>P–H.
Step 2: Reasoning.
PH$_3$ has almost pure p orbitals involved in bonding, with minimal hybridization. Hence, it has the lowest s character among the given molecules.
Step 3: Conclusion.
PH$_3$ has the smallest percent s character in the X–H bond.
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