Question:
The type of hybridisation of P atom in PCl\(_5\), PCl\(_4^+\) and PCl\(_6^-\) is (respectively)
The type of hybridisation of P atom in PCl\(_5\), PCl\(_4^+\) and PCl\(_6^-\) is (respectively)
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SN = 2 (sp), 3 (sp\(^2\)), 4 (sp\(^3\)), 5 (sp\(^3\)d), 6 (sp\(^3\)d\(^2\)).
Updated On: Apr 23, 2026
- sp\(^3\), sp\(^3\)d, sp\(^3\)d\(^2\)
- sp\(^3\)d, sp\(^3\), sp\(^3\)d\(^2\)
- sp\(^3\), sp\(^3\)d\(^2\), sp\(^3\)
- sp\(^3\)d\(^2\), sp\(^3\), sp\(^3\)d
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Steric number = number of bond pairs + lone pairs = hybridisation.
Step 2: Detailed Explanation:
PCl\(_5\): P has 5 bond pairs, 0 lone pairs \(\Rightarrow\) SN=5 \(\Rightarrow\) sp\(^3\)d.
PCl\(_4^+\): P has 4 bond pairs, 0 lone pairs \(\Rightarrow\) SN=4 \(\Rightarrow\) sp\(^3\).
PCl\(_6^-\): P has 6 bond pairs, 0 lone pairs \(\Rightarrow\) SN=6 \(\Rightarrow\) sp\(^3\)d\(^2\).
Step 3: Final Answer:
Thus, hybridisation: sp\(^3\)d, sp\(^3\), sp\(^3\)d\(^2\).
Steric number = number of bond pairs + lone pairs = hybridisation.
Step 2: Detailed Explanation:
PCl\(_5\): P has 5 bond pairs, 0 lone pairs \(\Rightarrow\) SN=5 \(\Rightarrow\) sp\(^3\)d.
PCl\(_4^+\): P has 4 bond pairs, 0 lone pairs \(\Rightarrow\) SN=4 \(\Rightarrow\) sp\(^3\).
PCl\(_6^-\): P has 6 bond pairs, 0 lone pairs \(\Rightarrow\) SN=6 \(\Rightarrow\) sp\(^3\)d\(^2\).
Step 3: Final Answer:
Thus, hybridisation: sp\(^3\)d, sp\(^3\), sp\(^3\)d\(^2\).
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