Question

The type of hybridisation of P atom in PCl\(_5\), PCl\(_4^+\) and PCl\(_6^-\) is (respectively)

• A. sp\(^3\), sp\(^3\)d, sp\(^3\)d\(^2\)
• B. sp\(^3\)d, sp\(^3\), sp\(^3\)d\(^2\)
• C. sp\(^3\), sp\(^3\)d\(^2\), sp\(^3\)
• D. sp\(^3\)d\(^2\), sp\(^3\), sp\(^3\)d

Hint:

SN = 2 (sp), 3 (sp\(^2\)), 4 (sp\(^3\)), 5 (sp\(^3\)d), 6 (sp\(^3\)d\(^2\)).

Answer 1

The Correct Option is B

Step 1: Concept:
Steric number = number of bond pairs + lone pairs = hybridisation.
Step 2: 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: Conclusion:
Thus, hybridisation: sp\(^3\)d, sp\(^3\), sp\(^3\)d\(^2\).