Q. The hybridisation state of central atom in $PCl_{5}$ is
A) $sp^{3}d$
B) $sp^{3}d^{2}$
C) $sp^{3}$
D) $d^{2}sp^{3}$
Explanation
(a)
Number of hybrid orbitals for neutral atom = $\frac{1}{2}$ [Number of valence electron in central atom + Number of monovalent atom]
Number of hybrid orbital $= \ \frac{5 + 5}{5}\ = \ 5$
Hence, hybridisation is $sp^{3}d.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.