JEE Main · 2025 · Shift-ImediumBOND-189

The molecules having square pyramidal geometry are:

Chemical Bonding · Class 11 · JEE Main Previous Year Question

Question

The molecules having square pyramidal geometry are:

Options
  1. a

    \ceBrF5\ce{BrF5} & \ceXeOF4\ce{XeOF4}

  2. b

    \ceSbF5\ce{SbF5} & \ceXeOF4\ce{XeOF4}

  3. c

    \ceSbF5\ce{SbF5} & \cePCl5\ce{PCl5}

  4. d

    \ceBrF5\ce{BrF5} & \cePCl5\ce{PCl5}

Correct Answera

\ceBrF5\ce{BrF5} & \ceXeOF4\ce{XeOF4}

Detailed Solution

🧠 Square pyramidal geometry: 5 bond pairs + 1 lone pair around the central atom (sp³d², with LP in equatorial position of the octahedral framework).

🗺️ Check each molecule:

\ceBrF5\ce{BrF5}: Br has 5 bonds + 1 LP = 6 e-pairs → sp³d² → square pyramidal

\ceXeOF4\ce{XeOF4}: Xe has 4 Xe–F + 1 Xe=O + 1 LP = 6 e-pairs → sp³d² → square pyramidal

\ceSbF5\ce{SbF5}: Sb has 5 bonds + 0 LP = 5 e-pairs → sp³d → trigonal bipyramidal

\cePCl5\ce{PCl5}: P has 5 bonds + 0 LP = 5 e-pairs → sp³d → trigonal bipyramidal

⚡ Square pyramidal requires exactly 5 bonds + 1 LP on the central atom. SbF₅ and PCl₅ have no lone pair → trigonal bipyramidal.

Answer: \ceBrF5 and \ceXeOF4\boxed{\text{Answer: } \ce{BrF5} \text{ and } \ce{XeOF4}}

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