JEE Main · 2020 · Shift-IImediumBOND-025

The molecular geometry of SF6 is octahedral. What is the geometry of SF4 (including lone pair(s) of electrons, if any)?

Chemical Bonding · Class 11 · JEE Main Previous Year Question

Question

The molecular geometry of SF6\mathrm{SF_6} is octahedral. What is the geometry of SF4\mathrm{SF_4} (including lone pair(s) of electrons, if any)?

Options
  1. a

    Tetrahedral

  2. b

    Trigonal bipyramidal

  3. c

    Pyramidal

  4. d

    Square planar

Correct Answerb

Trigonal bipyramidal

Detailed Solution

🧠 The question asks for the electron-pair geometry ("including lone pair") of \ceSF4\ce{SF4}, not the molecular (bond) shape.

\ceSF4\ce{SF4}: S has 6 valence electrons. 4 S–F bonds use 4e from S, leaving 2e = 1 lone pair. Total electron groups = 4 bonds + 1 LP = 5trigonal bipyramidal electron-pair geometry.

The molecular shape (positions of F atoms only) is see-saw, but when the question explicitly includes the lone pair, the full geometry is trigonal bipyramidal.

⚠️ Trap: Option (a) says tetrahedral — that would need 4 groups with 0 LP, which is \ceSF42+\ce{SF4^{2+}}, not \ceSF4\ce{SF4}.

Answer: (b) Trigonal bipyramidal\boxed{\text{Answer: (b) Trigonal bipyramidal}}

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