Question:

The shape of \(\mathrm{IF_6^-}\) is :

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Count the bond pairs and lone pairs on iodine; seven electron pairs give a pentagonal bipyramidal arrangement, but with one lone pair the shape is distorted.
Updated On: Oct 31, 2026
  • Trigonally distorted octahedron
  • Pyramidal
  • Octahedral
  • Square antiprism
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The Correct Option is A

Solution and Explanation

Iodine valence electrons: 7; add 6 for the six F atoms (bond pairs) and 1 for the negative charge: total = 7 + 6 + 1 = 14 electrons = 7 pairs.
So there are 6 bond pairs + 1 lone pair, steric number 7 (\(sp^3d^3\) hybridisation).
The lone pair distorts the octahedron of six F atoms along a threefold axis (\(C_{3v}\)), i.e. a trigonally distorted octahedron (it is not a regular octahedron, which would need no active lone pair, nor a pyramidal or antiprism shape).
Answer: (1) Trigonally distorted octahedron
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