Question:

The group of molecules having identical shape is :

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Count bond pairs and lone pairs on the central atom for each species and find the group with the same geometry.
Updated On: Nov 11, 2026
  • \(\mathrm{SF_4,\ XeF_4,\ CCl_4}\)
  • \(\mathrm{ClF_3,\ XeOF_2,\ XeF_3^+}\)
  • \(\mathrm{BF_3,\ PCl_3,\ XeO_3}\)
  • \(\mathrm{PCl_5,\ IF_5,\ XeO_2F_2}\)
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The Correct Option is B

Solution and Explanation

Option (1): \(\mathrm{SF_4}\) see-saw, \(\mathrm{XeF_4}\) square planar, \(\mathrm{CCl_4}\) tetrahedral: different.
Option (2): \(\mathrm{ClF_3}\) (3 bp + 2 lp, sp3d), \(\mathrm{XeOF_2}\) (2 F + 1 O bonded + 2 lp, sp3d) and \(\mathrm{XeF_3^+}\) (3 bp + 2 lp, sp3d) are all T-shaped: identical.
Option (3): \(\mathrm{BF_3}\) trigonal planar, \(\mathrm{PCl_3}\) pyramidal, \(\mathrm{XeO_3}\) pyramidal: different.
Option (4): \(\mathrm{PCl_5}\) trigonal bipyramidal, \(\mathrm{IF_5}\) square pyramidal, \(\mathrm{XeO_2F_2}\) see-saw: different.
Answer: (2) \(\mathrm{ClF_3,\ XeOF_2,\ XeF_3^+}\)
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