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NEET CHEMISTRYMedium

An isostructural compound with XeF2XeF_2 among the following is:

A

ICl₂⁻

B

SbCl₃

C

BaCl₂

D

TeF₂

Step-by-Step Solution

  1. Analyze XeF2XeF_2:
  • Central atom Xenon (Xe) has 8 valence electrons.
  • It forms 2 sigma bonds with Fluorine.
  • Remaining electrons = 82=68 - 2 = 6 (3 lone pairs).
  • Structure: 2 Bond Pairs + 3 Lone Pairs \rightarrow Linear geometry (Lone pairs occupy equatorial positions in trigonal bipyramidal electron geometry).
  1. Analyze Option A (ICl2ICl_2^-):
  • Central atom Iodine (I) has 7 valence electrons.
  • Negative charge adds 1 electron \rightarrow Total 8 valence electrons on the central shell context.
  • It forms 2 sigma bonds with Chlorine.
  • Remaining electrons = 82=68 - 2 = 6 (3 lone pairs).
  • Structure: 2 Bond Pairs + 3 Lone Pairs \rightarrow Linear geometry.
  • Since both have the same structure (Linear) and hybridization (sp3dsp^3d), they are isostructural.
  1. Analyze other options:
  • SbCl3SbCl_3: 3 Bond Pairs + 1 Lone Pair \rightarrow Trigonal Pyramidal.
  • BaCl2BaCl_2: Ionic/Linear (but typically considered distinct from covalent VSEPR analogs in this context).
  • TeF2TeF_2: 2 Bond Pairs + 2 Lone Pairs \rightarrow Bent/Angular.
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