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NEET CHEMISTRYChemical Bonding and Molecular StructureMedium

Question

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.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Chemical Bonding and Molecular Structure. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYChemical Bonding and Molecular Structureisostructuralcompoundfollowing

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