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MCQMediumJEE 2026VSEPR Theory & Shapes of Molecules

JEE Chemistry 2026 Question with Solution

Match List - I with List - II according to shape.

List - I A. XeO3\mathrm{XeO_3} B. XeF2\mathrm{XeF_2} C. XeO2F2\mathrm{XeO_2F_2} D. XeOF4\mathrm{XeOF_4}

List - II I. BrF5\mathrm{BrF_5} II. NH3\mathrm{NH_3} III. [I3]−\mathrm{[I_3]^-} IV. SF4\mathrm{SF_4}

Choose the correct answer from the options given below:

  • A

    A-II, B-III, C-I, D-IV

  • B

    A-II, B-I, C-III, D-IV

  • C

    A-II, B-III, C-IV, D-I

  • D

    A-III, B-II, C-IV, D-I

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Match the xenon compounds in List - I with the reference molecules in List - II according to shape.

Find: The correct matching.

Use VSEPR theory by counting bonding pairs and lone pairs around the central xenon atom.

  1. Analyze XeO3\mathrm{XeO_3}
  • Central atom Xe has 3 double bonds with O and 1 lone pair.
  • Shape: trigonal pyramidal, similar to NH3\mathrm{NH_3}.
A→II\text{A} \to \text{II}
  1. Analyze XeF2\mathrm{XeF_2}
  • Central atom Xe has 2 bonding pairs and 3 lone pairs.
  • Shape: linear, similar to [I3]−\mathrm{[I_3]^-}.
B→III\text{B} \to \text{III}
  1. Analyze XeO2F2\mathrm{XeO_2F_2}
  • Central atom Xe has 4 bonding pairs and 1 lone pair.
  • Shape: square pyramidal, similar to BrF5\mathrm{BrF_5}.
C→I\text{C} \to \text{I}
  1. Analyze XeOF4\mathrm{XeOF_4}
  • Central atom Xe has 5 bonding pairs and 1 lone pair.
  • Shape: distorted octahedral or square-pyramidal-like, matched here with SF4\mathrm{SF_4}.
D→IV\text{D} \to \text{IV}

Therefore, the final matching is

A→II,B→III,C→I,D→IV\text{A} \to \text{II}, \quad \text{B} \to \text{III}, \quad \text{C} \to \text{I}, \quad \text{D} \to \text{IV}

So, the correct option is A.

Shape Comparison Using Reference Molecules

Given: Xenon compounds and a list of reference molecules.

Find: Which reference shape matches each xenon compound.

The idea is to compare molecular shapes directly:

  • NH3\mathrm{NH_3} is trigonal pyramidal.
  • [I3]−\mathrm{[I_3]^-} is linear.
  • BrF5\mathrm{BrF_5} is square pyramidal.
  • SF4\mathrm{SF_4} is see-saw.

From the given analysis in the solution:

  • XeO3\mathrm{XeO_3} matches trigonal pyramidal ⇒\Rightarrow II
  • XeF2\mathrm{XeF_2} matches linear ⇒\Rightarrow III
  • XeO2F2\mathrm{XeO_2F_2} matches square pyramidal ⇒\Rightarrow I
  • XeOF4\mathrm{XeOF_4} is matched in the provided solution with IV

Hence,

A-II, B-III, C-I, D-IV\text{A-II, B-III, C-I, D-IV}

Therefore, the correct option is A.

Common mistakes

  • Treating every double bond as two separate electron domains. In VSEPR theory, a double bond counts as one electron domain for shape prediction. Count electron domains correctly before deciding the geometry.

  • Confusing electron-pair geometry with molecular shape. Lone pairs affect the observed molecular shape, so identify the positions of lone pairs before matching with reference molecules.

  • Matching compounds by formula appearance instead of by VSEPR structure. The correct approach is to count bonding pairs and lone pairs on xenon, then compare the resulting shape with the known reference molecules.

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