This Xenon Fluorides question repeats every year in NEET — See the pattern (XeF2, XeF4, XeF6)

PYQ + Trends

1. TITLE

This Xenon Fluorides question repeats every year in NEET — See the pattern (XeF2, XeF4, XeF6)

2. THE ACTUAL NEET QUESTION

NEET 2025 — Question
Match List-I with List-II

Choose the correct answer from the options given below:

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

3. OFFICIAL ANSWER

NEET Answer Key: (C)

4. COMPLETE SOLUTION

Concept Explanation:
To solve this, we use the hybridization formula: $H = \frac{1}{2} (V + M - C + A)$,.
* $V$ = Valence electrons of central atom (Xe = 8).
* $M$ = Number of monovalent atoms (F = 1, H = 1). Note: Oxygen is divalent and is not counted in $M$.
* $C/A$ = Cation/Anion charge.

Step-by-Step Breakdown:

1. $XeF_{2}$ (List-I B):

   • $H = \frac{1}{2} (8 + 2) = 5$. Hybridization: $sp^{3}d$.
   • Electron pairs = 5. Bond pairs (BP) = 2 (Xe-F). Lone pairs (LP) = $5 - 2 = 3$.
   • Geometry: Trigonal Bipyramidal.
   • Shape: Linear (3 Lone pairs occupy equatorial positions to minimize repulsion),.
   • Matches: I

2. $XeF_{6}$ (List-I D):

   • $H = \frac{1}{2} (8 + 6) = 7$. Hybridization: $sp^{3}d^{3}$.
   • Electron pairs = 7. Bond pairs = 6. Lone pairs = 1.
   • Shape: Distorted Octahedral,.
   • Matches: III

3. $XeO_{3}$ (List-I A):

   • $H = \frac{1}{2} (8 + 0) = 4$. (O is divalent, so M=0). Hybridization: $sp^{3}$.
   • Electron pairs = 4. Bond pairs = 3 (sigma bonds to O). Lone pairs = 1.
   • Shape: Pyramidal (Similar to Ammonia),.
   • Matches: II

4. $XeOF_{4}$ (List-I C):

   • $H = \frac{1}{2} (8 + 4) = 6$. (4 F atoms contribute to M). Hybridization: $sp^{3}d^{2}$.
   • Shape: Square Pyramidal.
   • Matches: IV

Conclusion: A-II, B-I, C-IV, D-III. Correct Option is C.

5. WHY NEET ASKED THIS

Examiner's Intent:
1. Testing Noble Gas Bonding: Xenon fluorides are the classic exception to the "inert gas" rule, and their structures are favorite test topics.
2. Hybridization Formula Application: They want to see if you remember that Oxygen does not count in the $H = \frac{1}{2}(V+M)$ formula for monovalent atoms, a common trap.
3. Lone Pair Repulsion: Specifically for $XeF_{2}$ and $XeF_{6}$, the shape is determined by lone pair placement (VSEPR theory),.

The "Trap":
Students often confuse the geometry (arrangement of electron pairs) with the shape (molecular structure). For example, $XeF_{4}$ has octahedral geometry but square planar shape. In this question, mixing up $XeF_{6}$ (Distorted Octahedral) with regular Octahedral is a potential error point.

6. TREND ANALYSIS

PATTERN ALERT:
Noble gas compounds are a high-yield sub-topic in Chemical Bonding.

Prediction for NEET 2026:
Expect a question focusing on Iso-structural pairs involving Xenon. For example: "Which molecule is isostructural with $XeF_{2}$?" (Answer: $I_{3}^{-}$ or $ICl_{2}^{-}$) or a deep dive into $XeOF_{4}$ specifically.

7. SIMILAR QUESTIONS TO PRACTICE

"If you understood this, you can solve:"
* NEET 2024 Q11: Match List-I ($XeF_{4}$) with List-II (Square Planar).
* NEET 2022 Q22: Amongst $ClF_{3}$, $IF_{5}$, $SF_{4}$, $XeF_{2}$, which has maximum 'lone pair - lone pair' repulsion? (Answer: $XeF_{2}$ with 3 lone pairs).
* NEET 2016 Q41: The correct geometry and hybridization for $XeF_{4}$ are? (Answer: Octahedral, $sp^{3}d^{2}$).

8. QUICK REVISION FOR THIS TYPE

For Xenon Fluoride questions, always memorize this table:

Watch out for:
* $XeF_{2}$ has 3 lone pairs (maximum repulsion cases).
* $XeF_{4}$ is Square Planar (not Tetrahedral).
* $XeF_{6}$ is Distorted Octahedral (due to 1 LP).

9. NCERT BASE

This question tests: NCERT Class 11, Chapter 4 (Chemical Bonding), Section 4.6 (Hybridisation) and Table 4.7 (Shapes of Molecules with Lone Pairs).
Also covered in Class 12, Group 18 Elements.

10. CTA

All PYQs from this chapter → Chemical Bonding & Molecular Structure Guide
Practice similar questions → Mock exams