Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option.
List I
List II
(P) XeF2 (1) Trigonal bipyramidal and two lone pair of electrons (Q) XeF4 (2) Tetrahedral and one lone pair of electrons (R) XeO3 (3) Octahedral and two lone pair of electrons (S) XeO3F2 (4) Trigonal bipyramidal and no lone pair of electrons (5) Trigonal bipyramidal and three lone pair of electrons
List I | List II | ||
| (P) | XeF2 | (1) | Trigonal bipyramidal and two lone pair of electrons |
| (Q) | XeF4 | (2) | Tetrahedral and one lone pair of electrons |
| (R) | XeO3 | (3) | Octahedral and two lone pair of electrons |
| (S) | XeO3F2 | (4) | Trigonal bipyramidal and no lone pair of electrons |
| (5) | Trigonal bipyramidal and three lone pair of electrons | ||
- P-5, Q-2, R-3, S-1
- P-5, Q-3, R-2, S-4
- P-4, Q-3, R-2, S-1
- P-4, Q-2, R-5, S-3
The Correct Option is B
Approach Solution - 1
Step 1: Determining the Geometry and Lone Pairs of Each Xenon Compound
- (P) XeF2
- Central atom: Xe
- Valence electrons: 8
- Bonding pairs: 2 (due to 2 F atoms)
- Lone pairs: \( 8 - 2 = 6 \) (3 lone pairs)
- Geometry: Trigonal bipyramidal (linear shape due to lone pairs in equatorial positions)
- Matches: (5)
- (Q) XeF4
- Central atom: Xe
- Valence electrons: 8
- Bonding pairs: 4 (due to 4 F atoms)
- Lone pairs: \( 8 - 4 = 4 \) (2 lone pairs)
- Geometry: Octahedral (square planar shape due to lone pairs at axial positions)
- Matches: (3)
- (R) XeO3
- Central atom: Xe
- Valence electrons: 8
- Bonding pairs: 3 (due to 3 O atoms)
- Lone pairs: \( 8 - 3 = 5 \) (1 lone pair)
- Geometry: Tetrahedral (pyramidal shape due to lone pair)
- Matches: (2)
- (S) XeOF4
- Central atom: Xe
- Valence electrons: 8
- Bonding pairs: 5 (3 O atoms and 2 F atoms)
- Lone pairs: \( 8 - 5 = 3 \) (0 lone pairs)
- Geometry: Trigonal bipyramidal (no lone pairs present)
- Matches: (4)
Step 2: Conclusion
- (P) → (5)
- (Q) → (3)
- (R) → (2)
- (S) → (4)
Thus, the correct answer is (B) P-5, Q-3, R-2, S-4.
Approach Solution -2
To solve the problem, we need to use the VSEPR model to determine the molecular geometries and lone pairs on xenon compounds and match them with the given options.
1. Xenon Difluoride (XeF2):
- Total electron pairs: 5 (2 bonding pairs, 3 lone pairs)
- Geometry: Trigonal bipyramidal
- Lone pairs: 3
- Corresponds to option (5) "Trigonal bipyramidal and three lone pairs".
2. Xenon Tetrafluoride (XeF4):
- Total electron pairs: 6 (4 bonding pairs, 2 lone pairs)
- Geometry: Octahedral
- Lone pairs: 2
- Corresponds to option (3) "Octahedral and two lone pairs".
3. Xenon Trioxide (XeO3):
- Total electron pairs: 4 (3 bonding pairs, 1 lone pair)
- Geometry: Tetrahedral
- Lone pairs: 1
- Corresponds to option (2) "Tetrahedral and one lone pair".
4. Xenon Oxyfluoride (XeO3F2):
- Total electron pairs: 5 (5 bonding pairs, 0 lone pairs)
- Geometry: Trigonal bipyramidal
- Lone pairs: 0
- Corresponds to option (4) "Trigonal bipyramidal and no lone pairs".
Final Matching:
P - 5, Q - 3, R - 2, S - 4
Final Answer:
Option (B)
Top Questions on Chemical bonding and molecular structure
- The number of molecules having lone pair of electrons on central atom in the following is: BF$_3$, SF$_4$, SiCl$_4$, XeF$_4$, NCl$_3$, XeF$_6$, PCl$_5$, HgCl$_2$, SnCl$_2$
- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
- Identify the pair of molecules which have the same hybridisation as the hybridisation in Xenon (II) fluoride.
- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
- The correct order of increasing bond lengths of C--H, O--H, C--C, and H--H is
- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
How many of the following molecules / ions have a trigonal planar structure?
\( \text{BO}_3^{3-}, \, \text{NH}_3, \, \text{PCl}_3, \, \text{BCl}_3, \, \text{ClF}_3, \, \text{XeO}_3 \)- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
- Consider the following
Assertion (A): Dipole moment of NF\(_3\) is lesser than NH\(_3\)
Reason (R): In NF\(_3\), the orbital dipole due to lone pair of electrons is in the opposite direction to the resultant dipole moment of the three N-F bonds.
The correct answer is:- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
Questions Asked in JEE Advanced exam
- At 25$^\circ$C, the concentration of H$^+$ ions in $ 1.00 \times 10^{-3} \, \text{M} $ aqueous solution of a weak monobasic acid having acid dissociation constant $ K_a = 4.00 \times 10^{-11} $ is $ X \times 10^{-7} \, \text{M} $. The value of $ X $ is ____. {Use: Ionic product of water $ K_w = 1.00 \times 10^{-14} $ at 25$^\circ$C
- JEE Advanced - 2025
- Ionic Equilibrium
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
- JEE Advanced - 2025
- Ray optics and optical instruments
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
