-
Engineering and Architecture
Exams
Colleges
Predictors
Resources
-
Computer Application and IT
Quick Links
Colleges
-
Pharmacy
Colleges
Resources
-
Hospitality and Tourism
Colleges
Resources
Diploma Colleges
-
Competition
Other Exams
Resources
-
School
Exams
Top Schools
Products & Resources
-
Study Abroad
Top Countries
Student Visas
-
Arts, Commerce & Sciences
Exams
Colleges
Upcoming Events
Resources
-
Management and Business Administration
Colleges & Courses
Predictors
-
Learn
Online Courses
Engineering Preparation
Medical Preparation
-
Online Courses and Certifications
Top Streams
Specializations
- Digital Marketing Certification Courses
- Cyber Security Certification Courses
- Artificial Intelligence Certification Courses
- Business Analytics Certification Courses
- Data Science Certification Courses
- Cloud Computing Certification Courses
- Machine Learning Certification Courses
- View All Certification Courses
Resources
-
Medicine and Allied Sciences
Colleges
Predictors
Resources
-
Law
Resources
Colleges
-
Animation and Design
Predictors & Articles
Colleges
Resources
-
Media, Mass Communication and Journalism
Colleges
Resources
-
Finance & Accounts
Top Courses & Careers
Colleges
Get Answers to all your Questions
- #Chemical Bonding and molecular structure
- #Medical
- #National Eligibility Cum Entrance Test
- #Chemistry
Q.81) Match List I with List - II
| List-I | List-II |
| A. $\mathrm{XeO}_3$ | I. $s p^3 d$; linear |
| B. $\mathrm{XeF}_2$ | II. $s p^3$, pyramidal |
| C. $\mathrm{XeOF}_4{ }^{-}$ | III. $\mathrm{sp}^3 \mathrm{~d}^3$; distorted octahedral |
| D. $\mathrm{XeF}_6$ | IV. $s p^3 d^2$; square pyramidal |
Choose the correct answer from the options given below:
A) A-IV, B-II, C-I, D-III
B) A-II, B-L, C-IV, D-III
C) A-II, B-I, C-III, D-IV
D) A-IV, B-II, C-III, D-I
Answers (1)
Q.81) A. $\mathrm{XeO}_3$
- Xenon has 8 valence electrons, forms $\mathbf{3} \mathrm{Xe}=\mathbf{O}$ double bonds and 1 lone pair
- Total regions $=\mathbf{3}$ bonds +1 lone pair $=\mathbf{4} \boldsymbol{\rightarrow} \mathrm{sp}^{\mathbf{3}}$ hybridization
- $\quad$ Shape $=$ Trigonal pyramidal
Match: II. $\mathbf{s p}^{\mathbf{3}}$; pyramidal
B. $\mathrm{XeF}_2$
- Xenon has 8 valence electrons, forms 2 bonds with $\mathbf{F}$, and has 3 lone pairs
- Total regions $=2$ bonds +3 lone pairs $=5 \rightarrow \mathbf{s p}^{\mathbf{3}} \mathrm{d}$ hybridization
- Electron geometry = trigonal bipyramidal; shape $=$ linear (3 lone pairs occupy equatorial positions)
Match: I. sp ${ }^3$ d; linear
C. $\mathrm{XeOF}_4^{-}$
- Think of parent: $\mathrm{XeOF}_4$ has $10,4 \mathrm{~F}=5$ bonds +1 lone pair $=6$ regions $\rightarrow \mathrm{sp}^3 \mathrm{~d}^2$
- The anion $\left(\mathrm{XeOF}_4\right)$ implies an extra electron, possibly altering geometry slightly, but overall still has 6 electron domains
- Likely shape = square pyramidal
Match: IV. sp ${ }^3{ }^2$; square pyramidal
D. $\mathrm{XeF}_6$
- Xenon forms $\mathbf{6}$ bonds with $F$, and has 1 lone pair $\rightarrow 6+1=7$ regions
- Hybridization $=s p^3 d^3$
- Geometry: distorted octahedral (due to lone pair)
Match: III. sp$^3$; distorted octahedral
Hence, the correct answer is option (2).
NEET 2024 Most scoring concepts
- Just Study 32% of the NEET syllabus and Score up to 100% marks
