
Which of the following inert gases liquefy easily?
A. Kr
B. He
C. Ne
D. Ar
Answer
160.8k+ views
Hint: Large van der waal’s attractive forces in a molecule are responsible for its easy liquefaction. Van der waals attractive forces increase with increase in size.
Complete Step by Step Answer:
Inert gases are present in group 18 of the periodic table. The elements in group 18 are Helium, Neon, Argon, Krypton, Xenon and Radon. The reactivity of these inert gases is very low due to their stable outer electron configuration and exists as monoatomic. But, these inert gases are liquefied at its critical temperature.
Liquification of gas is based on the attractive forces present between constituent molecules.
Inert gases are generally non-polar in nature. However, there are some attractive forces present in inert gas molecules. These attractive forces are responsible for their liquification . The Larger the magnitude of van der waal attractive forces, the more easily the gas will liquefy.
Along the group 18, the size of gases increases down the group due to addition of extra shells. Helium has the smallest size whereas Radon has maximum size. From the given inert gases, Kr (Krypton) is present in the fourth period and has a larger size than Helium, Neon and Argon. Therefore, the magnitude of van der waal’s attractive forces are more in Krypton than Helium, Neon and Argon. Hence, it can be easily liquefied.
So, option A is correct
Additional Information: Critical temperature of a gas is the temperature at which gas liquefies. Higher the critical temperature, higher the magnitude of attractive forces and more easily liquefy.
Note: In inert gases, Helium has the smallest size and small magnitude of van der waal’s attractive forces. It is more difficult to liquefy Helium gas in comparison to all other inert gases.
Complete Step by Step Answer:
Inert gases are present in group 18 of the periodic table. The elements in group 18 are Helium, Neon, Argon, Krypton, Xenon and Radon. The reactivity of these inert gases is very low due to their stable outer electron configuration and exists as monoatomic. But, these inert gases are liquefied at its critical temperature.
Liquification of gas is based on the attractive forces present between constituent molecules.
Inert gases are generally non-polar in nature. However, there are some attractive forces present in inert gas molecules. These attractive forces are responsible for their liquification . The Larger the magnitude of van der waal attractive forces, the more easily the gas will liquefy.
Along the group 18, the size of gases increases down the group due to addition of extra shells. Helium has the smallest size whereas Radon has maximum size. From the given inert gases, Kr (Krypton) is present in the fourth period and has a larger size than Helium, Neon and Argon. Therefore, the magnitude of van der waal’s attractive forces are more in Krypton than Helium, Neon and Argon. Hence, it can be easily liquefied.
So, option A is correct
Additional Information: Critical temperature of a gas is the temperature at which gas liquefies. Higher the critical temperature, higher the magnitude of attractive forces and more easily liquefy.
Note: In inert gases, Helium has the smallest size and small magnitude of van der waal’s attractive forces. It is more difficult to liquefy Helium gas in comparison to all other inert gases.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main Mock Test Series Class 12 Chemistry for FREE

Classification of Drugs Based on Pharmacological Effect, Drug Action

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

Solutions Class 12 Notes: CBSE Chemistry Chapter 1
