Problem 108
Question
The ease of liquefaction of noble gases increases in the order (a) \(\mathrm{Ar}>\mathrm{Kr}>\mathrm{Xe}>\mathrm{Ne}>\mathrm{He}\) (b) \(\mathrm{Xe}>\mathrm{Kr}>\mathrm{Ne}>\mathrm{Ar}>\mathrm{He}\) (c) \(\mathrm{Kr}>\mathrm{Xe}>\mathrm{He}>\mathrm{Ne}>\mathrm{Ar}\) (d) \(\mathrm{He}>\mathrm{Ne}>\mathrm{Ar}>\mathrm{Kr}>\mathrm{Xe}\)
Step-by-Step Solution
Verified Answer
The correct order is option (d): He > Ne > Ar > Kr > Xe.
1Step 1: Understand the Concept
The ease of liquefaction for noble gases is influenced by their intermolecular forces. The stronger the forces, the easier it is for a gas to liquefy. Van der Waals forces, which are dispersion forces, play a crucial role here. Larger atoms have more electrons, leading to stronger dispersion forces.
2Step 2: Identify Atomic Size Order
Noble gases in the periodic table are ordered by increasing atomic number as follows: He, Ne, Ar, Kr, Xe. With increasing atomic number, the atomic size increases.
3Step 3: Relate Atomic Size to Liquefaction
Since the ease of liquefaction increases with stronger intermolecular forces and these forces are stronger in larger atoms due to increased dispersion forces, larger noble gases are easier to liquefy.
4Step 4: Arrange According to Liquefaction Ease
Considering atomic sizes, the order of ease of liquefaction from smallest to largest is: He < Ne < Ar < Kr < Xe. Thus, larger atoms like Xe and Kr can be liquefied more easily than smaller ones like He and Ne.
5Step 5: Match With Given Options
Comparing our derived order with the given options, the correct order for increasing ease of liquefaction is option (d): dmathrm{He}>dmathrm{Ne}>dmathrm{Ar}>dmathrm{Kr}>dmathrm{Xe}.
Key Concepts
Intermolecular ForcesVan der Waals ForcesLiquefaction
Intermolecular Forces
Intermolecular forces are the interactions that occur between molecules. These forces are crucial for understanding how substances change states, like from gas to liquid. They are not as strong as the intramolecular forces which hold atoms together within a molecule, such as covalent bonds, but they play a significant role in the physical properties of substances.
There are several types of intermolecular forces:
There are several types of intermolecular forces:
- Dipole-Dipole Forces: Occur between polar molecules where partial positive and negative charges attract each other.
- Hydrogen Bonds: A special type of dipole-dipole interaction, stronger than other intermolecular forces, involving hydrogen atoms bonded to N, O, or F.
- London Dispersion Forces: Present in all molecules, especially significant in non-polar molecules and noble gases.
Van der Waals Forces
Van der Waals forces are the general term for the sum of the attractive or repulsive forces between molecules. These forces include both London dispersion forces and dipole-dipole interactions.
In noble gases, the most pertinent type of Van der Waals forces are the London dispersion forces. These occur because of temporary shifts in the electron cloud in atoms or molecules, creating a temporary dipole that can influence adjacent atoms or molecules, causing attraction. This is crucial for noble gases like helium and neon, which show these forces significantly due to their overall inert nature and lack of polarity.
The ease with which a gas can be liquefied correlates with the strength of these forces. As the atomic size increases, more electrons are involved, leading to stronger Van der Waals forces, thereby easing the liquefaction process of the noble gases.
In noble gases, the most pertinent type of Van der Waals forces are the London dispersion forces. These occur because of temporary shifts in the electron cloud in atoms or molecules, creating a temporary dipole that can influence adjacent atoms or molecules, causing attraction. This is crucial for noble gases like helium and neon, which show these forces significantly due to their overall inert nature and lack of polarity.
The ease with which a gas can be liquefied correlates with the strength of these forces. As the atomic size increases, more electrons are involved, leading to stronger Van der Waals forces, thereby easing the liquefaction process of the noble gases.
Liquefaction
Liquefaction is the process of turning a gas into a liquid. For noble gases, this transformation essentially depends on the ability of their atoms to stick together, which is influenced by intermolecular forces like Van der Waals forces.
The ease of liquefaction for noble gases increases with stronger intermolecular forces. Larger noble gases like krypton (Kr) and xenon (Xe), with more electrons, have stronger Van der Waals forces, making them easier to liquefy compared to smaller ones like helium (He) and neon (Ne).
When comparing the sequence of noble gases, starting from helium up to xenon, we see a clear pattern in which the capacity to transition from gaseous to liquid state becomes easier as you move from the lighter gases to the heavier ones. This is because the increase in atomic size contributes directly to the enhancement of the dispersion forces, thus facilitating liquefaction.
The ease of liquefaction for noble gases increases with stronger intermolecular forces. Larger noble gases like krypton (Kr) and xenon (Xe), with more electrons, have stronger Van der Waals forces, making them easier to liquefy compared to smaller ones like helium (He) and neon (Ne).
When comparing the sequence of noble gases, starting from helium up to xenon, we see a clear pattern in which the capacity to transition from gaseous to liquid state becomes easier as you move from the lighter gases to the heavier ones. This is because the increase in atomic size contributes directly to the enhancement of the dispersion forces, thus facilitating liquefaction.
Other exercises in this chapter
Problem 106
Noble gases do not react with other elements because (a) completely paired up electrons lead to stable electronic configuration (b) the size of their atoms is v
View solution Problem 107
The magnitude of enthalpy for formation of alkali metal halides decreases in the order (a) fluoride \(>\) chloride \(>\) iodide \(>\) bromide (b) iodide \(>\) b
View solution Problem 109
The elements which occupy the peaks of ionization energy curve are (a) \(\mathrm{Cl}, \mathrm{Br}, \mathrm{I}, \mathrm{F}\) (b) \(\mathrm{Na}, \mathrm{K}, \math
View solution Problem 111
The oxidation state of boron family shows which of the following trend for stable \(+1\) oxidation state? (a) \(\mathrm{Al}
View solution