What is the correct order of enthalpy of vapourization of noble gases?

He < Ne < Ar < Kr < Xe

The correct answer is option 1. He < Ne < Ar < Kr < Xe.

The enthalpy of vaporization, often denoted as ΔHvap, is the amount of heat energy required to vaporize one mole of a substance at its boiling point under standard conditions (constant temperature and pressure).  In the case of noble gases, the enthalpy of vaporization generally increases as the atomic size increases. This trend is primarily due to the increase in the strength of the London dispersion forces (also known as van der Waals forces) between the noble gas atoms as the atomic size increases.

Here is an explanation of the correct order:

Helium (He): Helium is the smallest noble gas with the smallest atomic size. It has weak London dispersion forces due to its small size, resulting in a relatively low enthalpy of vaporization compared to other noble gases.

Neon (Ne): Neon has a larger atomic size compared to helium, resulting in stronger London dispersion forces and a higher enthalpy of vaporization than helium.

Argon (Ar): Argon has a larger atomic size than neon, leading to stronger London dispersion forces and a higher enthalpy of vaporization compared to neon.

Krypton (Kr): Krypton has a larger atomic size than argon, resulting in stronger London dispersion forces and a higher enthalpy of vaporization compared to argon.

Xenon (Xe): Xenon has the largest atomic size among the noble gases listed, leading to the strongest London dispersion forces and the highest enthalpy of vaporization among the noble gases.

Therefore, the correct order of enthalpy of vaporization of noble gases is: \(He < Ne < Ar < Kr < Xe \)

This order reflects the trend of increasing atomic size and increasing strength of interatomic forces, resulting in higher enthalpies of vaporization as we move down the group of noble gases in the periodic table.