Which one has the highest boiling point?

We are asked to compare the boiling points of the noble gases $$\text{He},\; \text{Ne},\; \text{Kr},\; \text{Xe}.$$

First, we recall a key fact about intermolecular forces. For atoms or non-polar molecules, the only significant attractive force is the London dispersion (van der Waals) force. The empirical observation and theoretical treatment both show that

$$\text{Magnitude of dispersion force} \;\;\propto\;\; \alpha \times N_e^2,$$

where $$\alpha$$ is the polarisability of the electron cloud and $$N_e$$ is the number of electrons. In simpler words, a heavier atom with more electrons has a more easily distortable (more polarisable) electron cloud, leading to stronger attractive forces between neighbouring atoms.

Now, the boiling point is the temperature at which intermolecular attractions are overcome and the substance changes from liquid to gas. Stronger attractions mean more energy (higher temperature) is required, so

$$\text{Stronger dispersion forces} \;\;\Longrightarrow\;\; \text{Higher boiling point}.$$

Next, we list the atomic masses and electron counts in increasing order:

$$ \text{He}: 4\;(\text{u}),\; 2\;e^- \\ \text{Ne}: 20\;(\text{u}),\; 10\;e^- \\ \text{Kr}: 84\;(\text{u}),\; 36\;e^- \\ \text{Xe}: 131\;(\text{u}),\; 54\;e^- $$

Clearly, polarisability increases in the same sequence:

$$\alpha(\text{He}) < \alpha(\text{Ne}) < \alpha(\text{Kr}) < \alpha(\text{Xe}).$$

Therefore the strength of London dispersion forces follows exactly this order, and consequently the boiling points follow the same ascending trend:

$$T_b(\text{He}) < T_b(\text{Ne}) < T_b(\text{Kr}) < T_b(\text{Xe}).$$

The largest of the given boiling points thus belongs to xenon.

Hence, the correct answer is Option A.