'$$a$$' and '$$b$$' are van der Waals' constants for gases. Chlorine is more easily liquefied than ethane because:

The van der Waals equation for one mole of a real gas is
$$\left(P+\frac{a}{V_m^{\,2}}\right)\left(V_m-b\right)=RT$$
where
  • $$a$$ measures the magnitude of intermolecular attractions (greater $$a \Rightarrow$$ stronger attractive forces).
  • $$b$$ represents the effective excluded volume of one mole of molecules (larger $$b \Rightarrow$$ bulkier molecules, greater repulsion).

For a gas to liquefy easily it should
  1. possess strong intermolecular attractions → a large $$a$$ value,
  2. have comparatively small molecular size → a small $$b$$ value so that molecules can approach one another more closely.

Comparing chlorine ($$\text{Cl}_2$$) and ethane ($$\text{C}_2\text{H}_6$$):
(i) Chlorine atoms contain many more electrons than the atoms in ethane. This makes the electron cloud of $$\text{Cl}_2$$ highly polarisable, creating strong instantaneous dipole-induced-dipole (London dispersion) forces. Hence $$a_{\text{(Cl}_2)} \gt a_{\text{(C}_2\text{H}_6)}$$.
(ii) Although $$\text{Cl}_2$$ is heavier, its linear diatomic structure gives it a smaller excluded volume than the three-dimensional, zig-zag ethane molecule. Therefore $$b_{\text{(Cl}_2)} \lt b_{\text{(C}_2\text{H}_6)}$$.

Because $$\text{Cl}_2$$ has both a larger $$a$$ and a smaller $$b$$ than $$\text{C}_2\text{H}_6$$, chlorine condenses (liquefies) more readily.

Thus the correct statement is:
Option C which is: $$a$$ for $$\text{Cl}_2 \gt a$$ for $$\text{C}_2\text{H}_6$$ but $$b$$ for $$\text{Cl}_2 \lt b$$ for $$\text{C}_2\text{H}_6$$.