Why does small value of van der Waals gas constant "b" ensure easier liquefication? Larger value of b signifies that the size of the molecules is larger and hence the forces acting should become larger due to increased surface area.This should ensure easier liquefication. But my book says that smaller value of b corresponds to easier liquefication. Where am going wrong?

The book says "Chlorine is more easily liquefied than Ethane because 'a' for  chlorine is greater than 'a' for ethane but 'b' for fluorine is less than 'b' for ethane. Van der Waals constant 'a' is due to force of attraction and 'b' due to finite size of molecules. Thus, greater the value of 'a' and smaller the value of b, larger the liquefaction"
The $b$ in the Van der Waals equation of state signifies the volume that is effectively taken out per molecule by intermolecular repulsive interactions. You can imagine that a small $b$ will help the gas to liquefy, since in most ordinary liquids molecules are packed closer together than in a gas.

As written in Atkins Physical Chemistry: