The numerical value of $ a $, the Van der Waals constant is maximum for:
(A) $ He $
(B) $ {H_2} $
(C) $ {O_2} $
(D) $ N{H_3} $

Hint: Van der Waals constant $ a $ is the magnitude of intermolecular attractive forces with the gas. It depends on how strongly the molecules of gas attract each other, which in turn helps in the ease liquefaction of a gas. Higher the value of $ a $ , greater is the intermolecular force and easier the liquefaction.

Complete answer:
  $ N{H_3} $ is a polar molecule and exhibits three of the polar molecules: dipole-dipole attraction, induced attraction and London dispersion forces. Since $ H $ is bonded to $ N $ , it exhibits hydrogen bonding.
Due to the dipole-dipole attraction, $ N{H_3} $ is easily liquefiable compared to other molecules.
Here are the values of $ a $ for the given molecules in ascending order:
 $ He(0.0346) < {H_2}(0.2476) < {N_2}(1.370) < N{H_3}(4.225) $ .
Option (D) is correct.

Note:
Intermolecular interactions are electrostatic i.e. between delta negatively and delta positively charged atoms. In $ N{H_3} $ , the interaction is between delta positive $ H{\text{s}} $ and delta negative $ N $ which is attractive.
The intermolecular attractive forces can vary to some extent with temperature and pressure and are always minimum for inert gases.