Arrange the following in decreasing order of their van der Waals’ radii Cl, H, O, N.

The van der Waals’ radii increase as the number of energy shells increases and decreases as the nuclear charge increases. Since H has only one energy shell and Cl has three, therefore, the van der Waals’ radius of H is the smallest while that of Cl is the largest. Further both N and O have two energy shells but the nuclear charge on O (+8) is higher than that on N (+7), therefore, the van der Waals’ radius of N is bigger than that of O.

Thus, the overall decreasing order is Cl > N > O > H.