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Consider the elements N, P, O, S and arrange them in order of (a) increasing first ionization enthalpy, (b) increasing negative electron gain enthalpy, (c) increasing non-metallic character.

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Arranging all the given elements into different groups and periods in order of their increasing atomic numbers, we have, 

(a) Since ΔiH1 decreases down a group, therefore, ΔiH1 of N and O are higher than those of P and S. Further since N has more stable exactly half-filled electronic configuration in the 2p-subshell, therefore, it is more difficult to remove out an electron from N than from O even though O has higher nuclear charge. Similarly, P has exactly half-filled electronic configuration in the 3p-subshell. Therefore, ΔiH1 of P is higher than that of S. Thus, the overall increasing order of first ionization enthalpy of these elements follows the order: S < P < O < N. 

(b) Since adding an electron to smaller size 2p-orbital causes greater repulsion than adding an electron to larger 3p-orbital, therefore, electron gain enthalpies of P and S are more negative than those of N and O respectively. Further since S has higher nuclear charge but P has more stable exactly half-filled electronic configuration in the 3p-subshell, therefore, it is easier to add an electron to S than to P. In other words, electron gain enthalpy of S is more negative than that of P. Similarly, N has exactly half-filled electronic configuration in the 2p-subshell but O has higher nuclear charge. 

But the addition of an electron to N causes repulsions to such an extent that electron gain enthalpy of N is actually positive while that of O as expected in negative. Combining all the above results, the increasing order of negative electron gain enthalpy of these elements follows the order: N < P < O < S. 

(c) Since non-metallic character decreases down a group but increases along a period, therefore, O is the most non-metallic element while P is the least non-metallic element. The actual order of increasing non-metallic character is : P < S < N < O.

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