Arrange the elements N, P, O and S in the order of
i) increasing first ionisation enthalpy.
ii) increasing non-metallic character.
Give reason for the arrangement assigned.

Explanation

Arrange the elements N, P, O and S into different groups and periods in order of their increasing atomic numbers we have
`{:("Group " to ,,15,,16),("Period " 2,,N,,O),("Period " 3 ,,P ,,S):}`
(i) the electronic configuration of `N (1 s^(2) 2s^(2) 2p_(y)^(1) 2p_(y)^(1) 2p_(z)^(1))` in which 2p-orbitals are exactly half - filled is more stable than the electronic configuration of `O (1s^(2) 2s^(2) 2p_(x)^(2) 2p_(y)^(1) 2p_(z)^(1))` in which 2p- orbitals are neither half- filled nor completely filled. Therefore it is difficult to remove an electron from N and than from O in spites of the fact the O (+ 8) has highest nuclear charge than `N ( +7)` . As a result `Delta_(i) H_(1)` of N is higher thant that of O. Similarly the electronic configuration of `P (1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p_(x)^(1) 3p_(y)^(1) 3p_(z)^(1))` in which 3p- orbirtals are exactly half- filled is more stable than the electronic configuration of `S (1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p_(x)^(2) 3p_(y)^(1) 2p_(z)^(1))` in which 3p- orbitals are neither half - filled nor completely filled. Therefore is is difficult to remove on electron from P than from S in spite of the fact that S (+ 16) has higher nuclear charge than P (+ 15). Thus `Delta_(i) H_(1)` of P is higher than that S.
Further since `Delta_(i)H_(1)` decreases down a group therefore `Delta_(i)H_(i)` of N is greater than that of P and that of O is greater than that of S. Combining the two results together the first ionization enthalpies of N, P, O and increases in the order `: S lt P lt O lt N`
(ii) Since non metallic character increases along a period and decreases down a group , therefore non metallic character increases in the order `: P lt S lt N lt O`