Arrange the elements with the following electronic configuration of valence electron in decreasing order of `Delta_(eg)H^(ɵ)`.
(a) `3s^(2)3p^(4)` , (b) `2s^(2)2p^(4)`
( c) `2s^(2)2p^(3)` , (d) `2s^(2)2p^(5)`

Explanation

`c gt b gt a gt d (N gt O gt S gt F)`
Elements `(b), (c )` and `(d)` are of the `2nd` period namely `O, N` and `F` respectively, whereas element (a) is of the `3rd` period and is `S` atom.
Out of `F, O, N` and `S`, only `N` has positive `Delta_(eg)H^(ɵ)` because of its stable half-filled electronic configuration. `F, O` and `S` have negative `Delta_(eg)H^(ɵ)`. But `F` has the most negative `Delta_(eg)H^(ɵ)`, since it requires one electron to acquire the nearest inert gas configuration.
Out of `O` and `S, O` has the less negative `Delta_(eg)H^(ɵ)` than `S` because of electron-electron repulsion present in its small and compact `2p`-orbital.
Thus decreasing order of `Delta_(eg)H^(ɵ)` is
`c gt b gt a gt d (N gt O gt S gt F)`