Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Explanation

1. Electronic configuration: All the given elements have the same `ns^(2)np^(4) (n = 2-6)` valence shell electronic configuration. Hence, they are rightly placed in the group-16 of the periodic table.
`""_(8)O=[He]2s^(2)2p^(4),""_(16)S=[Ne]3s^(2)3p^(4),`
`""_(34)Se=[Ar]3d^(10)4s^(2)4p^(4),""_(52)Te=[Kr]4d^(10)5s^(2)5p^(4)` and `""_(84)Po=[Xe]4f^(14)5d^(10)6s^(2)6p^(4)`
2. Oxidation state: To acquire the nearest noble gas configuration, they need just 2 electrons and thus form a dinegative ion. O predominantly and S to some extent exhibit - 2 oxidation states. The other elements of the group are less electronegative than O and S and hence, do not exhibit negative oxidation states. These elements have 6 electrons in the valence shell. Therefore the highest oxidation state exhibited by these elements are +6 along with other states of +2 and +4.
However, oxygen cannot show +6 and +4 states due to the absence of d -orbitals. The various oxidation states shown by these elements justify their positioning in the group-16 of the periodic table.
Formation of hydrides: Elements of this group complete their respective octets by sharing two of their valence electrons with the 1s-orbitals of hydrogen, forming hydrides `(EH_(2))` such as `H_(2)O , H_(2)S, H_(2)Se, H_(2)Te` and `H_(2)Po` . Thus on the basis of their general formula of the hydrides `(EH_(2))`, these elements are justified to be placed in group-16 of the periodic table.