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

(i) Electronic configuration.
`""_(8)O=[He]2s^(2)sp^(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)`
All these elements have same `ns^(2) np^(4) (n = 2" to "6)` valence shell electronic configuration and hence are justified to be placed in Group - 16 of the periodic table.)
(ii) Oxidation states : They need two more electrons to form dinegative ions to acquire the nearest noble gas configuration. So, the minimum oxidation state of these elements should be `-2`. Oxygen predominantly and sulphur to some extent being electronegative show an oxidation state shown by these elements are `+2 and +4`. Although , oxygen due to the absence of d - orbitals does not show oxidation states of `+4 and +6`. Thus, on the basis of minimum and maximum oxidation states, these elements are justified to be placed in the same group, i.e., Group - 16 of the periodic table.
Formation of hydrides : All the elements complete their respective octets by sharing two of their valence electrons with 1s - orbital of hydrogen to form hydrides of the general formula `EH_(2)`, i.e., `H_(2)O, H_(2)S, H_(2)Se, H_(2) Te and H_(2)Po`. Therefore, on the basis of formation of hydrides of the general formula, `EH_(2)`, these elements are justified to be placed in Group 16 of the periodic table.