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: O, S, Se and Te have the same `ns^2np^4` valence shell electron configuration and hence are justified to be placed in group 16 of the periodic table.
`O(8)[He]2s^22p^4`
`S(16)[Ne]3s^23p^4`
`Se(34)[Ar]3d^(10)4s^24p^4`
`Te(52)[Kr]4d^(10)5s^25p^4`
(ii). Oxidation states: O, S, Se and Te need twio more electrons to form dinegative ion and acquire oxidation state of these elements shoud be -2. Oxygen predominantly and sulphur to some extent shows `O.S.-2`. Other elements of group 16, being less electronegative, have less stable `O.S.-2`. Since group 16 elements have 6 electrons in their valence shell, at maximum they can show `+6` O.S. other positive oxidation sattes shown by these elements is `+4` and `+6`. Thus on the basis of oxidation states shown by these elements they are justified to be placed in the same group i.e., group 16 of the periodic table.
(iii). Formation of hydrides: All the elements i.e., O,S Se and Te complete their respective octets by sharing of their valence electrons with 2 s-orbital of hydrogen to form hydrides of the type `EH_2` i.e., `H_2O,H_2S,H_2Se` and `H_2Te`. Thus on the basis of formation of hydrides of the general formula `EH_2` these elements are justified to be placed in the group ie., group 16 of the periodic table.