Q.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.

Step 1: Electronic configuration.

All five elements — O, S, Se, Te, Po — have the general outer electronic configuration $ns^2np^4$, i.e. six electrons in the valence shell, two short of achieving a stable noble-gas octet. This shared configuration is the fundamental basis for grouping them together.

Step 2: Oxidation states.

Because they are two electrons short of an octet, all these elements show a characteristic $-2$ oxidation state (by gaining 2 electrons, e.g. $O^{2-}, S^{2-}$) in compounds with more electropositive elements. In addition, S, Se, Te, Po (which can access d-orbitals, unlike O) also show positive oxidation states $+2, +4, +6$ in compounds with more electronegative elements like oxygen and the halogens (e.g. $SO_2, SO_3, SeO_3, TeF_6$).

Step 3: Hydride formation.

All five elements form covalent hydrides of the type $H_2E$ ($H_2O, H_2S, H_2Se, H_2Te, H_2Po$), reflecting their common valency of 2 towards hydrogen. The thermal stability of these hydrides decreases down the group ($H_2O$ most stable, $H_2Po$ least) as the E–H bond weakens with increasing atomic size, while their acidic strength and reducing character increase down the group.

Step 4: Overall justification. …

Justify the group placement using the three cited criteria in turn — shared ns²np⁴ configuration, common oxidation states, and a common H2E hydride-format …

• Do not just list oxidation states without linking them to the ns²np⁴ configuration (2 electrons short of octet) that is the actual justification.
• Oxygen does NOT show +4/+6 oxidation states (it lacks d-orbitals) — only S, Se, Te, Po do; stating all five show +2/+4/+6 uniformly is incorrect. …