Question

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.

Answer 1

Electronic configuration The electronic configuration of the above mentioned elements (group 16) is given below.

$\begin{array}{ccc}Element& ElectronicConfiguration& Oxidationstates\\ 8^O& \left[He\right]2s^{2}2p^4& -2,-1,+1\\ {}_{16}S& \left[Ne\right]3s^{2}3p^4& -2,+2,+4,+6\\ {}_{34}Se& \left[Ar\right]3d^{10}4s^{2}4p^4& -2,+2,+4,+6\\ {}_{52}Te& \left[Kr\right]4d^{10}5s^{2}5p^4& -2,+2,+4,+6\\ {}_{84}Po& \left[Xe\right]4f^{14}5d^{10}6s^{2}6p^4& +2,+4\end{array}$

All these elements have similar valence shell configuration $n{s}^2$ $n{p}^4$, hence their position in group 16 with each other is justified. Oxidation states are mentioned in the above table. As we observe that in all elements of group 16 oxidation states, polonium shows +4 oxidation state. Oxygen does not exhibit +4 and +6 oxidiation states due to absence of d- orbitals. Therefore, their position in group 16 is justified. Hydride formation All these elements form hydrides of general formula $H_{2}E$ e.g., $H_{2}O,H_{2}S,H_{2}Se,H_{2}Teand{H}_{2}Po$. All the hydrides with exception of $H_{2}O$, possess reducing properties, have acidic nature and are thermally not stable. On the basis of the above mentioned similarities these elements are correctly placed in group 16.