(1) Electronic configuration:
\(\begin{aligned}
&\mathrm{O} \text { (At. no. =8) }=[\mathrm{He}] 2 s^2 2 p^4 \\
&\mathrm{~S}(\text { At. no. }=16)=[\mathrm{Ne}] 3 s^2 3 p^4 \\
&\mathrm{Se}(\text { At. no. }=34)=[\mathrm{Ar}] 3 d^{10} 4 s^2 4 p^4 \\
&\text { Te (At. no. }=52)=[\mathrm{Kr}] 4 d^{10} 5 s^2 5 p^4 \\
&\mathrm{Po} \text { (At. no. }=84)=[\mathrm{Xe}] 4 f^{14} 5 d^{10} 6 s^2 6 p^4
\end{aligned}\)
Thus, all these elements have the same \(n s^2 n p^4\) ( \(n=2\) to 6 ) valence shell electronic configuration, hence are justified to be placed in group 16 of the Periodic Table.
(2) Oxidation state : Two more electrons are needed to acquire the nearest noble gas configuration. Thus, the minimum oxidation state of these elements should be \(-2\).
\(\mathrm{O}\) and to some extent \(\mathrm{S}\) show \(-2\) oxidation state. Other element being more electropositive than \(\mathrm{O}\) and \(\mathrm{S}\), do not show negative oxidation state. As these contain six electrons, thus, maximum oxidation state shown by them is \(+6\). Other oxidation state shown by them are \(+2\) and \(+4\). O do not show \(+4\) and \(+6\) oxidation state, due to the absence of d-orbitals.
Thus, on the basis of maximum and minimum oxidation states, these elements are justified to be placed in the same group 16 of the periodic table.
(3) Hydride formation: All these elements share two of their valence electrons with \(1 s-\) orbital of hydrogen to form hydrides of the general formula \(\mathrm{EH}_2\), i.e., \(\mathrm{H}_2 \mathrm{O}\), \(\mathrm{H}_2 \mathrm{~S}, \mathrm{H}_2 \mathrm{Se}, \mathrm{H}_2\) Te and \(\mathrm{H}_2 \mathrm{Po}\). Thus, on the basis of hydride formation, these elements are justified to be placed in the same group 16 of the Periodic Table.