1.

Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation states and hydride formation.

Answer» 1) Electronic configurations :
Oxygen (O) - [He] `2s^(2)2p^(2)`
Sulphur (S) - [Ne] `3s^(2)3p^(2)`
Selenium (Se) - [Ar] `3d^(10)4s^(2)4p^(4)`
Tellurium (Te) - [Kr] `4d^(10)5s^(2)sp^(4)`
Polonium (Po) - [Xe] `4f^(14)5d^(10)6s^(2)6p^(4)`
`to` All the above elements has general outer electronic configuration `ns^(2)np^(4)`.
2) Oxidation states :
All the gives elements (chalcogens) exhibits
common oxidation state of - 2
`to O^(-2),S^(-2,)Se^(-2)` etc.
3) Hydride formation :
All these elements (chalcogens) forms hydrides of type `EH_(2)` (E = chalcogen)
Eg : `H_(2)O,H_(2)S,H_(2)Se,H_(2)Te, H_(2)Po`.
The above mentioned concepts evident that the elements O, S, Se, Te and Po are present in the same group of periodic table.


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