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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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