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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. |
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Answer» Solution :(i) Electronic configuration : All these elements have the common `ns^(2) np^(4)` (n = 2 to 6) valence shell electronic configuration. `""_8O=[He] 2s^(2) 2p^4 , ""_(16)S = [Ne]3s^(2) 3p^4 , ""_16S= [ Ne ] 3s^(2) 3p^4 , ""_(34) Se = [Ar] 3d^(10) 4s^(2) 4p^(4)` `""_(52) Te = [Kr] 4d^(10)5s^(2) 5p^(4) and ""_(84) Po = [Xe] 4f^(14) 5d^(10) 6s^(2) 6p^(4)` Hence, it is justified to place them in Group 16 of the periodic table. (II) Oxidation states : They need two more electrons to form dinegative ions and acquire the nearest inert gas configuration. Thus, the minimum oxidation state of these elements should be -2. Oxygen and sulphur being electronegative show an oxidation state of -2. Other elements of this group, being more electropositive than O and S, do not show negative oxidation states. Since these elements have six electrons in the valence shell, THEREFORE, at the maximum they can show an oxidation state +6. Other positive oxidation states SHOWN by these elements are +2 and +4. However, due to the absence of d-orbtials, oxygen does not show oxidation states of +4 and +6. Thus, on the basis of minimum and maximum oxidation states, these elements are justified to be placed in the same group. (iii) Formation of hydrides : All the elements complete their respective octets by SHARING two of their valence electrons with 1s-orbital of hydrogen to form hydrides of the general formula `EH_2` i.e., `H_2O, H_2S , H_2 Se , H_2 Te and H_2Po`. Thus, on the basis of formation of hydrides of the general formula EH,, these elements are justified to be placed in Group 16 of the periodic table. |
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