NCERT Class XII Chemistry
Chapter - The p-Block Elements
Questions with Solutions
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Question : 17
Total: 74
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.
Solution:
(i) Electronic configuration :
8 O = [ H e ] 2 s 2 2 p 4 ; 16 S = [ N e ] 3 s 2 3 p 4 ;
34 S e = [ A r ] 3 d 10 4 s 2 4 p 4
52 T e = [ K r ] 4 d 10 5 s 2 5 p 4
and84 P O = [ X e ] 4 f 14 5 d 10 6 s 2 6 p 4
All these elements have same ns2np4(n = 2 to 6) valence shell electronicconfiguration and hence are justified to be placed in group 16 of theperiodic table.
(ii) Oxidation states : They need two more electronsto form dinegativeions by acquiring the nearest noble gas configuration. So, the minimumoxidation state of these elements should be –2. Oxygen predominantlyand sulphur to some extent being electronegative show an oxidationstate of –2. Since these elementshave six electrons in the valence shell,therefore, the maximum oxidation state they can show is, + 6. Otherpositive oxidation states shown by these elements are +2 and +4. Although,oxygen due to the absence of d-orbitals does not show oxidation statesof+4 and +6. Thus, on the basis of minimum and maximum oxidation states,these elements are justified to be placed in the same group i.e., group 16of the periodic table.
(iii)Formation of hydrides : All the elements complete their respectiveoctetsby sharing two of their valence electrons with 1s-orbital of hydrogen to form hydrides of the general formulaE H 2 i . e . , H 2 O , H 2 S , H 2 S e , H 2 T e H 2 P o E H 2 ,
and
All these elements have same ns2np4(n = 2 to 6) valence shell electronicconfiguration and hence are justified to be placed in group 16 of theperiodic table.
(ii) Oxidation states : They need two more electronsto form dinegativeions by acquiring the nearest noble gas configuration. So, the minimumoxidation state of these elements should be –2. Oxygen predominantlyand sulphur to some extent being electronegative show an oxidationstate of –2. Since these elementshave six electrons in the valence shell,therefore, the maximum oxidation state they can show is, + 6. Otherpositive oxidation states shown by these elements are +2 and +4. Although,oxygen due to the absence of d-orbitals does not show oxidation statesof+4 and +6. Thus, on the basis of minimum and maximum oxidation states,these elements are justified to be placed in the same group i.e., group 16of the periodic table.
(iii)Formation of hydrides : All the elements complete their respectiveoctetsby sharing two of their valence electrons with 1s-orbital of hydrogen to form hydrides of the general formula
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