SELECT * FROM question_mgmt as q WHERE id=1357 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=45 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

Compare the relative stability of the following species and indicate their magnetic properties:

O2,O2+,O2- (superoxide), O22-(peroxide)

Answer:

The stability of following species can be decided on the basis of bond order as follows:

O2 : KK

Bond order =1/2(Nb-Na)

= 1/2(8-4)

= 2 = Paramagnetic

 

Similarly, the electronic configuration of O2+ can be written as:

Bond order of  O2+  = 1/2(8-3)

= 2.5 = paramagnetic

 

Electronic configuration of O2- ion will be:

Bond order of = 1/2(8-5)

= 1.5 = paramagnetic

 

Electronic configuration of O22- ion will be:

Bond order of O22- =1/2(8-6) = 1  diamagnetic

 

The larger the bond order,greater is the stability

Therefore the stability decreases as O2+  > O2 > O2- > O22-.


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