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{:(,"Column I",,"Column II "), ((A),B_(2),(p),"Paramagnetic"),((B),N_(2),(q),"Undergoes oxidation "),((C),O_(2)^(-),(r),"Undergoes reduction"),((D),O_(2) ,(s),"Bond order"ge 2 ),(,,(t),"Mixing of s and p-orbitals"):} |
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Answer» It containsunpaired electrons . Hence, it is paramagnetic . Highest occupied molecular orbitals are bonding . Addition of electron increases stability . Hence , it undergoes reduction . BOND order = 1. As 2s and 2p orbitlas are close , mixing of these orbitals taken place. `(B) N_(2)= KK sigma (2s)^(2) sigma^(**) (2s)^(2) pi (2p_(x))^(2) pi(2p_(y))^(2) sigma(2p_(z))^(2)` All electrons are paired . Hence , it is diamagnetic . All orbitalsare completely filled . It is stable. Hence , it can neither gain nor lose electons/s, i.e., it does not undergo oxidition or reduction . Its bond order = 3.2 and 2p orbitals are close . Hence , there is 2s - 2p mixing . `(C) O_(2)^(-)= KK sigma (2s)^(2) sigma^(**) (2s)^(2)sigma (2p_(z))^(2) pi (2p_(x))^(2)` `pi(2p_(y))^(2) pi^(**)(2p_(x))^(2) pi^(**) (2p_(y))^(1)` Itis paramagnetic . Loss of electrons from antibonding molecular orbitals orbitals will increase stability. Hence, it undergoes OXIDATION . Bond order = 1.5 . 2s and 2p orbitals have LARGE DIFFERNECE in energy . There is no 2s - 2pmixing . `(D) O_(2)= KK sigma (2s)^(2) sigma^(**) (2s)^(2)sigma (2p_(z))^(2) pi (2p_(x))^(2)` `pi(2p_(y))^(2) pi^(**)(2p_(x))^(1) pi^(**) (2p_(y)^(1))` It is paramagnetic .Lose of electrons frm anti- bonding molecular orbitals will increase stability . Hence , it undergoes oxidation . Bond order 2. As 2s and 2p have large difference in energy , there is no 2s- 2p mixing |
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