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Using molecular orbital theory, compare the bond energy and magnetic character ofO_(2)^(+) and O_(2)^(-) species. |
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Answer» Solution :According to molecular ORBITAL theory electronic configurations of `O_(2)^(+) and O_(2)^(-)` species are as follows : `O_(2)^(+) : (sigma 1S)^(2) (sigma^(**) 1s)^(2) (sigma 2s)^(2) (sigma^(**)2s)^(2) (sigma 2p_(z))^(2) (pi 2p_(x)^(2) , pi 2p_(y)^(2)) (pi^(**) 2p_(x)^(2))` BOND ORDER of `O_(2)^(+) = (10 - 5)/(2) = (5)/(2) = 2.5 ` `O_(2)^(-) : (sigma 1s)^(2) (sigma^(**) 1s)^(2) (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)^(2))` Bond order of `O_(2)^(-) = (10 -7)/(2) = (3)/(2) = 1.5 ` As Higher bond order of `O_(2)^(+)` shows that it is more stable than `O_(2)^(-)` . Both the species have UNPAIRED electrons. So, both are paramagnetic in nature. |
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