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What is the effect of the following processes on the bond order in N_(2) and O_(2) ? (A) N_(2) rarr N_(2)^(+) + e^(-) "(B) " O_(2) rarr O_(2)^(+) + e^(-) |
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Answer» Solution :According to molecular orbital theory, electronic configurations and bond order of `N_(2) , N_(2)^(+) , O_(2), and O_(2)^(+)` species as follows `N_(2) (14 e^(-)) = SIGMA 1s^(2), overset(**)(sigma) 1s^(2) , sigma 2s^(2) , overset(**)(sigma) 2s^(2), (pi 2p_(x)^(2) = pi 2p_(y)^(2)),sigma 2p_(z)^(2)` Bond order = `(1)/(2) [ N_(b) - N_(a) ]= (1)/(2) (10 - 4) = 3 ` `N_(2)^(+) ( 13 e^(-)) = sigma 1s^(2),overset(*)(sigma) 1s^(2) , sigma 2s^(2), overset(**)(sigma) 2s^(2), (pi 2p_(x)^(2) = pi 2p_(y)^(2)), sigma 2p_(z)^(1)` Bond order = `(1)/(2) [ N_(b) - N_(a)] = (1)/(2) (9 - 4) = 2.5 ` `O_(2) (16 e^(-))= sigma 1s^(2) , overset(**)(sigma) 1s^(2), sigma 2s^(2) , overset(**)(sigma) 2s^(2), sigma 2p_(z)^(2), (pi 2p_(x)^(2) = pi 2p_(y)^(2)), (overset(**)(pi) 2p_(x)^(1) = overset(**)(pi) 2p_(y)^(1))` Bond order = `(1)/(2) [ N_(b) - N_(a) ]= (1)/(2) (10 - 6) = 2 ` `O_(2)^(+) (15 e^(-) ) = sigma 1s^(2) , overset(**)(sigma), sigma 2s^(2) , overset(**)(sigma) 2s^(2), sigma 2p_(z)^(2) , (pi 2p_(x)^(2), = pi 2p_(y)^(2)), (overset(**)(pi) 2p_(x)^(1) = overset(**)(pi ) 2p_(y))` Bond order = `(1)/(2) [N_(b) - N_(a)] = (1)/(2) (10 - 5) = 2.5 ` (a) `N_(2) rarr N_(2)^(+) + e^(-)` B.O = 3 B.O. = 2.5 Thus, bond order decreases. (b)` O_(2) rarr "" O_(2)^(+) + e^(-)` B.O. = 3B.O.= 2.5 Thus, bond order increases. |
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