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Give electron configuration, magnetic property bond order and energy diagram for Nitrogen (N_(2)) molecule. |
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Answer» Solution : `N_(2) (Z= 7) 1s^(2) 2S^(2) 2p^(3)`,total electron in `N_(2)` = 14 and affected valence electron = 10 Electron configuration in MO for `N_(2)` : KK `(SIGMA 2s)^(2) (sigma^(**) 2s)^(2) (pi 2p_(x))^(2) = (pi 2p_(y))^(2) (sigma 2p_(z))^(2) ` OR `(sigma 1s)^(2) (sigma^(**) 1s)^(2) (sigma 2s)^(2) (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) (8 - 2) = 3` OR BO `= (1)/(2) (N_(b) - N_(a)) = (1)/(2) (10 - 4) = 3 ` (Triple bond in `N_(2)`) Magnetic Property : All ELECTRONS are paired energy DIAGRAM for `N_(2)` Molecular 
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