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Give electron configuration,magnetic property bond order and energy diagram for oxygen (O_(2))molecule. |
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Answer» Solution :`O_(2) (Z = 8 ) 1 s^(2) 2s^(2) 2p^(4)` ,So total electron in `O_(2)`= 16 ,& 12 valence electrons are participate in bond. electron configuration in MO for `O_(2)` . KK `(sigma 2s)^(2) (sigma^(**)2s)^(2) (sigma 2p_(x))^(2) = (pi 2p_(y))^(2) (pi^(**) 2p_(x))^(1) (pi^(**) 2p_(y))^(1)` Bond order `= (1)/(2)(N_(b) - N_(a))` = `(1)/(2) (10 - 6)= 2("DOUBLE bond" O_(2))` in it unpair `bar(E) " in " pi_(2p_(x))^(**) and pi_(2p_(y))^(**) `, So paramagnetic Energy diagram for `O_(2)` molecule : 
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