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Which of the following pair is expected to have the same bond order ? |
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Answer» `O_(2), N_(2) ` On the base of molecular orbital THEORY we can calculate bond order of molecules ions as BO = `(1)/(2) (N_(B) - N_(a))` Molecular orbital electronic configuration (MOEC) of `N_(2)` is : `sigma 1s^(2), sigma(8*) 1s^(2), sigma 2S^(2), sigma^(**) 2s^(2), pi 2p_(X)^(2) = pi 2p_(y)^(2) , sigma 2p_(x)^(2) ` Bond order of `N_(2)^(+) = (1)/(2) (10 - 4) =3 ` MOEC of `N_(2)^(+) = sigma 1s^(2) , sigma^(**) 1s^(2) , sigma 2s^(2) , sigma^(**) 2s^(2), pi 2p_(x)^(2) = pi 2p_(y)^(2) , sigma 2p_(z)` BO of `N_(2)^(+) = (1)/(2) (9 - 4) = 2.5 ` MOEC of `N_(2)^(-) = sigma 1s^(2) , sigma^(**) 1s^(2), sigma 2s^(2), sigma^(**) 2s^(2),pi 2p_(x)^(2) = pi 2p_(y)^(2), sigma 2p_(z)^(2) , pi 2p_(x)^(1) = pi^(**) 2p_(y)` BO of `N_(2)^(-) = (1)/(2) ( 10 - 5)= 2.5` MOEC of `O_(2) =sigma 1s^(2), sigma^(**) 1s^(2) , sigma^(**) 2s^(2) , sigma 2p_(z)^(2) , pi 2p_(x)^(2) = pi 2p_(y)^(2) , pi^(**) 2p_(x)^(2) = pi^(**)2p_(y)^(1)` BO of `O_(2) = (1)/(2) (10 - 6) = 2 ` MOEC of `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)^(1)` BO of `O_(2)^(-) = (1)/(2) (10 - 7) =1.5` MOEC of `O_(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)` BO of `O_(2)^(+) = (1)/(2) (10 - 5) = 2.5 ` (A) Bond order of `O_(2) and N_(2)` are 2 and 3, respectively. (B) Bond order of both `O_(2)^(+) and N_(2)^(-) ` and 1.5 and 2.5 (C) Bond order of `O_(2)^(-) and N_(2)^(+) ` are 1.5 and 2.5, respectively. Bond order of `O_(2)^(-) and N_(2)^(-)` are 1.5 and 2.5 respectively. |
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