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Calculate the bond order ofN_(2) , O_(2), O_(2)^(+) and O_(2)^(-) |
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Answer» Solution :(A) BOND order of `N_(2)` : In `N_(2) ` Z = 7 , so total electron = 14 Electron configuration of in MO for `N_(2) :(sigma 1s)^(2) (sigma^(**)1s)^(2) (sigma 2s)^(2) (PI 2p_(x))^(2) (pi 2p_(y))^(2) (sigma 2p_(z))^(2)` BO = `(1)/(2) (N_(b) - N_(a)) = (1)/(2) (10 - 4) = 3` (Triple bond ) (B) Bond order of `O_(2)` : In `O_(2), Z = 8` So, Total electron = 16 Electron configuration in MO for `O_(2) : (sigma 1s)^(2) (sigma^(**) 1s)^(2) (sigma 2s)^(2) (sigma^(**) 2s)^(2) (sigma 2p_(x))^(2) (pi 2p_(x))^(2) (pi 2p_(y))^(2)(pi^(**) 2p_(x))^(1) (pi^(**) 2p_(y))^(1)` BO = `(1)/(2) (N_(b) - N_(a)) = (1)/(2) (10 - 6) =` (Double bond ) (C) Bond order of `O_(2)^(+)`: ,br> Total electron in `O_(2) = 16 ` and total electron in Electron configuration in MO for `O_(2) : (sigma_(1s))^(2) (sigma_(1s)^(**) )^(2) (sigma_(2s))^(2) (sigma_(2s)^(**))^(2) (sigma_(2p_(z)))^(2) (pi_(2p_(x)))^(2) (pi_(2pi_(y)))^(2)(pi_(2pi_(x))^(**))^(1)(pi_(2p_(y))^(**))^(0)` BO = `(1)/(2) (N_(b) - N_(a)) = (1)/(2) (10- 5) = 2.5 ` (D) Bond order of `O_(2)^(-)` : Total electron in `O_(2)^(-) = 16 + 1 = 17` Electron configuration in MO for`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 = `(1)/(2)(N_(b) - N_(a))` `= (1)/(2) (10 - 7) = (3)/(2) = 1.5 ` |
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