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The observed bond length of N_(2)^(+) is larger than N_(2) while the bond length in NO^(+) is less than in NO. Why? |
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Answer» Solution :(a) By molecular orbital theory , the bond order of both `N_(2)^(+)` is `2.5` where as `N_(2) " is " 3`. `N_(2)` has `5e^(-)` in the anti - bonding molecular where as `N_(2)^(+)` has `4e^(-)` in the anti - bonding molecular orbital so `N_(2)^(+)` will make a STRONGER and shorter bond length . More the bond order and bond STRENGTH and lesser will be the bond length. So we can easily CONCLUDE `N_(2)` has more bond length that `N_(2)`. Bond order in `= N_(2) = (N_(b) - N_(a))/2 = (10 - 4)/ 2= 6/2 = 3` Bond order in ` = N_(2) = (N_(b) - N_(a))/2 = (9 - 4)/ 2 = 5/2 = 2.5` So `N_(2)` is more stable that `N_(2)^(+)` but bond length `N_(2)^(+)` is greater than `N_(2)`. (b) ` NO^(+) & NO ` : Bond order of NO = `2.5` Bond order of `NO^(+) = 3` Due to lesser bond order in NO the bond length is greater than `NO^(+) `. So `NO^(+)` bond length is shorter than NO bond length. |
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