The observed bond length of N_(2)^(+) is larger than N_(2) while the bond length in NO^(+) is less than in NO. Why ?

The observed bond length of N_(2)^(+) is larger than N_(2) while the b

1.	The observed bond length of N_(2)^(+) is larger than N_(2) while the bond length in NO^(+) is less than in NO. Why ?
Answer» Solution :(a) (i) By molecular orbital theory, the bond order of both `N_(2)^(+)` is 2.5 whereas `N_(2)` is 3. (II) `N_(2) has 5e^(-)` in the antibonding molecular orbital whereas `N_(2)^(+) has 4E^(-)` in the antibonding molecular orbital. So `N_(2)^(+)` will make a stronger and shorter bond lenght. 9iii) 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 than `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)` ismore stable than `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 lenght is shorter than NO bond lenght.