1.

Which one of the following has highest bond order? N_(2), N_(2)^(+) or N_(2)^(-)?

Answer»

Solution :`N_(2)` (14 electrons)
Bond ORDER = 3, B.O. = `(N_(b)-N_(a))/(2)=(10-4)/(2)=3`
`N_(2)^(+)` (13 electrons)
Bond order = 2.5, B.O. `=(N_(b)-N_(a))/(2)=(9 - 4)/(2)=2.5`
`N_(2)^(-)` (15 electrons)
Bond order `= 2.5, B.O. = (N_(b)-N_(a))/(2)=(10 -5)/(2)=2.5`
So `N_(2)` has the HIGHEST bond order.


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