1.

Correct order of bond energy is :

Answer»

`N_2gtN_2^(+)gtN_2^(-)gtN_2^(2-)`
`N_2^(+)gtN_2^(-)gtN_2^(2-)gtN_2`
`N_2gtN_2^(-)=N_2^(+)gtN_2^(2-)`
`N_2^(-)gtN_2=N_2^(+)gtN_2^(2-)`

Solution :Bond order is directly PROPORTIONAL to the bond energy.
Bond order of `N_2=3, N_2^(+), N_2^(-)=2.5 " " N_2^(2-)=2`
But `N_2^(-)` has more ELECTRONS in antibonding MO's and THUS `N_2^(+)` is more stable than `N_2^-` .So correct order of bond energy will be `N_2 gt N_2^(+) gt N_2^(-) gt N_2^(2-)`
Therefore, (A) option is correct.


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