`N_(2)` and `O_(2)` converted to monopositive cations `N_(2)^(+)` and `O_(2)^(+)` respectively. Which is incorrect ?

Options

A. In `N_(2)^(+)`, the N-N bond is weakened
B. In `O_(2)^(+)`, the bond order increases
C. In `O_(2)^(+)`, paramagnetism decreases
D. `N_(2)^(+)` becomes diamagnetic

Answer

D

Explanation

MO configuration of `N_(2)` molecule is
`(sigma 1)^(2)(sigma ** 1s)^(2)(sigma2s)^(2)(sigma **2s)^(2)(pi2p_(x))^(2)(pi 2p_(y))^(2)(sigma 2p_(z))^(2)`
`therefore` Bond order of `N_(2) = (10-4)/(2) = 3`
MO configuration of `N_(2)^(+)` is
`(sigma 1)^(2)(sigma ** 1s)^(2)(sigma2s)^(2)(sigma **2s)^(2)(pi2p_(x))^(2)(pi 2p_(y))^(2)(sigma 2p_(z))^(2)`
Thus, `N_(2)^(+)` becomes paramagnetic.
`because` In `N_(2)^(+)`, the N-N bond is weakend.
MO configuration of `O_(2)` is
`(sigma 1)^(2)(sigma ** 1s)^(2)(sigma2s)^(2)(sigma **2s)^(2)(sigma2p_(z))^(2)(pi2p_(x))^(2)(pi 2p_(y))^(2)(pi ** 2p_(x))^(1)(pi ** 2p_(y))^(1)`
`therefore` Bond order `= (10 - 6)/(2) = 2`
MO configuration of `O_(2)^(+)` is
`(sigma 1)^(2)(sigma ** 1s)^(2)(sigma2s)^(2)(sigma **2s)^(2)(sigma2p_(z))^(2)(pi2p_(x))^(2)(pi 2p_(y))^(2)(pi ** 2p_(x))^(1)`
`therefore` Bond order `= (10 - 5)/(2) = 2.5`
hence, `O_(2)` and `O_(2)^(+)` both are paramagnetic, but in `O_(2)^(+)` paramagnetism decresaes.