In terms of the molecular orbital theory , which of the following species will most likely be the one to gain an electron to form thermodynamically more stable species?

Options

A. `CN`
B. `NO`
C. `O_(2)^(2+)`
D. `N_(2)`

Answer

1

Explanation

`(1)CN(6+7=13)=sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma 2p_(z)^(1),` Bond order `=(9-4)/(2)=2.5`
`CN^(-)(6+7+1=14)=sigma1s^(2)sigma^(**)1s^(2)sigma2s^(2)sigma^(**)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma2p_(z)^(2),` Bond order `=(10-4)/(2)=3`
`(2)NO(7+8=15)=sigma 1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma2p_(z)^(2) pi2p_(x)^(2)=pi2p_(y)^(2) pi^(*)2p_(x)^(1)=pi^(*)2p_(y)^(0),` Bond order `=(10-5)/(2)=2.5`
`NO^(-)(7+8+1=16)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi2p_(x)^(1)=pi^(*)2p_(y)^(1)`
Bond order `=(10-6)/(2)2.0`
`(3)O_(2)^(2+)(8+8-2=14)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi^(*)2p_(y)^(@)=pi^(*)2p_(y)^(@),` Bond order `=(10-4)/(2)=3.0`
`O_(2)^(+)(8+8-1=15)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)sigma^(**)2p_(z)^(2)pi2p_(x)^(2)=pi2p_(y)^(2)pi^(*)2p_(x)^(1)=pi^(*)2p_(y)^(@),` Bond order `=(10-5)/(2)=2.5`
`(4)N_(2)(7+7=14)sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma2p_(z)^(2),` Bond order `=(10-4)/(2)=3`
`N_(2)^(-)(7+7+1=15)=sigma1s^(2)sigma^(*)1s^(2)sigma2s^(2)sigma^(*)2s^(2)pi2p_(x)^(2)=pi2p_(y)^(2)sigma2p_(z)^(2)pi^(*)2p_(x)^(1),`Bond order `=(10-5)/(2)=2.5`