Explain the following :
(i) `O_(2)^(-)` is paramagnetic but `O_(2)^(2-)` is not.
(ii) `N_(2)` has higher bond order than `NO.`

Explanation

(i) `O_(2)^(-):KK(sigma2s)^(2)(sigma^(**)2s)^(2)(sigmap_(x))^(2)(pi2p_(x))^(2)(pi2p_(y))^(2)(pi^(**)2p_(x))^(2)(pi^(**)2p_(y))^(1)`
Paramagnetic due to the presence of 1 unpaired electron.
`O_(2)^(2-):KK(sigma2s)^(2)(sigma^(**)2s)^(2)(sigma2p_(z))^(2)(pi2p_(x))^(2)(pi2p_(y))^(2)(pi^(**)2p_(x))^(2)(pi^(**)2p_(y))^(2)`
It is diamagnetic.
(ii) `N_(2):KK(sigma2s)^(2)(sigma^(**)2s)^(2)(pi2p_(x))^(2)(sigma2p_(z))^(2)`
`"B.O."=(8-3)/(2)=3`
`NO:KK(sigma2s)^(2)(sigma^(**)2s)^(2)(pi2p_(x))^(2)(pi2p_(y))^(2)(sigma2p_(z))^(2)(pi^(**)2p_(x))^(1)`
`"B.O."=(8-3)/(2)=2.5`
`therefore N_(2)` has higher bond order than NO.