Diamagnetic species are those which contain no unapired electrons. Which among the following are diagmagnetic?

Options

A. `N_(2)`
B. `N_(2)^(2-)`
C. `F_(2)^(+)`
D. `O_(2)^(-)`

Answer

A, D

Explanation

(a,d)
a) Electronic configuration of `N_(2) = sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), pi2p_(x)^(2)=pi2p_(y)^(2), sigma2p_(z)^(2)`.
It has unpaired electron indicates diamagnetic species.
b) Electronic configuration of `N_(2)^(2-)` ion =`sigma1s^(2), sigma^(star)s^(2), sigma2s^(2), sigma^(star)2s^(2), pi2p_(x)^(2)=pi2_(y)^(2), sigma2p_(z)^(2), pi^(star)p_(x)^(1)=pi^(star)2p_(y)^(2)`
It has two unpaired electrons, paramagnetic in nature.
c) Electronic configuration of `O_(2)` = `sigma1s^(2), sigma^(star)2s^(2), pi2p_(s)^(2), sigma^(star)2s^(2), sigma2p_(z)^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi^(star)2p_(x)^(1)=pi^(star)2p_(y)^(2)`
The presence of two unpaired electrons shows its paramagnetic nature.
d) Electronic configuration of `O_(2)^(2-) = sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), sigma2p_(z)^(2), pi2p_(x)^(2)=pi2p_(y)^(2), pi^(star)2p_(x)^(2)=pi^(star)2p_(y)^(2)`
It contains no unpaired electron, therefore it is diamagnetic in nature.