1.

Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic ?

Answer»

`N_(2)`
`N_(2)^(2-)`
`O_(2)`
`O_(2)^(2-)`

Solution :(A,D)
Electronic configuration of `NO_(2) = sigma 1S^(2), sigma^(**) 1s^(2), sigma 2S^(2), sigma^(**) 2s^(2), pi 2p_(x)^(2) = pi 2p_(y)^(2) sigma 2p_(Z)^(2)`
It has no unpaired electron indicates DIAMAGNETIC species.
Electronic configuration of `N_(2)^(2-) " ion " = sigma 1s^(2), sigma^(**) 1s^(2), sigma 2s^(2), sigma^(**) 2s^(2), pi 2p_(x)^(2) = pi 2p_(y)^(2) sigma 2p_(z)^(2) pi^(**)2p_(x)^(1) = pi^(**) 2p_(x)^(1)`
It has two unpaired electrons, so it is PARAMAGNETIC in nature.
(C) Electronic configuration of `O_(2) = sigma 1s^(2), sigma^(**) 1s^(2) , sigma 2s^(2), sigma^(**) 2s^(2), sigma^(**) 2p_(z)^(2) , pi 2p_(x)^(2) = pi 2p_(y)^(2) , pi^(**) 2p_(x)^(1) = pi^(**) 2p_(y)^(1)`
The presence of two unpaired electrons shows it is paramagnetic nature.
(D) Electronic configuration of `O_(2)^(2-) "ion" = sigma 1s^(2), sigma^(**) 1s^(2) , sigma 2s^(2), sigma^(**) 2s^(2), sigma 2p_(z)^(2) , pi 2p_(x)^(2) = pi 2p_(y)^(2) , pi^(**) 2p_(x)^(1) = pi^(**) 2p_(y)^(1)`
It contains no unpaired electron, therefore, it is diamagnetic in nature.


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