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Q.

Which one is paramagnetic from the following

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a

$O_{2}^{-}$

b

$NO$

c

Both (a) and (b)

d

${CN}^{-}$

answer is C.

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Detailed Solution

Total number of electrons in $O_{2}^{-}$ is 17.
It is an odd-electron specie, this implies that it does not have paired electrons.
In MO diagram of $O_{2}^{-},$ there is one unpaired electron in $π_{2 py}^{*}$ .
This implies that $O_{2}^{-}$ is paramagnetic.
Total number of electrons in NO is 15. It is an odd-electron specie, hence it has at least one unpaired electron.
According to the MO diagram of NO, there is one unpaired electron present in $π_{2 py}^{*}$ .
This implies that NO is paramagnetic.
Total number of electrons in CN^- is 14. It is an even-electron specie.
This implies that all electrons present in CN^- are paired, thereby CN^- is diamagnetic.

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