Answer : For a species to be diamagnetic it needs to have no unpaired electrons and for paramagnetic the species needs t have unpaired electrons.
Answer 1) : \(C_{2} ^{+2}\)
Electronic configuration of carbon is \(1s^{2}2s^{2} 2p^{2}\)
So, electronic configuration of \(C_{2} ^{+2}\) will be \(1s^{2} 2s^{2}\), as it has given its two electrons present in the outermost valence shell.
This molecule of \(C_{2} ^{+2}\) will behave as diamagnetic species as there are no unpaired electrons in the orbital.
Answer 2) \(Li^{-2}\)
Explanation : The electronic configuration of Lithium is \(1s^{2} 2s^{1}\)
When lithium is in \(Li^{-2}\) state it will have electronic configuration as \(1s^{2} 2s^{2} 2p^{1}\)
This shows that Lithium has one electron which is available in 2p orbital, and it is also unpaired.
Therefore, the \(Li^{-2}\) species will behave like a paramagnetic substance.
Answer 3) \(B_{2}^{-2}\)
Boron has electronic configuration as \(1s^{2} 2s^{2} 2p^{1}\),
So, the excited state of \(B_{2}^{-2}\) will be \(1s^{2} 2s^{2} 2p^{3}\),
Here, Boron in excited state has one unpaired electron in 2p orbital rest 2 electrons are paired in the 2p orbital.
This will make \(B_{2}^{-2}\) as a paramagnetic substance.
