Question

which of the following atoms is/are paramagnetic in the ground electronic state? you may select more than one response. be o b c n

Explanation:

• Be (Beryllium): Its electron configuration is \(1s^{2}2s^{2}\). All electrons are paired (\(s\) - orbital can hold 2 electrons and they are paired here). So, it is diamagnetic.
• O (Oxygen): Electron configuration is \(1s^{2}2s^{2}2p^{4}\). In the \(2p\) - subshell (which has 3 orbitals), according to Hund's rule, electrons first occupy degenerate orbitals singly. For \(2p^{4}\), there are 2 unpaired electrons. So, it is paramagnetic.
• B (Boron): Electron configuration is \(1s^{2}2s^{2}2p^{1}\). The \(2p\) - orbital has 1 unpaired electron. So, it is paramagnetic.
• C (Carbon): Electron configuration is \(1s^{2}2s^{2}2p^{2}\). In the \(2p\) - subshell, there are 2 unpaired electrons (Hund's rule: electrons occupy degenerate orbitals singly first). So, it is paramagnetic.
• N (Nitrogen): Electron configuration is \(1s^{2}2s^{2}2p^{3}\). In the \(2p\) - subshell, there are 3 unpaired electrons (each \(2p\) - orbital has one electron as per Hund's rule). So, it is paramagnetic.

Answer:

O. Option O, B. Option B, C. Option C, N. Option N