Question

which of the following elements are paramagnetic in the ground state (for a free atom)? select all that apply however penalties will be applied for incorrect guesses. select one or more: ca zn c o be cl ne fe

Explanation:

Paramagnetic elements have unpaired electrons in their ground - state electron configurations.

• Ca (Calcium): Electron configuration \([Ar]4s^{2}\). All electrons are paired (\(4s^{2}\) has two paired electrons), so it is diamagnetic.
• Zn (Zinc): Electron configuration \([Ar]3d^{10}4s^{2}\). All \(3d\) and \(4s\) electrons are paired, so it is diamagnetic.
• C (Carbon): Electron configuration \([He]2s^{2}2p^{2}\). In the \(2p\) sub - shell, there are two unpaired electrons (Hund's rule: electrons occupy degenerate orbitals singly before pairing), so it is paramagnetic.
• O (Oxygen): Electron configuration \([He]2s^{2}2p^{4}\). In the \(2p\) sub - shell, there are two unpaired electrons (two of the three \(2p\) orbitals have paired electrons, and one has two unpaired electrons), so it is paramagnetic.
• Be (Beryllium): Electron configuration \([He]2s^{2}\). All electrons are paired (\(2s^{2}\)), so it is diamagnetic.
• Cl (Chlorine): Electron configuration \([Ne]3s^{2}3p^{5}\). In the \(3p\) sub - shell, there is one unpaired electron, so it is paramagnetic.
• Ne (Neon): Electron configuration \([He]2s^{2}2p^{6}\). All electrons are paired, so it is diamagnetic.
• Fe (Iron): Electron configuration \([Ar]3d^{6}4s^{2}\). In the \(3d\) sub - shell, there are four unpaired electrons (Hund's rule: electrons occupy \(3d\) orbitals singly as much as possible), so it is paramagnetic.

Answer:

C. C, D. O, F. Cl, H. Fe