Question

write ground - state electron configurations for the ions s^-, ca^2+, s^2 - and fe^2+. which do you expect will be paramagnetic due to the presence of unpaired electrons? (express your answer as a series of orbitals. for example, the electron configuration of li would be entered in complete form as 1s^2 2s^1 or in condensed form as he2s^1.)

Explanation:

Step1: Determine electron configuration for each ion

• For \(S^-\):
• Sulfur (\(S\)) has atomic number \(16\). \(S^-\) has \(17\) electrons.
• Using the Aufbau principle, electron configuration is \([Ne]3s^{2}3p^{4}\) (condensed form \([He]2s^{2}2p^{6}3s^{2}3p^{4}\)).
• For \(Ca^{2+}\):
• Calcium (\(Ca\)) has atomic number \(20\). \(Ca^{2+}\) has \(18\) electrons.
• Electron configuration is \([Ar]\) (condensed form \([He]2s^{2}2p^{6}3s^{2}3p^{6}\)).
• For \(S^{2 -}\):
• Sulfur (\(S\)) has atomic number \(16\). \(S^{2 -}\) has \(18\) electrons.
• Electron configuration is \([Ar]\) (condensed form \([He]2s^{2}2p^{6}3s^{2}3p^{6}\)).
• For \(Fe^{2+}\):
• Iron (\(Fe\)) has atomic number \(26\). \(Fe^{2+}\) has \(24\) electrons.
• Electron configuration is \([Ar]3d^{6}\) (condensed form \([He]2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}\)).

Step2: Determine paramagnetic nature

• For \(S^-\):
• In \(3p\) sub - shell, there are \(4\) electrons. According to Hund's rule, \(3p\) sub - shell has \(2\) unpaired electrons (\(3p^{4}\): two paired and two unpaired). So, it is paramagnetic.
• For \(Ca^{2+}\):
• Electron configuration \([Ar]\) has all paired electrons. So, it is not paramagnetic.
• For \(S^{2 -}\):
• Electron configuration \([Ar]\) has all paired electrons. So, it is not paramagnetic.
• For \(Fe^{2+}\):
• In \(3d\) sub - shell (\(3d^{6}\)), there are \(4\) unpaired electrons (using Hund's rule). So, it is paramagnetic.

Answer:

ion	electron configuration	paramagnetic
\(S^{2 -}\)	\([Ar]\)	✗
\(Ca^{2+}\)	\([Ar]\)	✗
\(S^-\)	\([Ne]3s^{2}3p^{4}\)	✓