Question

write ground - state electron configurations for the ions $co^{2 + }$, $al^{3 + }$, $p^{3 - }$ and $be^{ + }$. 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 \(Be^{+}\)

Beryllium (\(Be\)) has atomic number \(4\). \(Be^{+}\) has \(3\) electrons. The electron configuration is \(1s^{2}2s^{1}\) (or \([He]2s^{1}\)).

Step2: Determine electron configuration for \(P^{3 -}\)

Phosphorus (\(P\)) has atomic number \(15\). \(P^{3 -}\) has \(18\) electrons. The electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\) (or \([Ar]\)).

Step3: Determine electron configuration for \(Al^{3+}\)

Aluminum (\(Al\)) has atomic number \(13\). \(Al^{3+}\) has \(10\) electrons. The electron configuration is \(1s^{2}2s^{2}2p^{6}\) (or \([Ne]\)).

Step4: Determine electron configuration for \(Co^{2+}\)

Cobalt (\(Co\)) has atomic number \(27\). \(Co^{2+}\) has \(25\) electrons. The electron configuration is \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}\) (or \([Ar]3d^{7}\)).

Step5: Determine paramagnetic ions

Paramagnetic substances have unpaired electrons.

• \(Be^{+}\): \(2s^{1}\) has \(1\) unpaired electron (\(2s\) orbital), so it is paramagnetic.
• \(P^{3 -}\): All orbitals (\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\)) are fully - filled, no unpaired electrons, so it is diamagnetic.
• \(Al^{3+}\): \(1s^{2}2s^{2}2p^{6}\) (all orbitals filled), diamagnetic.
• \(Co^{2+}\): \(3d^{7}\) (using Hund's rule, \(d\) orbitals: \(↑↑↑↑↑↓↓\)), has \(3\) unpaired electrons, so it is paramagnetic.

Answer:

ion	electron configuration	paramagnetic
\(P^{3 -}\)	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}\) (or \([Ar]\))
\(Al^{3+}\)	\(1s^{2}2s^{2}2p^{6}\) (or \([Ne]\))
\(Co^{2+}\)	\(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7}\) (or \([Ar]3d^{7}\))