Are the following ions diamagnetic or paramagnetic? #Cr_3^+#, #Ca_2^+#, #Na^+#, #Cr#, #Fe_3^+#

1 Answer
Mar 29, 2016

Notice how #"Cr"# is not an ion. :)

Anyways, we can start from the electron configuration of the neutral atoms. I'm assuming you meant #"Cr"^(3+)#, #"Ca"^(2+)#, and #"Fe"^(3+)#...

#"Cr"#: #[Ar]3d^5 4s^1#

#"Ca"#: #[Ar]4s^2#

#"Na"#: #[Ne]3s^1#

#"Fe"#: #[Ar]3d^6 4s^2#

The rightmost orbitals listed here are highest in energy, so we ionize these atoms by booting off the highest-energy electrons. Thus:

#"Cr" -> "Cr"^(3+) + 3e^(-)#
#[Ar]3d^5 4s^1 -> color(blue)([Ar]3d^3)#

Since there are #5# #3d# orbitals, they are not all at least singly filled yet, and thus, all three electrons in the lowest-energy configuration are unpaired. So, this is paramagnetic.

The original atom is also paramagnetic. The #4s# subshell contains #1# electron (in one #4s# orbital) and the #3d# subshell contains #5# electrons, one in each #3d# orbital. No valence electrons are paired here.

That's in agreement with our expectations from Hund's Rule (generally, for the lowest-energy configuration, maximize parallel spins where possible by singly-filling all orbitals of very similar energies first and then doubling up afterwards).

#"Ca" -> "Ca"^(2+) + 2e^(-)#
#[Ar]4s^2 -> [Ar] => color(blue)(1s^2 2s^2 2p^6 3s^2 3p^6)#

This is a noble gas configuration, so no electrons are unpaired Thus, this is diamagnetic.

#"Na" -> "Na"^(+) + e^(-)#
#[Ne]3s^1 -> [Ne] => color(blue)(1s^2 2s^2 2p^6)#

This is a noble gas configuration, so no electrons are unpaired. Thus, this is diamagnetic.

#"Fe" -> "Fe"^(3+) + 3e^(-)#
#[Ar]3d^6 4s^2 -> color(blue)([Ar]3d^5)#

Since there are #5# #3d# orbitals, in accordance with Hund's Rule, all five electrons in the lowest-energy configuration are unpaired. So, this is paramagnetic.