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.