Question

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

Answer 1

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.