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

Mar 29, 2016

Notice how $\text{Cr}$ is not an ion. :)

Anyways, we can start from the electron configuration of the neutral atoms. I'm assuming you meant ${\text{Cr}}^{3 +}$, ${\text{Ca}}^{2 +}$, and ${\text{Fe}}^{3 +}$...

$\text{Cr}$: $\left[A r\right] 3 {d}^{5} 4 {s}^{1}$

$\text{Ca}$: $\left[A r\right] 4 {s}^{2}$

$\text{Na}$: $\left[N e\right] 3 {s}^{1}$

$\text{Fe}$: $\left[A r\right] 3 {d}^{6} 4 {s}^{2}$

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

${\text{Cr" -> "Cr}}^{3 +} + 3 {e}^{-}$
$\left[A r\right] 3 {d}^{5} 4 {s}^{1} \to \textcolor{b l u e}{\left[A r\right] 3 {d}^{3}}$

Since there are $5$ $3 d$ 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 $4 s$ subshell contains $1$ electron (in one $4 s$ orbital) and the $3 d$ subshell contains $5$ electrons, one in each $3 d$ 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).

${\text{Ca" -> "Ca}}^{2 +} + 2 {e}^{-}$
$\left[A r\right] 4 {s}^{2} \to \left[A r\right] \implies \textcolor{b l u e}{1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6}}$

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

${\text{Na" -> "Na}}^{+} + {e}^{-}$
$\left[N e\right] 3 {s}^{1} \to \left[N e\right] \implies \textcolor{b l u e}{1 {s}^{2} 2 {s}^{2} 2 {p}^{6}}$

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

${\text{Fe" -> "Fe}}^{3 +} + 3 {e}^{-}$
$\left[A r\right] 3 {d}^{6} 4 {s}^{2} \to \textcolor{b l u e}{\left[A r\right] 3 {d}^{5}}$

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