# How do you find electron configurations for ions?

Jun 11, 2018

By looking them up... or going off of neutral atoms. Some are tricky, and some are straightforward. The best you can do is know how to find the straightforward ones, and don't worry about the tricky ones.

[If your professor gives you a tricky example... (like vanadium or yttrium), they either didn't realize it or were being too tricksome and you should tell them.]

I'll show you two straightforward examples, and list some exceptions you don't need to know, but should recognize exist.

Except for the actinides, NIST has done a pretty good job tabulating these configurations (in a few years, we may be able to trust the high-oxidation state configurations for those, but not yet...).

STRAIGHTFORWARD EXAMPLE

1)

Magnesium has $12$ electrons, having atomic number $12$. So, its electron configuration, being in the $s$-block below neon, is

$\left[N e\right] 3 {s}^{2}$

Removing one $3 s$ electron (the only valence orbital, yay!) removes one $\left(-\right)$ charge, and thus we form ${\text{Mg}}^{+}$ (i.e. lost one $\left(-\right)$ $\implies$ gain one $\left(+\right)$), with configuration

overbrace([Ne] 3s^(color(red)(bb1)))^("Mg^(+)).

This is normal; for main-block elements, like $\text{P}$ and $\text{Ca}$, the electron removed is the one with the highest angular momentum (say, $5 p$ instead of $5 s$, or $5 s$ if no $5 p$ orbitals are filled).

2)

Titanium has $22$ electrons, having atomic number $22$. So, its electron configuration, being in the $d$-block below argon, is

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

Removing one $4 s$ electron removes one $\left(-\right)$ charge, and thus we form ${\text{Ti}}^{+}$ (i.e. lost one $\left(-\right)$ $\implies$ gain one $\left(+\right)$), with configuration

${\overbrace{\left[A r\right] 3 {d}^{2} 4 {s}^{\textcolor{red}{\boldsymbol{1}}}}}^{{\text{Ti}}^{+}}$.

Typically, it is not the $3 d$ that gets removed, but the $4 s$, as it has a farther radial extent than the $3 d$. Also, it is a bit higher in energy for this atom ($\text{1.90 eV}$).

TRICKY EXCEPTIONS (DON'T MEMORIZE)

The following are two of many $d$-block exceptions, in the sense that removing the first electron does not lead to the valence electron configuration it will end up having, or the first electron removed does not appear to be an $s$ electron.

(DISCLAIMER: this does not include $d$-block neutral transition metal atoms that have unusual electron configurations, which are $C r$, $C u$, $N b$, $M o$, $R u$, $R h$, $P d$, $A g$, $P t$, and $A u$. These can be verified here.)

"Exception A")

"V": 3d^3 4s^2 -> overbrace(3d^(color(red)(bb4)))^("V"^(+)) -> overbrace(3d^(color(red)(bb3)))^("V"^(2+)) -> overbrace(3d^(color(red)(bb2)))^("V"^(3+)) -> overbrace(3d^(color(red)(bb1)))^("V"^(4+)) -> overbrace(3d^(color(red)(bb0)))^("V"^(5+))

Here, it ends up that the remaining $4 s$ electron gets demoted to the $3 d$ orbital, because it is close enough in energy and is also lower in energy. Although the $3 d$ has a shorter radial extent, it does have three less radial nodes, so that may help reduce electron repulsions.

"Exception B")

"Y": 4d^1 5s^2 -> overbrace(5s^(color(red)(bb2)))^("Y"^(+)) -> overbrace(4d^(color(red)(bb1)))^("Y"^(2+)) -> overbrace(4d^(color(red)(bb0)))^("Y"^(3+))

This is an unusual transition metal that has $4 d$ orbitals a mere $\underline{\text{0.21 eV}}$ higher than the $5 s$ orbital, instead of lower (Appendix B.9). Again, this is strange, as $n s$ orbitals are usually higher in energy than the $\left(n - 1\right) d$.

A reasonable explanation would be to say that if a $5 s$ electron is removed, electron repulsions are reduced, lowering the energy of the $5 s$ orbital (which was already lower). This is apparently enough that the remaining $4 d$ electron gets bumped down to the $5 s$ to lower the energy of the atom more.