3
$\begingroup$

I am teaching my students how to write the electronic configuration of the elements of the periodic table.

For example, following the rules on how to fill the orbitals, the electronic configuration of iron is

$$\ce{Fe}:~\mathrm{(1s)^2(2s)^2(2p)^6(3s)^2(3p)^6(4s)^2(3d)^6}.$$

Thus, its abbreviated form is

$$\ce{Fe}:~[\ce{Ar}]~\mathrm{(4s)^2(3d)^6}.$$

However, I has not been easy to find the explanation on why in any periodic table it is written as

$$[\ce{Ar}]~\mathrm{(3d)^6(4s)^2}$$

instead of

$$[\ce{Ar}]~\mathrm{(4s)^2(3d)^6}.$$

Same case for many other elements. Why is that?

$\endgroup$
1
  • 4
    $\begingroup$ I have seen it written both ways. However, the '3d' orbitals in this case are going to be lower in energy than the 4s and therefore it is written in a way that indicates increasing energy from left to right. $\endgroup$ Commented Mar 19, 2015 at 13:48

2 Answers 2

3
$\begingroup$

I would recommend following the convention used by NIST which is to list the subshells in order of principle quantum number $(n),$ and in order of angular momentum quantum number $(\ell)$ for a given $n.$

See also ELECTRONIC STRUCTURE OF THE ELEMENTS (PDF).

The question refers to “rules on how to fill the orbitals”, but I don't think there are any rules, just memorization aids.

See The low-lying level structure of atomic lawrencium $(Z = 103):$ energies and absorption rates for an sample of how difficult it is to calculate the ground state configuration of an atom. The only rule is quantum electrodynamics.

Also, the configurations are really just designations or leading configurations, with the actually ground state being a mix of configuration. See for example the above lawrencium reference or A critical compilation of energy levels and spectral lines of neutral boron, where it explained “Configuration and term labels have little physical meaning for highly mixed levels”. Boron ground state is $95\%$ $\mathrm{(1s)^2(2s)^2(2p)}$ and $4\%$ $\mathrm{(1s)^2(2p)^3}.$ Lawrencium is calculated to be $86\%$ $\mathrm{(7s)^2(7p)}$ and $6\%$ $\mathrm{(6d)(7s)(7p)}.$

$\endgroup$
3
  • $\begingroup$ I guess, from a teaching perspective, listing the sub-shells order of increasing energy. My goal in teaching general chemistry is to help students make connections between quantum theory, the periodic table, and chemical reactivity. I teach my students to write electron configurations using the periodic table as a guide. However, I do tell them about exceptions and why they happen, but I don't want them to be more concerned about them. They can opt to take physical chemistry if they want to delve deeper. $\endgroup$
    – Tami M.
    Commented Mar 19, 2015 at 16:35
  • $\begingroup$ @TamiM. so would you have your students switch the order of 3d and 4s when you go from Fe to Fe+ since the 4s becomes higher energy, or do you always go by the neutral atom energy level order? $\endgroup$
    – DavePhD
    Commented Mar 19, 2015 at 17:18
  • $\begingroup$ I would probably accept either way. Honestly, I would accept either valid way of writing electron configurations, even for neutral atoms. But, when I teach general chemistry, I do teach them to use the periodic table to write electron configurations. If I were teaching a more advanced course, I would go much more in depth about mixed configurations. As it is, I go into more detail about these things then many general chemistry lecturers (I can't help it, I am a physical chemist). As I said before, I just try to focus on connecting theory to laws and chemical reactivity. $\endgroup$
    – Tami M.
    Commented Mar 19, 2015 at 18:53
1
$\begingroup$

In some textbooks that I have read, I came across electron configurations that listed the 3d electrons before the 4s electrons. I believe it is to keep the electrons with the same principal quantum number $(n)$ together.

For example, $[\ce{Ar}]~\mathrm{(3d)^6(4s)^2}$ in full would be $\mathrm{(1s)^2(2s)^2(2p)^6(3s)^2(3p)^6(3d)^6(4s)^2}.$ When I teach general chemistry, I opt to list the sub-shells in order of increasing energy instead, since it correlates with the structure of the periodic table.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.