(eli5) Why is the electron configuration of tin 2, 8, 18, 18, 4 and not 2,8,18,30,2 ?

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What is the rule for electron configuration, i do not understand it. Also isn’t the maximum number of electrons in shell 4, 32?

In: Chemistry
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Shells 4 and 5 actually overlap in their energy requirements. An electron always falls into the least energetic slot that’s open.

You are looking for the “Madelung Rule”, which describes the energy ordering of orbitals. It says orbitals are ordered by the sum of principal and azimuthal quantum numbers, *n* + *l*.

It is neatly captured by this diagram [here](https://en.m.wikipedia.org/wiki/Aufbau_principle#/media/File%3AAufbau_Principle.png). Orbitals are filled up in the order of the red arrows.

For tin, we are up to the 5p orbital, so the 4f has not been touched yet. And indeed, if you walk up the table to xenon, which has configuration [Kr]4d^10 5s^2 5p^6, the Madelung rule says that the next orbital should be 6s.

This should comport with our intuition, since xenon has a full final shell, so it is noble and stable, and caesium will have a dangling half-occupied 6s orbital making it volatile and reactive, as all good alkali metals are.

In short:

Orbitals are filled by the Aufbau principle in the order of energy. This ordering is described by *n*+*l*.

For tin specifically, this means that it should (and, indeed, does) have full 4d, 4p and 4s shells, a full 5s shell, a partially full 5p shell, and an empty 4f shell, resulting in an occupation 2, 8, 18, 18, 4

Because if you look _very_ closely at what the actual energy levels are in those 4 and 5 shells, then slots 19-22 in the 4 shell (the first four f slots, I believe) are all actually slightly higher than slots 1-4 in shell 5 (the two s slots and the first two p slots).

So those 4 electrons go into the lowest remaining 4 slots, which are all in shell 5.

–Dave, if you think that’s weird, don’t look closely at the lanthanides or actinides