The third shell must contain at least 18 electrons. But in iron there are 14 electrons in third and 2 electrons in 4th. Why are there 14 electrons in the third shell of iron atom?

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    $\begingroup$ The figure shows 12 in the 3rd and 2 in the 4th. The term shell is probably inappropriate as there was a model of Bohr which used the term "shell" to correspond to "periods" in the periodic table. As for the reason? There is nothing that requires an energy level to fill up before another one starts to fill. Its all about stability and energy. It depends on many factors. Transition metals will almost always have 1 or 2 electrons in the s sublevel above the d sublevel that corresponds to its place in the d block $\endgroup$ – Joseph Hirsch Dec 28 '16 at 6:26

Your doubt is - Total number of electrons in 3rd shell (i.e. electrons from 3s + electrons from 3p + electrons from 3d) is not yet 18 but still how we filled up the 4th shell?

You have a general misconception than atomic orbitals always follow the 2,8,18... electron filling rule. According to this 2,8,18.. electron filling rule, 2 electrons should get filled up in 1st shell, then 8 electrons in second shell, then 18 electrons in 3rd shell........

Remember this rule mentioned above is completely elementary! Just remember 2,8,18... electron filling rule is actually a waste. Simply I would recommend forget about that 2,8,18... rule. It is very elementary rule... simply forget it. In 2,8,18... electron filling rule, "lower energy level orbitals gets filled up first"- this statement is not considered in that elementary 2,8,18..rule. But this statement which that elementary rule doesn't consider is very important statement to be followed while filling up of electrons. So remember 2,8,18... electron filling rule is an elementary rule for beginners who just start to read about atoms and orbitals! Forget about that rule.

Its not necessary that 3rd shell has to have 18 electrons for 4th shell filling up to start. The electron filling depends upon the energy levels of the orbitals. Lower energy orbitals get filled up first.

Iron's configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d6. You can note from the configuration, 4s gets filled up before 3d . The reason behind this is lower energy orbitals are filled up first. Even though 4s belongs to 4th shell, still its energy is less than 3d which can be reasoned out by (n+l) rule. 4s energy is less than 3d, hence the two electrons which you expected to be going to 3d goes to 4s before 3d filling starts. Hence you end up having 14 electrons in 3rd shell and 2 electron in 4s. Remember 4s got filled up before 3d and the reason is 4s energy is less than that of 3d.

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    $\begingroup$ While the 2-8-18 shell model is elementary, the Aufbau filling is also a flawed concept. I won't vote down because Aufbau is consistent with 4s2 3d6, but the 4s electrons present in Iron are actually higher energy than the 3d electrons and will ionize first and enter last. It is still generally true that the 3d will "get full" after the 4s as you move across the periodic table, but it doesn't fill last in a given transition metal. Iron is now typically shown as having a configuration of 12s 2s2 2p6 3s2 3p6 3d6 4s2 tabulka.cz/english/elements/configuration.asp?id=26 and others $\endgroup$ – Joseph Hirsch Dec 29 '16 at 15:44
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    $\begingroup$ So "filling first" is ambiguous. In Iron, 4s is full while 3d is not. 4s tends to be occupied and or also full while 3d is not full as you move across the transition metals. The electrons that are IN the 4s orbitals of a given transition metal are higher than the electrons in the 3d orbital. chemguide.co.uk/atoms/properties/3d4sproblem.html Ultimately the reason is that the 4s and 3d sublevels are close to each other in energy and depending on the nuclear charge and the exact number of electrons, different orbitals become relatively more or less favorable for the next e-. $\endgroup$ – Joseph Hirsch Dec 29 '16 at 15:49
  • $\begingroup$ So do you mean to say in Iron 3d fills first and then 4s? Can you please elaborate where you think I went wrong. $\endgroup$ – Yb609 Dec 29 '16 at 16:08
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    $\begingroup$ Yes IN IRON (and almost all transition metals) the 3d gets its electrons first and then the 4s if you are literally looking at the energy of the electrons, or stripping electrons and putting them back. Modern electron configurations now show Iron as 3d6 4s2. This is not the same as saying that 3d FILLS first because 3d is not FULL for Iron, but the 6 3d electrons in Iron are lower energy, ionize later, refill preferrentially to the 2 4s electrons in Iron. The misconception comes from thinking that since Mg has 4s2 and no 3ds, that the 4s are lower energy for transition metals. $\endgroup$ – Joseph Hirsch Dec 29 '16 at 17:26
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    $\begingroup$ You answer the question but include additional information that is wrong, in part because the Aufbau is an oversimlification and in part because Aufbau is WRONG in terms of placing energy levels in a given element. The best way to correct it is to say that the 3d does not have to have all of its orbitals full for some electrons to go into the 4s. The ones that are in the 3d still are lower energy than the 4s present in transition metals. Some sources will still list the 4s below the 3d in Iron. It depends somewhat on what the order means to you-bookkeeping electrons or ordering energies. $\endgroup$ – Joseph Hirsch Dec 29 '16 at 17:40

The third shell must contain at least 18 electrons.

This statement is wrong. In fact, the third shell is only able to contain up to 18 electrons. This is a mathematical result of quantum chemistry that is, in principle, derived in every introductory quantum chemistry course. The reason being that there are three different subshells in the third shell (called s, p and d orbitals) which can contain 2, 6 and 10 electrons, respectively. After 18 electrons have been added to the third shell, there is no combination of quantum numbers to allow for a 19th electron within that same shell.

On to the question of why there are two electrons in the 4s orbitals (the fourth shell) although the third shell is not full yet. This is because the subshells in every atom with more than one electron have different energy levels. In general, the energy levels follow what is known as the aufbau principle: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, … However, this is more of a general observation. It works very well for general chemistry classes, but there are quite a number of edgy cases that these same classes will have to introduce special rules for.

A great general reference to check out would be this question. But I would also like to point you to this chemguide.co.uk guide, which explains the entire complication in more details. The gist is that the energy levels of 4s and 3d are so similar in an innocent case such as argon that depending on how many protons and additional electrons are added a different orbital may be the one lower in energy and leading to the more stable electronic configuration.

If, when coming from argon you add exactly one proton and exactly one electron (generating a potassium atom), the energy levels will shift relatively to each other in such a way that the 4s orbital is lower; thus, potassium’s electronic configuration is $\ce{[Ar] 4s^1}$. If you add another proton without an electron (generating $\ce{Ca+}$), the picture remains the same. Calcium(I) has the same electronic configuration. If you add a third proton to generate scandium(II) ($\ce{Sc^2+}$), the 3d orbitals will suddenly move below the 4s orbitals in energy so its configuration is $\ce{[Ar] 3d^1}$. (This is because the 3d orbitals are less diffuse or ‘closer to the nucleus’ and therfore get stabilised more.)

If you thought you could now easily add a second electron into another d orbital when generating scandium(I) ($\ce{Sc+}$), think again. Due to electron-electron repulsion (or more generally, the Hund rule), this additional electron is placed in the 4s orbital leading to $\ce{[Ar] 3d^1 4s^1}$. And the final electron to generate a neutral scandium atom would again be added into the 4s orbital to give $\ce{[Ar] 3d^1 4s^2}$.

(In reality, you do not simply ‘add electrons’ and all the electrons of an atom are indistinguishable. What you can do, however, is calculate or measure the electronic structure of all these atoms and ions to determine which levels are populated by how many electrons. And that will give you the result discussed above.)

Tl;dr: It’s complicated, but an easy rule of thumb would be to fill the next shell’s s electrons before filling up a given shell’s d electrons. Thus, for iron the 4s level is filled first, before an additional six d electrons are added. Read above as to why it is not that simple.


Electronic configuration of $\ce{Fe}$ : $\mathrm{1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^6 \ 3d^6 \ 4s^2}$

Add the electrons of 3rd shell $\mathrm{3s^2 + 3p^6 + 3d^6} = 14$ electrons

And 4th shell $\mathrm{4s^2}$ contains $2$ electrons.


According to the Bohr model of atoms, the maximum number of atoms which can be accommodated in the outermost shell is 8. That's why in the atomic shell diagram the 2 electrons are in the outermost shell.


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