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The periodic table has 7 periods and they have 2,8,8,18,18... elements respectively from 1 to 7. But from what I understand, the periods each state the number of electron shells that the elements in that period has. So if that is the case, shouldn't period 3 have more elements, since it can hold up to 18 electrons, and therefore it can have up to 18 more protons from the largest atomic number element in period 2? (Since period 3 has a M electron shell, it can also have a p orbital, and in total therefore can have 18 elements.)

The same could be said for elements in group 4, group 5, group 6 and so on since they can have f,g, and h orbitals as well. Am I missing something here?

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    $\begingroup$ There are a couple assumptions in your question that are really confusing me: (1) "shouldn't period 3...hold up to 18 electrons?" Why should you expect that? (2) "Since period 3 has a M electron shell" I haven't really been in a chemistry classroom for a while--what's an M electron shell? (3) " group 4, group 5, group 6...can have f,g, and h orbitals as well" No naturally-occurring element has electrons in anything higher than an f-shell. If you could explain these three points, I might be able to figure out where the confusion is coming from. $\endgroup$ – chipbuster Jul 30 '15 at 17:10
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    $\begingroup$ @chipbuster - M corresponds to principle quantum number 3 (K is 1, L is 2, and so on). It's old-style (or X-ray) notation. I would argue that f, g, h (and higher) atomic orbitals exist, but aren't populated for the elements in question. $\endgroup$ – Todd Minehardt Jul 30 '15 at 17:32
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    $\begingroup$ The standard periodic table isn't arranged according to electronic configuration; it's arranged so that elements with similar properties appear in the same group. There are alternative versions of the periodic table arranged in different ways, including by electronic configuration. $\endgroup$ – bon Jul 30 '15 at 17:44
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    $\begingroup$ @ToddMinehardt thanks, I had never heard x-ray notation used before. And agreed about the higher atomic orbitals. $\endgroup$ – chipbuster Jul 30 '15 at 17:52
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So if that is the case, shouldn't period 3 have more elements, since it can hold up to 18 electrons, and therefore it can have up to 18 more protons from the largest atomic number element in period 2?

Indeed, elements of the same period have the same number of electron shells, but the "problem" is that in accordance with the Madelung/Janet/Klechkowski rule, the $\mathrm{4s}$ orbital is occupied before the $\mathrm{3d}$ orbital. As a result, the $\mathrm{3d}$ orbital could be filled only when the $\mathrm{4s}$ orbital is already filled, i.e. only for elements of the 4th period.

Similarly, the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are occupied before the $\mathrm{4f}$ orbital. As a result, the $\mathrm{4f}$ orbital could be filled only when the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are already filled, i.e. only for elements of the 6th period.

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