Period 1 of the periodic table contains 2 elements ($1s^1$ and $1s^2$).

Period 2 contains 8 elements ($2s^1$, $2s^2$, $2p^1$, $2p^3$, ..., $2p^6$).

By the same argument, period 3 might contain 18 elements ($3s^1$, $3s^2$, $3p^1$, $3p^3$, ..., $3p^6$, $3d^1$, $3d^2$, $3d^{10}$). Why does period 3 of the periodic table contain 8 elements instead of 18?

I think that it could be explained by means of atomic stability, but I need help to understand what happens.

Thank you in advance.

  • 1
    $\begingroup$ The elements Sc through Zn have 3d electrons, but they also have a pair of 4s electrons, and Sc has one more proton than Ca, so it follows Ca in the periodic table. Therefore, Sc through Zn fit between Ca and Ga. The periodic table is ordered by atomic number. As for why the 4s orbitals get filled before the 3d orbitals, there is likely already an answer posted here, so I suggest searching for it. $\endgroup$ – Ed V Sep 1 '19 at 13:42
  • $\begingroup$ The energetics is nicely discussed here: chemistry.stackexchange.com/a/8426/79678 $\endgroup$ – Ed V Sep 1 '19 at 17:20

You could say that the period number tells you about the largest value of principal quantum number in which a electron is present.

Electrons are filled according to $n+l$ rule. It states as $n+l$ increases, the energy of the orbital increases. If $n+l$ is the same, then the bigger $n$ has larger energy.

So $\mathrm{4s}$ would be filled before $\mathrm{3d}$. Therefore elements containing $\mathrm{3d}$ electrons would have their $\mathrm{4s}$ orbital filled, and therefore, come in 4th period. There 3rd period contains elements having $\mathrm{3s}$ and $\mathrm{3p}$ electron as their last electron which are equal to 8.

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