In mine textbook, it states that the maximum number of electrons that have the right to fit in any type of given shell is given by 2n². This would typical 2 electrons could fit in the very first shell, 8 could fit in the 2nd shell, 18 in the third shell, and also 32 in the fourth shell.

However, ns was previously taught that the maximum number of electrons in the very first orbital is 2, 8 in the 2nd orbital, 8 in the 3rd shell, 18 in the fourth orbital, 18 in the fifth orbital, 32 in the 6th orbital. I am reasonably sure that orbitals and shells space the very same thing.

Which of these two methods is correct and also should be supplied to discover the number of electrons in one orbital?

I am in high school so please shot to leveling your answer and also use reasonably basic terms.

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Shells and orbitals space not the same. In terms of quantum numbers, electron in different shells will certainly have various values of principal quantum number n.

To answer your question...

In the very first shell (n=1), us have:

The 1s orbital

In the 2nd shell (n=2), us have:

The 2s orbitalThe 2p orbitals

In the 3rd shell (n=3), we have:

The 3s orbitalThe 3p orbitalsThe 3d orbitals

In the fourth shell (n=4), us have:

The 4s orbitalThe 4p orbitalsThe 4d orbitalsThe 4f orbitals

So an additional kind the orbitals (s, p, d, f) becomes obtainable as us go come a covering with higher n. The number in prior of the letter signifies which shell the orbital(s) space in. So the 7s orbital will certainly be in the 7th shell.

Now because that the different kinds the orbitalsEach sort of orbital has a different "shape", together you can see on the photo below. Girlfriend can likewise see that:

The s-kind has only one orbitalThe p-kind has actually three orbitalsThe d-kind has 5 orbitalsThe f-kind has seven orbitals


Each orbital deserve to hold two electrons. One spin-up and also one spin-down. This means that the 1s, 2s, 3s, 4s, etc., deserve to each organize two electrons due to the fact that they each have only one orbital.

The 2p, 3p, 4p, etc., can each hold six electrons due to the fact that they each have three orbitals, that can hold two electrons each (3*2=6).

The 3d, 4d etc., deserve to each host ten electrons, because they each have five orbitals, and also each orbital can hold two electrons (5*2=10).

Thus, to uncover the variety of electrons feasible per shell

First, us look at the n=1 shell (the an initial shell). It has:

The 1s orbital

An s-orbital stop 2 electrons. Therefore n=1 shell deserve to hold two electrons.

The n=2 (second) covering has:

The 2s orbitalThe 2p orbitals

s-orbitals have the right to hold 2 electrons, the p-orbitals can hold 6 electrons. Thus, the second shell have the right to have 8 electrons.

The n=3 (third) shell has:

The 3s orbitalThe 3p orbitalsThe 3d orbitals

s-orbitals deserve to hold 2 electrons, p-orbitals have the right to hold 6, and d-orbitals deserve to hold 10, because that a complete of 18 electrons.

Therefore, the formula $2n^2$ holds! What is the difference in between your two methods?

There"s critical distinction in between "the variety of electrons feasible in a shell" and also "the number of valence electrons possible for a duration of elements".

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There"s room for $18 \texte^-$ in the third shell: $3s + 3p + 3d = 2 + 6 + 10 = 18$, however, facets in the 3rd period only have up come 8 valence electrons. This is since the $3d$-orbitals aren"t filled until we obtain to elements from the fourth period - ie. Aspects from the third period don"t to fill the 3rd shell.

The orbitals space filled so that the persons of lowest energy are fill first. The energy is about like this: