In mine textbook, it claims that the maximum number of electrons that can fit in any kind of given shell is offered by 2n². This would median 2 electrons can fit in the very first shell, 8 can fit in the second shell, 18 in the third shell, and 32 in the fourth shell.

However, ns was previously taught the the maximum number of electrons in the an initial orbital is 2, 8 in the second orbital, 8 in the third shell, 18 in the 4th orbital, 18 in the fifth orbital, 32 in the sixth orbital. Ns am relatively sure the orbitals and also shells are the very same thing.

Which of this two approaches is correct and also should be provided to find the number of electrons in an orbital?

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

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Shells and also orbitals space not the same. In terms of quantum numbers, electrons in different shells will have different values of major quantum number n.

To answer your question...

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

The 1s orbital

In the second shell (n=2), we have:

The 2s orbitalThe 2p orbitals

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

The 3s orbitalThe 3p orbitalsThe 3d orbitals

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

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

So an additional kind the orbitals (s, p, d, f) becomes obtainable as we go to a covering with greater n. The number in former of the letter signifies which covering the orbital(s) room in. So the 7s orbital will be in the 7th shell.

Now for the various kinds the orbitalsEach sort of orbital has a various "shape", together you can see on the photo below. You can additionally see that:

The s-kind has only one orbitalThe p-kind has 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., can each host two electrons due to the fact that they each have only one orbital.

The 2p, 3p, 4p, etc., deserve to each host six electrons because they each have three orbitals, that deserve to hold two electrons every (3*2=6).

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

Thus, to discover the variety of electrons feasible per shell

First, us look at the n=1 covering (the very first shell). The has:

The 1s orbital

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

The n=2 (second) covering has:

The 2s orbitalThe 2p orbitals

s-orbitals have the right to hold 2 electrons, the p-orbitals have the right to hold 6 electrons. Thus, the second shell deserve 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 can hold 6, and also d-orbitals can hold 10, because that a full of 18 electrons.

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

There"s an essential distinction in between "the variety of electrons possible in a shell" and "the number of valence electrons feasible for a period of elements".

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There"s room for $18 \texte^-$ in the 3rd shell: $3s + 3p + 3d = 2 + 6 + 10 = 18$, however, elements in the third period only have actually up to 8 valence electrons. This is because the $3d$-orbitals aren"t filled till we get to facets from the 4th period - ie. Elements from the 3rd period don"t to fill the 3rd shell.

The orbitals room filled so the the ones of lowest energy are filled first. The energy is about like this: