# If each orbital can hold a maximum of 3 electrons, the number of elements in the 4th #bb"[period of the]"# periodic table (long form) is?

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*In brackets is a fix to the question.*

- Truong-Son

*In brackets is a fix to the question.
- Truong-Son*

##### 2 Answers

Well, this is kind of open to interpretation... but if I interpreted it correctly, I get

(If you already knew that the number of electrons allowed in a given orbital is derived from the properties of electrons, and NOT of the orbitals themselves, the answer quickly follows.)

**PRELIMINARY THINGS**

I think there's a typo in the question... I looked this up elsewhere, and it's probably...

If each orbital can hold a maximum of 3 electrons, the number of elements in the4th periodof the periodic table (long form) is...?

Also, note that this is **entirely theoretical**, as all electrons only have two possible spins (**two** electrons in real life.

That aside, when we suppose **three electrons** are "allowed" in a single orbital (assuming the other three quantum numbers are as normal), we suddenly "allow"

(It's not really *why* the periodic table was arranged historically, but... I suppose that's what the intent of this question was...)

**EXPANDING THE PERIOD...**

An **electron configuration** for the fourth period in a generalized manner is written as:

#4s^x 3d^x 4p^x# where:

#x = (2l+1) cdot N_(m_s)# is thetotalnumber of electrons in all the orbitals in a givensubshell.#l# is theangular momentum quantum number.#l = 0, 1, 2, 3, . . . # corresponds to#s, p, d, f, . . . # orbitals.#2l+1# is known as thedegeneracyof the subshell; it ishow many orbitalsare in that subshell.#N_(m_s)# is anarbitrarynumber of spinsthe electron could have, as it also then gives themaximum number of electrons per orbital. In this case we SUPPOSE that#N_(m_s) = 3# , but in real life it is just#2# .

Now, the number of allowed ** properties of the electron**, not of the orbitals themselves, so having more spins allowed does NOT change the orbital shapes or relative energies.

For the

#l = 0, => 2l+1 = 1#

For the

#l = 2, => 2l+1 = 5#

For the

#l = 1, => 2l+1 = 3#

Thus, the ** hypothetical** electron configuration we would then write is...

#4s^(1cdot3) 3d^(5 cdot 3) 4p^(3 cdot 3)#

#= 4s^3 3d^15 4p^9#

And that would apparently expand the fourth period of the periodic table from

#### Answer:

This is what I get.

#### Explanation:

We see that on the assumption that each orbital holds maximum of

Total number of electrons held in

If each orbital is allowed to hold

Total number of electrons held in this shell

This is the number of elements allowed in

.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Alternatively

Total number of electrons