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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Question

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Wyatt Atkins · Accepted Answer

Well, this is kind of open to interpretation... but if I interpreted it correctly, I get #27# elements, compared to the original #18# elements, in the 4th period of the periodic table.

(The answer comes easily if you already knew that the number of electrons permitted in a given orbital is determined by the characteristics of electrons, NOT by the orbitals themselves.)

PRIMINARY INFORMATION

I believe the question contains a typo; after checking it out elsewhere, it's most likely...

In the long form of the periodic table, how many elements are there in the fourth period if each orbital can only contain three electrons?

Also, note that this is entirely theoretical, as all electrons only have two possible spins (#m_s = pm1/2#) in a given orbital, and no two electrons can share the same quantum state (Pauli Exclusion Principle); that restricts each orbital to contain only 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" #50%# more elements in a given quantum level.

(I suppose that was the purpose of this question; it's not really why the periodic table was arranged historically.)

LARGER TIME PERIOD...

In a generalized way, an electron configuration for the fourth period is expressed as follows:

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

where:

Now, the number of allowed #m_s# values derives from the 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 #4s# orbital:

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

For the #3d# orbitals:

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

For the #4p# orbitals:

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

Therefore, the fictitious electron configuration that we would subsequently 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 #18# elements to #color(blue)(27)# elements.

Wyatt Atkins · Answer

This is what I get.

We see that on the assumption that each orbital holds maximum of #2# electrons, for #4th# period, #n=4# allowed orbitals are #s,p,d andf#

#s# orbital holds #2# electrons

#p# orbital holds #6# electrons

#d# orbital holds #10# electrons

#f# orbital holds #14# electrons

Total number of electrons held in #n=4# shell, #=32# electrons.

If each orbital is allowed to hold #3# electrons.

Total number of electrons held in this shell #=32/2xx3=48# electrons.

This is the number of elements allowed in #4th# period.
.-.-.-.-.-.-.-.-.-.-.-.-.-.-

Alternatively

#s# orbital holds #3# electrons

#p# orbital holds #9# electrons

#d# orbital holds #15# electrons

#f# orbital holds #21# electrons

Total number of electrons #=3+9+15+21=48#

Serenity Allen · Answer

undefined There are eighteen elements in the long form of the periodic table's fourth period, and each orbital can only contain two electrons at most, not three.

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?

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

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