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

2 Answers
Sep 4, 2017

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.

(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 the 4th period of the periodic table (long form) is...?

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.

(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 the total number of electrons in all the orbitals in a given subshell.
  • #l# is the angular momentum quantum number. #l = 0, 1, 2, 3, . . . # corresponds to #s, p, d, f, . . . # orbitals.
  • #2l+1# is known as the degeneracy of the subshell; it is how many orbitals are in that subshell.
  • #N_(m_s)# is an arbitrary number of spins the electron could have, as it also then gives the maximum 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 #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#

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 #18# elements to #color(blue)(27)# elements.

Sep 13, 2017

Answer:

This is what I get.

Explanation:

chemhume.co.uk

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#