# 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

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 (${m}_{s} = \pm \frac{1}{2}$) 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:

$4 {s}^{x} 3 {d}^{x} 4 {p}^{x}$

where:

• $x = \left(2 l + 1\right) \cdot {N}_{{m}_{s}}$ 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.
• $2 l + 1$ 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 $4 s$ orbital:

$l = 0 , \implies 2 l + 1 = 1$

For the $3 d$ orbitals:

$l = 2 , \implies 2 l + 1 = 5$

For the $4 p$ orbitals:

$l = 1 , \implies 2 l + 1 = 3$

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

$4 {s}^{1 \cdot 3} 3 {d}^{5 \cdot 3} 4 {p}^{3 \cdot 3}$

$= 4 {s}^{3} 3 {d}^{15} 4 {p}^{9}$

And that would apparently expand the fourth period of the periodic table from $18$ elements to $\textcolor{b l u e}{27}$ elements.

Sep 13, 2017

This is what I get.

#### Explanation: We see that on the assumption that each orbital holds maximum of $2$ electrons, for $4 t h$ period, $n = 4$ allowed orbitals are $s , p , d \mathmr{and} f$

$s$ 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 $= \frac{32}{2} \times 3 = 48$ electrons.

This is the number of elements allowed in $4 t h$ 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 $= 3 + 9 + 15 + 21 = 48$