# If each orbital can hold a maximum of 3 electrons

• xiphoid

## Homework Statement

If each orbital can hold a maximum of 3 electrons, the number of elements in the 4th period of long form of periodic table will be?

?

## The Attempt at a Solution

I am not able to decide from where should I begin?

hi xiphoid! i don't understand the question …

there are no orbitals which hold a maximum of 3 electrons (the maximum would always be even, anyway … spin-ups matching spin-downs)

what exactly is the question, and the background to it?​

I bet it is a hypothetical question. That doesn't mean it makes sense.

This question is just to develop the thinking skill of the student!
It does not/ never is related to the original/factual happening around or perhaps inside the nucleus.
From where should i start.
I guess, the moment i get the completely filled configuration of noble gas, the number of elements preceding to them would be the answer, but again from where to start?
By taking the default configuration of element in the mentioned period?
I bet it is a hypothetical question. That doesn't mean it makes sense.

Yep this IS hypothetical. 4th Period means that electrons are being filled in specifically, 4s, 3d, 4p orbitals.

Calculate the number of electrons you can fill. Try.

• honey
Here is a hint, the first period of the long form periodic table contains two elements. This is because each s orbital can only contain 2 electrons. Thus, there can only be two elements; one with Z=1 and the other with Z=2.

In your problem you can have 3 electrons per orbital. So this version of the periodic table must have three elements in the first period with Z=1, Z=2 and Z=3 and the element corresponding to Z=3 is the first noble gas. You can do the same analysis for the second, third and fourth period.