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I came across the following question:

If each orbital can hold a max. of 3 e– what is the number of elements in the 4th period of the periodic table?

I was unable to even start the thought process of answering the question because I'm rather horrible at discerning the period number using the electronic configuration, the number of elements that can be accommodated in a period etc. If someone could please explain it to me, that would be so great, especially since I've googled this too many times already and haven't been able to wrap my head around it yet!

Thanks in advance :) Cheers!

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In the 4th period of the periodic table there is 1 s orbital, 3 p orbitals, and 5 d orbitals. That's a total of 9 orbitals. $9 \times 3=27$

That's 27 elements.

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  • $\begingroup$ Since when there's supposed to be 3 electrons per orbital? $\endgroup$
    – Mithoron
    Commented Aug 18, 2016 at 10:57
  • $\begingroup$ It's a hypothetical question. $\endgroup$
    – GeeJay
    Commented Aug 18, 2016 at 11:05
  • $\begingroup$ It's not hypothetical, but sarcastic - there can be only two... $\endgroup$
    – Mithoron
    Commented Aug 18, 2016 at 11:07
  • $\begingroup$ Yes, and in the "there can only..." stream of thought, the double slit experiment, GR, and the quantum model of the atom were attempts at sarcasm. $\endgroup$
    – GeeJay
    Commented Aug 18, 2016 at 12:39
  • $\begingroup$ en.wikipedia.org/wiki/Spin_quantum_number - seems you're somehow not getting my point. It's 9*2 = 18 as spin quantum number can have only two different values, orbitals can be occupied only by 2 electrons. Geez you can count elements in periodic table to see there are 18 elements in this period. $\endgroup$
    – Mithoron
    Commented Aug 18, 2016 at 14:35