Suppose an orbital may accommodate three electrons, then estimate the number of elements in 2nd period:

In the second period of the periodic table, there is a filling of 1s,2s,2p orbitals. So the total number of orbitals to be filled,

 $=1\left(1s\right)+1\left(2s\right)+2\left(2p\right)=4$

If each orbital can hold a maximum of 3 electrons, the number of elements,

$=3\times 4=12$