Question

How many elements would be in the third period of the periodic table if the spin quantum number $m_s$ could have the value $-\frac{1}{2}$, $0$ and $+\frac{1}{2}$?

Answer 1

If $m_s$ could have three values then ${3e}^{-}$ could be accomodated in one orbital

s orbital no. of electrons=$3\left(3s\right)$

p orbital no. of electrons=$9\left(3p\right)$

d orbital no. of electrons=$15\left(3d\right)$

Therefore no. of elements in third period=$3+9+15=27$elements