Question

How many elements would be in the $I{I}^{nd}$ period of the Periodic Table if the spin quantum numbers could have the value $+\frac{1}{2},0,-\frac{1}{2}$?

• A. 18
• B. 12
• C. 10
• D. None of these

Answer 1

The correct option is B 12

For $I{I}^{nd}$ period

$n=2$ hence

$\begin{array}{ccc}l& m& s\\ 0& 0& \frac{1}{2}+0-\frac{1}{2}\\ 1& -1& \frac{1}{2}+0-\frac{1}{2}\\ & 0& \frac{1}{2}+0,-\frac{1}{2}\\ & +1& \frac{1}{2}+,0,-\frac{1}{2}\end{array}$

Hence, total number of elements (total value of spin quantum number) = 12