• # question_answer How many elements would be in the second period of the Periodic Table if the spin quantum number ${{m}_{s}}$ could have the value $-\frac{1}{2},\,0,\,+\frac{1}{2}$? A)  8                                 B)  10               C)  12                               D)  18

For second period n = 2 hence I           m         ${{m}_{s}}$ 0          0          $+\frac{1}{2},\,\,0,\,-\frac{1}{2}$ 1          - 1        $+\frac{1}{2},\,0,\,-\frac{1}{2}$ 0          $+\frac{1}{2},\,0,\,-\frac{1}{2}$ $+1$  $+\frac{1}{2},\,0,\,-\frac{1}{2}$  Hence, total number of elements = 12 (= total values of spin quantum number)