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. $8$

Answer 1

Answer: (B)

$12$

Number of orbit in second period ($n$) $=2$ and orbitals ($n^2$) $=4$

Since, possible values of spin quantum number are $+\frac{1}{2},0,-\frac{1}{2}$ i.e. 3.

Therefore, each orbital can occupy 3 electrons with different spin and a total of $4\times 3$ i.e. 12 electrons in the second period.

Therefore option B is the correct answer.