Question

Suppose there were three values possible $(-\frac{1}{2},0,+\frac{1}{2})$ for the spin magnetic quantum number $\left(m\right)$ . Which of the following statement(s) would be correct regarding a hypothetical periodic table based on this condition?

• A. The first period would have only $2$ vertical columns
• B. The second period would have $12$ elements
• C. The periodic table would contain a total of $27$ groups
• D. The third period would have $12$ elements

Answer 1

Answer: (B), (C), (D)

The correct options are
B The third period would have $12$ elements
C The periodic table would contain a total of $27$ groups
D The second period would have $12$ elements

Since three values are possible for the spin quantum number $\left(m_3\right)$ in a hypothetical periodic table, each orbital can, therefore, accommodate $3$ electrons.

$\left(A\right)$ The first period will have $3$ vertical columns as $1s$ orbital will accommodate $3$ electrons. Thus, option A is incorrect.

$\left(B\right)$ The second period would have $12$ elements as $2s$ orbital will contain $3$ electrons and $2p$ orbital will contain $9$ electrons. Hence, option B is correct.

$\left(C\right)$ The periodic table would contain $27$ groups. This is because $s,p$ and $d$ orbitals will contain $3,9$ and $15$ electrons respectively. Hence, option C is correct.

$\left(D\right)$ The third period would have $12$ elements as $3s$ orbital will contain $3$ electrons and $3p$ orbital will contain $9$ electrons. Hence, option D is correct.

Hence, options $B,C$ and $D$ are correct.