Question

If each orbital can hold a maximum of 3 electrons. The number of elements in $2^{nd}$ period of the periodic table (long form) will be:

• A. 27
• B. 9
• C. 18
• D. 12

Answer 1

The correct option is D 12

First we need to calculate:

i) No. of subshells ii) No. of orbitals iii) No. of elements

Formula to calculate no. of subshell

1) No. of period is even : $\frac{(n+2)}{2}$

2) No. of period is odd : $\frac{(n+1)}{2}$

$\therefore \frac{(n+2)}{2}\Rightarrow n=2$

Thus no. of orbitals in various subshell are:

For n=2 we have only $s$ & $p$ orbitals

$s$ has $1$ orbital

$p$ has $3$ orbital

$\therefore$ no. of orbital $=4$

It is given that each orbital can have a maximum of 3 electrons.

Thus maximum no. of electrons $=4\times 3=12$

Hence, the no. of elements = 12
Option D is correct.