Question

What is the maximum number of electrons in a \[3p\] subshell?

Answer 1

Hint: You must know the concept of Pauli exclusion principle as more than two electrons may not inhabit the same orbital and To answer this question, we need to figure how many sets of lobes are present in the given subshell.

Complete answer:
The subshell of $p$ will always have $3$ orbitals and a maximum of $2$ electrons each orbital can accommodate. Therefore, the p subshell can accommodate a maximum of 6 electrons. So, \[Xe\] will also have 6 electrons which belong to the subshell of $3p$.
By having "3" before the subshell-'p' we don’t want to do anything because it represents only the number of shells. Now coming to the p subshell which contains three orbitals that are px, py, pz due to its shape, representing p orbitals arranged in three dimensions. Now as we know that maximum of $2$ electrons each orbital can contain so there will be a $6$ electron totally in the subshell of p. which is $\left( {2 \times 3} \right) = 6$ electrons in p subshell.
Therefore $6$ electrons can fit in the orbital of $3p$.
But as you know that in a text book in any given shell is given by $2{n^2}$, that can fit a maximum of the electron. It would mean in the first shell, $2$ electrons could fit. In the second shell, $8$ electrons could fit. In the third shell, $18$ electrons can fit. In the fourth shell, $32$ electrons can fit. As I said before in the first orbital the maximum number of orbital is $2$. Surely, they are the same.

Note:
There are three sets of lobes in the subshell of $3p$that each carries two electrons.
By the Pauli exclusion principle.
Within an atom, no two electrons may have the same n, l, ml, or ms. Therefore more than two electrons may not inhabit the same orbital.