Answer
The electron configuration of $P^{3-}$ is $$[Ne]3s^23p^6=[Ar]$$
$P^{3-}$ possesses noble-gas configuration of $Ar$.
Work Step by Step
*RULES FOR WRITING ELECTRON CONFIGURATION FOR ANIONS:
1) Write the electron configuration of the neutral atom.
2) Find out the number of electrons to be added to the neutral atom to create the anion.
3) Adding the electrons according to the following rules:
- The electrons will occupy the empty or partially occupied subshell with the lowest value of $n$ first
1) The electron configuration of a neutral $P$ atom is $$[Ne]3s^23p^3$$
2) To make $P^{3-}$, 3 electrons need to be added to a neutral $P$ atom.
3) All the inner subshells as well as subshell $3s$ have been fulled occupied.
So subshell $3p$ is the partially occupied subshell with the lowest value of $n$ $(n=3)$. Electrons will be added to here first.
Subshell $3p$ is occupied by 3 electrons in a neutral $P$ atom, while its maximum number of electrons that can occupy subshell $3p$ is 6.
Therefore, to make $P^{3-}$, 3 electrons would fill exactly 3 empty spots in subshell $3p$.
So, the electron configuration of $P^{3-}$ is $$[Ne]3s^23p^6=[Ar]$$
The above configuration is the electron configuration of noble gas $Ar$. So $P^{3-}$ possesses noble-gas configuration of $Ar$.
