# What is the electron configuration for a phosphorous anion with a charge of -2?

Nov 6, 2015

"P"^(2-): ["Ne"] 3s^2 3p^5

#### Explanation:

Your starting point here will be the electron configuration of a neutral phosphorus atom.

Phosphorus, $\text{P}$, is located in period 3, group 15 of the periodic table, and has an atomic number equal to $15$. This means that a neutral phosphorus atom will have $15$ electrons surrounding its nucleus.

Therefore, the electron configuration of a neutral phosphorus atom will show $15$ electrons

$\text{P: } {1.2}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{3}$

Now, in order for the phosphorus atom to become a phosphorus anion, it must gain two electrons. Since the 3p-subshell only holds $3$ electrons in the neutral atom, these incoming electrons will be placed there.

Therefore, the electron configuration of the ${\text{P}}^{2 -}$ anion will be

${\text{P}}^{2 -} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{5}$

Using the noble gas shorthand notation, you will get

"P"^(2-): ["Ne"] 3s^2 3p^5