# What is the electron configuration for a phosphide ion?

$\left[\text{Ne}\right] 3 {s}^{2} 3 {p}^{6}$
The electron configuration of phosphorus is $\left[\text{Ne}\right] 3 {s}^{2} 3 {p}^{3}$
When the phosphide ion forms, the phosphorus atom gains three electrons. Thus, the electron configuration becomes $\left[\text{Ne}\right] 3 {s}^{2} 3 {p}^{6}$ because the phosphorus atom gains three electrons that are added to the $3 p$ subshell.