# What is the full ground state electron configuration of O^+?

Nov 20, 2016

${O}^{+} = 1 {s}^{2} 2 {s}^{2} 2 {p}^{3}$

#### Explanation:

Typically, an atom of $O$ has 8 electrons, so based on the electron configuration system that would be $1 {s}^{2} 2 {s}^{2} 2 {p}^{4}$, or $2 + 2 + 4 = 8 {e}^{-}$.

However, ${O}^{+}$ means you've lost an electron, hence the positive charge. Thus, you subtract an electron:
$8 \setminus \textcolor{t e a l}{- 1} = \setminus \textcolor{red}{7} {e}^{-}$
$1 {s}^{2} 2 {s}^{2} 2 {p}^{4 \setminus \textcolor{t e a l}{- 1}} \setminus \Rightarrow 1 {s}^{2} 2 {s}^{2} 2 {p}^{\setminus \textcolor{red}{3}}$

Electron configuration rule: