# What is the electron configuration (arrangement) by spdf notation of a sodium ion?

Feb 24, 2018

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$

#### Explanation:

Sodium loses 1 electron from its outermost shell to get
the configuration of neon, or form the sodium ion.

• The first shell has $1 {s}^{2}$
• The second shell has $2 {s}^{2} , 2 {p}^{6}$
Feb 24, 2018

$1 {s}^{2} \setminus 2 {s}^{2} \setminus 2 {p}^{6}$ or just $\left[N e\right]$
A neutral sodium atom has an electron configuration of $\left[N e\right] 3 {s}^{1}$.
Since it has $11$ electrons, it would want to lose one electron to remove its third shell, while keeping the other $2$ electron shells full. So, in a sodium ion, it becomes $N {a}^{+}$ and has only $10$ electrons.
So, it will remove $1$ electron, and lose its $3 s$ orbital. The electron configuration therefore just becomes $\left[N e\right]$ or if you want the full configuration, it is $1 {s}^{2} \setminus 2 {s}^{2} \setminus 2 {p}^{6}$.