# What is the shorthand electron configuration for Al?

Apr 18, 2016

Given the periodic table: We see aluminum is atomic number $13$.

When we write our electron configurations, we utilize the valence orbitals relative to each period to construct it, and write a superscript for the maximum number of electrons in each orbital for each given period.

So the full configuration is:

$\textcolor{g r e e n}{\stackrel{\text{Period 1")overbrace(1s^2) stackrel("Period 2")overbrace(2s^2 2p^6) stackrel("Period 3}}{\overbrace{3 {s}^{2} 3 {p}^{1}}}}$

To write the noble gas shorthand configuration, you simply find the previous noble gas, which is neon, and determine which orbitals in the full configuration you've already accounted for by shortening the notation.

$\left[N e\right] = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$.

Therefore, we just have:

cancel(stackrel("Period 1")overbrace(1s^2) stackrel("Period 2")overbrace(2s^2 2p^6))^([Ne]) stackrel("Period 3")overbrace(3s^2 3p^1)

$\to \textcolor{b l u e}{\left[N e\right] 3 {s}^{2} 3 {p}^{1}}$