Write the electronic configuration of aluminum

Aluminum (Al) has an atomic number ($Z$) of 13. This means a neutral aluminum atom has 13 electrons.

• The $1s$ orbital can hold 2 electrons: $1s^2$ (Remaining electrons: $13 - 2 = 11$)
• The $2s$ orbital can hold 2 electrons: $2s^2$ (Remaining electrons: $11 - 2 = 9$)
• The $2p$ orbital can hold 6 electrons: $2p^6$ (Remaining electrons: $9 - 6 = 3$)
• The $3s$ orbital can hold 2 electrons: $3s^2$ (Remaining electrons: $3 - 2 = 1$)
• The $3p$ orbital can hold up to 6 electrons. The remaining 1 electron goes into the $3p$ orbital: $3p^1$ (Remaining electrons: $1 - 1 = 0$)

Combining the filled orbitals:

The noble gas preceding Aluminum is Neon (Ne), which has an electronic configuration of $1s^2 2s^2 2p^6$. We can replace this core configuration with $[\text{Ne}]$.

The electronic configuration of aluminum is:
$\boxed{\text{1s}^2 \text{2s}^2 \text{2p}^6 \text{3s}^2 \text{3p}^1}$ or $\boxed{[\text{Ne}] \text{3s}^2 \text{3p}^1}$