# Question b3c29

Jan 4, 2016

""_13Al:1s^(2)2s^(2)2p^(6)3s^(2)3p^1

#### Explanation:

The electron configuration of Aluminum is:

""_13Al:1s^(2)2s^(2)2p^(6)3s^(2)3p^1#

Jul 4, 2016

$\left[N e\right] 3 {s}^{2} 3 {p}^{1}$

#### Explanation:

Atomic number is 13. The last noble gas with atomic number below 13 is Neon (atomic number 10).

Configuration of Neon is $1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$ so this leaves 3 more electrons for Aluminium. Two electrons will go into the 3s, and the remaining one into the 3p.

Jul 4, 2016

$\left[N e\right] 3 {s}^{2} 3 {p}^{1}$

#### Explanation:

The electron configuration for Aluminum is

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{1}$

Aluminum is found in the $3 r d$ energy level $1 s t$ column of the $p$ block.

The short hand or noble gas notation for this would use

$\left[N e\right] 3 {s}^{2} 3 {p}^{1}$

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

Jul 25, 2016

$\text{Al: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{1}$

#### Explanation:

For starters, grab a periodic table and look for aluminium, $\text{Al}$. You'll find it listed in period 3, group 13. Notice that the element has an atomic number equal to $13$.

This tells you that a neutral aluminium atom contains $13$ protons inside its nucleus and $13$ electrons surrounding its nucleus. You can thus say that the electron configuration for aluminium must account for a total of $13$ electrons.

As you know, electron configurations are written in accordance with the Aufbau Principle. The available empty orbitals in order of increasing energy look like this Your goal now will be to start adding electrons to these orbitals. The first energy level contains a single orbital, so the first two electrons go in the $1 s$ orbital

$\text{Al: } \textcolor{b l u e}{1 {s}^{2}}$

You're now down to $11$ electrons. The second energy level contains a total of $4$ orbitals. The first orbital to be filled on this energy level is the $2 s$ orbital

$\text{Al: } \textcolor{b l u e}{1 {s}^{2}} \textcolor{red}{2 {s}^{2}}$

You're down to $9$ electrons. The next $6$ electrons are distributed in the $2 p$ orbitals

$\text{Al: } \textcolor{b l u e}{1 {s}^{2}} \textcolor{red}{2 {s}^{2} 2 {p}^{6}}$

You're down to $3$ electrons. The first orbital to be filled on the third energy level is the $3 s$ orbital

$\text{Al: } \textcolor{b l u e}{1 {s}^{2}} \textcolor{red}{2 {s}^{2} 2 {p}^{6}} \textcolor{g r e e n}{3 {s}^{2}}$

Finally, the last electron is added to one of the $3 p$ orbitals

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{Al: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

And there you have it, the complete electron configuration for aluminium.