Question #b3c29

4 Answers
Jan 4, 2016

Answer:

#""_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

Answer:

#[Ne]3s^2 3p^1#

Explanation:

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

Configuration of Neon is #1s^2 2s^2 2p^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

Answer:

#[Ne] 3s^2 3p^1#

Explanation:

The electron configuration for Aluminum is

#1s^2 2s^2 2p^6 3s^2 3p^1#

Aluminum is found in the #3rd# energy level #1st# column of the #p# block.

The short hand or noble gas notation for this would use

#[Ne] 3s^2 3p^1#

The #[Ne]# represents #1s^2 2s^2 2p^6#

Jul 25, 2016

Answer:

#"Al: " 1s^2 2s^2 2p^6 3s^2 3p^1#

Explanation:

For starters, grab a periodic table and look for aluminium, #"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

http://www.chemguide.co.uk/atoms/properties/atomorbs.html

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 #1s# orbital

#"Al: " color(blue)(1s^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 #2s# orbital

#"Al: " color(blue)(1s^2) color(red)(2s^2)#

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

#"Al: " color(blue)(1s^2) color(red)(2s^2 2p^6)#

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

#"Al: " color(blue)(1s^2) color(red)(2s^2 2p^6) color(green)(3s^2)#

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

#color(green)(|bar(ul(color(white)(a/a)color(black)("Al: " 1s^2 2s^2 2p^6 3s^2 3p^1)color(white)(a/a)|)))#

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