Question

Question #b3c29

Answer 1

`""_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`

Answer 2

`[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.

Answer 3

`[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`

Answer 4

`"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

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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.