Question #b3c29

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Question #b3c29

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Answer 1 · 2016-01-04T16:48:07

$_{13} A l : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$ $""_13Al:1s^(2)2s^(2)2p^(6)3s^(2)3p^1$

Explanation:

The electron configuration of Aluminum is:

$_{13} A l : 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}$ $""_13Al:1s^(2)2s^(2)2p^(6)3s^(2)3p^1$

Answer 2 · 2016-07-04T15:31:31

$[N e] 3 s^{2} 3 p^{1}$ $[Ne]3s^2 3p^1$

Explanation:

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

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

Answer 3 · 2016-07-04T15:34:21

$[N e] 3 s^{2} 3 p^{1}$ $[Ne] 3s^2 3p^1$

Explanation:

The electron configuration for Aluminum is

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

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

The short hand or noble gas notation for this would use

$[N e] 3 s^{2} 3 p^{1}$ $[Ne] 3s^2 3p^1$

The $[N e]$ $[Ne]$ represents $1 s^{2} 2 s^{2} 2 p^{6}$ $1s^2 2s^2 2p^6$

Answer 4 · 2016-07-25T00:29:50

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

Explanation:

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

This tells you that a neutral aluminium atom contains $13$ $13$ protons inside its nucleus and $13$ $13$ electrons surrounding its nucleus. You can thus say that the electron configuration for aluminium must account for a total of $13$ $13$ 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 $1 s$ $1s$ orbital

$Al: 1 s^{2}$ $"Al: " color(blue)(1s^2)$

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

$Al: 1 s^{2} 2 s^{2}$ $"Al: " color(blue)(1s^2) color(red)(2s^2)$

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

$Al: 1 s^{2} 2 s^{2} 2 p^{6}$ $"Al: " color(blue)(1s^2) color(red)(2s^2 2p^6)$

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

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

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

$∣ ∣ ∣ ∣ ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ \frac{a}{a} Al: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1} \frac{a}{a} ∣ ∣ - -------------------- -$ $color(green)(|bar(ul(color(white)(a/a)color(black)("Al: " 1s^2 2s^2 2p^6 3s^2 3p^1)color(white)(a/a)|)))$

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

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