# Writing electronic configurations

> Sublevel by sublevel, and the short form with [Ne] or [Ar]


## Objectives

- Write an atom's configuration sublevel by sublevel, in the filling order, so the electrons add up to Z.
- Write the short form with a noble-gas core such as [Ne] or [Ar].

## 1. Electronic configurations and oxidation states




## 2. A long line of electrons

Predict: which part of this line is exactly the configuration of argon, $\ce{_{18}Ar}$?

- $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6$ ✓
- $1s^2\ 2s^2\ 2p^6$
- $4s^2$

$1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6$ holds 2 + 2 + 6 + 2 + 6 = 18 electrons, exactly the configuration of argon. So calcium is argon's configuration plus $4s^2$.

[ar-EG] الجزء $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6$ فيه 2 + 2 + 6 + 2 + 6 = 18 electrons، وده configuration الـ argon بالظبط. يعني الـ calcium هو configuration الـ argon ومعاه $4s^2$.
## 3. The rules for writing

Fill the sublevels in the filling order you studied, starting with 1s. Fill one sublevel before you start the next. Chapter 1 shows two exceptions, Cr and Cu. Filling order, with the maximum number of electrons in each sublevel: s holds 2, p holds 6, d holds 10 Quick check: the exponents must add up to the atomic number, Z.


## 4. Worked example: S (Z = 16)

Sulphur has 16 electrons. Let's place them, step by step. Step 1 $1s^2\ 2s^2\ 2p^6$ uses 10 electrons. Why? Here, a sublevel is filled to its maximum before the next one starts. Step 2 6 electrons are left. 3s comes next and takes its 2. Then 3p takes the remaining 4. Check: 2 + 2 + 6 + 2 + 4 = 16 Step 3: the short form noble-gas core The configuration of the noble gas just before the element, written as its symbol in square brackets. Example: [Ne] = $1s^2\ 2s^2\ 2p^6$. Neon (Z = 10) is the noble gas just before sulphur


## 5. Your turn: K (Z = 19)

Potassium has 19 electrons. The first part is done for you. Full: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6$ 4s___ (type the exponent). In the short form, the noble-gas core is [___].


## 6. On your own

Which is the short-form configuration of phosphorus, $\ce{_{15}P}$?

- $[\text{Ne}]\ 3s^2\ 3p^3$ ✓
- $[\text{Ne}]\ 3p^5$
- $[\text{Ne}]\ 3s^2\ 3p^5$
- $[\text{Ar}]\ 3s^2\ 3p^3$

Phosphorus has 15 electrons, and the noble gas just before it is neon (10). So the short form is $[\text{Ne}]\ 3s^2\ 3p^3$: 10 + 2 + 3 = 15, with 3s filled before 3p.

[ar-EG] الـ phosphorus فيه 15 electrons، والـ noble gas اللي قبله على طول هو الـ neon (10). فالـ short form هو $[\text{Ne}]\ 3s^2\ 3p^3$: 10 + 2 + 3 = 15، والـ 3s بيتملي قبل الـ 3p.
## 7. Summary

Fill the sublevels in the filling order: $1s\ 2s\ 2p\ 3s\ 3p\ 4s\ 3d$ … s holds 2, p holds 6, d holds 10. Check: the exponents add up to Z. Short form: [the noble gas just before the element] + the electrons after it, e.g. $\ce{_{20}Ca}$: $[\text{Ar}]\ 4s^2$. The examples of this lesson: full configuration and short form Chapter 1 starts right after Ca, from Sc (Z = 21). Its short forms start with [Ar]. Can you explain how to check in a few seconds that $[\text{Ne}]\ 3s^2\ 3p^4$ is sulphur?



Chapter: Transition Elements (https://yaani.ai/ar/learn/chemistry/chem-s3-en.ch1)
Source: https://yaani.ai/ar/lesson/chem-s3-en.lesson-atom.electron-configuration-writing