# Ions: 4s electrons are lost first

> Which electrons leave when a transition metal forms an ion


## Objectives

- Write the configuration of a first-series positive ion: remove the 4s electrons first, then 3d electrons one by one.
- Explain why: 4s is in the outermost level (n = 4).
- Handle Cr and Cu ions, and identify an element from its ion's configuration.
- State that all elements of the series have oxidation state (+2), except scandium.

## 1. Electronic configurations and oxidation states




## 2. Which electrons leave first?

Predict: which two electrons leave?

- One 4s electron and one 3d electron
- Two 3d electrons, because 3d was filled last
- The two 4s electrons ✓

The two 4s electrons leave, because 4s is the outermost level. So $\ce{Fe^{2+}}$ is $[\text{Ar}]\ 3d^6$.

[ar-EG] الاتنين 4s electrons هما اللي بيطلعوا، عشان الـ 4s هو المستوى الخارجي. فالـ $\ce{Fe^{2+}}$ بيبقى $[\text{Ar}]\ 3d^6$.
## 3. 4s is in the outermost level

Fe: 4s is in level n = 4, 3d is in level n = 3 In Fe, the 4s electrons are in the outermost level (n = 4). The 3d electrons are in n = 3, nearer the nucleus. So a first-series atom loses its 4s electrons first. For higher charges, 3d electrons then leave one by one. $\ce{Fe^{2+}}$ is $[\text{Ar}]\ 4s^2\ 3d^4$: the electrons that went in last come out first.


## 4. Worked: Fe²⁺, then your turn: Co²⁺

Step 1: Write the atom Step 2: Remove the 4s electrons first 4s is empty; 3d keeps its 6 electrons Check: 18 + 6 = 24 electrons = 26 − 2. Your turn: Co²⁺ (Z = 27) $\ce{_{27}Co}$ is $[\text{Ar}]\ 4s^2\ 3d^7$. So $\ce{Co^{2+}}$ = [Ar] 4s___ 3d___


## 5. Strip a vanadium atom

Vanadium is $[\text{Ar}]\ 4s^2\ 3d^3$. Remove its electrons one at a time. Which sublevel empties first? electrons removed $\ce{V}$: $4s^2\ 3d^3$ (the atom) $\ce{V+}$: $4s^1\ 3d^3$ $\ce{V^{2+}}$: $4s^0\ 3d^3$: 4s is empty $\ce{V^{3+}}$: $4s^0\ 3d^2$ $\ce{V^{4+}}$: $4s^0\ 3d^1$ $\ce{V^{5+}}$: $4s^0\ 3d^0$ Tried all five? Tap to see the pattern 4s empties first: V to $\ce{V^{2+}}$. Then 3d electrons leave one by one: $\ce{V^{3+}}$ to $\ce{V^{5+}}$. The book shows the same chain for vanadium on p.10.


## 6. Ni²⁺ and a mystery ion

Which is the configuration of $\ce{_{28}Ni^{2+}}$? (Ni: $[\text{Ar}]\ 4s^2\ 3d^8$)

- $[\text{Ar}]\ 4s^1\ 3d^7$
- $[\text{Ar}]\ 4s^2\ 3d^8$
- $[\text{Ar}]\ 4s^2\ 3d^6$
- $[\text{Ar}]\ 3d^8$ ✓

$\ce{Ni^{2+}}$: $[\text{Ar}]\ 3d^8$. Check: 18 + 8 = 26 = 28 − 2.

[ar-EG] $\ce{Ni^{2+}}$: $[\text{Ar}]\ 3d^8$. اتأكد: 18 + 8 = 26 = 28 − 2.
## 7. Cr and Cu ions

Cr and Cu atoms have only one 4s electron. So for a 2+ ion, the second electron must come from 3d. $\ce{Cr^{2+}}$: the 4s electron and one 3d electron are gone $\ce{Cr^{2+}}$ is $[\text{Ar}]\ 4s^1\ 3d^3$: the ion keeps the atom's $4s^1$.


## 8. +2 for all, except scandium

Book point 3: all elements of the first series have oxidation state (+2) after losing their 4s electrons, except scandium. Oxidation state For a simple metal ion like $\ce{Fe^{2+}}$, it equals the ion's charge (+2). The next lesson explains it fully. Scandium $\ce{_{21}Sc}$: $[\text{Ar}]\ 4s^2\ 3d^1$. The book's table gives Sc only +3: $\ce{Sc^{3+}}$ is $[\text{Ar}]$. $\ce{Sc^{3+}}$: $[\text{Ar}]$, 4s and 3d are empty Zinc $\ce{_{30}Zn}$ loses its two 4s electrons: $\ce{Zn^{2+}}$ is $[\text{Ar}]\ 3d^{10}$, oxidation state +2. $\ce{Zn^{2+}}$: $[\text{Ar}]\ 3d^{10}$


## 9. Copper, titanium

Which is the configuration of $\ce{Cu^{2+}}$? ($\ce{_{29}Cu}$: $[\text{Ar}]\ 4s^1\ 3d^{10}$)

- $[\text{Ar}]\ 3d^9$ ✓
- $[\text{Ar}]\ 4s^2\ 3d^7$
- $[\text{Ar}]\ 4s^1\ 3d^8$
- $[\text{Ar}]\ 3d^{10}$

Copper loses its single 4s electron first, then one 3d electron, so $\ce{Cu^{2+}}$ is $[\text{Ar}]\ 3d^9$: 18 + 9 = 27 = 29 − 2.

[ar-EG] الـ copper بيفقد الـ 4s electron الوحيد بتاعه الأول، وبعده 3d electron واحد، فالـ $\ce{Cu^{2+}}$ يبقى $[\text{Ar}]\ 3d^9$: 18 + 9 = 27 = 29 − 2.
## 10. Summary

A first-series atom loses its 4s electrons first: 4s is in the outermost level (n = 4). Cr and Cu have only $4s^1$, so their 2+ ions also lose one 3d electron: $\ce{Cr^{2+}}$ is $[\text{Ar}]\ 3d^4$. All have oxidation state (+2), except Sc (only +3). All elements of the first series have **oxidation state (+2)** after **losing the 4s electrons first** (except **scandium**). In higher states they **lose 3d electrons in sequence**. Back to the hook: Fe loses its two 4s electrons, so $\ce{Fe^{2+}}$ is $[\text{Ar}]\ 3d^6$. Can you explain why $\ce{Mn^{2+}}$ is $[\text{Ar}]\ 3d^5$ and not $[\text{Ar}]\ 4s^2\ 3d^3$? Mn is $[\text{Ar}]\ 4s^2\ 3d^5$. Its 4s electrons are in the outermost level (n = 4), so they leave first and 3d keeps its 5 electrons: $\ce{Mn^{2+}}$ is $[\text{Ar}]\ 3d^5$. Removing two 3d electrons instead would give $[\text{Ar}]\ 4s^2\ 3d^3$. $\ce{Mn^{2+}}$: 4s is empty; 3d keeps its 5 electrons



Chapter: Transition Elements (https://yaani.ai/en/learn/chemistry/chem-s3-en.ch1)
Source: https://yaani.ai/en/lesson/chem-s3-en.lesson-te.ion-configuration