Ions: 4s electrons are lost first
Which electrons leave when a transition metal forms an ion
1. Electronic configurations and oxidation states
2. Which electrons leave first?
Iron is [Ar]\ 4s²\ 3d⁶. To form Fe²⁺, it loses two electrons.
Pick the right answer
Predict: which two electrons leave?
- A. The two 4s electrons
- B. Two 3d electrons, because 3d was filled last
- C. One 4s electron and one 3d electron
Explanation
The two 4s electrons leave, because 4s is the outermost level. So Fe²⁺ is [Ar]\ 3d⁶.
3. 4s is in the outermost level
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.
4. Worked: Fe²⁺, then your turn: Co²⁺
Step 1: Write the atom
₂₆Fe:\ [Ar]\ 4s²\ 3d⁶
Step 2: Remove the 4s electrons first
Fe²⁺:\ [Ar]\ 3d⁶
4s
↑↓↑↑↑↑
3dCheck: 18 + 6 = 24 electrons = 26 − 2.
Your turn: Co²⁺ (Z = 27)
Fill in the blanks
₂₇Co is [Ar]\ 4s²\ 3d⁷. So Co²⁺ = [Ar] 4s_____ 3d_____
Explanation
Co²⁺: [Ar]\ 4s⁰\ 3d⁷, written [Ar]\ 3d⁷. Check: 18 + 7 = 25 = 27 − 2.
Need a hint?
- The charge is 2+, so 2 electrons leave.
- They leave from the outermost level first.
- 4s holds 2 electrons; 3d is not touched.
5. Strip a vanadium atom
Vanadium is [Ar]\ 4s²\ 3d³. Remove its electrons one at a time. Which sublevel empties first?
electrons removed 0
↑↓
4s↑↑↑
3dTried all five? Tap to see the pattern
- 4s empties first: V to V²⁺.
- Then 3d electrons leave one by one: V³⁺ to V⁵⁺.
- The book shows the same chain for vanadium on p.10.
6. Ni²⁺ and a mystery ion
Pick the right answer
Which is the configuration of ₂₈Ni²⁺? (Ni: [Ar]\ 4s²\ 3d⁸)
- A. [Ar]\ 4s²\ 3d⁶
- B. [Ar]\ 3d⁸
- C. [Ar]\ 4s¹\ 3d⁷
- D. [Ar]\ 4s²\ 3d⁸
Explanation
Ni²⁺: [Ar]\ 3d⁸. Check: 18 + 8 = 26 = 28 − 2.
Fill in the blanks
An ion X³⁺ is [Ar]\ 3d³. It has _____ electrons, so the atom X has Z = _____.
Explanation
18 + 3 = 21 electrons. X lost 3, so Z = 21 + 3 = 24: X is Cr, and the ion is Cr³⁺.
Need a hint?
- [Ar] stands for 18 electrons.
- A 3+ ion has 3 electrons fewer than its atom.
- Z of the atom = electrons in the ion + 3.
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.
₂₄Cr:\ [Ar]\ 4s¹\ 3d⁵ \;\; Cr²⁺:\ [Ar]\ 3d⁴
4s
↑↑↑↑
3d8. +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 Fe²⁺, it equals the ion's charge (+2). The next lesson explains it fully.
Scandium
₂₁Sc: [Ar]\ 4s²\ 3d¹. The book's table gives Sc only +3: Sc³⁺ is [Ar].
4s
3d
Zinc
₃₀Zn loses its two 4s electrons: Zn²⁺ is [Ar]\ 3d¹⁰, oxidation state +2.
4s
↑↓↑↓↑↓↑↓↑↓
3d9. Copper, titanium
Pick the right answer
Which is the configuration of Cu²⁺? (₂₉Cu: [Ar]\ 4s¹\ 3d¹⁰)
- A. [Ar]\ 4s¹\ 3d⁸
- B. [Ar]\ 3d⁹
- C. [Ar]\ 4s²\ 3d⁷
- D. [Ar]\ 3d¹⁰
Explanation
Copper loses its single 4s electron first, then one 3d electron, so Cu²⁺ is [Ar]\ 3d⁹: 18 + 9 = 27 = 29 − 2.
True or false?
Titanium (₂₂Ti), like scandium, has no +2 oxidation state.
Explanation
Ti²⁺: [Ar]\ 3d². The book's table lists Ti as 2, 3, 4.
10. Summary
- A first-series atom loses its 4s electrons first: 4s is in the outermost level (n = 4).
- Cr and Cu have only 4s¹, so their 2+ ions also lose one 3d electron: Cr²⁺ is [Ar]\ 3d⁴.
- All have oxidation state (+2), except Sc (only +3).
Back to the hook: Fe loses its two 4s electrons, so Fe²⁺ is [Ar]\ 3d⁶.
Can you explain why Mn²⁺ is [Ar]\ 3d⁵ and not [Ar]\ 4s²\ 3d³?
Reveal
Mn is [Ar]\ 4s²\ 3d⁵. Its 4s electrons are in the outermost level (n = 4), so they leave first and 3d keeps its 5 electrons: Mn²⁺ is [Ar]\ 3d⁵. Removing two 3d electrons instead would give [Ar]\ 4s²\ 3d³.
4s
↑↑↑↑↑
3d