Electronic configuration of the first transition series
From Sc to Zn: the series where the 3d sublevel is filled successively
1. Electronic configurations and oxidation states
2. Where does the extra electron go?
From Sc to Zn there are ten elements side by side. Each one has one electron more than the element before it. Where does that electron go?
Element table
Ca, Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn
- Ca: [Ar] 4s², before the series
↑↓
4s3d
Pick the right answer
Predict: where does electron number 21 of Sc go?
- A. 4s
- B. 3d
- C. 4p
Explanation
Electron 21 goes into 3d. In the filling order 4s comes before 3d, and 4s is already full at Ca, so Sc is [Ar] 4s²\ 3d¹.
3. The general pattern
[Ar] is short for the first 18 electrons. After it comes 4s², which was filled at Ca, and then 3d.
₂₁Sc:\ 1s²\ 2s²\ 2p⁶\ 3s²\ 3p⁶\ 4s²\ 3d¹
₂₁Sc:\ [Ar]\ 4s²\ 3d¹
↑↓
4s↑
3d[Ar]\ 4s²\ 3d^\,Z-20
4. Worked example: Fe (Z = 26)
Step 1: Take away the [Ar]
26 - 18 = 8 electrons are left after [Ar], because [Ar] = the first 18 electrons.
Step 2: Share them between 4s and 3d
The first 2 go to 4s (just like Ca), and the other 6 go to 3d.
₂₆Fe:\ [Ar]\ 4s²\ 3d⁶
Step 3: Draw the boxes
Hund's rule: the first five go in singly, and the sixth pairs with the first one.
↑↓
4s↑↓↑↑↑↑
3d5. Your turn: Co (Z = 27)
Same steps. Step 1 is done for you: 27 - 18 = 9 electrons after [Ar].
Fill in the blanks
The first _____ electrons go to 4s, and the rest (_____) go to 3d. So 3d has _____ unpaired electrons.
Explanation
₂₇Co: [Ar]\ 4s²\ 3d⁷. 3d is [][][][][]: 3 unpaired electrons.
Need a hint?
- 4s has only one orbital.
- 3d has 5 orbitals: fill them one at a time first.
- 7 electrons = 5 singly + 2 that make pairs.
Show the boxes
↑↓
4s↑↓↑↓↑↑↑
3d6. The whole series in your hands
Move the slider from Sc to Zn. Watch the 3d boxes and the number of unpaired electrons.
Z (atomic number) 21
↑↓
4s↑
3dTried all ten? Tap to see the pattern
- The five 3d orbitals are filled singly up to Mn (3d⁵).
- After Mn, pairing takes place up to Zn (3d¹⁰).
- Cr and Cu are the exceptions: 4s¹, as the book writes them.
- So 4s can change too. Compare Cr and Mn: both are 3d⁵, but 4s goes from 1 to 2.
7. Why no early pairing?
Mn (Z = 25)
↑↓
4s↑↑↑↑↑
3dFe (Z = 26)
↑↓
4s↑↓↑↑↑↑
3dThe book says it clearly: the five 3d orbitals are filled singly up to Mn (3d⁵). After Mn, pairing takes place up to Zn (3d¹⁰).
8. Draw Ni
Pick the right answer
₂₈Ni: which diagram of 4s and 3d is correct?
- A. 4s\,[] 3d\,[][][][][\ \ ]
- B. 4s\,[] 3d\,[][][][][]
- C. 4s\,[\ \ ] 3d\,[][][][][]
- D. 4s\,[] 3d\,[][][][][]
Explanation
₂₈Ni: [Ar]\ 4s²\ 3d⁸, so 2 unpaired electrons in 3d.
Need a hint?
- Start with 28 - 18.
- 4s takes 2, like the rest of the series (except Cr and Cu).
- 3d: 5 singly first, the rest as pairs.
9. Write the configuration
Pick the right answer
Which configuration is correct for ₂₃V?
- A. [Ar]\ 3d³
- B. [Ar]\ 3d⁵
- C. [Ar]\ 4s²\ 3d³
- D. [Ar]\ 4s¹\ 3d⁴
Explanation
General pattern [Ar]\ 4s²\ 3d^\,Z-20: 23 - 20 = 3.
Fill in the blanks
Full configuration of Ti (Z = 22): 1s²\ 2s²\ 2p⁶\ 3s²\ 3p⁶ 4s_____ 3d_____ (type the exponents)
Explanation
₂₂Ti: 1s²\ 2s²\ 2p⁶\ 3s²\ 3p⁶\ 4s²\ 3d², which is [Ar]\ 4s²\ 3d². Quick check: the exponents add up to 22.
Need a hint?
- Start with 22 - 18.
- 4s takes 2, like Ca.
- 3d takes the rest. The exponents must add up to 22.
10. Summary
- From Sc to Zn the 3d sublevel is filled successively. General pattern: [Ar]\ 4s²\ 3d^\,Z-20.
- 3d follows Hund's rule: singly up to Mn (3d⁵), then pairing up to Zn (3d¹⁰).
- In an atom, never leave out 4s: the electrons must add up to Z.
- Two exceptions: ₂₄Cr: [Ar]\ 4s¹\ 3d⁵ and ₂₉Cu: [Ar]\ 4s¹\ 3d¹⁰ (the reason is in its own lesson).
Can you explain why Fe has 4 unpaired electrons in 3d, not just 2?
Reveal
₂₆Fe: [Ar]\ 4s²\ 3d⁶. By Hund's rule the first five 3d electrons go in singly, and only the sixth pairs with the first one, so 4 electrons stay unpaired.
↑↓
4s↑↓↑↑↑↑
3d