d- and f-Block Elements · Lesson 1 of 12
Electronic Structure of the d-Block
“Locate the transition series, explain configuration exceptions and derive configurations of metal ions.”
• Distinguish a d-block element from a transition element using atoms and ions. • Write the ground-state configurations of the first transition series and recognise heavier-series exceptions. • Remove electrons in the correct order to obtain ionic d-electron counts. • Explain why configuration patterns are useful without treating them as universal stability rules.
Why a block of metals sits between the main groups
Iron conducts electricity, copper forms blue solutions, and zinc forms many colourless salts. All three occupy the broad central region of the periodic table, yet their atoms and ions do not have identical electronic structures. A block identifies the subshell involved in the characteristic filling pattern; a transition element is identified more specifically by an incomplete d subshell in its atom or a chemically accessible cation. This distinction explains why membership of the central block does not guarantee every characteristic transition-metal property.
An element with a partially occupied d subshell in its atom, or one capable of forming a cation with a partially occupied d subshell.
Groups 3–12 form the d-block in the conventional long-form arrangement. The series involve 3d, 4d, 5d and 6d orbitals. For period number n, the characteristic d subshell has principal quantum number n−1: in period 4, 3d fills while 4s is the outer s subshell. The 4f and 5f series are usually displayed below the main table to keep its width manageable. Their physical position on a printed chart does not imply weaker chemical importance.
| Series | Elements in the conventional sequence | Orbitals involved |
|---|---|---|
| 3d | Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn | 3d and 4s |
| 4d | Y, Zr, Nb, Mo, Tc, Ru, Rh, Pd, Ag, Cd | 4d and 5s |
| 5d | La, then Hf through Hg after insertion of the 4f series | 5d and 6s |
| 6d | Ac, then Rf through Cn after insertion of the 5f series | 6d and 7s |
Configurations are outcomes of competing energies
A useful general pattern is (n−1)d¹–¹⁰ns⁰–². It is a pattern, not an instruction to force every atom into ns². The d and s levels are close enough that electron attraction, repulsion and exchange contributions can change which occupation has the lowest total energy. Chromium is [Ar]3d⁵4s¹ rather than a simple 3d⁴4s² prediction; copper is [Ar]3d¹⁰4s¹ rather than 3d⁹4s². Half-filled and filled subshells help explain these cases, but “every atom prefers a half-filled subshell” would predict many wrong configurations.
| Atom | Z | Beyond [Ar] | Common M²⁺ d count |
|---|---|---|---|
| Sc | 21 | 3d¹4s² | d¹ (Sc³⁺ is the usual stable ion) |
| Ti | 22 | 3d²4s² | d² |
| V | 23 | 3d³4s² | d³ |
| Cr | 24 | 3d⁵4s¹ | d⁴ |
| Mn | 25 | 3d⁵4s² | d⁵ |
| Fe | 26 | 3d⁶4s² | d⁶ |
| Co | 27 | 3d⁷4s² | d⁷ |
| Ni | 28 | 3d⁸4s² | d⁸ |
| Cu | 29 | 3d¹⁰4s¹ | d⁹ |
| Zn | 30 | 3d¹⁰4s² | d¹⁰ |
Heavier rows show additional exceptions. Beyond [Kr], Y through Cd have respectively 4d¹5s², 4d²5s², 4d⁴5s¹, 4d⁵5s¹, 4d⁵5s², 4d⁷5s¹, 4d⁸5s¹, 4d¹⁰5s⁰, 4d¹⁰5s¹ and 4d¹⁰5s². Palladium is especially useful because its empty 5s subshell defeats the assumption that every neutral d-block atom must retain at least one outer s electron. In the 5d row, Pt is 5d⁹6s¹ and Au is 5d¹⁰6s¹. Configurations of very heavy atoms may be theoretical assignments affected by relativity; they should not be presented as measurements of equal certainty.
Forming ions: remove outer s electrons first
Writing 4s before 3d while constructing an approximate filling sequence does not mean 4s remains lower in energy in every atom and ion. Once d orbitals are occupied, their energies and radial distributions change. For the common first-row cations, remove 4s electrons before 3d electrons. Count the electrons actually present; do not remove two 4s electrons from chromium or copper when their neutral atoms possess only one.
Problem
Find Cr³⁺, Cu⁺, Co²⁺ and Mn²⁺ configurations.
- 1.Cr begins [Ar]3d⁵4s¹. Remove its one 4s electron, then two 3d electrons: Cr³⁺=[Ar]3d³.
- 2.Cu begins [Ar]3d¹⁰4s¹. Removal of the one 4s electron gives Cu⁺=[Ar]3d¹⁰; a second removal would give Cu²⁺=[Ar]3d⁹.
- 3.Co begins [Ar]3d⁷4s² and loses both 4s electrons: Co²⁺=[Ar]3d⁷. Mn similarly gives Mn²⁺=[Ar]3d⁵.
- 4.Check electron totals: chromium has 24−3=21 electrons and copper(I) has 29−1=28. The core plus remaining d electrons gives the same totals.
For example, Fe is in group 8: Fe²⁺ has d⁶ and Fe³⁺ has d⁵. The oxidation state is a formal electron-accounting quantity; it does not assert that a covalent compound contains an isolated free ion of that charge. For most simple cation problems here the shortcut agrees with removing s and then d electrons explicitly.
Why scandium qualifies and zinc does not
Problem
Compare Sc, Zn, Cu and Ag as transition elements.
- 1.Sc has a 3d¹ atom, so it qualifies even though its usual Sc³⁺ ion is d⁰. The atom OR cation criterion matters.
- 2.Zn has a 3d¹⁰ atom and its normal Zn²⁺ chemistry retains d¹⁰; it is a d-block element without the usual transition-element classification.
- 3.Cu has a d¹⁰ atom but forms Cu²⁺, d⁹. Ag has a d¹⁰ atom and familiar Ag⁺, d¹⁰, but also forms Ag²⁺, d⁹. Both therefore qualify.
- 4.Cd and Hg are conventionally grouped with Zn because their ordinary ionic chemistry retains filled d subshells. Avoid claiming that no unusual species involving higher states can ever be made.
Electronic configuration supplies a foundation for later explanations of oxidation states, colour and magnetism. It does not supply a complete energy balance. A d⁵ ion can be favoured in one medium while another charge is favoured with a different ligand. The next lessons add bonding, hydration, size and redox potentials to the picture.
Quick Quiz
Quiz
Why is scandium a transition element?
Explanation: The atom itself satisfies the incomplete-d criterion, despite d⁰ Sc³⁺.
What is the configuration of Cr³⁺?
Explanation: Remove one 4s and then two 3d electrons from Cr.
Which atom has an empty outer s subshell in its ground configuration?
Explanation: Pd has 4d¹⁰5s⁰ beyond [Kr].
Which cation establishes that copper is a transition element?
Explanation: Cu²⁺ is formed by removing one s and one d electron.
What is the usual d count of Fe³⁺?
Explanation: Group 8 minus oxidation state 3 gives five d electrons.
Practice Problems
- Derive Ti³⁺, V²⁺, Fe²⁺, Fe³⁺ and Ni²⁺ configurations.
- Why is silver classified as a transition element although Ag and Ag⁺ have filled d subshells?
- Explain the Cr and Cu exceptions without claiming a universal half-filled rule.
- Compare the valence configurations of Cr, Mo and W.
- A group-7 metal forms a +6 oxoanion. Find its formal d count and explain the limit of this statement.
Write the neutral atoms first, remove both 4s electrons, then remove further d electrons when needed. The results are [Ar]3d¹, [Ar]3d³, [Ar]3d⁶, [Ar]3d⁵ and [Ar]3d⁸ respectively. Check each total against Z minus ionic charge.
Key Takeaways
• The central block is characterised by d-subshell filling. • An incomplete d subshell in either the atom or an accessible cation establishes transition-element character. • Cr and Cu illustrate exceptions caused by closely competing total energies. • Outer s electrons are removed before d electrons in common d-metal cations. • Sc qualifies through its atom; Cu and Ag also qualify through their cations. • Zn, Cd and Hg retain filled d subshells in their normal ionic chemistry. • Configuration is a starting point for chemical reasoning, not a complete stability calculation.
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Metallic Bonding, Size and Ionisation Trends