Concept Core
3d transition metals — electronic configurations, properties, and important compounds.
Electronic Configurations of 3d Transition Metals
General configuration: [Ar] 3dˢ 4sˢ. Remember the two exceptions:
| Element | Z | Expected | Actual | Reason |
|---|
| Chromium (Cr) | 24 | [Ar]3d⁴4s² | [Ar]3d⁵4s¹ | Half-filled 3d⁵ stability |
| Copper (Cu) | 29 | [Ar]3d⁹4s² | [Ar]3d¹⁰4s¹ | Fully-filled 3d¹⁰ stability |
These two exceptions appear in EAPCET every year. All other 3d elements fill in expected order.
General Properties of Transition Metals
1. Variable oxidation states: due to comparable energies of 3d and 4s orbitals (e.g., Fe: +2, +3; Mn: +2,+3,+4,+6,+7)
2. Coloured ions: d-d transitions absorb visible light (ions with unpaired d electrons are coloured)
3. Magnetic properties: paramagnetic due to unpaired d electrons
4. Catalytic activity: variable valency allows electron transfer (Fe in Haber, V₂O₅ in Contact process)
5. Alloy formation: similar atomic radii → form alloys (e.g., stainless steel = Fe+Cr+Ni)
6. Complex formation: empty d orbitals accept lone pairs from ligands
Potassium Permanganate (KMnO₄)
Mn is in +7 oxidation state. Deep purple colour. Strong oxidising agent.
In acidic medium: MnO₄⁻ → Mn²⁺ (colourless) — gains 5e⁻. Half-reaction:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (acidic)
In alkaline medium: MnO₄⁻ → MnO₂ (brown ppt) — gains 3e⁻
Uses: oxidation of Fe²⁺ to Fe³⁺, oxalic acid titration, water purification.
Potassium Dichromate (K₂Cr₂O₇)
Cr is in +6 oxidation state. Orange colour. Strong oxidising agent in acidic medium.
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (acidic)
(colour change: orange → green)
Uses: chrome alum preparation, dyeing, tanning leather.
Interconversion: Cr₂O₇²⁻ (orange, acidic) ⇌ 2CrO₄²⁻ (yellow, alkaline)
Lanthanide Contraction
Going from La to Lu (4f filling), atomic radius DECREASES despite increasing Z. Called lanthanide contraction.
Cause: 4f electrons have poor shielding → effective nuclear charge increases → radius decreases.
Consequence: 5d elements (Hf, Ta, W…) have nearly the same atomic radii as their 4d counterparts (Zr, Nb, Mo…), making them difficult to separate.
Actinides vs Lanthanides
| Property | Lanthanides | Actinides |
|---|
| f-orbital filling | 4f | 5f |
| Common oxidation state | +3 | +3, +4, +5, +6 (more variable) |
| Radioactivity | Non-radioactive (mostly) | All radioactive |
| Availability | Found in nature | Many are synthetic |
Formula Vault
Key d-block facts, oxidation state half-reactions, and electronic configurations.
Cr Configuration
[Ar] 3d⁵ 4s¹
Not 3d⁴4s²; half-filled stability
Cu Configuration
[Ar] 3d¹⁰ 4s¹
Not 3d⁹4s²; fully-filled stability
Mn Oxidation States
+2,+3,+4,+6,+7
Most variable in 3d series
Fe Oxidation States
+2 (Fe²⁺), +3 (Fe³⁺)
+2 reducing, +3 more stable
KMnO₄ in Acid
MnO₄⁻+8H⁺+5e⁻ → Mn²⁺+4H₂O
Purple → colourless
K₂Cr₂O₇ in Acid
Cr₂O₇²⁻+14H⁺+6e⁻ → 2Cr³⁺+7H₂O
Orange → green
Dichromate-Chromate
Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O
Orange (acid) ↔ Yellow (alkali)
Paramagnetic Formula
μ = √(n(n+2)) BM
n = number of unpaired electrons
Worked Examples
5 problems — oxidation states, electronic config, half-reactions, colour, and a magnetic moment trap.
EasyFind oxidation state of Mn in KMnO₄▾
Find the oxidation state of Mn in KMnO₄.
1
K = +1, O = −2 (×4 = −8). Let Mn = x.
2
+1 + x + 4(−2) = 0 → 1 + x − 8 = 0 → x = +7
✓ Mn = +7 in KMnO₄
EasyFind oxidation state of Cr in K₂Cr₂O₇▾
Find the oxidation state of Cr in K₂Cr₂O₇.
1
2(+1) + 2x + 7(−2) = 0 → 2 + 2x − 14 = 0 → 2x = 12 → x = +6
✓ Cr = +6 in K₂Cr₂O₇
MediumWrite the half-reaction for KMnO₄ in acidic medium and balance▾
Balance the half-reaction for MnO₄⁻ reduction in acidic medium.
1
Mn goes from +7 to +2: gain of 5 electrons.
2
Balance O using H₂O: 4H₂O on right.
3
Balance H using H⁺: 8H⁺ on left.
4
Final: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
5
Check: Mn: 1=1 ✓; O: 4=4 ✓; H: 8=8 ✓; Charge: −1+8−5 = +2 ✓
✓ MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
EAPCET LevelHow many unpaired electrons in Cr³⁺? What is its magnetic moment?▾
Cr has Z=24, configuration [Ar]3d⁵4s¹. Cr³⁺ loses 3 electrons. Find unpaired electrons and magnetic moment μ.
1
Cr: [Ar]3d⁵4s¹. Cr³⁺ loses 4s¹ and two 3d electrons.
2
Cr³⁺: [Ar]3d³. Three d electrons, one each in three d orbitals (Hund's rule) → 3 unpaired electrons
3
μ = √(n(n+2)) = √(3×5) = √15 ≈ 3.87 BM
✓ 3 unpaired electrons; μ = √15 ≈ 3.87 BM
Trap QuestionZn²⁺ is coloured since Zn is a d-block element — True or False?▾
A student says: 'Zn²⁺ must be coloured because zinc is a transition metal with d orbitals.' Evaluate.
1
Transition metals are coloured due to d-d transitions — absorption of visible light by electrons moving between split d-orbitals.
2
For d-d transitions to occur, the d orbitals must be PARTIALLY FILLED.
3
Zn: [Ar]3d¹⁰4s². Zn²⁺: [Ar]3d¹⁰ — fully filled d orbitals.
4
With all d orbitals filled, no d-d transition is possible → Zn²⁺ is colourless.
5
Similarly, Sc³⁺ (3d⁰) is colourless — no d electrons to transition.
✓ False — Zn²⁺ has d¹⁰ configuration (fully filled); no d-d transition possible → colourless
Mistake DNA
4 d-block errors from EAPCET distractor analysis.
🔵
All Transition Metal Ions are Coloured
Only ions with PARTIALLY FILLED d orbitals are coloured. d⁰ (Sc³⁺) and d¹⁰ (Zn²⁺, Cu⁺) are colourless.
❌ Wrong
Zn²⁺ (d¹⁰): coloured ✗
Sc³⁺ (d⁰): coloured ✗
✓ Correct
d¹⁰ (Zn²⁺, Cu⁺): colourless ✓
d⁰ (Sc³⁺, Ti⁴⁺): colourless ✓
Partially filled d → coloured ✓
Colour requires d-d transitions. For this, there must be electrons in some d orbitals and empty d orbitals to jump to. Completely full (d¹⁰) or completely empty (d⁰) = no transitions = colourless.
🔄
KMnO₄ in Alkaline Medium: Mn goes to +2
In alkaline medium, KMnO₄ reduces to MnO₂ (brown precipitate, Mn = +4), NOT to Mn²⁺. The +2 state is the acidic medium product.
❌ Wrong
KMnO₄ in alkali → Mn²⁺
(colourless) ✗
✓ Correct
KMnO₄ in acid → Mn²⁺ ✓
KMnO₄ in alkali → MnO₂ ✓
(brown precipitate, Mn = +4)
Memorise: Acidic KMnO₄ → Mn²⁺ (colourless, +2, gains 5e⁻). Alkaline KMnO₄ → MnO₂ (brown, +4, gains 3e⁻). Neutral KMnO₄ → MnO₂ also.
🟡
Chromate and Dichromate: Colour Confusion
CrO₄²⁻ is yellow (chromate, alkaline). Cr₂O₇²⁻ is orange (dichromate, acidic). Students sometimes swap these.
❌ Wrong
CrO₄²⁻: orange (acidic) ✗
Cr₂O₇²⁻: yellow (alkaline) ✗
✓ Correct
CrO₄²⁻: yellow (alkaline) ✓
Cr₂O₇²⁻: orange (acidic) ✓
Add acid → orange;
add alkali → yellow
Memory: Cr₂O₇²⁻ (Orange, acidic) — remember O as in Orange and Oxidising. CrO₄²⁻ (Yellow, alkaline). Interconversion: add H⁺ → dichromate, add OH⁻ → chromate.
📐
Lanthanide Contraction: Radius Increases Across Lanthanides
Atomic radius DECREASES across the lanthanide series (La to Lu) due to poor shielding by 4f electrons.
❌ Wrong
Across lanthanides (La→Lu):
radius increases ✗
(like normal atomic radius
trend!?)
✓ Correct
La→Lu: radius decreases ✓
(lanthanide contraction) ✓
4f poor shielding →
Z_eff increases → r decreases
Unlike main group elements, 4f electrons shield very poorly. So as Z increases across lanthanides, effective nuclear charge increases significantly, pulling electrons inward and decreasing radius.
Chapter Intelligence
d-block covers quantitative (OS calculations) and qualitative (properties, compounds) questions.
EAPCET Weightage (2019–2024)
Electronic configurations (Cr, Cu exceptions)~8 Oxidation states and half-reactions~7 Colours and magnetic properties~6 KMnO₄ and K₂Cr₂O₇ chemistry~5 High-Yield PYQ Patterns
OS of Mn/Cr in compoundsCr/Cu configuration exceptionsZn²⁺ / Sc³⁺: coloured or not?Balance KMnO₄ half-reactionDichromate ↔ chromate pH dependenceMagnetic moment from unpaired e⁻Lanthanide contraction consequence
Exam Strategy
- Always check Cr (Z=24) and Cu (Z=29) electronic configurations — they are exceptions and appear every year.
- Oxidation state in compound: set up charge balance equation. KMnO₄: 1+x−8=0 → x=+7. K₂Cr₂O₇: 2+2x−14=0 → x=+6.
- Colour: only partially filled d orbitals (d¹ to d⁹) give coloured ions. d⁰ and d¹⁰ are colourless.
- KMnO₄ reactions: acidic → Mn²⁺ (colourless, 5e⁻ change); alkaline → MnO₂ (brown, 3e⁻ change).
- This chapter connects to Coordination Compounds (complex formation uses d-block metals) and Atomic Structure (electronic configurations).