Transition Metals — Electronic Configuration and General Properties

• How to write electronic configurations of transition metals using the [Ar] core notation
• Why Cr and Cu are exceptions to the expected [Ar]3d^n 4s^2 pattern
• Why transition metals exhibit variable oxidation states
• The cause of colour in transition metal compounds (d-d transitions)
• How to calculate magnetic moments using the spin-only formula
• Why transition metals are excellent catalysts

What are Transition Metals?

Transition metals (d-block elements) are elements in which the last electron enters a d-orbital. They occupy groups 3–12 of the periodic table. The first transition series (3d series) spans from Sc (Z = 21) to Zn (Z = 30).

Definition (IUPAC): An element is a transition metal if it has a partially filled d-subshell in any of its common oxidation states.

Note: Zinc (Zn), Cadmium (Cd), and Mercury (Hg) do NOT qualify as transition metals because Zn²⁺ has a completely filled d¹⁰ configuration.

Electronic Configurations

The general configuration for 3d series is: [Ar] 3d^(1–10) 4s^(0–2)

Exceptions: Cr and Cu

• Chromium (Cr): Expected [Ar]3d⁴4s², actual [Ar]3d⁵4s¹
• Copper (Cu): Expected [Ar]3d⁹4s², actual [Ar]3d¹⁰4s¹

Reason: Half-filled (3d⁵) and completely-filled (3d¹⁰) d-subshells have extra stability due to:

1. Symmetrical electron distribution (lower inter-electron repulsion)
2. Exchange energy — more electrons with parallel spins means greater exchange energy stabilisation

So one electron is "promoted" from 4s to 3d to achieve these stable configurations.

Variable Oxidation States

Transition metals show multiple oxidation states because:

1. The energy difference between 3d and 4s orbitals is small
2. Both 3d and 4s electrons can participate in bonding
3. Loss of successive electrons is energetically feasible

Examples:

• Iron: +2 (Fe²⁺) and +3 (Fe³⁺)
• Manganese: +2, +4, +6, +7 (MnO₄⁻)
• Chromium: +2, +3, +6 (CrO₄²⁻, Cr₂O₇²⁻)

The highest oxidation states generally correspond to loss of all d and s electrons.

Colour in Transition Metal Compounds

Most transition metal ions are coloured. The colour arises from d-d transitions:

1. In a complex, ligands split the d-orbitals into energy levels (crystal field splitting, Δ)
2. An electron absorbs visible light and jumps from a lower d-level to a higher d-level
3. The complementary colour of the absorbed light is what we observe

Condition: The metal must have at least one d-electron AND at least one vacancy in d-orbitals. This is why:

• Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) are colourless
• Cu²⁺ (3d⁹) is blue; Fe³⁺ (3d⁵) is pale yellow

Magnetic Properties

Magnetic behaviour depends on unpaired electrons:

• Paramagnetic: Has unpaired electrons (attracted to magnetic field)
• Diamagnetic: No unpaired electrons (repelled by magnetic field)

Spin-only magnetic moment formula: $\mu = \sqrt{n(n+2)} \text{ BM (Bohr Magnetons)}$

where n = number of unpaired electrons

Catalytic Behaviour

Transition metals are excellent catalysts because:

1. Variable oxidation states — can form intermediate compounds with reactants, lowering activation energy
2. Large surface area — can adsorb reactants on metal surface (heterogeneous catalysis)

Examples:

• Fe in Haber process (N₂ + H₂ → NH₃)
• V₂O₅ in Contact process (SO₂ → SO₃)
• Ni in hydrogenation of oils (Sabatier reaction)
• MnO₂ as catalyst in decomposition of KClO₃
• Pt in catalytic converters (oxidises CO and NOₓ)

Why 4s Electrons are Lost Before 3d in Ionisation

Although 4s fills before 3d (Aufbau), during ionisation electrons are removed from 4s first. This is because:

• In the presence of 3d electrons, the effective nuclear charge experienced by 4s electrons decreases
• The 3d orbitals contract and shield 4s; 4s actually has higher energy in the ionised state
• This is confirmed by the ground-state electronic configuration of cations (e.g., Fe = [Ar]3d⁶4s², Fe²⁺ = [Ar]3d⁶)

Crystal Field Theory Basics

When ligands approach a metal ion, they interact electrostatically and split the d-orbitals into two sets:

• t₂g set: dxy, dyz, dxz (lower energy in octahedral field)
• eg set: dx²-y², dz² (higher energy in octahedral field)

The energy gap between these sets is Δo (crystal field splitting energy). If Δo equals the energy of visible light, the complex absorbs that colour.

Standard Electrode Potentials

The reduction potential (E°) of transition metals varies non-uniformly due to:

1. Ionisation energies
2. Hydration enthalpies
3. Sublimation energies

Important E° values (M²⁺/M):

• Mn²⁺/Mn: −1.18 V (Mn is a good reducing agent as M)
• Fe²⁺/Fe: −0.44 V
• Cu²⁺/Cu: +0.34 V (Cu is least reactive; does not dissolve in dilute acids)
• Zn²⁺/Zn: −0.76 V

Higher Oxidation States in Oxyanions

In high oxidation states, transition metals form oxoanions:

• Cr(VI): CrO₄²⁻ (yellow, chromate) and Cr₂O₇²⁻ (orange, dichromate)
• Mn(VII): MnO₄⁻ (purple, permanganate)

These are powerful oxidising agents because the metal is in a high OS and tends to be reduced.

Example 1: Calculating Magnetic Moment of Fe³⁺

Problem: Calculate the magnetic moment of Fe³⁺ ion.

Step 1: Electronic configuration of Fe (Z=26): [Ar] 3d⁶ 4s²
Step 2: Fe³⁺ loses 3 electrons (2 from 4s, 1 from 3d):
         Fe³⁺ = [Ar] 3d⁵

Step 3: Count unpaired electrons in 3d⁵:
         3d:  ↑  ↑  ↑  ↑  ↑   → 5 unpaired electrons (Hund's rule, half-filled)

Step 4: Apply spin-only formula:
         μ = √(n(n+2)) BM = √(5 × 7) = √35 ≈ 5.92 BM

Answer: 5.92 BM — Fe³⁺ is strongly paramagnetic

Example 2: Identifying the Exception — Why Cr is [Ar]3d⁵4s¹

Problem: Write the electronic configuration of Cr (Z=24) and explain the exception.

Expected (Aufbau): [Ar] 3d⁴ 4s²

Actual: [Ar] 3d⁵ 4s¹

Explanation:
- Half-filled d-subshell (3d⁵) has all five d-orbitals singly occupied
- This gives maximum exchange energy and symmetric electron distribution
- The gain in stability from 3d⁵ > energy cost of moving one 4s electron to 3d
- Hence the actual configuration [Ar]3d⁵4s¹ is more stable

Memory tip: "Cr and Cu are rebels — they prefer symmetry over rules"
Cr: d⁵s¹ (half-filled d)
Cu: d¹⁰s¹ (fully-filled d)

1. Write the electronic configuration of Cu²⁺ (Z = 29).
2. Calculate the magnetic moment of Mn²⁺. How many unpaired electrons does it have?
3. Why is TiO₂ white while CuSO₄·5H₂O is blue?
4. Which transition metal is used as a catalyst in the Haber process, and why?
5. (Stretch) The E°(Cu²⁺/Cu) = +0.34 V while E°(Zn²⁺/Zn) = −0.76 V. Why does copper not dissolve in dilute H₂SO₄ but zinc does? What does this tell us about their relative reducing power?

• NCERT Class 12 Chemistry, Chapter 8 — The d and f Block Elements
• Sections 8.2 (Position in PT), 8.3 (Electronic Configurations), 8.4 (General Properties)
• NEET/JEE: 2–3 questions typically from electronic config exceptions, magnetic moments, and catalysis
• Common MCQ formats: identify incorrect configuration, calculate μ, predict colour from d-count

Study strategy: Master the 10 elements Sc–Zn with their d-configurations cold. Memorise Cr and Cu exceptions with the reason (not just the fact). Practice the spin-only formula until it's automatic.

Practice MCQs on transition metal configurations and magnetic moments in the D-Block Transition Metals topic bank — try Easy first, then Medium for JEE Foundation level.

Stuck on why electrons are removed from 4s before 3d during ionisation? Ask the Drishti AI tutor to explain with orbital energy diagrams.

Complete the Quick Check questions and mark them in your Drishti progress tracker. Aim for 4/5 before moving to the next note.

• Read: d-Block Properties — trends, interstitial compounds, alloys, E° values
• Then: f-Block Elements — lanthanoids, actinoids, lanthanoid contraction
• Practice: Mixed d-block MCQs (Medium difficulty)