Properties of Transition Elements

• Trends in ionisation energy, atomic radius, density, and melting point across the 3d series
• What interstitial compounds are and why transition metals form them
• What alloys are and why transition metals readily form them
• How transition metals form coordination complexes (ligands, coordination number)
• Standard electrode potentials (E°) and what they mean chemically
• Why transition metals are such effective catalysts

Trends Across the 3d Series (Sc to Zn)

Atomic Radius:

• Decreases from Sc to Cr due to increasing nuclear charge (poor shielding by d-electrons)
• Then nearly constant from Cr to Cu (d-electrons shield each other effectively)
• Slight increase at Zn (d¹⁰ — maximum electron-electron repulsion)

Ionisation Energy:

• Generally increases across the series (Sc → Zn) due to increasing nuclear charge
• Irregularities at Cr and Cu (half-filled and fully-filled d subshells — extra stable)
• Mn shows slightly higher first IE than expected; Fe slightly lower (anomaly due to d⁵ stability)

Density:

• Increases across the series (more nuclear mass in similar/smaller volume)
• Osmium (5d series) is the densest natural element
• Zn has lower density than Cu (d¹⁰ 4s² — slightly expanded radius)

Melting Point:

• Generally high across the series (strong metallic bonding involving d-electrons)
• Mn and Tc have anomalously low melting points within their series (unusual crystal structures)
• Cr and W have exceptionally high melting points
• Zn has a low melting point (419°C) — closed d-shell, weaker metallic bonding

Interstitial Compounds

Transition metals form compounds by trapping small non-metal atoms (H, C, N, B) in the interstitial spaces (holes) of their metallic lattice.

Key features:

• Non-stoichiometric composition (e.g., TiH₁.₇₃, not TiH₂ exactly)
• Metal lattice structure is largely retained
• Properties: very hard, high melting point, chemically inert

Examples:

Why transition metals form these? Large metallic radius creates gaps that small atoms (H, C, N radius < 100 pm) can fit into.

Alloy Formation

An alloy is a homogeneous mixture of two or more metals (or a metal and a non-metal). Transition metals readily form alloys because:

1. Similar atomic radii — atoms of different metals can substitute for each other in the lattice
2. Same crystal structure tendencies

Important alloys:

Complex Formation

Transition metals have a strong tendency to form coordination compounds (complexes):

Why? They have:

1. Small, highly charged ions (high charge density)
2. Empty d-orbitals to accept electron pairs from ligands
3. Ability to show variable oxidation states

Key terms:

• Ligand: A molecule or ion that donates a lone pair to the metal centre (Lewis base)
• Coordination number (CN): Number of ligand donor atoms bonded to the metal
• Complex ion examples: [Fe(CN)₆]⁴⁻, [Cu(NH₃)₄]²⁺, [Cr(H₂O)₆]³⁺

Standard Electrode Potentials (E°)

E° values tell us the tendency of a metal ion to be reduced (positive E° = oxidising agent; negative E° = reducing agent as the metal).

3d series M²⁺/M E° values (approximate):

Key point: Cu has positive E°, meaning Cu²⁺ is easily reduced and Cu metal is less reactive than H — copper does NOT dissolve in dilute HCl or H₂SO₄.

Catalytic Properties

Transition metals are exceptional catalysts because:

1. Variable oxidation states allow them to form intermediate compounds with reactants, creating alternative reaction pathways with lower activation energy

2. Adsorption on metal surfaces — metals can adsorb reactant molecules on their surfaces (heterogeneous catalysis), weakening bonds and increasing concentration locally

Key industrial examples:

Why Mn Has Anomalously Low E° (More Negative Than Cr)

Despite Mn coming after Cr (higher nuclear charge), its E°(Mn²⁺/Mn) = −1.18 V is more negative than Cr's −0.91 V. This is because:

• Mn²⁺ has a half-filled 3d⁵ configuration — extra stability (difficult to be formed, meaning the forward reduction is thermodynamically less favoured)
• Actually the metal Mn has lower enthalpy of atomisation than Cr (weaker metallic bonding, anomalous bcc structure)
• Net result: the overall cycle (Born-Haber for E°) favours Mn²⁺ being MORE stable (harder to reduce)

E° for Fe²⁺/Fe³⁺ Equilibrium

E°(Fe³⁺/Fe²⁺) = +0.77 V means Fe²⁺ is oxidised to Fe³⁺ by common oxidising agents like Cl₂ but not always by O₂ in acidic solution. This is why Fe²⁺ can be used as a reducing agent in titrations (e.g., with KMnO₄).

Stability of Oxidation States

• Higher oxidation states are stabilised by:
  • Electronegative ligands (F, O) — e.g., MnO₄⁻ (Mn in +7)
  • Oxyanion formation (e.g., CrO₄²⁻, Cr₂O₇²⁻)
• Lower oxidation states are stabilised by:
  • Large, polarisable (soft) ligands like CN⁻, CO
  • Conditions reducing conditions (lower pH)

Mn stability trend: Mn²⁺ is most stable (half-filled d⁵); Mn³⁺ is a good oxidising agent (rapidly oxidises to Mn²⁺).

Electrode Potential and Predicting Reactions

Use E° values to predict if a reaction is spontaneous:

• E°cell = E°cathode − E°anode
• If E°cell > 0, reaction is spontaneous

Example: Will Fe dissolve in CuSO₄ solution?

• Fe²⁺/Fe: E° = −0.44 V (anode)
• Cu²⁺/Cu: E° = +0.34 V (cathode)
• E°cell = 0.34 − (−0.44) = +0.78 V → spontaneous → Fe displaces Cu from CuSO₄

Interstitial Compounds vs Ionic/Covalent Compounds

Example 1: Predicting Whether a Metal Dissolves in Acid

Problem: Will copper dissolve in dilute sulphuric acid?

Given: E°(Cu²⁺/Cu) = +0.34 V
       E°(H⁺/H₂) = 0.00 V

For reaction: Cu → Cu²⁺ + 2e⁻ (oxidation, anode)
              2H⁺ + 2e⁻ → H₂ (reduction, cathode)

E°cell = E°cathode − E°anode = 0.00 − (+0.34) = −0.34 V

Negative E°cell → non-spontaneous → Copper does NOT dissolve in dilute H₂SO₄.

(Copper DOES dissolve in hot concentrated H₂SO₄ or HNO₃ because they are oxidising acids.)

Example 2: Identifying an Interstitial Compound

Problem: Which of the following is an interstitial compound?
(a) CuZn  (b) TiH₁.₇  (c) NiSO₄  (d) CrCl₃

Answer: (b) TiH₁.₇

Reason:
- TiH₁.₇ is non-stoichiometric (not exactly TiH₂)
- H atoms occupy interstitial voids in the Ti metal lattice
- The Ti metallic structure is retained
- It has high hardness and melting point typical of interstitial compounds

CuZn = alloy (substitutional), NiSO₄ = ionic salt, CrCl₃ = covalent/ionic compound

1. Why is stainless steel more resistant to corrosion than ordinary steel?
2. List the small atoms that can occupy interstitial sites in transition metal lattices.
3. Name the catalyst used in the Contact process and write the reaction it catalyses.
4. Which transition metal has a positive E° for M²⁺/M, and what does this mean for its reactivity?
5. (Stretch) E°(Mn³⁺/Mn²⁺) = +1.51 V and E°(Fe³⁺/Fe²⁺) = +0.77 V. Which ion (Mn³⁺ or Fe³⁺) is a stronger oxidising agent in aqueous solution? Explain in terms of d-orbital stability.

• NCERT Class 12, Chapter 8, Sections 8.4.2–8.4.7 (general properties of transition elements)
• JEE Main: E° values, catalysts, alloy composition — direct recall questions
• JEE Advanced: Multi-step reasoning on E° trends, stability of oxidation states
• NEET: Catalysts (Haber/Contact process), alloy composition, interstitial compound examples

Study strategy: Make a one-page table with Sc→Zn showing: configuration, common OS, E°(M²⁺/M), colour of common ion. Reviewing this table takes 2 minutes and covers 80% of exam questions.

Work through D-Block Properties question sets — start with Easy (alloys, interstitial basics) then progress to Medium (E° values, catalysts). The Hard set includes JEE assertion-based questions on E° reasoning.

Ask the Drishti AI: "Why does Mn have a more negative E° than Cr despite being to its right in the periodic table?" for a step-by-step thermodynamic cycle explanation.

Log your Quick Check score in your Drishti dashboard. If you missed Q5, revisit the E° section and try the Medium MCQ set before moving on.

• Read: f-Block Elements — lanthanoids and actinoids, lanthanoid contraction
• Revise: Transition metals electronic configurations (previous note)
• Practice: Full mixed d-block MCQ set at Medium → Hard difficulty