IB DP Chemistry - S2.3.3 Transition elements and delocalized d-electrons- Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S2.3.3 Transition elements and delocalized d-electrons- Study Notes – New Syllabus
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- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
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Structure 2.3.3 — Transition Elements: Metallic Bonding, Melting Points, Conductivity, and Chemical Properties
Structure 2.3.3 — Transition Elements: Metallic Bonding, Melting Points, Conductivity, and Chemical Properties
Transition Elements
Transition metals are d-block elements that form at least one ion with a partially filled d-subshell. Typical examples include iron (Fe), copper (Cu), and chromium (Cr).
Metallic Bonding in Transition Metals
Like other metals, transition elements have metallic bonding – the electrostatic attraction between positive metal ions and a ‘sea’ of delocalized electrons.
However, in transition metals, the delocalized electrons include both s and d electrons, making the metallic bonds much stronger than in s-block metals.
- More delocalized electrons → stronger electrostatic forces of attraction.
- Stronger bonding → higher melting points and greater hardness.
Physical Properties Explained:
- High melting point: Strong metallic bonding due to additional delocalized d-electrons. This requires more energy to break.
- Electrical conductivity: Delocalized electrons (from both s and d orbitals) move freely through the metallic lattice, carrying charge.
Chemical Properties of Transition Elements:
Transition metals exhibit unique chemical behavior compared to s-block elements due to the presence of partially filled d orbitals.
- Variable oxidation states: They can lose different numbers of d and s electrons.
- Formation of colored compounds: Due to d–d electronic transitions.
- Catalytic activity: Many act as catalysts (e.g. Fe in Haber process, MnO₂ in decomposition of H₂O₂).
- Complex ion formation: Transition metals can coordinate with ligands to form complex ions.
Common Oxidation States of Selected Transition Elements:
| Element | Common Oxidation States | Example Compounds |
|---|---|---|
| Fe | +2, +3 | FeCl₂, FeCl₃ |
| Cu | +1, +2 | Cu₂O, CuO |
| Mn | +2, +4, +7 | MnO, MnO₂, KMnO₄ |
| Cr | +3, +6 | CrCl₃, Cr₂O₇²⁻ |
Example
Explain why iron has a higher melting point than sodium, even though both are metals.
▶️Answer/Explanation
Iron has a higher melting point because it is a transition metal with delocalized d and s electrons, leading to stronger metallic bonding.
Sodium, an s-block metal, has only one delocalized s-electron per atom.
Stronger bonding in Fe → more energy required to overcome → higher melting point.
Example
Why do many transition metals act as catalysts?
▶️Answer/Explanation
Transition metals can easily change oxidation states and provide a surface for reactions.
This allows them to form intermediate compounds with reactants, lowering activation energy.
For example, Fe catalyzes the Haber process: \( \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \)
Example
The table shows melting points of several metals:
| Element | Group | Melting Point (°C) |
|---|---|---|
| Na | 1 | 98 |
| Mg | 2 | 650 |
| Fe | Transition | 1538 |
| Cu | Transition | 1085 |
Use the data to explain the trend in melting points from Na to Fe.
▶️Answer/Explanation
- Melting point increases from Na to Fe.
- Na has only 1 delocalized electron per atom and weak metallic bonding.
- Mg has 2 delocalized electrons → stronger bonding → higher melting point.
- Fe has delocalized s and d electrons → much stronger metallic bonding → highest melting point.
Example
Determine the oxidation state of chromium in the ion \( \text{Cr}_2\text{O}_7^{2-} \).
▶️Answer/Explanation
Let the oxidation state of Cr be \( x \). The oxidation state of O is -2.
\( 2x + 7(-2) = -2 \Rightarrow 2x – 14 = -2 \Rightarrow 2x = 12 \Rightarrow x = +6 \)
So each Cr has an oxidation state of +6 in \( \text{Cr}_2\text{O}_7^{2-} \).
Example
The melting point of titanium is 1668°C, while zinc melts at only 419.5°C. Both are transition metals. Explain the difference.
▶️Answer/Explanation
- Titanium has more delocalized d-electrons, which contribute to stronger metallic bonding.
- Zinc has a full d-subshell (3d10) and only 2 valence electrons, so fewer electrons are available for metallic bonding.
- Therefore, Zn has weaker bonding and a lower melting point than Ti.