d- and f-Block Elements · Lesson 2 of 12
Metallic Bonding, Size and Ionisation Trends
“Explain metallic strength, irregular size trends and ionisation energies using electron structure and measured data.”
• Relate atomisation enthalpy and melting behaviour to metallic bonding. • Explain atomic and same-charge ionic size trends without forcing strict monotonicity. • Connect lanthanoid contraction to the similarity of 4d and 5d radii. • Use configurations to interpret first, second and third ionisation enthalpies.
Strong metallic bonding and its exceptions
A metal crystal is a collective bonded structure, not a pile of separate neutral atoms. Mobile electrons allow electrical and thermal conduction; non-directional metallic bonding permits many metals to be drawn into wires and hammered into sheets. Transition metals often have strong bonding because both s and d valence states contribute to cohesion. Separating the solid into isolated gaseous atoms therefore needs substantial energy.
The enthalpy change for producing one mole of gaseous atoms from the element in its standard state. For a solid metal, M(s) → M(g), it is positive.
Melting and atomisation are related to cohesion but describe different changes. Melting disrupts long-range crystal order while retaining interactions in a liquid; atomisation separates atoms. High atomisation enthalpy often accompanies high melting and boiling temperatures, but the quantities do not have identical rankings. Crystal structure, entropy and electronic band filling also matter. Mn has an anomalously low melting point within its row despite its half-filled atomic d shell; a count of unpaired electrons alone is not a quantitative theory of a solid.
| Metal | Atomisation / kJ mol⁻¹ | Metallic radius / pm | Density / g cm⁻³ |
|---|---|---|---|
| Sc | 326 | 164 | 3.43 |
| Ti | 473 | 147 | 4.1 |
| V | 515 | 135 | 6.07 |
| Cr | 397 | 129 | 7.19 |
| Mn | 281 | 137 | 7.21 |
| Fe | 416 | 126 | 7.8 |
| Co | 425 | 125 | 8.7 |
| Ni | 430 | 125 | 8.9 |
| Cu | 339 | 128 | 8.9 |
| Zn | 126 | 137 | 7.1 |
Cohesion tends to be large near the middle of a transition row and smaller near its end. Zinc’s filled 3d shell contributes less to the strong d-band bonding characteristic of many preceding metals, and its atomisation enthalpy is particularly low. This does not mean zinc has no metallic bonds: its 4s-derived valence electrons still hold a conducting metal together. Heavier 4d and 5d metals often have stronger metal–metal interactions than their 3d counterparts. W, for example, is exceptionally refractory.
Size: increased attraction competes with shielding
Across a row, each additional proton attracts electrons more strongly, but each additional d electron partly screens that charge. Screening is incomplete, so the effective attraction generally increases. Same-charge ions usually contract across a series. Metallic radii show a smaller, less regular change: they decrease early, become nearly level and can rise toward the filled-shell end. Comparing a divalent ion with a trivalent ion indiscriminately would mix the effect of charge with the across-row trend.
The 4d atoms are larger than their 3d counterparts because an additional main shell is involved. Moving from 4d to 5d does not produce an equally large increase: filling the 4f series raises nuclear charge while 4f electrons screen it poorly. The resulting lanthanoid contraction largely compensates for the extra shell. Representative metallic radii are Zr 160 pm and Hf 159 pm. Their chemistry is consequently unusually similar, making their separation difficult. Radius numbers must be compared on the same definition; metallic and ionic radii are not interchangeable.
Problem
Use the idea of mass per atomic volume to explain the broad increase from Ti to Cu.
- 1.Density is mass divided by volume. Across much of the row atomic mass increases while the characteristic metallic radius decreases.
- 2.A smaller characteristic radius tends to reduce atomic volume; in a fixed packing type the volume scale varies roughly as r³.
- 3.Both changes favour higher density. The actual crystal structure and packing still matter, so this explains a broad trend rather than a strict prediction for every neighbour.
Ionisation is a change of a particular gaseous species
First ionisation enthalpy refers to M(g) → M⁺(g)+e⁻. Second ionisation starts with M⁺, and third with M²⁺. Because each removal leaves a more positively charged ion, successive removals from the same element generally require increasing energy. Across different elements, however, the initial and final configurations also change. These two comparisons must not be confused.
| Metal | I₁ | I₂ | I₃ (all kJ mol⁻¹) |
|---|---|---|---|
| Sc | 631 | 1235 | 2393 |
| Ti | 656 | 1309 | 2657 |
| V | 650 | 1414 | 2833 |
| Cr | 653 | 1592 | 2990 |
| Mn | 717 | 1509 | 3260 |
| Fe | 762 | 1561 | 2962 |
| Co | 758 | 1644 | 3243 |
| Ni | 736 | 1752 | 3402 |
| Cu | 745 | 1958 | 3556 |
| Zn | 906 | 1734 | 3837 |
The first ionisation values rise only broadly across 3d because inner d electrons provide some screening and orbital energies reorganise on ionisation. Exchange stabilisation lowers the energy of arrangements with suitable parallel spins in degenerate orbitals. Losing favourable exchange stabilisation costs energy; it does not make the ion more stable. For Cr⁺, the next electron must leave d⁵, while Mn⁺ still has one 4s electron that can be removed to form d⁵. This helps explain why I₂(Cr)>I₂(Mn).
Problem
Why is I₃(Mn)=3260 kJ mol⁻¹ larger than I₃(Fe)=2962 kJ mol⁻¹?
- 1.The initial ions are Mn²⁺, d⁵, and Fe²⁺, d⁶. The question concerns these ions, not the neutral atoms.
- 2.Mn²⁺ loses a singly occupied d electron and becomes d⁴, disrupting the favourable half-filled arrangement.
- 3.Fe²⁺ loses one electron from a paired occupation and becomes d⁵. This favourable final arrangement makes its third removal comparatively easier.
- 4.The measured difference is 3260−2962=298 kJ mol⁻¹. Neither number alone predicts every solution oxidation state, because hydration and compound bonding are additional contributions.
Cu⁺ is d¹⁰, so its second ionisation is costly because it begins to break into a filled d shell. Zn⁺ still has a 4s electron; removing it yields d¹⁰ Zn²⁺. Removing a third electron from Zn²⁺ then becomes very costly. These examples explain irregularities while also showing why “stable configuration” must specify which initial and final species are being compared.
Crystal packing is an additional variable
Common metallic structures include body-centred cubic (bcc), hexagonal close-packed (hcp) and cubic close-packed (ccp, also called face-centred cubic). At ordinary conditions, Sc and Ti are hcp, V and Cr bcc, Fe bcc, Co hcp and Ni and Cu ccp. Co can also occur in ccp modifications. Mn has a more complicated structure; Zn and Cd are distorted hcp metals, while Hg is liquid near room temperature. Structures can change with temperature, so a table listing several forms should not imply that every form is simultaneously the stable ambient phase. Crystal packing helps explain why simple atomic electron counts do not perfectly predict melting behaviour.
| Metal | M²⁺ radius / pm | M³⁺ radius / pm |
|---|---|---|
| Sc | — | 73 |
| Ti | — | 67 |
| V | 79 | 64 |
| Cr | 82 | 62 |
| Mn | 82 | 65 |
| Fe | 77 | 65 |
| Co | 74 | 61 |
| Ni | 70 | 60 |
| Cu | 73 | — |
| Zn | 75 | — |
These tabulated ionic radii use the selected source convention and available entries; a dash is not a claim that an ion is impossible. Coordination number and spin state can alter radii, so the small irregularities are not errors to smooth away. Compare like charge and environment before attributing a difference solely to nuclear charge.
Quick Quiz
Quiz
Which change defines metal atomisation?
Explanation: Atomisation separates the element into isolated gaseous atoms.
Why are Zr and Hf nearly the same size?
Explanation: Poor 4f screening increases effective attraction before the 5d series.
Why is I₃ of Mn unusually high?
Explanation: Third ionisation begins with gaseous Mn²⁺, whose d⁵ arrangement is favoured.
Which statement about metallic radii is sound?
Explanation: Nuclear attraction, shielding and solid-state bonding jointly determine the measured size.
Why is the atomisation enthalpy of Zn low within 3d?
Explanation: Zinc remains a metal, but does not gain the same d-bonding cohesion as many central members.
Practice Problems
- Distinguish melting point, atomisation enthalpy and first ionisation enthalpy.
- Explain why the Ti-to-Cu density trend has a structural basis.
- Use Cr⁺ and Mn⁺ configurations to compare their second ionisations.
- Why does Zn resist a third ionisation much more than a second?
- A student predicts Mn has the highest melting point because its atom is d⁵. Diagnose the error.
- Explain why a same-charge ionic radius comparison is more useful than mixing M²⁺ and M³⁺ data.
Melting changes solid to liquid, atomisation creates isolated gaseous atoms, and ionisation removes an electron from a gaseous atom. Their energetic pathways differ, so correlated trends are not numerical equivalences.
Key Takeaways
• Metallic properties arise from collective bonding and mobile electrons. • Atomisation and melting probe different physical changes. • Size trends across d rows are small and include meaningful exceptions. • Poor 4f shielding explains the similarity of corresponding 4d and 5d sizes. • Ionisation enthalpy must specify the initial gaseous ion. • Exchange stabilisation is lost at an energetic cost when favourable occupations are disrupted. • Hydration and compound bonding must be added before predicting solution stability.