This MCQ module is based on: Properties Trends
Properties Trends
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Properties Trends
Why d-Block Chemistry Is Endlessly Varied
What makes a transition metal a transition metal? Five interlocking properties — variable oxidation states, paramagnetism, coloured ions, catalysis and complex formation — all flow from a single architectural fact: the (n-1)d sub-shell is partly filled and lies energetically close to the ns sub-shell. In Part 2 we examine how that single fact ripples through the periodic trends of the 3d series.
8.5 Variation in Atomic and Ionic Sizes
As we move across a transition series, the nuclear charge rises by one each step, and a new electron enters the inner d sub-shell. Inner d-electrons shield the outer ns electrons rather poorly, so the effective nuclear charge felt by the outermost shell rises — and the atomic / ionic radius shrinks. The shrinkage, however, is much smaller than across a typical s- or p-block period because the d-electrons themselves screen each other to some extent.
| Element | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn |
|---|---|---|---|---|---|---|---|---|---|---|
| Atomic radius / pm | 164 | 147 | 135 | 129 | 137 | 126 | 125 | 125 | 128 | 137 |
| M²⁺ ionic radius / pm | — | — | 79 | 82 | 82 | 77 | 74 | 70 | 73 | 75 |
| M³⁺ ionic radius / pm | 73 | 67 | 64 | 62 | 65 | 65 | 61 | 60 | — | — |
| Density / g cm⁻³ | 3.43 | 4.10 | 6.07 | 7.19 | 7.21 | 7.80 | 8.70 | 8.90 | 8.90 | 7.10 |
8.5.1 The "Plateau" After Mn — and the d-Block Contraction
From Sc to Cr the radius drops smoothly. Then from Mn to Cu the values flatten near 125–129 pm. Why? In the second half, electrons start to pair up in the d-orbitals; the rising electron–electron repulsion partly cancels the rising nuclear pull, so the radius barely changes. At Zn, the now-full 3d¹⁰ shell repels the 4s electrons outwards, pushing the radius back up to 137 pm.
8.5.2 Comparing the Three Series — Lanthanoid Contraction
If you compare the 3d, 4d and 5d series at the same group, you find that the radii grow on going 3d → 4d (more shells), but then 4d ≈ 5d (Zr 160 pm vs Hf 159 pm; Nb 146 pm vs Ta 146 pm). The reason is a one-time event embedded between the second and third series — the filling of the entire 4f sub-shell of the lanthanoids, called the lanthanoid contraction. The 4f electrons shield very poorly, so the 5d series begins after a hidden 14-element shrinkage that almost exactly cancels the size increase you would expect on going from Period 5 to Period 6.
8.6 Ionisation Enthalpies
The first ionisation enthalpy of the 3d series rises slowly from 631 kJ mol⁻¹ (Sc) to 906 kJ mol⁻¹ (Zn) — a much shallower climb than across a typical s/p period. Why? Because the new electrons enter the inner 3d sub-shell, which screens the outer 4s electron quite well; the effective nuclear charge experienced by 4s rises only slightly.
The trend is irregular at d⁵ (Mn) and d¹⁰ (Zn): ionising Mn (which would expose its half-filled 3d⁵) costs less than expected after an inflection at Cr, while ionising Zn (4s² → 4s¹ leaving 3d¹⁰) costs more than expected.
| Element | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn |
|---|---|---|---|---|---|---|---|---|---|---|
| ΔᵢH₁ / kJ mol⁻¹ | 631 | 656 | 650 | 653 | 717 | 762 | 758 | 736 | 745 | 906 |
| ΔᵢH₂ / kJ mol⁻¹ | 1235 | 1309 | 1414 | 1592 | 1509 | 1561 | 1644 | 1752 | 1958 | 1734 |
| ΔᵢH₃ / kJ mol⁻¹ | 2393 | 2657 | 2833 | 2990 | 3260 | 2962 | 3243 | 3402 | 3556 | 3837 |
Anomalies: the ΔᵢH₂ of Cr and Cu is unusually high — removing the second electron destroys the freshly-stable d⁵ (Cr⁺) and d¹⁰ (Cu⁺) configurations. The ΔᵢH₃ of Mn is unusually high — removing it from Mn²⁺ wrecks the half-filled d⁵.
8.7 Variable Oxidation States
Because (n-1)d and ns electrons are very close in energy, transition metals can lose different numbers of electrons in different reactions — giving an enormous variety of oxidation states.
| Element | Common oxidation states (most stable in bold) |
|---|---|
| Sc | +3 |
| Ti | +2, +3, +4 |
| V | +2, +3, +4, +5 |
| Cr | +2, +3, +4, +5, +6 |
| Mn | +2, +3, +4, +5, +6, +7 |
| Fe | +2, +3, +4, +6 |
| Co | +2, +3, +4 |
| Ni | +2, +3, +4 |
| Cu | +1, +2 |
| Zn | +2 |
8.7.1 Patterns to Remember
- The greatest range of oxidation states occurs in the middle of the series — Mn shows +2 to +7. Ends of the series show fewer states (too few electrons in Sc/Ti; too many in Cu/Zn for higher valence).
- Maximum oxidation state up to Mn equals (s + d) electrons: Ti⁴⁺ (3d⁰), V⁵⁺ (in VO₂⁺), Cr⁶⁺ (in CrO₄²⁻), Mn⁷⁺ (in MnO₄⁻).
- Beyond Mn the maximum oxidation state drops abruptly: Fe(II,III), Co(II,III), Ni(II), Cu(I,II), Zn(II).
- Within a group of d-block elements, heavier members favour the higher oxidation state — opposite to the p-block. Cr(VI) is a strong oxidant; Mo(VI) and W(VI) are stable and weakly oxidising.
- Very low oxidation states (0, -1) appear when ligands like CO, NO can π-accept electrons. Example: Ni(CO)₄ — Ni is in the 0 state.
Q (Ex 8.5). Suggest the most stable oxidation state for atoms with d⁻electron configurations 3d³, 3d⁵, 3d⁸, 3d⁴.
- 3d³ → +3 (e.g. V, V³⁺ stable)
- 3d⁵ → +2 (Mn²⁺) or +3 (Fe³⁺) — both half-filled and very stable
- 3d⁸ → +2 (Ni²⁺ stable in solution)
- 3d⁴ → not particularly stable (no half- or fully-filled bonus); higher state is preferred (e.g. Cr in +3 not +4)
8.8 Standard Electrode Potentials and Reactivity
The key data for the 3d series are tabulated below.
| Couple | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn |
|---|---|---|---|---|---|---|---|---|---|
| E°(M²⁺/M) / V | −1.63 | −1.18 | −0.90 | −1.18 | −0.44 | −0.28 | −0.25 | +0.34 | −0.76 |
| E°(M³⁺/M²⁺) / V | −0.37 | −0.26 | −0.41 | +1.57 | +0.77 | +1.97 | — | — | — |
- Cu has positive E°(Cu²⁺/Cu) = +0.34 V — Cu cannot liberate H₂ from non-oxidising acids. The high ΔₐH and modest ΔₕydH of Cu are not balanced enough to give a negative potential.
- The Mn²⁺/Mn and Zn²⁺/Zn potentials are more negative than the smooth trend predicts — the products Mn²⁺ (d⁵) and Zn²⁺ (d¹⁰) are unusually stable.
- Cr²⁺ is a strong reducing agent: changing d⁴ (Cr²⁺) → d³ (Cr³⁺) gives a half-filled t₂g level. Mn³⁺ is a strong oxidising agent: changing d⁴ (Mn³⁺) → d⁵ (Mn²⁺) gives the half-filled d⁵.
Q. Why is Cr²⁺ reducing while Mn³⁺ is oxidising, although both have d⁴?
The product of Cr²⁺ oxidation is Cr³⁺ (d³ — half-filled t₂g, extra-stable in octahedral field). The product of Mn³⁺ reduction is Mn²⁺ (d⁵ — half-filled d-shell, extra-stable). Each species moves to a more stable d-configuration.
8.9 Magnetic Properties
A substance is paramagnetic if it is attracted into a magnetic field — caused by unpaired electrons (each electron is a tiny magnet). It is diamagnetic if it is feebly repelled — all electrons are paired. Ferromagnetism (Fe, Co, Ni at room temperature) is an extreme cooperative form of paramagnetism.
For most 3d ions the orbital contribution is "quenched", so the spin-only formula gives the magnetic moment to good accuracy:
One unpaired electron gives μ = √3 = 1.73 BM. Five unpaired electrons (high-spin d⁵ like Mn²⁺) give μ = √35 = 5.92 BM.
| Ion | Configuration | n | μ (calc.) / BM | μ (obs.) / BM |
|---|---|---|---|---|
| Sc³⁺ | 3d⁰ | 0 | 0 | 0 |
| Ti³⁺ | 3d¹ | 1 | 1.73 | 1.75 |
| V³⁺ | 3d² | 2 | 2.84 | 2.76 |
| Cr³⁺ | 3d³ | 3 | 3.87 | 3.86 |
| Mn²⁺ | 3d⁵ | 5 | 5.92 | 5.96 |
| Fe²⁺ | 3d⁶ | 4 | 4.90 | 5.3–5.5 |
| Ni²⁺ | 3d⁸ | 2 | 2.84 | 2.9–3.4 |
| Cu²⁺ | 3d⁹ | 1 | 1.73 | 1.8–2.2 |
| Zn²⁺ | 3d¹⁰ | 0 | 0 | 0 |
Q. Calculate the spin-only magnetic moment of an M²⁺ ion (Z = 25) in aqueous solution.
Z = 25 → Mn. Mn²⁺ has [Ar]3d⁵ → n = 5 unpaired electrons.
Interactive: Magnetic Moment Calculator + Oxidation-State Predictor L3 Apply
Pick a 3d-series ion. The simulator returns the d-electron count, the number of unpaired electrons (high-spin), and the spin-only magnetic moment.
8.10 Coloured Ions of the d-Block
Most d-block ions are coloured in the solid state and in solution. The colour arises because the five d-orbitals of a transition-metal ion split into two sets of slightly different energies under the influence of surrounding ligands. An electron in the lower set absorbs a photon of visible light to jump to the upper set — a "d-d transition" — and the colour we see is the complementary colour of the wavelength absorbed.
Configurations with no d-electrons (d⁰: Sc³⁺, Ti⁴⁺) and configurations with all d-orbitals filled (d¹⁰: Zn²⁺, Cu⁺, Cd²⁺) cannot undergo d-d transitions — they are colourless.
| Ion | Config | Colour (aqueous) |
|---|---|---|
| Sc³⁺ / Ti⁴⁺ | 3d⁰ | Colourless |
| Ti³⁺ | 3d¹ | Purple |
| V³⁺ | 3d² | Green |
| V²⁺ | 3d³ | Violet |
| Cr³⁺ | 3d³ | Violet |
| Cr²⁺ | 3d⁴ | Blue |
| Mn²⁺ | 3d⁵ | Pale pink |
| Fe³⁺ | 3d⁵ | Yellow |
| Fe²⁺ | 3d⁶ | Pale green |
| Co²⁺ | 3d⁷ | Pink |
| Ni²⁺ | 3d⁸ | Green |
| Cu²⁺ | 3d⁹ | Blue |
| Zn²⁺ | 3d¹⁰ | Colourless |
8.11 Catalytic Properties
Transition metals and their compounds are master catalysts. Two factors are responsible: (i) they can shuttle between multiple oxidation states and (ii) they offer surface sites at which reacting molecules can adsorb. Some industrially crucial examples:
| Catalyst | Industrial Process |
|---|---|
| V₂O₅ | Contact process (SO₂ → SO₃ for H₂SO₄) |
| Finely divided Fe | Haber–Bosch (N₂ + 3 H₂ → 2 NH₃) |
| Ni | Catalytic hydrogenation of unsaturated oils |
| Pd or Pt | Catalytic converters (CO/NOₓ → CO₂/N₂) |
| TiCl₄ + Al(C₂H₅)₃ | Ziegler–Natta polymerisation of ethylene |
| PdCl₂ | Wacker process (ethene → ethanal) |
A homogeneous example: Fe³⁺ catalyses the oxidation of I⁻ by S₂O₈²⁻:
The Fe³⁺/Fe²⁺ couple shuttles electrons faster than the direct (uncatalysed) reaction.
8.12 Complex Formation
Transition metals form an extraordinary range of complex compounds with ligands such as CN⁻, NH₃, H₂O, Cl⁻, CO, en (ethylenediamine). Three structural features make this possible:
- Small ionic size + high ionic charge – strong electric field draws ligands in.
- Vacant low-energy d-orbitals – room to accept lone pairs from ligands.
- Variable oxidation states – different ligand fields stabilise different states.
Examples: [Fe(CN)₆]³⁻, [Fe(CN)₆]⁴⁻, [Cu(NH₃)₄]²⁺, [PtCl₄]²⁻. The detailed treatment is in Chapter 9 (Coordination Compounds).
8.13 Interstitial Compounds and Alloy Formation
Interstitial compounds form when small atoms (H, B, C, N) get trapped in the holes between metal atoms in the lattice. Examples: TiC, Mn₄N, Fe₃H, VH₀.₅₆, TiH₁.₇. They are typically non-stoichiometric. Properties:
- Higher melting points than the parent metals.
- Extremely hard — some borides rival diamond (e.g., TiB₂).
- Retain metallic conductivity.
- Chemically inert.
Alloys are solid solutions of two or more metals. Because transition-metal radii sit within ~15% of each other and they share similar bonding, they form alloys readily. Examples: stainless steel (Fe + Cr + Ni), brass (Cu + Zn), bronze (Cu + Sn), nichrome (Ni + Cr).
Aim: Use the spin-only formula to predict μ for any 3d-series ion given its atomic number and oxidation state.
Procedure:
- Take the atomic number of the metal. Subtract the number of electrons removed (the charge) — but always remove the 4s electrons first.
- Count the d-electrons in the ion. Apply Hund's rule: fill all five d-orbitals singly first, then pair up.
- Count the unpaired electrons (n).
- Compute μ = √[n(n+2)] BM.
Predict: For Co²⁺ (Z = 27), how many unpaired electrons does it have, and what is the spin-only magnetic moment?
Co (Z = 27) = [Ar] 3d⁷4s². Remove two electrons (4s² first) → Co²⁺ = [Ar] 3d⁷.
Distribution: ↑↓ ↑↓ ↑ ↑ ↑ → n = 3 unpaired electrons.
μ = √[3(3+2)] = √15 = 3.87 BM (observed value 4.4–5.2 BM, the discrepancy due to small orbital contribution).
Q (In-text 4.9). Why is Cu⁺ ion not stable in aqueous solutions?
Cu⁺ undergoes disproportionation in water:
The very large hydration enthalpy of Cu²⁺ (smaller, doubly-charged) drives the equilibrium far to the right; the energy released compensates for the second ionisation cost of Cu.
Competency-Based Questions L3 L4
Q1. (MCQ) The spin-only magnetic moment of Fe³⁺(d⁵, high spin) in BM is approximately:
Q2. (MCQ) Which ion is colourless in aqueous solution?
Q3. (SA) Predict (with reasoning) which of the following ions are coloured in aqueous solution: Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺, Co²⁺.
Q4. (LA) Why is Fe²⁺ a stronger reducing agent than Cu²⁺ in aqueous solution? Give thermodynamic reasoning.
Q5. (HOT) A student claims the high catalytic activity of Pt in catalytic converters is due to "Pt's variable oxidation states alone". Critique this statement.
Assertion–Reason Questions L4 L5
Choose: A) Both A and R true and R explains A · B) Both true but R does not explain A · C) A true, R false · D) A false, R true.
Assertion (A): The atomic radii of 5d-series elements are almost identical to those of the 4d series in the same group.
Reason (R): The lanthanoid contraction cancels the size increase expected on going from Period 5 to Period 6.
Assertion (A): Mn²⁺ has a magnetic moment of about 5.92 BM.
Reason (R): Mn²⁺ has the configuration 3d⁵ with all five electrons paired.
Assertion (A): Cr²⁺ is a stronger reducing agent than Fe²⁺.
Reason (R): Oxidation of Cr²⁺ to Cr³⁺ converts d⁴ to the half-filled t₂g d³ configuration, which is more stable than the d⁵ obtained on oxidising Fe²⁺ to Fe³⁺.
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