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Properties Trends

🎓 Class 12 Chemistry CBSE Theory Ch 4 – The d- and f-Block Elements ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Properties Trends

આ મૂલ્યાંકન આના પર આધારિત હશે: 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.

Roadmap: Atomic / ionic radii → ionisation enthalpies → standard electrode potentials → oxidation states → magnetic properties → coloured ions → complex / interstitial / alloy formation → catalysis.

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.

ElementScTiVCrMnFeCoNiCuZn
Atomic radius / pm164147135129137126125125128137
M²⁺ ionic radius / pm7982827774707375
M³⁺ ionic radius / pm7367646265656160
Density / g cm⁻³3.434.106.077.197.217.808.708.908.907.10

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.

120 140 160 180 200 Atomic radius / pm Group → 3 4 5 6 7 8 9 10 11 12 5d (La...Hg) 4d (Y...Cd) 3d (Sc...Zn) 4d ≈ 5d due to lanthanoid contraction
Fig 8.4: Atomic radii across the three d-series. The 5d curve almost exactly overlies the 4d curve.

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.

ElementScTiVCrMnFeCoNiCuZn
ΔᵢH₁ / kJ mol⁻¹631656650653717762758736745906
ΔᵢH₂ / kJ mol⁻¹1235130914141592150915611644175219581734
ΔᵢH₃ / kJ mol⁻¹2393265728332990326029623243340235563837

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.

ElementCommon 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.
Disproportionation: When a particular oxidation state becomes less stable than both a higher and a lower one, it splits in two. Example in acid solution:
3 MnO₄²⁻ + 4 H⁺ → 2 MnO₄⁻ + MnO₂ + 2 H₂O
Mn(VI) disproportionates to Mn(VII) (in MnO₄⁻) and Mn(IV) (in MnO₂).
Worked Example 8.4 L4 Analyse

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.

CoupleTiVCrMnFeCoNiCuZn
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.
Worked Example 8.5 L4 Analyse

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:

μ = √[n(n+2)] BM, where n = number of unpaired electrons

One unpaired electron gives μ = √3 = 1.73 BM. Five unpaired electrons (high-spin d⁵ like Mn²⁺) give μ = √35 = 5.92 BM.

IonConfigurationnμ (calc.) / BMμ (obs.) / BM
Sc³⁺3d⁰000
Ti³⁺3d¹11.731.75
V³⁺3d²22.842.76
Cr³⁺3d³33.873.86
Mn²⁺3d⁵55.925.96
Fe²⁺3d⁶44.905.3–5.5
Ni²⁺3d⁸22.842.9–3.4
Cu²⁺3d⁹11.731.8–2.2
Zn²⁺3d¹⁰000
Worked Example 8.6 L3 Apply

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.

μ = √[5(5+2)] = √35 = 5.92 BM

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.

Pick an ion to begin.

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.

IonConfigColour (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:

CatalystIndustrial Process
V₂O₅Contact process (SO₂ → SO₃ for H₂SO₄)
Finely divided FeHaber–Bosch (N₂ + 3 H₂ → 2 NH₃)
NiCatalytic hydrogenation of unsaturated oils
Pd or PtCatalytic converters (CO/NOₓ → CO₂/N₂)
TiCl₄ + Al(C₂H₅)₃Ziegler–Natta polymerisation of ethylene
PdCl₂Wacker process (ethene → ethanal)

A homogeneous example: Fe³⁺ catalyses the oxidation of I⁻ by S₂O₈²⁻:

Step 1: 2 Fe³⁺ + 2 I⁻ → 2 Fe²⁺ + I₂ Step 2: 2 Fe²⁺ + S₂O₈²⁻ → 2 Fe³⁺ + 2 SO₄²⁻ Net: 2 I⁻ + S₂O₈²⁻ → I₂ + 2 SO₄²⁻

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).

Activity 8.2 — Predict the Magnetic Moment from Atomic NumberL3 Apply

Aim: Use the spin-only formula to predict μ for any 3d-series ion given its atomic number and oxidation state.

Procedure:

  1. Take the atomic number of the metal. Subtract the number of electrons removed (the charge) — but always remove the 4s electrons first.
  2. Count the d-electrons in the ion. Apply Hund's rule: fill all five d-orbitals singly first, then pair up.
  3. Count the unpaired electrons (n).
  4. 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).

Worked Example 8.7 L4 Analyse

Q (In-text 4.9). Why is Cu⁺ ion not stable in aqueous solutions?

Cu⁺ undergoes disproportionation in water:

2 Cu⁺(aq) → Cu²⁺(aq) + Cu(s) (E°ₚᵣₒc > 0)

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

Stimulus: A laboratory chemist tabulates magnetic moments measured for several first-series transition-metal ions and notices that the values match those predicted by the spin-only formula closely for d¹–d⁴ but deviate for d⁶–d⁹. She also notes that Cr²⁺ acts as a reducing agent while Mn³⁺ acts as an oxidant, even though both are d⁴.

Q1. (MCQ) The spin-only magnetic moment of Fe³⁺(d⁵, high spin) in BM is approximately:

  • (a) 1.73
  • (b) 2.84
  • (c) 4.90
  • (d) 5.92
(d) 5.92 BM. Fe³⁺ = d⁵, n = 5 → μ = √35 ≈ 5.92 BM.

Q2. (MCQ) Which ion is colourless in aqueous solution?

  • (a) Cu²⁺
  • (b) Zn²⁺
  • (c) Ni²⁺
  • (d) Fe³⁺
(b) Zn²⁺ is 3d¹⁰ — no d-d transition possible, so colourless.

Q3. (SA) Predict (with reasoning) which of the following ions are coloured in aqueous solution: Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺, Co²⁺.

Coloured (have unpaired d-electrons): Ti³⁺ (d¹), V³⁺ (d²), Mn²⁺ (d⁵), Fe³⁺ (d⁵), Co²⁺ (d⁷). Colourless (d⁰ or d¹⁰): Cu⁺ (d¹⁰) and Sc³⁺ (d⁰).

Q4. (LA) Why is Fe²⁺ a stronger reducing agent than Cu²⁺ in aqueous solution? Give thermodynamic reasoning.

E°(Fe³⁺/Fe²⁺) = +0.77 V is moderately positive while E°(Cu²⁺/Cu⁺) = +0.16 V is small. More importantly, Fe²⁺ → Fe³⁺ moves to the very stable d⁵ configuration, so the oxidation is favourable. Cu²⁺ has no half-/full-shell driving force to oxidise further, and Cu²⁺/Cu has E° = +0.34 V — already on the noble side. Hence Fe²⁺ tends to release electrons (be oxidised) whereas Cu²⁺ does not.

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.

The claim is incomplete. Pt does have multiple oxidation states (0, +2, +4), but its catalytic role in converters is mainly heterogeneous: CO and NO adsorb on Pt surface atoms via partial bonds with Pt 5d/6s electrons. This (i) increases reactant concentration on the surface, (ii) weakens the C=O / N=O bonds, and (iii) lowers the activation energy. Variable oxidation states matter more in homogeneous catalysis (e.g., Fe³⁺/Fe²⁺ shuttling electrons in solution).

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.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. e.g. Zr 160 pm vs Hf 159 pm.

Assertion (A): Mn²⁺ has a magnetic moment of about 5.92 BM.

Reason (R): Mn²⁺ has the configuration 3d⁵ with all five electrons paired.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: C. A is correct, but R is false — the 5 d-electrons of Mn²⁺ are all unpaired (Hund), not paired. Pairing would give μ = 0.

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⁺.

  • A. Both true, R explains A.
  • B. Both true, R does not explain A.
  • C. A true, R false.
  • D. A false, R true.
Answer: A. The CFSE of half-filled t₂g (d³) in an aqueous (octahedral) field exceeds the symmetry stabilisation of d⁵, so Cr²⁺ is the stronger reductant.

Frequently Asked Questions - Properties Trends

What is the main concept covered in Properties Trends?
In NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements), "Properties Trends" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is Properties Trends useful in real-life or applied chemistry?
Real-life applications of "Properties Trends" from NCERT Class 12 Chemistry Chapter 4 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for Properties Trends?
Key reactions in "Properties Trends" (NCERT Class 12 Chemistry Chapter 4 The d- and f-Block Elements) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 4?
NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements) is structured so each part builds chemical understanding sequentially. "Properties Trends" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Properties Trends?
CBSE board questions from "Properties Trends" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Properties Trends" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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Chemistry Class 12 Part I – NCERT (2025-26)
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