Class 12 Chemistry | Chapter 4
The d- and f-Block Elements
Transition Metals • Lanthanoids • Actinoids • Magnetic Properties • Catalytic Activity
1. The d-Block (Transition) Elements
Transition Elements: Elements which have incompletely filled d-orbitals in their ground
state or in any of their common oxidation states.
- Groups 3 to 12. Positioned between s- and p-blocks.
- Zn, Cd, Hg: Have completely filled d-orbitals (d10) in ground state and in their common oxidation states. Hence, they are NOT considered transition elements.
- General Electronic Configuration: (n−1)d1−10 ns1−2.
1.1 Four Transition Series
- 3d series (Sc to Zn): 4th period (Z = 21 to 30).
- 4d series (Y to Cd): 5th period (Z = 39 to 48).
- 5d series (La, Hf to Hg): 6th period (Z = 57, 72 to 80).
- 6d series (Ac, Rf to Cn): 7th period (Incomplete).
⚠️ EXCEPTIONS: Cr (Z=24) is [Ar] 3d5 4s1 and Cu (Z=29)
is [Ar] 3d10 4s1. This is due to the extra stability of exactly half-filled and
completely filled d-orbitals.
2. General Properties of Transition Elements
2.1 Atomic and Ionic Radii
- In a series, atomic radii decrease from left to right initially, remain almost constant in the middle, and increase slightly at the end (due to increased inter-electronic repulsion in d-orbitals).
- Radii of 4d & 5d elements are almost identical due to Lanthanoid Contraction (poor shielding of 4f electrons). Example: Zr and Hf have very similar radii and properties.
2.2 Ionisation Enthalpies
Increase generally from left to right. Irregular trend in 1st IE is due to altered stability of configurations (e.g., half-filled). However, the 2nd and 3rd IEs show significant jumps when removing an electron from stable d5 or d10 configurations.
- Zn, Cd, Hg have very high I.E. due to completely filled (n-1)d10 ns2 configuration.
2.3 Oxidation States
Transition elements show variable oxidation states because energy difference between (n-1)d and ns orbitals is very small. Both ns and (n-1)d electrons can participate in bonding.
- Maximum number of oxidation states is shown by elements in the middle of series (e.g., Mn shows +2 to +7).
- Sc shows only +3. Zn shows only +2.
- Highest oxidation states are found in fluorides and oxides (e.g., OsO4 and RuO4 have +8) because F and O are highly electronegative and small in size.
2.4 Melting and Boiling Points
They have high M.P. and B.P. due to strong metallic bonds arising from the participation of large number of electrons from both (n-1)d and ns orbitals.
- Melting point is maximum near the middle of series (e.g., Cr, Mo, W have maximum M.P. due to maximum number of unpaired electrons resulting in very strong metallic bonds).
- Zn, Cd, Hg have lowest M.P. (Hg is liquid) due to absence of unpaired d-electrons resulting in very weak metallic bonds.
3. Magnetic and Colour Properties
3.1 Magnetic Properties
Substances with unpaired electrons are paramagnetic (attracted by magnetic field). Substances with all paired electrons are diamagnetic (repelled by magnetic field).
Magnetic Moment (μ):
μ = √[n(n+2)] B.M. (Bohr Magneton)
Where 'n' is the number of unpaired electrons.
μ = √[n(n+2)] B.M. (Bohr Magneton)
Where 'n' is the number of unpaired electrons.
Greater the number of unpaired d-electrons, greater the magnetic moment. Mn2+ (3d5) has max n=5, so max μ ≈ 5.92 B.M.
3.2 Formation of Coloured Ions
Most transition metal compounds are coloured in solid state and solution. Colour is due to d-d transitions.
- When an electron from a lower energy d-orbital is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed within the visible region. The observed colour is complementary to the colour of the light absorbed.
- Ions with d0 (e.g., Sc3+, Ti4+) or d10 (e.g., Cu+, Zn2+) configurations are colourless because d-d transitions are not possible.
4. Other Characteristic Properties
4.1 Catalytic Properties
Transition metals and their compounds are known for their catalytic activity. Examples: V2O5 (Contact Process), Fe (Haber's Process), Ni (Catalytic Hydrogenation).
Reasons:
- Ability to adopt multiple oxidation states.
- Ability to form intermediate complexes.
- Provision of large surface area with free valencies (when in finely divided solid form).
4.2 Formation of Interstitial Compounds
Small non-metal atoms (like H, C, or N) get trapped inside the crystal lattices of metals. Examples: TiC, Mn4N, Fe3H.
- They are non-stoichiometric and typically neither ionic nor covalent.
- Properties: Very hard (some approach diamond in hardness), high melting points (higher than pure metals), retain metallic conductivity, chemically inert.
4.3 Alloy Formation
Alloys are solid solutions of two or more metals. Because the atomic radii of transition metals are very similar (differing by < 15%), atoms of one metal can readily substitute for atoms in the crystal lattice of another metal.
- Examples: Brass (Cu+Zn), Bronze (Cu+Sn). They are hard and often have high melting points. Ferrous alloys are the most common.
4.4 Tendency to form Complex Compounds
They form a large number of complex compounds. E.g., [Fe(CN)6]3−, [PtCl4]2−.
Reasons:
- Small sizes of the metal ions.
- High ionic charges.
- Availability of vacant d-orbitals of suitable energy to accept lone pairs of electrons donated by ligands.
5. Some Important Compounds of Transition Elements
5.1 Potassium Dichromate (K2Cr2O7)
Orange coloured crystals. Powerful oxidizing agent in acidic medium.
- Preparation is from Chromite ore (FeCr2O4). Involves converting chromite to sodium chromate, then to sodium dichromate, then to potassium dichromate.
- Chromate-Dichromate interconversion: Depends on pH. They exist in equilibrium.
In acidic medium: 2CrO42− (yellow) + 2H+ → Cr2O72− (orange) + H2O
In basic medium: Cr2O72− (orange) + 2OH− → 2CrO42− (yellow) + H2O - Oxidizing action (Acidic): Cr2O72− +
14H+ + 6e− → 2Cr3+ + 7H2O (E° = +1.33
V).
It oxidizes I− to I2, Fe2+ to Fe3+, Sn2+ to Sn4+, H2S to S.
5.2 Potassium Permanganate (KMnO4)
Dark purple (almost black) crystals. Very strong oxidizing agent.
- Prepared from pyrolusite (MnO2). Fused with KOH in presence of O2 to form potassium manganate (K2MnO4 - green), which is then electrolytically oxidized to KMnO4.
- Oxidizing action depends on pH:
- Acidic medium: MnO4− + 8H+ + 5e− → Mn2+ + 4H2O (E° = +1.52 V). Oxidizes oxalate to CO2, Fe2+ to Fe3+, nitrites to nitrates.
- Neutral/faintly alkaline medium: MnO4− + 2H2O + 3e− → MnO2(s) + 4OH−. Oxidizes iodide to iodate (IO3−), and thiosulphate to sulphate.
6. The f-Block Elements (Inner Transition Elements)
6.1 The Lanthanoids
14 elements following lanthanum (Z=58 to 71). General configuration: [Xe] 4f1-14 5d0-1 6s2.
- Lanthanoid Contraction: the steady decrease in atomic and ionic sizes of lanthanoid elements with increasing atomic number. Very crucial consequence: Radii of 4d and 5d transition series (e.g., Zr and Hf) become nearly identical.
- Reason for Contraction: Poor shielding effect of the 4f electrons. As nuclear charge increases, the pull on electrons increases, causing size reduction.
- Oxidation States: The most common and stable oxidation state is +3. Some elements show +2 (e.g., Eu2+, [Xe]4f7) and +4 (e.g., Ce4+, [Xe]4f0) to attain stable f0, f7, or f14 configurations. Ce4+ is a good analytical oxidizing agent.
- Mischmetall: An alloy of lanthanoid metals (~95%), iron (~5%), and traces of S, C, Ca, Al. Used making bullets, shells, and lighter flints.
6.2 The Actinoids
14 elements following actinium (Z=90 to 103). General configuration: [Rn] 5f0-14 6d0-2 7s2.
- All actinoids are radioactive. Early actinoids have relatively long half-lives, later ones have very short half-lives.
- Actinoid Contraction: Similar to lanthanoid contraction but greater from element to element due to even poorer shielding by 5f electrons compared to 4f.
- Oxidation States: Show a greater range of oxidation states (up to +7 for Np and Pu) compared to lanthanoids because 5f, 6d, and 7s levels are of comparable energies. Common O.S. is +3, but +4 is most stable for Th.
- Most actinoids are artificially prepared (Transuranium elements).
6.3 Lanthanoids vs Actinoids
| Lanthanoids | Actinoids |
|---|---|
| Maximum oxidation state is +4 (Ce) | Show higher oxidation states (+5, +6, +7) |
| Except Promethium (Pm), they are non-radioactive | All are radioactive |
| Compounds are less basic | Compounds are more basic |
| Do not form oxocations | Form oxocations (e.g., UO22+, PuO2+) |
🎓 NEET Previous Year Questions
Q1. [NEET 2022] The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9
BM. The suitable ligand for this complex is:
Answer μ = 5.9 BM indicates there are 5 unpaired
electrons (√[5(5+2)] = √35 ≈ 5.91). Mn2+ is 3d5. To have 5
unpaired electrons, the complex MUST be high-spin (weak field ligand). Among common ligands, NCS⁻
or H2O are weak field. (CN⁻ is strong field and would cause pairing).
Q2. [NEET 2021] Zr (Z=40) and Hf (Z=72) have similar atomic and ionic radii because
of:
Answer Lanthanoid contraction. The 14
lanthanoid elements lying between La and Hf have poorly shielding 4f electrons, leading to a steady
decrease in size that offsets the expected increase from the 4d to 5d series.
Q3. [NEET 2020] Identify the incorrect statement.
(a) Cr2+ (d4) is a stronger reducing agent than Fe2+ (d6) in water.
(b) The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states.
(c) Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
(d) The oxidation states of chromium in CrO42- and Cr2O72- are not the same.
(a) Cr2+ (d4) is a stronger reducing agent than Fe2+ (d6) in water.
(b) The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states.
(c) Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
(d) The oxidation states of chromium in CrO42- and Cr2O72- are not the same.
Answer Incorrect is (d). The
oxidation state of Cr is +6 in both CrO42- and
Cr2O72-.
Q4. [NEET 2019] Because of lanthanoid contraction, which of the following pairs of
elements have nearly same atomic radii? (Numbers in parenthesis are atomic numbers)
Answer Zr (40) and Hf (72). (Other
classic pairs: Nb & Ta, Mo & W).
Q5. [NEET 2018] Which of the following ions exhibits d-d transition and paramagnetism
as well? (a) CrO42- (b) Cr2O72- (c)
MnO4- (d) MnO42-
Answer In CrO42-, Cr is +6
(3d0). In Cr2O72-, Cr is +6 (3d0). In
MnO4-, Mn is +7 (3d0). In MnO42- (manganate), Mn
is +6 (3d1). Since MnO42- has 1 unpaired electron, it exhibits d-d
transition and paramagnetism. Correct answer is (d) MnO42-.
💡 Rapid Revision
- Zinc, Cadmium, Mercury are NOT considered transition elements because their d-orbitals are full in both atomic and ionic states.
- Highest oxidation state is shown by Mn (+7) in 3d series, and Os/Ru (+8) overall.
- Colour in transition metal ions is typically strictly due to d-d transitions. If d0 or d10, it is typically colourless. (Intense colour of KMnO4 is due to charge transfer spectra, not d-d!).
- Lanthanoid contraction makes the properties of 4d and 5d series elements very similar, making them hard to separate in nature.
- Cerium (Ce) commonly shows a +4 oxidation state, giving a stable empty f-subshell. Eu shows +2, for a stable half-filled f7 shell.
CLASS 12 CHEMISTRY | NCERT SOLUTIONS
Chapter 4 — The d- and f-Block Elements
22 Solved Questions — Reasoning & Calculations
Note: This chapter primarily features reasoning-based questions (why transition metals form
complexes, act as catalysts, show variable oxidation states) and magnetic moment calculations (μ =
√[n(n+2)] B.M.) rather than heavy numericals.
📝 Magnetic Moment Calculations (Q1 – Q5)
2 MarksQ1. Calculate the 'spin only' magnetic moment
of M²⁺ (aq) ion (Z = 27).
✓ Solution
For Z = 27 (Cobalt), the electronic configuration is [Ar] 3d⁷ 4s².
For M²⁺ ion, two electrons are removed from the 4s orbital. Co²⁺ is [Ar] 3d⁷.
Arrangement in 3d: ↑↓ ↑↓ ↑ ↑ ↑. This gives 3 unpaired electrons (n=3).
Magnetic moment (μ) = √[n(n+2)] B.M. = √[3(3+2)] = √15 B.M.
For Z = 27 (Cobalt), the electronic configuration is [Ar] 3d⁷ 4s².
For M²⁺ ion, two electrons are removed from the 4s orbital. Co²⁺ is [Ar] 3d⁷.
Arrangement in 3d: ↑↓ ↑↓ ↑ ↑ ↑. This gives 3 unpaired electrons (n=3).
Magnetic moment (μ) = √[n(n+2)] B.M. = √[3(3+2)] = √15 B.M.
μ = 3.87 B.M.
2 MarksQ2. Calculate the magnetic moment of a
divalent ion in aqueous solution if its atomic number is 25.
✓ Solution
For Z = 25 (Manganese), the configuration is [Ar] 3d⁵ 4s².
Divalent ion (Mn²⁺) configuration is [Ar] 3d⁵.
There are 5 unpaired electrons (n = 5).
μ = √[n(n+2)] = √[5(5+2)] = √35 B.M.
For Z = 25 (Manganese), the configuration is [Ar] 3d⁵ 4s².
Divalent ion (Mn²⁺) configuration is [Ar] 3d⁵.
There are 5 unpaired electrons (n = 5).
μ = √[n(n+2)] = √[5(5+2)] = √35 B.M.
μ = 5.92 B.M.
3 MarksQ3. The magnetic moment of a transition metal
ion is 5.92 BM. Predict the number of unpaired electrons and given an example of such an ion.
✓ Solution
Given μ = √[n(n+2)] = 5.92 B.M. Squaring both sides:
n(n+2) = (5.92)² ≈ 35
n² + 2n - 35 = 0 ⇒ (n+7)(n-5) = 0
Since n cannot be negative, n = 5 unpaired electrons.
Given μ = √[n(n+2)] = 5.92 B.M. Squaring both sides:
n(n+2) = (5.92)² ≈ 35
n² + 2n - 35 = 0 ⇒ (n+7)(n-5) = 0
Since n cannot be negative, n = 5 unpaired electrons.
An example of an ion with 5 unpaired electrons is
Mn²⁺ or Fe³⁺.
2 MarksQ4. Explain why Cu⁺ ion is diamagnetic
whereas Cu²⁺ ion is paramagnetic.
✓ Solution
Cu (Z=29) configuration: [Ar] 3d¹⁰ 4s¹.
Cu⁺: [Ar] 3d¹⁰. All electrons in the 3d subshell are paired (n=0). Hence it is repelled by magnetic fields (diamagnetic).
Cu (Z=29) configuration: [Ar] 3d¹⁰ 4s¹.
Cu⁺: [Ar] 3d¹⁰. All electrons in the 3d subshell are paired (n=0). Hence it is repelled by magnetic fields (diamagnetic).
Cu²⁺: [Ar] 3d⁹. It has 1 unpaired electron
(n=1) in the 3d orbital. Hence it is attracted by magnetic fields (paramagnetic).
2 MarksQ5. Write down the electronic configuration of
Cr³⁺ and calculate its magnetic moment.
✓ Solution
Cr (Z=24) configuration: [Ar] 3d⁵ 4s¹.
Cr³⁺ configuration: [Ar] 3d³.
Number of unpaired electrons, n = 3.
μ = √[3(3+2)] = √15 B.M.
Cr (Z=24) configuration: [Ar] 3d⁵ 4s¹.
Cr³⁺ configuration: [Ar] 3d³.
Number of unpaired electrons, n = 3.
μ = √[3(3+2)] = √15 B.M.
μ = 3.87 B.M.
💡 Transition Metals Reasoning (Q6 – Q14)
2 MarksQ6. Why are Zn, Cd and Hg not considered as
transition elements?
✓ Solution
A transition element must have incompletely filled d-orbitals in its ground state or in any of its common oxidation states.
A transition element must have incompletely filled d-orbitals in its ground state or in any of its common oxidation states.
Zn (3d¹⁰ 4s²), Cd (4d¹⁰ 5s²), and Hg
(5d¹⁰ 6s²) have completely filled d-orbitals in their ground state as well as in
their common oxidation state (+2). Hence, they are excluded from transition elements.
3 MarksQ7. Give reasons: (i) Transition metals and
many of their compounds show paramagnetic behaviour. (ii) The enthalpies of atomisation of the
transition metals are high. (iii) The transition metals generally form coloured compounds.
✓ Solution
(i) They have unpaired electrons in their (n-1)d orbitals.
(ii) They have a large number of unpaired electrons in their atoms, which results in strong interatomic metallic bonding. Hence, high energy is required to break these bonds (high enthalpy of atomisation).
(i) They have unpaired electrons in their (n-1)d orbitals.
(ii) They have a large number of unpaired electrons in their atoms, which results in strong interatomic metallic bonding. Hence, high energy is required to break these bonds (high enthalpy of atomisation).
(iii) They have incomplete d-orbitals. Unpaired electrons can
undergo d-d transitions by absorbing energy from visible light. The transmitted/reflected light
shows the complementary colour.
2 MarksQ8. What are interstitial compounds? Why are
such compounds well known for transition metals?
✓ Solution
Interstitial compounds are formed when small atoms like H, C, or N get trapped inside the empty spaces (interstices) of the crystal lattices of metals.
Interstitial compounds are formed when small atoms like H, C, or N get trapped inside the empty spaces (interstices) of the crystal lattices of metals.
Transition metals are well known for forming these because their crystal
lattices contain appropriately sized voids that can accommodate these small atoms without
substantially disturbing the host lattice.
2 MarksQ9. Explain why transition elements form
complex compounds.
✓ Solution
Transition elements form complex compounds due to:
1. Considerably small size and high effective nuclear charge of their ions.
2. High ionic charge density.
Transition elements form complex compounds due to:
1. Considerably small size and high effective nuclear charge of their ions.
2. High ionic charge density.
3. Availability of vacant d-orbitals of compatible energy to accept lone pairs
of electrons donated by the ligands.
3 MarksQ10. Name a member of the lanthanoid series
which is well known to exhibit +4 oxidation state. Why does it show +4 oxidation state?
✓ Solution
The member is Cerium (Ce, Z = 58).
The member is Cerium (Ce, Z = 58).
It exhibits a +4 oxidation state because by losing 4 electrons it acquires a
very stable empty orbital (noble gas) configuration: Ce⁴⁺ is [Xe] 4f⁰ 5d⁰
6s⁰.
2 MarksQ11. Why does copper not replace hydrogen from
acids?
✓ Solution
For a metal to displace H₂ from an acid, it must have a negative standard reduction potential (E°).
For a metal to displace H₂ from an acid, it must have a negative standard reduction potential (E°).
The standard reduction potential of Copper (E° Cu²⁺/Cu) is
+0.34 V, which is positive. Being a weaker reducing agent than hydrogen (E° = 0.00 V), Cu
cannot reduce H⁺ ions to H₂ gas.
3 MarksQ12. What is lanthanoid contraction? Name an
important alloy which contains some of the lanthanoid metals.
✓ Solution
Lanthanoid Contraction: The steady, gradual decrease in atomic and ionic radii along the lanthanoid series (from Ce to Lu) due to the poor shielding effect of 4f electrons against the increasing nuclear charge.
Lanthanoid Contraction: The steady, gradual decrease in atomic and ionic radii along the lanthanoid series (from Ce to Lu) due to the poor shielding effect of 4f electrons against the increasing nuclear charge.
Important Alloy: Mischmetall (consists of roughly 95%
lanthanoids, 5% iron, and traces of S, C, Al, Ca). Used making lighter flints.
2 MarksQ13. Zinc has the lowest enthalpy of
atomisation in the 3d series. Why?
✓ Solution
The enthalpy of atomisation depends on the strength of metallic bonding, which in turn depends on the number of unpaired d-electrons.
The enthalpy of atomisation depends on the strength of metallic bonding, which in turn depends on the number of unpaired d-electrons.
Zinc (3d¹⁰ 4s²) has completely filled d-orbitals and zero
unpaired electrons. Hence, the metallic bonding is very weak, leading to the lowest enthalpy of
atomisation in its series.
3 MarksQ14. Why is the highest oxidation state of a
metal exhibited in its oxide or fluoride only?
✓ Solution
Oxygen and Fluorine have:
1. Small atomic size.
2. High electronegativity. (Fluorine is the most, Oxygen is the second most).
Oxygen and Fluorine have:
1. Small atomic size.
2. High electronegativity. (Fluorine is the most, Oxygen is the second most).
Because of these properties, they are strong oxidizing agents and can remove
the maximum number of electrons from the transition metals, stabilizing them in their highest
oxidation states. (Oxygen can also form multiple bonds).
📈 Balancing Redox Equations (Q15 – Q18)
4 MarksQ15. Write balanced ionic equations for: (i)
Oxidation of iodide by permanganate ion in neutral/faintly alkaline medium. (ii) Oxidation of
Fe²⁺ by dichromate ion in acidic medium.
✓ Solution
(i) Neutral/Faintly alkaline medium (Permanganate + Iodide):
Permanganate reduces to MnO₂, Iodide oxidizes to Iodate (IO₃⁻).
2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻
(ii) Acidic medium (Dichromate + Fe²⁺):
Dichromate reduces to Cr³⁺, Fe²⁺ oxidizes to Fe³⁺.
(i) Neutral/Faintly alkaline medium (Permanganate + Iodide):
Permanganate reduces to MnO₂, Iodide oxidizes to Iodate (IO₃⁻).
2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻
(ii) Acidic medium (Dichromate + Fe²⁺):
Dichromate reduces to Cr³⁺, Fe²⁺ oxidizes to Fe³⁺.
Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ →
2Cr³⁺ + 6Fe³⁺ + 7H₂O
3 MarksQ16. Complete the following equation:
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → ?
✓ Solution
This is the oxidation of oxalate ion by permanganate in acidic medium.
Permanganate (MnO₄⁻) reduces to Mn²⁺.
Oxalate (C₂O₄²⁻) oxidizes to CO₂.
This is the oxidation of oxalate ion by permanganate in acidic medium.
Permanganate (MnO₄⁻) reduces to Mn²⁺.
Oxalate (C₂O₄²⁻) oxidizes to CO₂.
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ →
2Mn²⁺ + 10CO₂ + 8H₂O
3 MarksQ17. Complete the following equation:
Cr₂O₇²⁻ + 3H₂S + 8H⁺ → ?
✓ Solution
This is the oxidation of Hydrogen Sulphide (H₂S) by dichromate in acidic medium.
Dichromate (Cr₂O₇²⁻) reduces to Cr³⁺.
Sulphide (S²⁻) from H₂S oxidizes to elemental Sulphur (S).
This is the oxidation of Hydrogen Sulphide (H₂S) by dichromate in acidic medium.
Dichromate (Cr₂O₇²⁻) reduces to Cr³⁺.
Sulphide (S²⁻) from H₂S oxidizes to elemental Sulphur (S).
Cr₂O₇²⁻ + 3H₂S + 8H⁺ →
2Cr³⁺ + 3S + 7H₂O
3 MarksQ18. How is potassium dichromate prepared from
chromite ore? Mention the steps.
✓ Solution
Step 1: Roasting chromite ore with Na₂CO₃ in air to form Sodium chromate.
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
Step 2: Conversion of sodium chromate to sodium dichromate by acidification.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
Step 1: Roasting chromite ore with Na₂CO₃ in air to form Sodium chromate.
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
Step 2: Conversion of sodium chromate to sodium dichromate by acidification.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
Step 3: Conversion of sodium dichromate to potassium dichromate by adding KCl
(K₂Cr₂O₇ is less soluble, so it crystallizes out).
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
🔥 Actinoids and Lanthanoids (Q19 – Q22)
3 MarksQ19. Describe actinoid contraction. How does
it differ from lanthanoid contraction?
✓ Solution
Actinoid Contraction: It is the steady decrease in atomic/ionic size with an increase in atomic number across the actinoid series.
Difference: The actinoid contraction is greater from element to element than the lanthanoid contraction. This is because 5f electrons provide even poorer shielding from nuclear charge than 4f electrons. Thus, the effective nuclear charge felt by outer electrons increases more sharply in actinoids.
Actinoid Contraction: It is the steady decrease in atomic/ionic size with an increase in atomic number across the actinoid series.
Difference: The actinoid contraction is greater from element to element than the lanthanoid contraction. This is because 5f electrons provide even poorer shielding from nuclear charge than 4f electrons. Thus, the effective nuclear charge felt by outer electrons increases more sharply in actinoids.
2 MarksQ20. Why is the chemistry of all the
lanthanoids so identical?
✓ Solution
Lanthanoids involve the filling of inner 4f orbitals. Since the outermost shell (6s) and penultimate shell (5s, 5p, 5d) remain virtually identical for all these elements, their chemical properties are very similar.
Lanthanoids involve the filling of inner 4f orbitals. Since the outermost shell (6s) and penultimate shell (5s, 5p, 5d) remain virtually identical for all these elements, their chemical properties are very similar.
Due to the lanthanoid contraction, their atomic sizes are also very similar,
which further makes their separation in chemistry very difficult.
2 MarksQ21. The first ionization enthalpies of 5d
elements are higher than those of 3d and 4d elements. Why?
✓ Solution
Because of the lanthanoid contraction.
Because of the lanthanoid contraction.
The filling of 4f orbitals before the 5d series results in poor shielding. As
a result, the effective nuclear charge pulling on the outermost 6s electrons is unusually high in 5d
elements. This makes it much harder to remove an electron, causing higher ionization enthalpies.
2 MarksQ22. Compare the chemistry of actinoids with
that of the lanthanoids with special reference to: (i) oxidation state (ii) chemical reactivity.
✓ Solution
(i) Oxidation state: Lanthanoids mostly show +3 (some +2, +4). Actinoids show a much wider range of oxidation states (+3, +4, +5, +6, +7) because the 5f, 6d, and 7s levels have comparable energies.
(i) Oxidation state: Lanthanoids mostly show +3 (some +2, +4). Actinoids show a much wider range of oxidation states (+3, +4, +5, +6, +7) because the 5f, 6d, and 7s levels have comparable energies.
(ii) Chemical Reactivity: Actinoids are generally more
reactive than lanthanoids, especially when finely divided. This is largely because the 5f electrons
extend further from the nucleus than 4f, making them more available for bonding.
✍ Score Guide — 22 Questions
All questions from NCERT Exercises & CBSE Previous Year Question Papers.
All questions from NCERT Exercises & CBSE Previous Year Question Papers.
High-Yield Facts & Formulas: d- and f- Block
Transition Elements
Elements with partially filled d-orbitals in ground state or oxidation state. (Zn, Cd, Hg are not transition metals).
Elements with partially filled d-orbitals in ground state or oxidation state. (Zn, Cd, Hg are not transition metals).
d-Block Configuration
General: (n-1)d1-10 ns1-2.
General: (n-1)d1-10 ns1-2.
Highest Oxidation State
+7 shown by Manganese (Mn) in the 3d series. Osmium (Os) and Ruthenium (Ru) show +8.
+7 shown by Manganese (Mn) in the 3d series. Osmium (Os) and Ruthenium (Ru) show +8.
Spin-only Magnetic Moment
μ = √[n(n+2)] Bohr Magnetons (BM).
μ = √[n(n+2)] Bohr Magnetons (BM).
Ionization Enthalpy Trend
Increases along the series due to increase in nuclear charge, but irregularly because of d-orbital filling.
Increases along the series due to increase in nuclear charge, but irregularly because of d-orbital filling.
Color of Ions
Transition metal ions are usually colored due to d-d transitions. (Exceptions: d0 like Sc3+, d10 like Zn2+).
Transition metal ions are usually colored due to d-d transitions. (Exceptions: d0 like Sc3+, d10 like Zn2+).
Alloy Formation
Metals of similar radii can replace each other in crystal lattice. (e.g., Brass, Bronze, Stainless steel).
Metals of similar radii can replace each other in crystal lattice. (e.g., Brass, Bronze, Stainless steel).
Interstitial Compounds
Non-stoichiometric compounds like TiC, ZrH1.9, Mn4N. They are hard and conduct electricity.
Non-stoichiometric compounds like TiC, ZrH1.9, Mn4N. They are hard and conduct electricity.
Lanthanoid Contraction
Steady decrease in size of atoms/ions from La to Lu due to poor shielding of 4f electrons.
Steady decrease in size of atoms/ions from La to Lu due to poor shielding of 4f electrons.
Actinoid Contraction
Steady decrease in size from Ac to Lr. It is greater than lanthanoid contraction due to poorer shielding by 5f.
Steady decrease in size from Ac to Lr. It is greater than lanthanoid contraction due to poorer shielding by 5f.
KMnO4 Structure
Tetrahedral MnO4- ion. Mn is d0 (purple color is due to charge transfer).
Tetrahedral MnO4- ion. Mn is d0 (purple color is due to charge transfer).
Potassium Dichromate color
Orange in acidic medium, Yellow (chromate) in basic medium.
Orange in acidic medium, Yellow (chromate) in basic medium.
Density Trend
Density increases from Ti to Cu because atomic mass increase is more dominant than radius change.
Density increases from Ti to Cu because atomic mass increase is more dominant than radius change.
Enthalpy of Atomization
High for transition metals due to strong metallic bonding (participation of (n-1)d electrons).
High for transition metals due to strong metallic bonding (participation of (n-1)d electrons).
V2O5 catalyst
Used in the Contact Process for making H2SO4.
Used in the Contact Process for making H2SO4.
TiCl4/Al(Et)3
Ziegler-Natta catalyst used for polymerization of ethene.
Ziegler-Natta catalyst used for polymerization of ethene.
Liquid metal
Mercury (Hg) is liquid because of very weak metallic bonding (no unpaired d-electrons).
Mercury (Hg) is liquid because of very weak metallic bonding (no unpaired d-electrons).
Paramagnetism
Property of being attracted by a magnetic field. Increases with number of unpaired electrons.
Property of being attracted by a magnetic field. Increases with number of unpaired electrons.
Diamagnetism
Property of being weakly repelled by a magnetic field. Shown by d0 and d10 species.
Property of being weakly repelled by a magnetic field. Shown by d0 and d10 species.
Eu2+ and Tb4+
Lanthanoid ions showing anomalous oxidation states (+2 and +4) for stable f7 configuration.
Lanthanoid ions showing anomalous oxidation states (+2 and +4) for stable f7 configuration.
Separation of Lanthanoids
Done by ion-exchange method based on slight differences in basicity.
Done by ion-exchange method based on slight differences in basicity.
Actinoid Basicity
Actinoid hydroxides are more basic than lanthanoid hydroxides.
Actinoid hydroxides are more basic than lanthanoid hydroxides.
Chromite Ore
FeCr2O4. Raw material for K2Cr2O7.
FeCr2O4. Raw material for K2Cr2O7.
Pyrolusite Ore
MnO2. Raw material for KMnO4 manufacture.
MnO2. Raw material for KMnO4 manufacture.
Disproportionation of
MnO42-
In acidic medium: Manganate ions disproportionate to permanganate and MnO2.
In acidic medium: Manganate ions disproportionate to permanganate and MnO2.
Inner Transition Elements Position
Placed separately at bottom to avoid horizontal expansion of periodic table.
Placed separately at bottom to avoid horizontal expansion of periodic table.
Third Transition Series Radii
Almost identical to second series due to Lanthanoid contraction (e.g., Zr ≈ Hf).
Almost identical to second series due to Lanthanoid contraction (e.g., Zr ≈ Hf).
Complexation ability
Transition ions form complexes with ligands like NH3, H2O, CN-.
Transition ions form complexes with ligands like NH3, H2O, CN-.
Noble metals
Pt and Au are very resistant to oxidation (low reactivity).
Pt and Au are very resistant to oxidation (low reactivity).
CuSO4.5H2O
Blue Vitriol. Contains [Cu(H2O)4]2+ SO42-.H2O.
Blue Vitriol. Contains [Cu(H2O)4]2+ SO42-.H2O.
Ferromagnetism
Permanent magnetism shown by Fe, Co, Ni.
Permanent magnetism shown by Fe, Co, Ni.
Coinage Metals
Group 11 elements (Cu, Ag, Au).
Group 11 elements (Cu, Ag, Au).
f-Block group
All lanthanoids and actinoids belong to Group 3 of the periodic table.
All lanthanoids and actinoids belong to Group 3 of the periodic table.
Malleability and Ductility
Typical properties of transition metals due to metallic bonds.
Typical properties of transition metals due to metallic bonds.
Ionic vs Covalent bonds
Lower oxidation states form ionic compounds; higher ones form covalent compounds (Fajans' Rule).
Lower oxidation states form ionic compounds; higher ones form covalent compounds (Fajans' Rule).
Transition metal hydrides
Often non-stoichiometric and hydrogen deficient.
Often non-stoichiometric and hydrogen deficient.
Magnetic susceptibility
Increases with temperature for antiferromagnetic substances and decreases for paramagnetic.
Increases with temperature for antiferromagnetic substances and decreases for paramagnetic.
Oxoanions of metals
Chromate (CrO42-), Permanganate (MnO4-).
Chromate (CrO42-), Permanganate (MnO4-).
Chromium +6 toxicity
Highly toxic and carcinogenic compared to Cr3+.
Highly toxic and carcinogenic compared to Cr3+.
Nickel in Hydrogenation
Finely divided Ni is used as a catalyst for hydrogenating oils.
Finely divided Ni is used as a catalyst for hydrogenating oils.
Titanium durability
"Space age metal" due to its high strength, low weight, and corrosion resistance.
"Space age metal" due to its high strength, low weight, and corrosion resistance.
Silver reactivity
Gets tarnished in air due to H2S, forming Ag2S.
Gets tarnished in air due to H2S, forming Ag2S.
Copper's uniqueness
The only metal in 3d series with positive E0 (reduction potential), hence does not release H2 from acids.
The only metal in 3d series with positive E0 (reduction potential), hence does not release H2 from acids.
Steel components
Mainly Fe + C; Alloyed with Cr, Ni, Mo to change properties.
Mainly Fe + C; Alloyed with Cr, Ni, Mo to change properties.
Zn, Cd, Hg soft metals
Low enthalpies of atomization due to completed d-orbitals.
Low enthalpies of atomization due to completed d-orbitals.
Platinum black
Finely divided Pt with very high catalytic activity.
Finely divided Pt with very high catalytic activity.
Cerium anomaly
Ce shows +4 oxidation state (noble gas configuration [Xe]).
Ce shows +4 oxidation state (noble gas configuration [Xe]).
Th, Pa, U
Elements after which 5f filling becomes dominant.
Elements after which 5f filling becomes dominant.
Manganese +2 stability
High stability in aqueous solution due to half-filled d5 subshell.
High stability in aqueous solution due to half-filled d5 subshell.
Catalyst in Haber's process
Iron mixed with Mo (as promoter) or Al2O3/K2O.
Iron mixed with Mo (as promoter) or Al2O3/K2O.
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App