d- and f-Block Elements

What to remember

• Transition elements have partly filled d orbitals in the atom or in a common ion. Zn, Cd and Hg are d-block elements but not transition elements, because they have d¹⁰ in the atom and in the +2 ion.
• Variable oxidation states, coloured ions (d–d transitions), paramagnetism, catalytic activity, interstitial compounds and alloy formation are the hallmarks of the d-block. The highest oxidation state is +7, shown by Mn.
• Spin-only magnetic moment μ = √(n(n+2)) BM, where n is the number of unpaired electrons. n = 1, 2, 3, 4, 5 gives 1.73, 2.83, 3.87, 4.90, 5.92 BM.
• Lanthanoid contraction (poor shielding by 4f electrons) makes 4d and 5d elements almost equal in size. +3 is the common state of lanthanoids; actinoids show a wider range of states.

Position and electronic configuration

The d-block lies in Groups 3 to 12 and has four series: 3d (Sc to Zn), 4d (Y to Cd), 5d (La, Hf to Hg) and 6d (Ac, Rf onwards). The general outer configuration is (n–1)d¹⁻¹⁰ ns¹⁻². The 4s orbital fills before 3d, but when ions form, 4s electrons are lost first.

Element	Configuration	Element	Configuration
Sc (21)	3d¹ 4s²	Fe (26)	3d⁶ 4s²
Ti (22)	3d² 4s²	Co (27)	3d⁷ 4s²
V (23)	3d³ 4s²	Ni (28)	3d⁸ 4s²
Cr (24)	3d⁵ 4s¹	Cu (29)	3d¹⁰ 4s¹
Mn (25)	3d⁵ 4s²	Zn (30)	3d¹⁰ 4s²

Cr and Cu are the two exceptions in the 3d series. A half-filled (d⁵) or completely filled (d¹⁰) subshell has extra stability from symmetry and exchange energy, so one 4s electron moves into 3d.

Common ions: Ti³⁺ (3d¹), V³⁺ (3d²), Cr³⁺ (3d³), Mn²⁺ (3d⁵), Fe³⁺ (3d⁵), Fe²⁺ (3d⁶), Co²⁺ (3d⁷), Ni²⁺ (3d⁸), Cu²⁺ (3d⁹), Cu⁺ (3d¹⁰), Zn²⁺ (3d¹⁰). Sc³⁺ is 3d⁰.

Metallic character, size and ionisation enthalpy

• All are metals with high tensile strength, ductility, malleability, and good heat and electrical conduction. Both the ns and the unpaired (n–1)d electrons take part in metallic bonding, so melting points and enthalpies of atomisation are high. Zn, Cd and Hg have no unpaired d electrons, so their atomisation enthalpies and melting points are low (Hg is a liquid). Mn is also lower than its neighbours.
• Atomic and ionic radii decrease from Sc to about Cr or Mn because the nuclear charge rises while the added d electrons shield poorly. The radii stay nearly constant in the middle (Fe, Co, Ni), then rise slightly at Cu and Zn because d–d repulsion in d¹⁰ outweighs the nuclear pull.
• Down a group radius increases from 3d to 4d, but 5d elements have about the same radius as 4d elements (Zr 160 pm, Hf 159 pm). The reason is the lanthanoid contraction, caused by the 4f electrons that are filled before 5d.
• Ionisation enthalpies rise slowly across the series because the nuclear charge increases and the d electrons shield poorly. The second ionisation enthalpies of Cr and Cu are unusually high because the ion Cr⁺ (d⁵) and Cu⁺ (d¹⁰) have stable configurations. The third ionisation enthalpy of Mn and of Zn is high for the same reason (Mn²⁺ is d⁵, Zn²⁺ is d¹⁰).

Property	Trend across 3d series
Atomic radius	Decreases, then almost constant, then slight rise at Cu and Zn
First ionisation enthalpy	Increases slowly
Enthalpy of atomisation	High; lowest for Zn
Density	Generally increases (small volume, rising mass)

Variable oxidation states and electrode potentials

The (n–1)d and ns energies are close, so many electrons can be used for bonding. The oxidation states of one element usually differ by one unit (Fe +2, +3; Mn +2 to +7). In p-block elements the states differ by two units.

Element	Common states	Highest
Sc	+3	+3
Ti	+3, +4	+4
V	+2, +3, +4, +5	+5
Cr	+2, +3, +6	+6
Mn	+2, +3, +4, +6, +7	+7
Fe	+2, +3	+6 (rare)
Co	+2, +3	+4
Ni	+2	+4
Cu	+1, +2	+2
Zn	+2	+2

Up to Mn, the highest state equals the total number of 3d and 4s electrons (group number). After Mn the highest state falls. The highest states appear in oxides and oxoanions (CrO₄²⁻, Cr₂O₇²⁻, MnO₄⁻), where fluorine and oxygen stabilise them. In a group, the higher oxidation states become more stable on going down (Mo(VI) and W(VI) are more stable than Cr(VI)). Lower oxides are basic (CrO), intermediate are amphoteric (Cr₂O₃) and highest are acidic (CrO₃, Mn₂O₇).

Standard electrode potentials. E°(M²⁺/M) values are: Ti –1.63, V –1.18, Cr –0.90, Mn –1.18, Fe –0.44, Co –0.28, Ni –0.25, Cu +0.34, Zn –0.76 V.

• Cu is the only element in the series with a positive E°(M²⁺/M). The energy to atomise and ionise copper is not offset by the hydration enthalpy of Cu²⁺, so Cu does not liberate H₂ from dilute acids.
• Mn and Zn are more negative than the general trend because Mn²⁺ is d⁵ (half-filled) and Zn²⁺ is d¹⁰ (completely filled), which makes the ions unusually stable.
• E°(M³⁺/M²⁺) values: Mn +1.57, Fe +0.77, Co +1.97, Cr –0.41, V –0.26. Mn³⁺ and Co³⁺ are strong oxidising agents (Mn²⁺ is stable d⁵). A low value for Cr³⁺/Cr²⁺ means Cr³⁺ is stable (d³, t₂g³) and Cr²⁺ is a strong reducing agent. The value for Fe³⁺/Fe²⁺ (+0.77 V) is intermediate.
• Cu⁺ is unstable in water and disproportionates: 2Cu⁺ → Cu²⁺ + Cu. The high hydration enthalpy of Cu²⁺ makes this favourable.

Magnetic, colour, catalytic, interstitial and alloy properties

Magnetism. A substance with unpaired electrons is paramagnetic; one with all electrons paired is diamagnetic. The spin-only moment is μ = √(n(n+2)) BM.

Unpaired electrons n	1	2	3	4	5
μ (BM)	1.73	2.83	3.87	4.90	5.92

Worked example 1: Fe³⁺ is 3d⁵, so n = 5 and μ = √(5 × 7) = √35 = 5.92 BM. Worked example 2: Ni²⁺ is 3d⁸, with t₂g⁶ eg² giving n = 2, so μ = √(2 × 4) = √8 = 2.83 BM. Cu²⁺ (d⁹) has n = 1 and μ = 1.73 BM. Zn²⁺, Sc³⁺ and Cu⁺ are diamagnetic.

Colour. Colour arises from d–d transitions: an electron in a lower d level absorbs visible light and jumps to a higher d level, and the complementary colour is seen. Ions with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) have no d–d transition and are colourless.

Ion	Configuration	Colour in water
Ti³⁺	d¹	Purple
V³⁺	d²	Green
Cr³⁺	d³	Violet
Mn²⁺	d⁵	Pink
Fe²⁺	d⁶	Green
Fe³⁺	d⁵	Yellow
Co²⁺	d⁷	Pink
Ni²⁺	d⁸	Green
Cu²⁺	d⁹	Blue
Zn²⁺	d¹⁰	Colourless

Catalysis. Variable oxidation states and the ability to adsorb reactants make them good catalysts: V₂O₅ in the contact process (SO₂ → SO₃); finely divided Fe in the Haber process; Ni in hydrogenation of oils; Pt in the Ostwald process; MnO₂ in the decomposition of KClO₃.

Interstitial compounds. Small atoms such as H, C and N occupy the voids of the metal lattice, giving compounds like TiC, Fe₃H and Fe₃C. They are very hard, have high melting points, keep metallic conductivity and are chemically inert. They are less malleable than the pure metal.

Alloys. Similar atomic sizes let one metal replace another in the lattice. Examples: brass (Cu + Zn), bronze (Cu + Sn), steel (Fe + C and other metals). Misch metal (about 95% lanthanoids, 5% Fe, traces of S, C, Ca, Al) is used in Mg-based alloys for bullets, shells and lighter flints.

Potassium dichromate, K₂Cr₂O₇

Preparation from chromite ore (FeCr₂O₄).

• 1. Fuse with Na₂CO₃ in excess air: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂.
• 2. Filter and acidify the yellow sodium chromate: 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O.
• 3. Treat the orange sodium dichromate with KCl: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl. Orange crystals of K₂Cr₂O₇ separate because it is less soluble.

Chromate–dichromate equilibrium. 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. In acid (pH below about 4) the orange dichromate ion dominates. In alkali it turns yellow (chromate). Chromium is +6 in both, so this is not a redox change. The chromate ion is tetrahedral; the dichromate ion has two tetrahedra sharing one corner oxygen (Cr–O–Cr angle about 126°).

Oxidising action in acid. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (orange to green).

• With Fe²⁺: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.
• With iodide: Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O.
• With H₂S: Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O.

Worked example 3: In acid, 1 mol Cr₂O₇²⁻ takes 6 mol electrons and oxidises 6 mol Fe²⁺. So 0.01 mol K₂Cr₂O₇ oxidises 0.06 mol FeSO₄.

Potassium permanganate, KMnO₄

Preparation from pyrolusite (MnO₂).

• 1. Fuse MnO₂ with KOH in air or with KNO₃: 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O (green potassium manganate).
• 2. Oxidise manganate to permanganate by electrolysis (MnO₄²⁻ → MnO₄⁻ + e⁻) or by acid disproportionation: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O.

KMnO₄ forms dark purple crystals. On heating at 513 K: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂. The MnO₄⁻ ion is tetrahedral and diamagnetic (Mn +7, d⁰); its purple colour is due to charge transfer from O to Mn, not d–d transition. MnO₄²⁻ (Mn +6, d¹) is paramagnetic.

Oxidising action depends on the medium.

Medium	Half-reaction	Product of Mn
Acidic	MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	Mn²⁺ (colourless/pale pink)
Neutral or weakly alkaline	MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻	MnO₂ (brown)
Strongly alkaline	MnO₄⁻ + e⁻ → MnO₄²⁻	MnO₄²⁻ (green)

Acidic examples: 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O; MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O; 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O. Neutral example: 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻. Dilute H₂SO₄ is used for acidification; HCl is avoided because it is oxidised to Cl₂.

Worked example 4: 25 mL of 0.02 M KMnO₄ contains 5 × 10⁻⁴ mol MnO₄⁻. In acid each MnO₄⁻ oxidises 5 Fe²⁺, so Fe²⁺ = 2.5 × 10⁻³ mol.

Lanthanoids (inner transition elements, 4f series)

The 14 elements Ce (58) to Lu (71) follow La (57). General configuration: [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s². Examples: La 5d¹ 6s²; Ce 4f¹ 5d¹ 6s²; Gd 4f⁷ 5d¹ 6s²; Lu 4f¹⁴ 5d¹ 6s².

• Oxidation states: +3 is the most common. +2 and +4 appear where they give a stable empty, half-filled or full 4f shell: Ce⁴⁺ (4f⁰, strong oxidising agent, Ce⁴⁺ → Ce³⁺), Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴) (both strong reducing agents, going to +3). Tb⁴⁺ is 4f⁷.
• Colour: many Ln³⁺ ions are coloured because of f–f transitions; La³⁺ (f⁰) and Lu³⁺ (f¹⁴) are colourless.
• Lanthanoid contraction: a steady decrease in atomic and ionic radii from La to Lu (ionic radius of Ln³⁺ falls from about 106 pm to about 85 pm). The 4f electrons shield one another poorly, so the effective nuclear charge rises.
• Consequences: (1) 4d and 5d elements of a group have nearly equal radii (Zr–Hf, Nb–Ta) and similar chemistry, so they are hard to separate; (2) lanthanoids are difficult to separate from one another; (3) basic character of Ln(OH)₃ decreases from La to Lu because covalent character rises as the ion becomes smaller; (4) 5d elements have high ionisation enthalpies and densities.
• Pm is the only radioactive lanthanoid. Lanthanoids are silvery, fairly reactive, and are used in misch metal.

Actinoids (5f series)

The 14 elements Th (90) to Lr (103) follow Ac (89). General configuration: [Rn] 5f⁰⁻¹⁴ 6d⁰⁻¹ 7s². All are radioactive.

• Oxidation states: +3 is common, but +4, +5, +6 and +7 also occur (for example U +3 to +6, Np and Pu up to +7). The 5f, 6d and 7s energies are close, and 5f electrons are less tightly held, so more electrons take part in bonding.
• Actinoid contraction is larger than lanthanoid contraction because 5f electrons shield even more poorly.
• The elements are very reactive, and their ions are generally coloured.

Point	Lanthanoids	Actinoids
Filling orbital	4f	5f
Common oxidation state	+3 (also +2, +4)	+3 with +4, +5, +6, +7
Radioactivity	Only Pm	All
4f/5f electrons in bonding	Rarely	Frequently
Contraction	Lanthanoid contraction	Greater actinoid contraction

NEET traps

• 1. Zn, Cd and Hg are d-block but not transition elements (d¹⁰ in atom and in +2 ion). Sc is counted as a transition element because its atom has 3d¹.
• 2. Cr is 3d⁵ 4s¹ and Cu is 3d¹⁰ 4s¹; Mn is 3d⁵ 4s² (no exception).
• 3. In ions, 4s electrons leave before 3d electrons.
• 4. E°(Cu²⁺/Cu) is positive; Mn and Zn are more negative than expected because of d⁵ and d¹⁰.
• 5. Co³⁺/Co²⁺ (+1.97 V) is the highest; Mn³⁺/Mn²⁺ (+1.57 V) is also high. Cr³⁺/Cr²⁺ is negative.
• 6. Chromate to dichromate in acid is not redox (Cr stays +6).
• 7. KMnO₄ gives Mn²⁺ in acid, MnO₂ in neutral medium, MnO₄²⁻ in strong alkali.
• 8. Highest oxidation state of Mn is +7 (Mn₂O₇ is acidic); after Mn the highest state falls, so Fe, Co, Ni and Cu do not reach their group number.
• 9. The 4d and 5d size similarity is due to lanthanoid contraction, not to the d orbitals themselves.
• 10. Ce⁴⁺ is an oxidising agent; Eu²⁺ and Yb²⁺ are reducing agents.

One-liners

• 1. The 3d series has 10 elements, Sc to Zn.
• 2. Cr: 3d⁵ 4s¹ and Cu: 3d¹⁰ 4s¹ are the two exceptions.
• 3. Zn has the lowest enthalpy of atomisation in the 3d series.
• 4. Spin-only moment for 5 unpaired electrons is 5.92 BM.
• 5. Mn shows oxidation states +2 to +7, with +7 the highest.
• 6. Cu⁺ disproportionates in water to Cu²⁺ and Cu.
• 7. TiC, Fe₃C and Fe₃H are interstitial compounds.
• 8. V₂O₅ catalyses the contact process; Fe catalyses the Haber process.
• 9. Chromite is FeCr₂O₄; pyrolusite is MnO₂.
• 10. Orange Cr₂O₇²⁻ turns yellow CrO₄²⁻ in alkali.
• 11. Cr₂O₇²⁻ in acid takes 6 electrons and forms Cr³⁺.
• 12. MnO₄⁻ in acid takes 5 electrons and forms Mn²⁺.
• 13. Lanthanoids are Ce to Lu; most common state is +3.
• 14. Misch metal is mainly lanthanoids with about 5% iron.
• 15. All actinoids are radioactive; among lanthanoids only Pm is.

Practice questions

1. Assertion (A): In acid solution orange K₂Cr₂O₇ turns green when it oxidises Fe²⁺. Reason (R): Cr₂O₇²⁻ is reduced to green Cr³⁺ in acidic solution.

   1. Both A and R are true and R is the correct explanation of A
   2. Both A and R are true but R is not the correct explanation of A
   3. A is true but R is false
   4. A is false but R is true
   Answer

   A. Both A and R are true and R is the correct explanation of A

   Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O; Cr³⁺ ions are green.

2. Given below are two statements: Statement I: The most common oxidation state of actinoids is +2. Statement II: Actinoids show a wider range of oxidation states than lanthanoids. Choose the correct answer.

   1. Both Statement I and Statement II are correct
   2. Both Statement I and Statement II are incorrect
   3. Statement I is correct but Statement II is incorrect
   4. Statement I is incorrect but Statement II is correct
   Answer

   D. Statement I is incorrect but Statement II is correct

   The common state is +3; the 5f, 6d and 7s energies are close, so more oxidation states occur.

3. Match the catalyst in List I with its process in List II. List I: a. V₂O₅ b. Fe c. Ni d. MnO₂. List II: i. Haber process ii. Hydrogenation of oils iii. Contact process iv. Decomposition of KClO₃.

   1. a-i, b-iii, c-ii, d-iv
   2. a-iii, b-i, c-ii, d-iv
   3. a-iii, b-ii, c-i, d-iv
   4. a-iv, b-i, c-ii, d-iii
   Answer

   B. a-iii, b-i, c-ii, d-iv

   V₂O₅ contact process, Fe Haber process, Ni hydrogenation, MnO₂ KClO₃ decomposition.

4. What is the spin-only magnetic moment of the green MnO₄²⁻ ion?

   1. 0 BM
   2. 3.87 BM
   3. 1.73 BM
   4. 2.83 BM
   Answer

   C. 1.73 BM

   Mn is +6 in MnO₄²⁻, so Mn⁶⁺ is 3d¹ with n = 1; μ = √3 = 1.73 BM (MnO₄⁻ with Mn⁷⁺, d⁰, is diamagnetic).

5. What is the ground-state electronic configuration of chromium (Z = 24)?

   1. [Ar] 3d⁴ 4s²
   2. [Ar] 3d³ 4s² 4p¹
   3. [Ar] 3d⁵ 4s¹
   4. [Ar] 3d⁶ 4s⁰
   Answer

   C. [Ar] 3d⁵ 4s¹

   Half-filled d⁵ gives extra stability, so one 4s electron shifts to 3d.

6. Match the lanthanoid species in List I with the property in List II. List I: a. Ce⁴⁺ b. Eu²⁺ c. Lu³⁺ d. Pm. List II: i. Radioactive element ii. Colourless ion iii. Oxidising agent iv. Reducing agent.

   1. a-iv, b-iii, c-ii, d-i
   2. a-iii, b-iv, c-ii, d-i
   3. a-ii, b-iv, c-iii, d-i
   4. a-iii, b-iv, c-i, d-ii
   Answer

   B. a-iii, b-iv, c-ii, d-i

   Ce⁴⁺ oxidising, Eu²⁺ reducing, Lu³⁺ (4f¹⁴) colourless, Pm the radioactive lanthanoid.

7. In acidic solution, how many moles of Fe²⁺ are oxidised by 0.02 mol of KMnO₄?

   1. 0.02 mol
   2. 0.20 mol
   3. 0.06 mol
   4. 0.10 mol
   Answer

   D. 0.10 mol

   MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, so each MnO₄⁻ oxidises 5 Fe²⁺: 0.02 × 5 = 0.10 mol.

8. Assertion (A): Cu²⁺ compounds are paramagnetic. Reason (R): Aqueous Cu²⁺ salts are blue.

   1. Both A and R are true and R is the correct explanation of A
   2. Both A and R are true but R is not the correct explanation of A
   3. A is true but R is false
   4. A is false but R is true
   Answer

   B. Both A and R are true but R is not the correct explanation of A

   Cu²⁺ (3d⁹) has one unpaired electron; the blue colour comes from d–d transition and does not explain paramagnetism.

9. A metal ion M²⁺ has the configuration 3d⁵. What are the atomic number of M and the spin-only magnetic moment of the ion?

   1. 25 and 5.92 BM
   2. 26 and 4.90 BM
   3. 24 and 3.87 BM
   4. 27 and 5.92 BM
   Answer

   A. 25 and 5.92 BM

   M has 5 + 2 = 7 electrons beyond argon (3d⁵ 4s²), so Z = 25 (Mn); n = 5, μ = √35 = 5.92 BM.

10. Given below are two statements: Statement I: In acidic solution MnO₄⁻ is reduced to Mn²⁺. Statement II: In neutral solution MnO₄⁻ is reduced to MnO₂. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Acidic: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Neutral: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻.

11. Given below are two statements: Statement I: Zn, Cd and Hg are regarded as typical transition elements. Statement II: The atomic radii increase steadily from Sc to Zn. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    They have d¹⁰ in atom and ion, so they are not typical transition elements. Radii decrease, stay nearly constant, then rise only slightly at Cu and Zn.

12. Which of the following statements are correct? 1. The spin-only moment of free Fe²⁺ is 4.90 BM. 2. The spin-only moment of Cu²⁺ is 1.73 BM. 3. Zn²⁺ is paramagnetic with two unpaired electrons.

    1. 1, 2 and 3
    2. 1 and 2 only
    3. 2 and 3 only
    4. 1 and 3 only
    Answer

    B. 1 and 2 only

    Fe²⁺ (d⁶, n = 4) gives √24 = 4.90 BM; Cu²⁺ (d⁹, n = 1) gives √3 = 1.73 BM; Zn²⁺ (d¹⁰) is diamagnetic.

13. 20 mL of 0.05 M KMnO₄ is just decolourised in acid by a solution of sodium oxalate. How many moles of oxalate are present?

    1. 4.0 × 10⁻⁴ mol
    2. 1.0 × 10⁻³ mol
    3. 5.0 × 10⁻³ mol
    4. 2.5 × 10⁻³ mol
    Answer

    D. 2.5 × 10⁻³ mol

    MnO₄⁻ = 0.020 × 0.05 = 1.0 × 10⁻³ mol. 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → ... so oxalate = 1.0 × 10⁻³ × 5/2 = 2.5 × 10⁻³ mol.

14. Which of the following is NOT regarded as a transition element?

    1. Ni
    2. Mn
    3. Zn
    4. Cu
    Answer

    C. Zn

    Zn has 3d¹⁰ in the atom and in Zn²⁺, so it has no partly filled d subshell. Cu²⁺ is 3d⁹.

15. Which of the following lanthanoid ions is a strong oxidising agent?

    1. Yb²⁺
    2. Eu²⁺
    3. Ce⁴⁺
    4. La³⁺
    Answer

    C. Ce⁴⁺

    Ce⁴⁺ goes to the more common Ce³⁺; Eu²⁺ and Yb²⁺ are reducing agents; La³⁺ is already the stable +3 state.

16. Given below are two statements: Statement I: Lanthanoid contraction is the steady decrease in ionic radii from La³⁺ to Lu³⁺. Statement II: The basic strength of Ln(OH)₃ increases from La to Lu. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Basic strength decreases from La(OH)₃ to Lu(OH)₃ as the ions get smaller and covalent character rises.

17. Green K₂MnO₄ is acidified and a purple solution with a brown precipitate forms. What is the brown precipitate?

    1. MnO
    2. Mn(OH)₂
    3. Mn₂O₃
    4. MnO₂
    Answer

    D. MnO₂

    3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O; MnO₂ is brown.

18. How many moles of Fe²⁺ are oxidised to Fe³⁺ by one mole of Cr₂O₇²⁻ in acidic solution?

    1. 3
    2. 5
    3. 6
    4. 2
    Answer

    C. 6

    Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O; Cr(VI) to Cr(III) takes 2 × 3 = 6 electrons.

19. Match the step in List I with the product in List II. List I: a. Fusion of FeCr₂O₄ with Na₂CO₃ in air b. Acidification of sodium chromate c. Fusion of MnO₂ with KOH in air d. Acidification of manganate. List II: i. Na₂Cr₂O₇ ii. Na₂CrO₄ iii. K₂MnO₄ iv. KMnO₄ and MnO₂.

    1. a-i, b-ii, c-iii, d-iv
    2. a-ii, b-i, c-iv, d-iii
    3. a-iii, b-i, c-ii, d-iv
    4. a-ii, b-i, c-iii, d-iv
    Answer

    D. a-ii, b-i, c-iii, d-iv

    Chromite gives Na₂CrO₄, which gives Na₂Cr₂O₇ with acid; MnO₂ gives K₂MnO₄, which disproportionates in acid to MnO₄⁻ and MnO₂.

20. Which is the correct increasing order of spin-only magnetic moment of Ti³⁺, Ni²⁺, Cr³⁺ and Fe³⁺?

    1. Ti³⁺ < Ni²⁺ < Cr³⁺ < Fe³⁺
    2. Ti³⁺ < Cr³⁺ < Ni²⁺ < Fe³⁺
    3. Fe³⁺ < Cr³⁺ < Ni²⁺ < Ti³⁺
    4. Ni²⁺ < Ti³⁺ < Cr³⁺ < Fe³⁺
    Answer

    A. Ti³⁺ < Ni²⁺ < Cr³⁺ < Fe³⁺

    Unpaired electrons: Ti³⁺ 1 (1.73), Ni²⁺ 2 (2.83), Cr³⁺ 3 (3.87), Fe³⁺ 5 (5.92 BM).

21. Yellow sodium chromate solution turns orange on adding dilute H₂SO₄ and turns yellow again on adding NaOH. Which statement is correct?

    1. Cr(VI) is reduced to Cr(III) in acid and re-oxidised by alkali
    2. Cr(VI) is reduced to Cr(II) in acid
    3. The colour change is due to a change in the oxidation state of oxygen
    4. Chromium stays in the +6 state; only the pH-dependent equilibrium CrO₄²⁻ ⇌ Cr₂O₇²⁻ shifts
    Answer

    D. Chromium stays in the +6 state; only the pH-dependent equilibrium CrO₄²⁻ ⇌ Cr₂O₇²⁻ shifts

    2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O; both ions have Cr(VI), so no redox change takes place.

22. Match the medium for KMnO₄ in List I with the manganese product in List II. List I: a. Acidic b. Neutral c. Strongly alkaline d. Heated at 513 K. List II: i. MnO₂ ii. Mn²⁺ iii. MnO₄²⁻ iv. K₂MnO₄ + MnO₂ + O₂.

    1. a-ii, b-i, c-iv, d-iii
    2. a-ii, b-i, c-iii, d-iv
    3. a-i, b-ii, c-iii, d-iv
    4. a-ii, b-iii, c-i, d-iv
    Answer

    B. a-ii, b-i, c-iii, d-iv

    Acidic gives Mn²⁺, neutral MnO₂, strongly alkaline MnO₄²⁻; heating gives K₂MnO₄, MnO₂ and O₂.

23. Assertion (A): Copper does not liberate hydrogen from dilute hydrochloric acid. Reason (R): The standard electrode potential E°(Cu²⁺/Cu) is positive.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    A positive E° means Cu is a poorer reducing agent than H₂, so it cannot reduce H⁺.

24. Which 3d-series metal has a positive standard electrode potential E°(M²⁺/M)?

    1. Fe
    2. Cu (+0.34 V)
    3. Ni
    4. Zn
    Answer

    B. Cu (+0.34 V)

    Cu is the only one with a positive value; the others are negative (Fe –0.44, Ni –0.25, Zn –0.76 V).

25. Which 3d-series element shows the highest oxidation state?

    1. Fe (+6)
    2. V (+5)
    3. Cr (+6)
    4. Mn (+7)
    Answer

    D. Mn (+7)

    Mn has seven electrons in 3d and 4s, giving +7 in MnO₄⁻ and Mn₂O₇.

26. How many unpaired electrons are present in the Fe²⁺ ion (Z = 26)?

    1. 2
    2. 5
    3. 4
    4. 6
    Answer

    C. 4

    Fe²⁺ is 3d⁶: one orbital holds a pair and four orbitals hold one electron each, so n = 4.

27. Which product is formed when chromite ore is fused with sodium carbonate in excess air?

    1. Na₂CrO₄
    2. K₂CrO₄
    3. Na₂Cr₂O₇
    4. Cr₂O₃
    Answer

    A. Na₂CrO₄

    4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂; the dichromate forms only after acidification.

28. What is the spin-only magnetic moment of Mn²⁺ (Z = 25)?

    1. 4.90 BM
    2. 2.83 BM
    3. 5.92 BM
    4. 3.87 BM
    Answer

    C. 5.92 BM

    Mn²⁺ is 3d⁵, n = 5; μ = √(5×7) = √35 = 5.92 BM.

29. Given below are two statements: Statement I: Adding an alkali to orange dichromate solution gives orange chromate. Statement II: The conversion of dichromate to chromate is a redox reaction in which Cr(VI) changes to Cr(III). Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Alkali gives yellow chromate, not orange; chromium stays +6, so no redox change occurs.

30. Assertion (A): Eu²⁺ is a strong oxidising agent. Reason (R): Eu²⁺ has a stable half-filled 4f⁷ configuration.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    Eu²⁺ readily changes to Eu³⁺, so it is a reducing agent; Eu²⁺ is 4f⁷, which is true.

31. Given below are two statements: Statement I: Chromium has the configuration 3d⁵ 4s¹ and copper has the configuration 3d¹⁰ 4s¹. Statement II: Both are exceptions to the regular filling order because half-filled and completely filled d subshells are extra stable. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both statements are true; Cr (d⁵) and Cu (d¹⁰) take one electron from 4s.

32. Given below are two statements: Statement I: Cu⁺ is stable in aqueous solution. Statement II: Cu⁺ in water undergoes disproportionation: 2Cu⁺ → Cu²⁺ + Cu. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The high hydration enthalpy of Cu²⁺ makes Cu⁺ unstable in water, so Statement I is wrong.

33. Match List I (spin-only moment) with List II (ion). List I: a. 1.73 BM b. 2.83 BM c. 3.87 BM d. 5.92 BM. List II: i. Mn²⁺ ii. Cr³⁺ iii. Ni²⁺ iv. Cu²⁺.

    1. a-iii, b-iv, c-ii, d-i
    2. a-i, b-iii, c-ii, d-iv
    3. a-iv, b-iii, c-ii, d-i
    4. a-iv, b-ii, c-iii, d-i
    Answer

    C. a-iv, b-iii, c-ii, d-i

    Cu²⁺ n = 1 (1.73), Ni²⁺ n = 2 (2.83), Cr³⁺ n = 3 (3.87), Mn²⁺ n = 5 (5.92 BM).

34. In acid, 50 mL of 0.1 M K₂Cr₂O₇ is used to oxidise Fe²⁺ completely. How many moles of Fe²⁺ are oxidised?

    1. 0.005 mol
    2. 0.03 mol
    3. 0.025 mol
    4. 0.015 mol
    Answer

    B. 0.03 mol

    Cr₂O₇²⁻ = 0.050 L × 0.1 M = 0.005 mol; each takes 6 electrons so oxidises 6 Fe²⁺: 0.005 × 6 = 0.03 mol.

35. A 3d-series metal ion in the +3 state has a spin-only magnetic moment of 3.87 BM. The ion is

    1. Cr³⁺
    2. Ti³⁺
    3. V³⁺
    4. Fe³⁺
    Answer

    A. Cr³⁺

    μ = 3.87 BM means n = 3. Cr³⁺ is d³ (n = 3); V³⁺ d² (2.83), Fe³⁺ d⁵ (5.92), Ti³⁺ d¹ (1.73).

36. Which of the following is the only radioactive lanthanoid?

    1. Ce
    2. Pm
    3. Gd
    4. Lu
    Answer

    B. Pm

    Promethium (Z = 61) is radioactive; all other lanthanoids are non-radioactive.

37. Given below are two statements: Statement I: Mn²⁺ is particularly stable because it has a half-filled d⁵ configuration. Statement II: The value of E°(Mn³⁺/Mn²⁺) is a large negative number, about –1.57 V. Choose the correct answer.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    The value is +1.57 V (positive): Mn³⁺ is a strong oxidising agent because Mn²⁺ is stable.

38. The nearly equal atomic radii of Zr and Hf are due to poor shielding by the electrons of which subshell?

    1. 4f
    2. 4d
    3. 3d
    4. 5d
    Answer

    A. 4f

    The 4f electrons filled before 5d shield poorly, so the lanthanoid contraction offsets the usual increase in size.

39. What is the spin-only magnetic moment of the free Co²⁺ ion (Z = 27)?

    1. 1.73 BM
    2. 2.83 BM
    3. 4.90 BM
    4. 3.87 BM
    Answer

    D. 3.87 BM

    Co²⁺ is 3d⁷ with 3 unpaired electrons; μ = √(3×5) = √15 = 3.87 BM.

40. Assertion (A): Transition metals and their compounds are good catalysts. Reason (R): Transition metals have completely filled d orbitals.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    They are good catalysts because of variable oxidation states and available d orbitals; their d orbitals are partly filled.

41. Match List I with List II. List I: a. Ti³⁺ b. Cr³⁺ c. Cu²⁺ d. Mn²⁺. List II: i. Blue ii. Pink iii. Violet iv. Purple.

    1. a-ii, b-iii, c-i, d-iv
    2. a-iv, b-iii, c-ii, d-i
    3. a-iii, b-iv, c-i, d-ii
    4. a-iv, b-iii, c-i, d-ii
    Answer

    D. a-iv, b-iii, c-i, d-ii

    Ti³⁺ purple, Cr³⁺ violet, Cu²⁺ blue, Mn²⁺ pink.

42. Assertion (A): Zinc is not regarded as a transition element. Reason (R): Zinc has a low melting point.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Zn is not a transition element because its d subshell is full (d¹⁰) in atom and ion; the low melting point is true but is not the reason.

43. The oxidation state of manganese in the green potassium manganate, K₂MnO₄, is

    1. +6
    2. +4
    3. +7
    4. +5
    Answer

    A. +6

    2(+1) + x + 4(–2) = 0, so x = +6.

44. How many electrons does one MnO₄⁻ ion gain when it is reduced in neutral solution?

    1. 1
    2. 5
    3. 6
    4. 3
    Answer

    D. 3

    MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (Mn +7 to +4).