d- and f-Block Elements — Class 12 Chemistry practice questions

Question 1

A research chemist is studying two coordination compounds formed by manganese in different oxidation states. She observes that aqueous KMnO₄ solution is deep purple, while MnSO₄ solution is nearly colourless. She also notes that when KMnO₄ is added to an acidified FeSO₄ solution, the purple colour disappears. Using your knowledge of d-block chemistry, answer the following:

(a) Account for the deep purple colour of KMnO₄, even though Mn in KMnO₄ is in the +7 oxidation state (i.e., has no d electrons). [2]

(b) Why is MnSO₄ solution nearly colourless? Give one reason. [1]

(c) Write the balanced ionic equation for the reaction of MnO₄⁻ with Fe²⁺ in acidic medium. [1]

Accepted Answer

(a) The deep purple colour of KMnO₄ is NOT due to d–d transitions, since Mn in the +7 state has a d⁰ configuration (no d electrons available for such transitions). Instead, the intense purple colour arises due to charge transfer transitions (metal-to-ligand or ligand-to-metal charge transfer). Electrons are transferred from the oxygen ligands to the Mn(VII) centre under the influence of visible light, absorbing in the visible region and producing the characteristic deep purple/violet colour. Such charge transfer absorptions are much more intense than d–d transitions and do not require the presence of d electrons.

∴ Colour of KMnO₄ is due to charge transfer transitions, not d–d transitions. [2 marks — 1 mark for correctly stating d⁰/no d–d transition; 1 mark for charge transfer explanation]

(b) Mn²⁺ has the electronic configuration [Ar] 3d⁵. The 3d⁵ configuration is half-filled, and d–d transitions in this configuration are spin-forbidden (all five 3d orbitals singly occupied; any transition would require spin flip and violates the spin selection rule). As a result, d–d transitions occur with extremely low probability, and MnSO₄ solution absorbs very little visible light.

∴ Mn²⁺ (3d⁵) is nearly colourless because d–d transitions are spin-forbidden due to the half-filled 3d⁵ configuration. [1 mark]

(c) In acidic medium, MnO₄⁻ is reduced to Mn²⁺.

Reduction half-reaction:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation half-reaction:
Fe²⁺ → Fe³⁺ + e⁻

Multiplying oxidation half by 5 and adding:

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

∴ Balanced ionic equation: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O [1 mark — equation must be balanced for both mass and charge]

Question 2

A research chemist is studying the oxidising behaviour of permanganate ions in three different reaction media. She observes that when KMnO₄ is added to separate flasks containing acidified FeSO₄ solution, neutral FeSO₄ solution, and alkaline FeSO₄ solution, the colour changes and products formed are distinctly different in each flask.

(a) Identify the manganese-containing product formed when KMnO₄ reacts with FeSO₄ in (i) acidic medium and (ii) alkaline medium. Give the oxidation state of Mn in each product. (2 marks)

(b) Write the balanced ionic equation for the reaction of MnO₄⁻ with Fe²⁺ in acidic medium. (1 mark)

(c) The chemist notes that KMnO₄ acts as a self-indicator in acidic medium but not in alkaline medium. Justify this observation. (1 mark)

Accepted Answer

(a)
(i) Acidic medium:
Product: Mn²⁺ (as MnSO₄ / manganous ion — colourless solution).
Oxidation state of Mn in Mn²⁺ = +2.

(ii) Alkaline medium:
Product: MnO₂ (brown/black precipitate).
Oxidation state of Mn in MnO₂ = +4.

(½ mark for each correct product + ½ mark for each correct oxidation state = 2 marks)

(b) Balanced ionic equation for MnO₄⁻ reacting with Fe²⁺ in acidic medium:

Reduction half-reaction:
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Oxidation half-reaction:
Fe²⁺ → Fe³⁺ + e⁻   (× 5)

Net balanced ionic equation:
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

(1 mark for the fully balanced ionic equation with correct mass and charge balance)

(c) In acidic medium, MnO₄⁻ is reduced to Mn²⁺, which is nearly colourless (very pale pink). The intense purple/violet colour of KMnO₄ disappears sharply at the equivalence point — the first permanent pale pink colour marks the end-point. Thus KMnO₄ acts as its own (self) indicator in acidic medium.

In alkaline medium, MnO₄⁻ is reduced to MnO₂ (brown precipitate) rather than the colourless Mn²⁺. The brown precipitate persists throughout the reaction and masks any colour change, so a clear end-point cannot be observed. Hence KMnO₄ cannot serve as a self-indicator in alkaline medium.

(1 mark for the correct justification covering: Mn²⁺ colourless in acid → clear colour change / MnO₂ brown precipitate in alkali → no clear end-point)

Question 3

A research chemist is analysing two unknown transition metal complexes, X and Y, separated from an industrial effluent sample. The following observations are recorded:

(i) Complex X is intensely purple-coloured in aqueous solution and decolourises acidified potassium permanganate solution rapidly upon addition of excess oxalic acid.

(ii) Complex Y contains a metal ion M²⁺ with the electronic configuration [Ar] 3d⁵. When Y is dissolved in water, the resulting solution shows NO visible colour despite M²⁺ being a d-block ion.

(iii) A third ion, Z³⁺, is known to have a higher second ionisation enthalpy than its neighbours in Period 4 of the d-block, yet Z³⁺ is one of the most stable +3 ions among all first-row transition metals.

Based on the above observations, answer the following:

(a) Identify the metal ion responsible for the intense purple colour in Complex X. Write the balanced ionic equation for its reaction with oxalic acid (H₂C₂O₄) in acidic medium. (2 marks)

(b) Identify metal ion M²⁺ in Complex Y and explain why its aqueous solution is colourless, even though it belongs to the d-block. (1 mark)

(c) Identify ion Z³⁺ and give ONE reason for its exceptional stability. (1 mark)

Accepted Answer

(a) The intense purple colour of Complex X is due to the permanganate ion, MnO₄⁻ (Mn in +7 oxidation state). The metal ion responsible for the purple colour is Mn⁷⁺ (present as MnO₄⁻).

In acidic medium, MnO₄⁻ oxidises oxalic acid (H₂C₂O₄) and is itself reduced to the colourless Mn²⁺ ion.

Balanced ionic equation:

2MnO₄⁻(aq) + 5H₂C₂O₄(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 10CO₂(g)↑ + 8H₂O(l)

∴ The decolourisation confirms reduction of MnO₄⁻ (purple) → Mn²⁺ (colourless).

[Value points: correct identification of MnO₄⁻/Mn⁷⁺ — ½ mark; balanced ionic equation with correct products, charges and state symbols — 1½ marks]

(b) The electronic configuration [Ar] 3d⁵ with a charge of +2 corresponds to Mn²⁺ (atomic number of Mn = 25; ground state [Ar] 3d⁵ 4s²; removing 2 electrons gives [Ar] 3d⁵).

Despite belonging to the d-block and having five 3d electrons, Mn²⁺ is colourless in aqueous solution because its 3d subshell is exactly half-filled (3d⁵, all five orbitals singly occupied). Any d–d electronic transition within this configuration would require an electron to move to an orbital of the same energy (all 3d orbitals are degenerate in a free ion, and in an octahedral field the t₂g³ e_g² arrangement of high-spin Mn²⁺ makes all spin-allowed d–d transitions spin-forbidden). Since no spin-allowed d–d transition is possible, no visible light is absorbed and the solution appears colourless.

∴ M²⁺ = Mn²⁺; colourless because all spin-allowed d–d transitions are forbidden in the half-filled 3d⁵ configuration.

[Value points: correct identification as Mn²⁺ — ½ mark; correct reason (half-filled 3d⁵, all d–d transitions spin-forbidden / no spin-allowed d–d transition) — ½ mark]

(c) Ion Z³⁺ is Fe³⁺ (Iron, atomic number 26).

Fe has a higher second ionisation enthalpy than its neighbours in Period 4 d-block because the removal of the second electron from Fe⁺ would disturb the stable half-filled 3d⁵ configuration of Fe²⁺. However, Fe³⁺ (configuration [Ar] 3d⁵) is exceptionally stable precisely because it achieves the half-filled 3d⁵ configuration, which has extra stability due to exchange energy.

Reason for exceptional stability: Fe³⁺ has the electronic configuration [Ar] 3d⁵ — a exactly half-filled d subshell — which confers maximum exchange energy and high symmetry, making Fe³⁺ one of the most stable +3 ions among first-row transition metals.

∴ Z³⁺ = Fe³⁺; stable due to the extra stability of the half-filled 3d⁵ configuration (maximum exchange energy).

[Value points: correct identification as Fe³⁺ — ½ mark; correct reason (half-filled 3d⁵ / exchange energy / symmetrical configuration) — ½ mark]

Question 4

A school chemistry club is investigating transition metals for use in industrial catalysts and battery electrodes. They observe the following facts about chromium and manganese compounds:

• Fact 1: A deep green solution turns purple on acidification.
• Fact 2: A purple solution decolourises when excess iron(II) sulphate solution is added in acidic medium.
• Fact 3: Manganese shows a higher number of oxidation states than chromium in its compounds.

Based on the above observations, answer the following:
(a) Identify the ionic species responsible for the deep green colour and the purple colour mentioned in Fact 1. Write the balanced ionic equation for the conversion that occurs on acidification. [2]
(b) Write the balanced ionic equation for the reaction described in Fact 2, showing the change in colour. [1]
(c) Justify Fact 3: Why does manganese exhibit more oxidation states than chromium? (Atomic numbers: Cr = 24, Mn = 25) [1]

Accepted Answer

(a) The deep green species is manganate ion, MnO₄²⁻ (Mn in +6 state), and the purple species is permanganate ion, MnO₄⁻ (Mn in +7 state).

On acidification, manganate undergoes disproportionation (Mn moves from +6 to both +7 and +4):

3MnO₄²⁻(aq) + 4H⁺(aq) → 2MnO₄⁻(aq) + MnO₂(s) + 2H₂O(l)

[1 mark for correct identification of both species; 1 mark for the balanced ionic equation — mass and charge both balanced]

(b) In Fact 2, MnO₄⁻ (purple) oxidises Fe²⁺ to Fe³⁺ in acidic medium and is itself reduced to Mn²⁺ (colourless/very pale pink), causing decolourisation:

MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

[1 mark for the balanced ionic equation with correct products and charge balance]

(c) Electronic configurations:
Cr (Z = 24): [Ar] 3d⁵ 4s¹
Mn (Z = 25): [Ar] 3d⁵ 4s²

Transition metals show variable oxidation states because the energies of the (n−1)d and ns orbitals are close, allowing different numbers of electrons to participate in bonding.

Manganese has one extra electron (4s²) compared to chromium (4s¹), giving it more electrons available for involvement in oxidation states. Mn can therefore exhibit oxidation states from +2 up to +7 (using all five 3d and both 4s electrons progressively), whereas Cr's half-filled 3d⁵ configuration is extra stable, limiting its common higher oxidation states. ∴ Mn displays a wider range of oxidation states (+2, +3, +4, +6, +7) than Cr (+2, +3, +6).

[1 mark for citing the close energy of 3d and 4s orbitals AND correctly linking Mn's electronic configuration to its wider range of oxidation states]

Question 5

A school laboratory has four unlabelled aqueous solutions, each containing one of the following ions: Mn²⁺, Fe³⁺, Cu²⁺, and Zn²⁺. A student is asked to identify each solution using simple observations and chemical tests. Using your knowledge of d-block elements, answer the following:

(a) Which solution will appear colourless? Name the ion responsible and give ONE reason why it is colourless, even though it is a d-block ion. (2 marks)

(b) The student adds excess aqueous NaOH to two of the remaining solutions. One gives a blue precipitate insoluble in excess NaOH, and the other gives a rust-brown precipitate insoluble in excess NaOH. Identify the ion responsible for each observation. (1 mark)

(c) The student adds aqueous KMnO₄ to a dilute H₂SO₄ solution and then bubbles SO₂ gas through it. The purple colour of KMnO₄ is discharged. Write the half-reaction showing the change undergone by MnO₄⁻ in acidic medium. (1 mark)

Accepted Answer

(a) The Zn²⁺ solution will appear colourless.

Reason: Zn²⁺ has the electronic configuration [Ar] 3d¹⁰. Since the 3d subshell is completely filled, no d–d electronic transitions are possible. Absorption of visible light cannot occur, so the ion appears colourless.

(Award 1 mark for correct identification of Zn²⁺ with its configuration / completely filled 3d¹⁰; 1 mark for the reason — no d–d transitions possible.)

(b) Blue precipitate (insoluble in excess NaOH) → Cu²⁺ ion.

Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)  (blue precipitate)

Rust-brown precipitate (insoluble in excess NaOH) → Fe³⁺ ion.

Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s)  (rust-brown precipitate)

(Award ½ mark for each correct identification — Cu²⁺ and Fe³⁺ — total 1 mark. Accept ionic equations as supporting evidence.)

(c) In acidic medium MnO₄⁻ is reduced to Mn²⁺ (colourless).

Reduction half-reaction:

MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)

∴ The purple colour of KMnO₄ is discharged because Mn²⁺ is nearly colourless in aqueous solution.

(Award 1 mark for the correctly balanced half-reaction — mass and charge must both be balanced. Deduct if electrons, H⁺ or H₂O are missing or incorrect.)

Question 6

A metallurgist is studying two industrial catalysts used in different manufacturing processes. Catalyst X is finely divided iron (Fe) used in the Haber process for ammonia synthesis. Catalyst Y is platinum (Pt) used in the Contact process for sulphuric acid manufacture. Both are transition metals.

Based on the above context, answer the following:
(a) Why are transition metals generally good catalysts? Give one reason.  (1 mark)
(b) The metallurgist notes that Fe has a higher melting point than Zn, even though both are in the 4th period. Account for this observation.  (1 mark)
(c) Complete the following ionic equation:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ →  (1 mark)
(d) Out of the two half-reactions given below, identify which one acts as the oxidation half-reaction and which acts as the reduction half-reaction when KMnO₄ oxidises Fe²⁺ in acidic medium. Also state the colour change observed.
(i) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
(ii) Fe²⁺ → Fe³⁺ + e⁻  (1 mark)

Accepted Answer

(a) Due to the availability of vacant d-orbitals, transition metals can provide a suitable surface for adsorption of reactants and can also exhibit variable oxidation states, thereby forming intermediate compounds with reactants. This lowers the activation energy and increases the rate of reaction. / They can act as good catalysts because of their ability to provide a reaction surface and exhibit variable oxidation states.

[Award 1 mark for any one valid reason — variable oxidation states OR surface adsorption / available vacant d-orbitals.]

(b) Due to the presence of a large number of unpaired d-electrons in Fe (electronic configuration: [Ar] 3d⁶ 4s²), it forms strong metallic bonds involving both 3d and 4s electrons. Zn (electronic configuration: [Ar] 3d¹⁰ 4s²) has completely filled 3d orbitals and only 4s electrons participate in metallic bonding. Therefore, metallic bonding is much stronger in Fe than in Zn, resulting in a higher melting point for Fe.

[Award 1 mark for the reason — Zn has completely filled 3d¹⁰ so only 4s electrons contribute to metallic bonding / Fe has more unpaired d-electrons contributing to stronger metallic bonds.]

(c) The balanced ionic equation is:

MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

[Award 1 mark for the fully balanced equation with correct products. Deduct the mark if the equation is unbalanced for mass or charge.]

(d) Reaction (i): MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O is the REDUCTION half-reaction (Mn goes from +7 to +2, gain of electrons).

Reaction (ii): Fe²⁺ → Fe³⁺ + e⁻ is the OXIDATION half-reaction (Fe goes from +2 to +3, loss of electrons).

Colour change observed: The deep purple / violet colour of KMnO₄ (due to MnO₄⁻) is discharged and the solution becomes nearly colourless (Mn²⁺ is almost colourless / very pale pink).

[Award 1 mark: ½ mark for correct identification of both half-reactions AND ½ mark for the colour change — purple/violet to colourless/pale pink.]

Question 7

A researcher in an industrial laboratory is studying the chemistry of transition metal compounds for use in manufacturing processes. She observes the following:

• A deep purple solution of compound A (acidified with H₂SO₄) decolourises completely when excess iron(II) sulphate solution is added.
• An orange solution of compound B, on adding dilute NaOH, gives a yellow solution (compound C).
• Compound D is a black oxide of manganese used widely in dry cells.
• A lanthanoid alloy (compound E), consisting of miscellaneous lanthanoid metals, is used in cigarette lighters.

On the basis of this information, answer the following questions:

(i) Identify compound A and write the balanced ionic equation for its reaction with Fe²⁺ ions in acidic medium. [2]
(ii) Identify compounds B and C. Write the equation for the interconversion and state the condition under which it occurs. [1]
(iii) Give the formula of compound D and state why MnO₂ acts as a catalyst in the decomposition of KClO₃. [1]

Accepted Answer

Marking Scheme:
(i) Compound A — 1 mark (identification + balanced ionic equation)
(ii) Compounds B and C — 1 mark (identification + equation + condition)
(iii) Compound D — 1 mark (formula + reason for catalytic action)
(Bonus note — Compound E = Misch metal; not separately marked here but may be asked as extension)

─────────────────────────────────────
(i) Compound A = KMnO₄ (potassium permanganate) — deep purple, acidic medium.

In acidic medium, MnO₄⁻ is reduced to Mn²⁺ (colourless), while Fe²⁺ is oxidised to Fe³⁺.

Reduction half-reaction:
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Oxidation half-reaction:
Fe²⁺ → Fe³⁺ + e⁻  (×5)

Balanced net ionic equation:
MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

(The deep purple colour disappears because Mn²⁺ is nearly colourless.)

─────────────────────────────────────
(ii) Compound B = K₂Cr₂O₇ (potassium dichromate) — orange solution.
      Compound C = K₂CrO₄ (potassium chromate) — yellow solution.

Interconversion:
Cr₂O₇²⁻(aq) + 2OH⁻(aq) → 2CrO₄²⁻(aq) + H₂O(l)

Condition: Addition of dilute NaOH (alkaline medium) converts dichromate (orange) to chromate (yellow).

[Note: The reverse occurs in acidic medium — 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O]

─────────────────────────────────────
(iii) Compound D = MnO₂ (manganese dioxide).

MnO₂ acts as a catalyst in the decomposition of KClO₃ because transition metal oxides such as MnO₂ can exhibit variable oxidation states (+2, +3, +4), allowing them to form intermediate compounds with the reactant. This lowers the activation energy of the reaction and speeds up the decomposition:

2KClO₃(s) →(MnO₂, Δ) 2KCl(s) + 3O₂(g)↑

∴ MnO₂ provides an alternative reaction pathway of lower activation energy due to its ability to change oxidation state during the catalytic cycle.

Question 8

A chemistry teacher demonstrates an experiment to her class. She prepares three beakers, each containing a different solution of a transition metal ion: Beaker 1 has a pale pink solution of Mn²⁺ ions, Beaker 2 has a deep blue solution of Cu²⁺ ions, and Beaker 3 has a colourless solution of Zn²⁺ ions. She tells the class: 'One of these ions will not show d-d transitions, one has a half-filled d subshell, and one has a variable oxidation state that makes it useful in electroplating.'

Based on this scenario, answer the following:
(a) Identify which beaker contains the ion that CANNOT show d-d transitions. Give ONE reason for this. (2 marks)
(b) Which ion has a half-filled d subshell? Write its electronic configuration of the ion. (1 mark)
(c) Name ONE specific use of the ion present in Beaker 2 that is based on its variable oxidation state or electrochemical property. (1 mark)

Accepted Answer

(a) Beaker 3 (Zn²⁺) cannot show d-d transitions.

Zn²⁺ has the electronic configuration [Ar] 3d¹⁰ — a completely filled d subshell. Since all five d orbitals are fully occupied, no d-d transition is possible (there is no vacant d orbital of the same subshell for an electron to jump into). Hence Zn²⁺ solutions are colourless.

[Award 1 mark for correct identification of Beaker 3 / Zn²⁺; 1 mark for the reason — completely filled 3d¹⁰ configuration / no vacant d orbital available for d-d transition]

(b) Beaker 1 — Mn²⁺ — has a half-filled d subshell.

Electronic configuration of Mn²⁺: [Ar] 3d⁵

(Mn: [Ar] 3d⁵ 4s²; loss of two electrons from 4s gives Mn²⁺: [Ar] 3d⁵)

[Award 1 mark for correct identification with configuration]

(c) Cu²⁺ (Beaker 2) is used in electroplating / copper electroplating of metals and articles.

(Cu²⁺/Cu electrode is used as the cathode in electroplating baths; Cu²⁺ is reduced to Cu metal at the cathode, depositing a shiny, corrosion-resistant copper layer on the object.)

[Award 1 mark for any ONE correct use: electroplating / use in electrochemical cells / Daniel cell / copper refining by electrolysis]

Question 9

In which oxidation state is manganese most stable, and why?

Accepted Answer

(A)

Explanation: Mn²⁺ has the electronic configuration [Ar] 3d⁵ — a half-filled d-subshell, which is exceptionally stable due to symmetrical distribution of electrons and maximum exchange energy. This makes +2 the most stable oxidation state of manganese.

Question 10

Which of the following ions is colourless?
(A) Cu²⁺
(B) Zn²⁺
(C) Cr³⁺
(D) Fe³⁺
(Atomic numbers: Cu = 29, Zn = 30, Cr = 24, Fe = 26)

Accepted Answer

(B) Zn²⁺

Explanation: Colour in transition metal ions arises from d–d electronic transitions, which require partially filled d orbitals. Zn²⁺ has the electronic configuration [Ar] 3d¹⁰ — a completely filled d subshell — so no d–d transition is possible and it absorbs no visible light. Hence Zn²⁺ is colourless. In contrast, Cu²⁺ (3d⁹), Cr³⁺ (3d³) and Fe³⁺ (3d⁵) all have partially filled d orbitals and are therefore coloured.

Question 11

Which of the following ions is colourless in aqueous solution?
(Atomic numbers: Sc = 21, Ti = 22, V = 23, Cr = 24)

Accepted Answer

(D) Sc³⁺

Explanation: Colour in transition-metal ions arises from d–d transitions, which require partially filled d-orbitals. Sc³⁺ has the electronic configuration [Ar] 3d⁰; it has NO d-electrons and therefore cannot undergo any d–d transition. As a result, it does not absorb visible light and appears colourless in aqueous solution. The other ions — V³⁺ (3d²), Cr³⁺ (3d³) and Ti³⁺ (3d¹) — all have partially filled d-orbitals and are therefore coloured.

Question 12

Account for the following:
Zn²⁺ compounds are white (colourless), whereas Cu²⁺ compounds are blue.

Accepted Answer

Zn²⁺ has the electronic configuration [Ar] 3d¹⁰ — the 3d sub-shell is completely filled. No d–d transition is possible, so no visible light is absorbed, and Zn²⁺ compounds appear white (colourless).

Cu²⁺ has the electronic configuration [Ar] 3d⁹ — the 3d sub-shell is incompletely filled. d–d transitions are possible; visible light is absorbed, and the complementary colour (blue) is transmitted/reflected, making Cu²⁺ compounds appear blue.

Question 13

Account for the following: Mn²⁺ compounds are more stable than Fe²⁺ compounds towards further oxidation.

Accepted Answer

Due to the extra stability of the half-filled 3d⁵ configuration, Mn²⁺ (3d⁵) is highly resistant to further oxidation to Mn³⁺. Fe²⁺ (3d⁶), on the other hand, can be oxidised to Fe³⁺ (3d⁵), which itself attains the stable half-filled configuration. Hence Fe²⁺ is more easily oxidised than Mn²⁺.

Question 14

Complete and balance the following chemical equations:
(a) MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) →
(b) Cr₂O₇²⁻(aq) + 3Sn²⁺(aq) + 14H⁺(aq) →

Accepted Answer

(a) MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

MnO₄⁻ acts as the oxidising agent; Mn is reduced from +7 to +2 (acidic medium → Mn²⁺, colourless solution). Each Fe²⁺ is oxidised to Fe³⁺. Mass and charge are balanced: charge on left = −1 + 5(2+) + 8(1+) = −1 + 10 + 8 = +17; charge on right = 2+ + 5(3+) = 2 + 15 = +17. ✓

(b) Cr₂O₇²⁻(aq) + 3Sn²⁺(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 3Sn⁴⁺(aq) + 7H₂O(l)

Cr₂O₇²⁻ acts as the oxidising agent; each Cr is reduced from +6 to +3 (gain of 3e⁻ per Cr, 6e⁻ total). Each Sn²⁺ is oxidised to Sn⁴⁺ (loss of 2e⁻ per Sn, 6e⁻ total). Electrons balance (6e⁻ transferred). Charge check: left = −2 + 3(2+) + 14(1+) = −2 + 6 + 14 = +18; right = 2(3+) + 3(4+) = 6 + 12 = +18. ✓

Question 15

Why is Zn²⁺ colourless in its aqueous solution?

Accepted Answer

Due to completely filled 3d¹⁰ configuration, Zn²⁺ has no vacant d-orbital available for d-d transitions, and hence it does not absorb visible light. Therefore, Zn²⁺ appears colourless in aqueous solution.

Question 16

Account for the following:
(a) The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺, although both Mn and Cr are transition metals.
(b) Zn²⁺ salts are white, whereas Cu²⁺ salts are coloured.
(c) The atomic radii of Zr (4d series) and Hf (5d series) are almost identical, even though Hf lies one period below Zr.

Accepted Answer

(a) The E° value for Mn³⁺/Mn²⁺ is highly positive (+1.57 V) compared to Cr³⁺/Cr²⁺ (+0.41 V).
Due to the extra stability of the half-filled 3d⁵ configuration of Mn²⁺ (3d⁵), it strongly resists oxidation to Mn³⁺. Therefore, a high oxidising potential (large positive E°) is required to bring about the Mn²⁺ → Mn³⁺ change, making E°(Mn³⁺/Mn²⁺) unusually high.
In contrast, Cr²⁺ (3d⁴) does not possess any extra half-filled or fully-filled stability, so oxidation to Cr³⁺ (3d³, stable half-filled t₂g³) is relatively easier, giving a much lower E° value.

(b) Colour in transition-metal ions arises from d–d transitions, which require at least one unpaired d-electron and a partially filled d-orbital set.
Zn²⁺ has the configuration 3d¹⁰ — all d-orbitals are completely filled. No d–d transition is possible, so Zn²⁺ salts are white (colourless).
Cu²⁺ has the configuration 3d⁹ — it has one unpaired electron and the d-subshell is partially filled. d–d transitions absorb visible light, imparting a characteristic blue colour to Cu²⁺ salts.

(c) Although Hf belongs to the 5d series and lies one full period below Zr (4d series), the two elements have nearly identical atomic radii (Zr: 160 pm; Hf: 159 pm).
This is due to lanthanoid contraction — the steady decrease in size of the 14 lanthanoid elements (Ce to Lu) that precede Hf in the periodic table. The poor shielding of nuclear charge by 4f electrons causes the outer electrons of each successive lanthanoid to be pulled in more strongly, resulting in a progressive contraction. By the time Hf is reached, this contraction has almost exactly offset the expected increase in size on going from the 4d to the 5d series, so Zr and Hf end up with virtually the same atomic radius and very similar chemical properties.

Question 17

A chemistry student is investigating the properties of two transition metal ions, Cr³⁺ and Cu²⁺, in a laboratory experiment. She notices the following:

(a) An aqueous solution of Cr³⁺ appears violet/green, while Cu²⁺ solution appears blue. However, when she prepares a solution of Zn²⁺, it is colourless. Explain why Zn²⁺ is colourless while Cr³⁺ and Cu²⁺ are coloured. (2 marks)

(b) The student also observes that CrO₄²⁻ (chromate) and Cr₂O₇²⁻ (dichromate) can be inter-converted by changing the pH of the solution. Write the balanced chemical equation for the conversion of CrO₄²⁻ to Cr₂O₇²⁻ in acidic medium, and give the oxidation state of Cr in both ions. (1 mark)

(c) The student reads that KMnO₄ acts as an oxidising agent in acidic, neutral, and alkaline media. Write the balanced ionic equation for the reaction of KMnO₄ with FeSO₄ in acidic medium (H₂SO₄), identifying the colour change observed. (1 mark)

Accepted Answer

(a) Colour in transition metal ions arises due to d–d transitions. When visible light falls on a transition metal ion, electrons in lower-energy d orbitals absorb certain wavelengths and get excited to higher-energy d orbitals; the complementary colour is transmitted and observed.

Cr³⁺ has the electronic configuration [Ar] 3d³ — three unpaired electrons are present. Since the d orbitals are partially filled (d³), d–d transitions are possible, and Cr³⁺ appears coloured (violet/green).

Cu²⁺ has the configuration [Ar] 3d⁹ — one unpaired electron is present. Since the d orbitals are partially filled (d⁹), d–d transitions are possible, and Cu²⁺ appears blue.

Zn²⁺ has the configuration [Ar] 3d¹⁰ — the d orbitals are completely filled. No d–d transitions are possible because there are no vacant d orbitals for electrons to be excited into. ∴ Zn²⁺ is colourless.

(Award 1 mark for correct explanation of d–d transition principle; 1 mark for correctly applying it to Zn²⁺ with 3d¹⁰ configuration.)

(b) In acidic medium, chromate (CrO₄²⁻) is converted to dichromate (Cr₂O₇²⁻):

2 CrO₄²⁻(aq) + 2H⁺(aq) → Cr₂O₇²⁻(aq) + H₂O(l)

Oxidation state of Cr in CrO₄²⁻: Let x + 4(−2) = −2 ∴ x = +6.
Oxidation state of Cr in Cr₂O₇²⁻: Let 2x + 7(−2) = −2 ∴ x = +6.

∴ Cr is in the +6 oxidation state in both CrO₄²⁻ and Cr₂O₇²⁻. (This is not a redox change — it is an acid–base equilibrium.)

(Award 1 mark for the balanced equation with correct charges and water.)

(c) In acidic medium, MnO₄⁻ is reduced to Mn²⁺ (colourless) — this is a 5-electron reduction.

Reduction half-reaction:
MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l)

Oxidation half-reaction:
Fe²⁺(aq) → Fe³⁺(aq) + e⁻  [× 5]

Balanced net ionic equation:
MnO₄⁻(aq) + 5Fe²⁺(aq) + 8H⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)

Colour change observed: The purple/violet colour of KMnO₄ (MnO₄⁻) is discharged — the solution turns nearly colourless (pale pink → colourless) as Mn²⁺ is formed.

(Award 1 mark for the correctly balanced ionic equation with correct charges, stoichiometry, and identification of the colour change.)

Question 18

A chemistry teacher demonstrates the following observations to her class during a laboratory session:

(i) When excess ammonia solution is added to a pale blue aqueous solution of copper(II) sulphate, the solution turns deep blue.
(ii) A student notices that zinc sulphate solution is colourless even though zinc is a d-block element.
(iii) The teacher shows that manganese(II) ion solution is almost colourless/very pale pink, but when KMnO₄ is added to the same solution in acidic medium, the purple colour of KMnO₄ disappears.
(iv) The teacher points out that chromium has the electronic configuration [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s².

Account for each of the above observations.

Accepted Answer

(i) When excess NH₃ is added to Cu²⁺(aq), the water ligands in [Cu(H₂O)₄]²⁺ are replaced by the stronger-field NH₃ ligands to form the deep blue tetraamminecopper(II) complex:

[Cu(H₂O)₄]²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺ + 4H₂O

Due to a change in the crystal-field splitting (Δo) caused by the stronger ligand NH₃, the d-d transition absorbs a different wavelength of visible light, resulting in the observed deep blue colour.  (1 mark)

(ii) Zinc (Zn) has the electronic configuration [Ar] 3d¹⁰ 4s². The Zn²⁺ ion has the configuration 3d¹⁰ (completely filled d orbitals). Since all d orbitals are fully occupied, d-d electronic transitions are not possible. Consequently, Zn²⁺ does not absorb visible light and its solution appears colourless.  (1 mark)

(iii) In acidic medium, MnO₄⁻ (purple) acts as a strong oxidising agent and is reduced to Mn²⁺ (colourless/very pale pink). The balanced ionic equation is:

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

(In the context of the demonstration, MnO₄⁻ oxidises Mn²⁺ or any reducing species present in acidic medium, getting reduced to Mn²⁺, hence the purple colour disappears.)

Or, generally: The purple colour of KMnO₄ disappears in acidic medium because MnO₄⁻ is reduced to the almost colourless Mn²⁺ ion:  (1 mark)

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

(iv) Chromium has the electronic configuration [Ar] 3d⁵ 4s¹ and not [Ar] 3d⁴ 4s² because a half-filled d subshell (3d⁵) has extra stability due to symmetrical distribution of electrons and maximum exchange energy. Therefore, one electron from the 4s orbital is shifted to the 3d orbital to attain this stable half-filled configuration.  (1 mark)

Question 19

Answer the following questions about d-block elements:
(a) Why does Mn²⁺ exhibit greater stability compared to Fe²⁺ in aqueous solution? Give reason.
(b) Write the balanced chemical equation for the reaction of K₂Cr₂O₇ with KI in acidic medium.
(c) The E° (M²⁺/M) value for Cu is positive (+0.34 V) whereas for Zn it is negative (−0.76 V). Explain why Cu does not dissolve in dilute H₂SO₄ but Zn does.

Accepted Answer

(a) Mn²⁺ has the electronic configuration [Ar] 3d⁵ — a half-filled d sub-shell. This configuration is exceptionally stable due to exchange energy and symmetrical electron distribution. Oxidation of Mn²⁺ to Mn³⁺ would disturb this half-filled arrangement, requiring a large amount of energy. Therefore, Mn²⁺ resists further oxidation and is more stable than Fe²⁺, which has the configuration [Ar] 3d⁶ and is readily oxidised to Fe³⁺ ([Ar] 3d⁵), also attaining half-filled stability.

(b) In acidic medium, K₂Cr₂O₇ acts as a strong oxidising agent and oxidises I⁻ to I₂:

K₂Cr₂O₇ + 6KI + 7H₂SO₄ → Cr₂(SO₄)₃ + 4K₂SO₄ + 3I₂ + 7H₂O

Ionic form:
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O

(Cr is reduced from +6 to +3; I is oxidised from −1 to 0.)

(c) A metal dissolves in dilute H₂SO₄ only if it can reduce H⁺ ions, i.e., its E°(M²⁺/M) must be negative (metal is a stronger reducing agent than H₂).

• For Zn: E°(Zn²⁺/Zn) = −0.76 V < 0. ∴ Zn can reduce H⁺ to H₂ and dissolves readily in dilute H₂SO₄:
Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂↑

• For Cu: E°(Cu²⁺/Cu) = +0.34 V > 0. ∴ Cu cannot reduce H⁺ ions (H₂ would be more easily oxidised than Cu). Hence Cu does not dissolve in dilute H₂SO₄.

Question 20

Give reasons for the following:
(i) The standard electrode potential (E°) for the Cu²⁺/Cu couple is positive (+0.34 V), unlike most other transition metals which have negative E° values.
(ii) Mn²⁺ compounds are more stable towards oxidation than Fe²⁺ compounds.
(iii) Zn²⁺ salts are white/colourless, whereas Cu²⁺ salts are blue.

Accepted Answer

(i) Due to the relatively high sum of first and second ionisation enthalpies of copper, its E° would be expected to be negative. However, the exceptionally high enthalpy of atomisation of copper is more than offset by its very high hydration enthalpy of Cu²⁺ ions. The high hydration enthalpy of Cu²⁺ arises from its small ionic size and high charge density, making the net energy change favourable for Cu²⁺/Cu — hence E° is positive (+0.34 V). [1 mark]

(ii) Mn²⁺ has the electronic configuration [Ar] 3d⁵ — a half-filled d-subshell, which is extra-stable due to exchange energy and symmetrical electron distribution. Oxidation of Mn²⁺ to Mn³⁺ would disrupt this stable half-filled configuration, requiring a large amount of energy. Therefore, Mn²⁺ resists oxidation and is more stable compared to Fe²⁺ (configuration [Ar] 3d⁶), which is readily oxidised to Fe³⁺ (configuration [Ar] 3d⁵, attaining the stable half-filled state). [1 mark]

(iii) Zn²⁺ has the electronic configuration [Ar] 3d¹⁰ — a completely filled d-subshell. There are no vacant d-orbitals of appropriate energy available for d–d electronic transitions. Since colour in transition metal ions arises from d–d transitions (absorption of visible light), Zn²⁺ compounds are colourless/white. In contrast, Cu²⁺ has the configuration [Ar] 3d⁹, which has one unpaired electron and a vacancy in the d-subshell. The d–d transition in Cu²⁺ absorbs red light from the visible spectrum, and the complementary colour — blue — is transmitted, imparting a blue colour to Cu²⁺ salts. [1 mark]

d- and f-Block Elements: Class 12 Chemistry Practice Questions

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