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Chapter 4 · Class 12 Chemistry
d- and f-Block Elements
1 exercises10 questions solved
Exercise 4.1Transition Elements — Properties and Compounds
Q1
Write down the electronic configuration of: (i) Cr³⁺ (ii) Cu⁺ (iii) Co²⁺ (iv) Mn²⁺ (v) Pm³⁺ (vi) Ce⁴⁺ (vii) Lu²⁺ (viii) Th⁴⁺
Solution
Electronic configurations (using [Ar] = 1s²2s²2p⁶3s²3p⁶, [Kr] = ...4s²4p⁶, [Xe] = ...5s²5p⁶):
(i) Cr: [Ar]3d⁵4s¹ → Cr³⁺: [Ar]3d³
(ii) Cu: [Ar]3d¹⁰4s¹ → Cu⁺: [Ar]3d¹⁰
(iii) Co: [Ar]3d⁷4s² → Co²⁺: [Ar]3d⁷
(iv) Mn: [Ar]3d⁵4s² → Mn²⁺: [Ar]3d⁵
(v) Pm: [Xe]4f⁵6s² → Pm³⁺: [Xe]4f⁴
(vi) Ce: [Xe]4f¹5d¹6s² → Ce⁴⁺: [Xe]
(vii) Lu: [Xe]4f¹⁴5d¹6s² → Lu²⁺: [Xe]4f¹⁴5d¹ (or [Xe]4f¹⁴ if 5d electron removed)
(viii) Th: [Rn]6d²7s² → Th⁴⁺: [Rn]
Q2
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state?
Solution
Mn²⁺ configuration: [Ar]3d⁵ (half-filled d subshell)
Fe²⁺ configuration: [Ar]3d⁶
Mn²⁺ has extra stability due to:
• Half-filled 3d subshell (3d⁵) — all five d orbitals singly occupied
• Exchange energy is maximum for half-filled configuration
• High exchange energy → greater stability → resistance to oxidation
Oxidation of Mn²⁺ to Mn³⁺ would give 3d⁴ (less stable)
Oxidation of Fe²⁺ to Fe³⁺ gives 3d⁵ (half-filled, stable)
Therefore Fe²⁺ is more readily oxidised to Fe³⁺ (because Fe³⁺ achieves the stable half-filled d⁵ configuration), while Mn²⁺ resists oxidation to Mn³⁺.
Mn²⁺ is more stable than Fe²⁺ towards oxidation.
Q3
Explain briefly how +2 state becomes more and more stable in the first row transition elements with increasing atomic number.
Solution
In 3d transition series (Sc to Zn), the +2 oxidation state becomes progressively more stable as atomic number increases because:
1. As we move along the series, nuclear charge increases and 3d electrons are added
2. The 3d electrons are increasingly poor at shielding each other
3. Both 4s electrons are removed for +2 state; 3d electrons remain
4. The ionisation energies for removing the third electron (to reach +3 state) increase significantly from left to right
5. Particularly at d⁵ (Mn²⁺) and d¹⁰ (Zn²⁺), extra stability makes +2 the dominant state
For elements like Zn (d¹⁰), Cu (d¹⁰ after ionisation), the +2 state is stable because:
• Removing a third electron would disrupt stable d configurations
• IE₃ values are very high for the later elements
Thus, from Mn onwards, +2 state becomes increasingly the common stable state.
Q4
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate with examples.
Solution
Electronic configurations play a major role in determining stability of oxidation states:
1. Cr shows +3 most commonly (d³: half-filled t₂g in octahedral field)
Cr²⁺ (d⁴) is less stable; Cr⁶⁺ (d⁰) is stable in CrO₄²⁻
2. Mn shows +2 most commonly (d⁵: half-filled d subshell)
KMnO₄ (Mn⁷⁺, d⁰) also stable; MnO₂ (Mn⁴⁺) is stable intermediate
3. Fe²⁺ (d⁶) easily oxidised to Fe³⁺ (d⁵, half-filled)
Fe³⁺ is more stable than Fe²⁺
4. Cu²⁺ (d⁹) is more common than Cu⁺ (d¹⁰) in aqueous solution
Despite d¹⁰ stability, hydration energy favours Cu²⁺
5. Zn²⁺ (d¹⁰) is the only stable oxidation state
Full d subshell provides stability
Conclusion: Extra stability at d⁰, d⁵, and d¹⁰ configurations strongly influences which oxidation states are most stable.
Q5
What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d³, 3d⁵, 3d⁸ and 3d⁴?
Solution
3d³ → This corresponds to V (vanadium): [Ar]3d³4s²
Stable oxidation state: +5 (d⁰), +4 (d¹), +3 (d²), +2 (d³)
Most stable: +5 (VO₄³⁻, VO₂⁺) and +4
Actually for d³ ground state configuration, the ion with 3d³ is V²⁺ or Cr³⁺
Cr³⁺ (3d³) is highly stable — most stable oxidation state for Cr is +3
3d⁵ → Mn (3d⁵4s²); Mn²⁺ has 3d⁵
Most stable oxidation state: +2 (due to half-filled d⁵ stability)
3d⁸ → Ni (3d⁸4s²); Ni²⁺ has 3d⁸
Most stable oxidation state: +2
3d⁴ → Cr (3d⁵4s¹ actual), but for 3d⁴: would be Cr²⁺ or Mn³⁺ (d⁴ is less stable)
For an atom with 3d⁴ ground state → most stable oxidation state: +3 (giving d³ half-filled t₂g)
Q6
Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Solution
Oxometal anions where oxidation state = group number:
1. Chromate: CrO₄²⁻ — Cr is in Group 6, oxidation state = +6
(Also dichromate: Cr₂O₇²⁻, same oxidation state +6)
2. Permanganate: MnO₄⁻ — Mn is in Group 7, oxidation state = +7
3. Vanadate: VO₄³⁻ — V is in Group 5, oxidation state = +5
(also VO₂⁺ where V = +5)
These are the main oxometal anions where the transition metal's oxidation state equals its group number.
Q7
What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.
Solution
Disproportionation is a redox reaction in which a single substance simultaneously undergoes both oxidation and reduction, producing two different products in different oxidation states.
In other words, one species acts both as an oxidising agent and a reducing agent.
Examples in aqueous solution:
1. Copper(I) ion disproportionates:
2Cu⁺(aq) → Cu(s) + Cu²⁺(aq)
Cu⁺ → Cu (reduction, +1 to 0)
Cu⁺ → Cu²⁺ (oxidation, +1 to +2)
2. Manganese(VI) disproportionates in neutral/acidic conditions:
3MnO₄²⁻(aq) + 4H⁺(aq) → 2MnO₄⁻(aq) + MnO₂(s) + 2H₂O(l)
Mn(+6) → Mn(+7) (oxidation) and Mn(+6) → Mn(+4) (reduction)
Q8
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Solution
Copper (Cu) exhibits the +1 oxidation state most frequently among the first series transition metals.
Reason:
Electronic configuration of Cu: [Ar]3d¹⁰4s¹
Cu⁺ configuration: [Ar]3d¹⁰ (completely filled d subshell)
The +1 state is stable because:
• Loss of one 4s electron gives the fully filled 3d¹⁰ configuration
• Completely filled d subshell provides extra stability (similar to noble gas-like arrangement)
Note: In aqueous solution, Cu²⁺ is more stable than Cu⁺ due to high hydration enthalpy compensating the higher IE. However, in solid compounds (e.g., CuCl, Cu₂O, CuI), Cu⁺ is stable.
Q9
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO₂ (iii) oxalic acid? Write the ionic equations for the reactions.
Solution
Preparation of KMnO₄:
Step 1 — Fusion with KOH:
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
(Mn is oxidised from +4 to +6; green manganate formed)
Step 2 — Electrolytic oxidation or disproportionation:
In neutral/acidic solution: 3K₂MnO₄ + 2CO₂ → 2KMnO₄ + MnO₂ + 2K₂CO₃
Or electrolytic: MnO₄²⁻ → MnO₄⁻ + e⁻
Deep purple KMnO₄ crystallises on cooling.
Reactions of acidified KMnO₄ (in H₂SO₄):
(i) With Fe²⁺ ions:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
(ii) With SO₂:
2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺
(iii) With oxalic acid (C₂O₄²⁻):
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
Q10
Describe the oxidising action of potassium dichromate and write the ionic equation for its reaction with (i) iodide (ii) iron(II) solution and (iii) H₂S.
Solution
K₂Cr₂O₇ is a strong oxidising agent, especially in acidic medium.
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (E° = +1.33 V)
Cr is reduced from +6 to +3 (orange → green colour change).
(i) With iodide:
Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O
(I⁻ oxidised to I₂)
(ii) With Fe²⁺ solution:
Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(Fe²⁺ oxidised to Fe³⁺)
(iii) With H₂S:
Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O
(H₂S oxidised to S)
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