Chapter 4 — Transition and Inner Transition Elements Exercise Solutions
Most answers in this chapter need a property plus its reason. Never write only the property.
Key Concepts Quick Recap
| Concept | Key Fact |
|---|
| Transition elements | d-block elements with partially filled d orbitals (Groups 3–12) |
| Variable oxidation states | ns and (n−1)d electrons have similar energies — both can be used in bonding |
| Colour | d-d electronic transitions in partially filled d orbitals absorb visible light |
| Magnetic behaviour | Due to presence of unpaired electrons in d orbitals |
| Catalytic activity | Variable oxidation states allow electron transfer; surface adsorption also helps |
| Complex formation | High charge density (small size + high charge) attracts ligands |
| Lanthanide contraction | Steady decrease in atomic/ionic radii from La to Lu due to poor shielding of 4f electrons |
| Lanthanides | 14 elements: La (57) to Lu (71) — 4f series; all +3 in compounds |
| Actinides | 14 elements: Ac (89) to Lr (103) — 5f series; more variable oxidation states than lanthanides |
| Spin-only formula | μ = √(n(n+2)) BM; n = number of unpaired electrons |
| Anomalous configs | Cr = [Ar]3d⁵4s¹; Cu = [Ar]3d¹⁰4s¹ — half-filled/fully-filled d sub-shell stability |
I. Choose the Best Answer (MCQs)
1. Which of the following is NOT a characteristic of transition metals? (a) Variable oxidation states (b) Coloured compounds (c) Fixed valency (d) Catalytic activity
Answer(c) Fixed valency Transition metals show variable oxidation states — a defining characteristic. Fixed valency is a feature of s-block metals like Na (+1) and Mg (+2).
2. The electronic configuration of Cr (Z=24) is: (a) [Ar]3d⁴4s² (b) [Ar]3d⁵4s¹ (c) [Ar]3d⁶4s⁰ (d) [Ar]3d³4s²
Answer(b) [Ar]3d⁵4s¹ Half-filled d sub-shell (3d⁵) is more stable. One electron from 4s transfers to 3d, giving [Ar]3d⁵4s¹.
3. The spin-only magnetic moment of Fe³⁺ is: (a) 3.87 BM (b) 4.90 BM (c) 5.92 BM (d) 2.83 BM
Answer(c) 5.92 BM Fe³⁺ = [Ar]3d⁵ — 5 unpaired electrons. μ = √(5×7) = √35 ≈ 5.92 BM.
4. Lanthanide contraction is due to: (a) Increase in nuclear charge (b) Poor shielding by 4f electrons (c) Increase in atomic mass (d) d-d transitions
Answer(b) Poor shielding by 4f electrons The 4f electrons shield the outer electrons poorly from the increasing nuclear charge (f orbitals are diffuse), causing atomic radii to decrease steadily across the lanthanide series.
5. Which transition metal is the best catalyst for the Haber process? (a) Cu (b) Ni (c) Fe (d) Pt
Answer(c) Fe Iron (Fe) with promoters Al₂O₃ and K₂O is the catalyst for N₂ + 3H₂ → 2NH₃. Variable oxidation states allow it to adsorb and activate reactant molecules.
6. The highest oxidation state of manganese (Mn) is: (a) +4 (b) +5 (c) +6 (d) +7
Answer(d) +7 Mn shows +2, +3, +4, +6, +7 oxidation states. The highest (+7) is seen in KMnO₄ (potassium permanganate) and Mn₂O₇.
7. Zn²⁺ is colourless because: (a) It has a completely filled d subshell (b) It has empty d orbitals (c) It is diamagnetic (d) It is a d-block element
Answer(a) Completely filled d subshell Zn²⁺ = [Ar]3d¹⁰ — all d orbitals are filled. No d-d transitions are possible, so no visible light is absorbed → colourless.
II. Short Answer Questions
Q1. Explain why transition metals show variable oxidation states.
Answer
In transition metals, the energies of ns and (n−1)d orbitals are very close. So both ns electrons and some (n−1)d electrons can participate in bonding. As different numbers of d electrons are used, different oxidation states are possible.
Example for Mn:
- +2 → uses only 4s²
- +4 → uses 4s² + 2 of 3d electrons
- +7 → uses 4s² + all 5 of 3d electrons
Q2. Why do transition metal compounds show colour?
Answer
When white light falls on a transition metal complex, electrons in the lower d orbitals absorb specific wavelengths of visible light and jump to higher d orbitals (d-d transition). The remaining wavelengths are transmitted or reflected as colour.
Example:
- [Cu(H₂O)₆]²⁺ appears blue — absorbs red/orange light (~630 nm).
- KMnO₄ is purple — Mn⁷⁺ absorbs green light.
Compounds of Zn²⁺, Cu⁺ (d¹⁰) and Sc³⁺ (d⁰) are colourless — no d-d transitions possible.
Q3. What is lanthanide contraction? State two consequences.
Answer
Lanthanide contraction: The steady decrease in atomic and ionic radii across the lanthanide series (La to Lu) despite increasing atomic number.
Cause: The 14 electrons added across the 4f orbitals shield very poorly (f orbitals are diffuse). Nuclear charge increases by 1 each step, but the shielding barely increases → effective nuclear charge increases → radii decrease.
Consequences: 1. The 5d transition metals (Hf, Ta, W…) have nearly the same radii as their 4d counterparts (Zr, Nb, Mo…), making them very difficult to separate. 2. Similar ionic radii of lanthanides (Ln³⁺) makes them difficult to separate from each other (ion exchange methods needed).
Q4. Why do transition metals form complex compounds?
Answer
Transition metals can form complexes because: 1. They have high charge density (small size, high charge) — attract electron-donating ligands. 2. They have vacant d orbitals that can accept lone pairs from ligands. 3. They can show variable oxidation states, allowing stable complexes of different charges.
Example: [Fe(CN)₆]³⁻, [Co(NH₃)₆]³⁺, [Cu(NH₃)₄]²⁺.
Q5. Compare lanthanides and actinides.
Answer
| Property | Lanthanides | Actinides |
|---|
| Series | 4f (Ce–Lu) | 5f (Th–Lr) |
| Common oxidation state | +3 | +3, but more variable |
| Radioactivity | Mostly stable | All radioactive |
| Complex formation | Less | More (due to 5f availability) |
| Extraction | Ion exchange | Solvent extraction |
| Occurrence | Naturally available | Mostly synthetic |
Worked Examples
Example 1 — Spin-only magnetic moment
Find μ for the following ions:
Fe²⁺ = [Ar]3d⁶ → 4 unpaired e⁻ → μ = √(4×6) = √24 ≈ 4.90 BM
Fe³⁺ = [Ar]3d⁵ → 5 unpaired e⁻ → μ = √(5×7) = √35 ≈ 5.92 BM
Ti³⁺ = [Ar]3d¹ → 1 unpaired e⁻ → μ = √(1×3) = √3 ≈ 1.73 BM
Cu²⁺ = [Ar]3d⁹ → 1 unpaired e⁻ → μ = √(1×3) = √3 ≈ 1.73 BM
Zn²⁺ = [Ar]3d¹⁰→ 0 unpaired e⁻ → μ = 0 BM (diamagnetic)
Example 2 — Find number of unpaired electrons from μ
A transition metal ion has μ = 4.90 BM. Find the number of unpaired electrons.
μ = √(n(n+2)) = 4.90
n(n+2) = 24
Try n = 4: 4×6 = 24 ✓
Answer: 4 unpaired electrons (e.g. Fe²⁺)
Example 3 — Anomalous electronic configurations
Expected Cr: [Ar]3d⁴4s² Actual: [Ar]3d⁵4s¹
Expected Cu: [Ar]3d⁹4s² Actual: [Ar]3d¹⁰4s¹
Reason: Half-filled (3d⁵) and fully-filled (3d¹⁰) sub-shells are extra stable
due to symmetric electron distribution and maximum exchange energy.
Example 4 — Oxidation states of Mn in its compounds
MnO → Mn is +2 (MnO → MnO + ... wait, MnO is Mn²⁺)
MnO₂ → Mn is +4 (most stable common compound)
K₂MnO₄ → Mn is +6 (manganate — green)
KMnO₄ → Mn is +7 (permanganate — purple, strong oxidising agent)
Example 5 — Why Zn, Cd, Hg are not true transition elements
True transition elements have partially filled d orbitals in their ground state
OR in their common ionic state.
Zn = [Ar]3d¹⁰4s² → Zn²⁺ = [Ar]3d¹⁰ (completely filled)
Cd = [Kr]4d¹⁰5s² → Cd²⁺ = [Kr]4d¹⁰ (completely filled)
Hg = [Xe]4f¹⁴5d¹⁰6s² → Hg²⁺ = [Xe]4f¹⁴5d¹⁰ (completely filled)
Therefore, Zn, Cd, Hg are d-block elements but NOT true transition elements.
They show fixed +2 oxidation state and form colourless compounds.
Common Mistakes
- Calling Zn, Cd, Hg typical transition elements without noting they have complete d¹⁰ subshell.
- Forgetting to count unpaired electrons correctly for ions (not atoms).
- Mixing lanthanides (4f) and actinides (5f) series.
- Saying colour is due to f-f transitions — it is d-d transitions for transition metal compounds.
- Not including both property AND reason in answers.
Exam Checklist
- Electronic configuration written for the ion (not the neutral atom) when finding unpaired electrons.
- Property and reason are both present in short answers.
- Magnetic moment formula applied correctly with correct n.
- Lanthanide vs actinide differences are clear.
- Anomalous configurations of Cr and Cu are justified.
