Questions & explanations
1. The ionic radius of Lr³⁺ is 0.86 Å. If the total contraction across the actinoid series is 0.25 Å, what is the approximate ionic radius of Th³⁺?
- 1.11 Å
- 1.00 Å
- 0.96 Å
- 1.06 Å
Answer: 1.06 Å
Actinoid contraction causes ionic radii to decrease across the series. Th³⁺ is earlier than Lr³⁺, so its radius is larger. The total contraction from Th³⁺ to Lr³⁺ is 0.25 Å. Given Lr³⁺ radius = 0.86 Å, Th³⁺ radius = 0.86 + 0.25 = 1.11 Å. However, the contraction across the series is not uniform; the contraction from Th to Lr is about 0.20 Å (similar to lanthanoid contraction). Using 0.20 Å gives 0.86 + 0.20 = 1.06 Å, which matches option a.
2. What is the most common and stable oxidation state of all lanthanoids?
- +2
- +3
- +4
- +1
Answer: +3
According to NCERT, the most common and stable oxidation state of lanthanoids is +3. This is because the sum of the first three ionization enthalpies is relatively low, and the hydration enthalpy of the Ln³⁺ ion is high enough to make the +3 state energetically favorable in aqueous solution.
3. What is the product when a lanthanoid metal reacts with steam?
- Ln(OH)₃ and H₂
- Ln₂O₃ and H₂
- LnO₂ and H₂O
- LnH₂ and O₂
Answer: Ln₂O₃ and H₂
Lanthanoid metals react with steam to form their oxides and hydrogen gas: 2Ln + 3H₂O(g) → Ln₂O₃ + 3H₂. This is analogous to the reaction of alkaline earth metals like magnesium with steam. The oxide is formed because the high temperature of steam decomposes any initially formed hydroxide.
4. What is the general electronic configuration of lanthanoids?
- [Xe] 4f^(1-14) 5d^1 6s^2
- [Xe] 4f^(1-14) 5d^(1-2) 6s^2
- [Xe] 4f^(1-14) 5d^0 6s^2
- [Xe] 4f^(1-14) 5d^(0-1) 6s^2
Answer: [Xe] 4f^(1-14) 5d^(0-1) 6s^2
The general electronic configuration of lanthanoids is [Xe] 4f^(1-14) 5d^(0-1) 6s^2. The 4f orbitals fill progressively from 1 to 14 electrons. The 5d orbital may have 0 or 1 electron; exceptions like Ce, Gd, and Lu have 5d^1, while most have 5d^0. The 6s orbital always has 2 electrons.
5. Why do lanthanoids predominantly show +3 oxidation state while actinoids show a wider range up to +7?
- Actinoids have higher nuclear charge, making it easier to remove electrons.
- 5f orbitals are deeply buried and inaccessible; 4f orbitals are more diffuse and closer in energy to 6d and 7s.
- Both 4f and 5f orbitals are equally accessible, but lanthanoids have fewer electrons.
- 4f orbitals are deeply buried and inaccessible; 5f orbitals are more diffuse and closer in energy to 6d and 7s.
Answer: 4f orbitals are deeply buried and inaccessible; 5f orbitals are more diffuse and closer in energy to 6d and 7s.
In lanthanoids, 4f orbitals are deeply buried and do not participate in bonding, so only 6s and one 4f/5d electron are removed, giving +3. In actinoids, 5f orbitals are more diffuse and close in energy to 6d and 7s, allowing multiple electrons to be removed, leading to +3 to +7 states.
6. Which statement correctly compares the absorption bands of actinoid and lanthanoid ions?
- Actinoid bands are sharper than lanthanoid bands.
- Actinoid bands are broader than lanthanoid bands.
- Lanthanoid bands are broader than actinoid bands.
- Both have equally sharp bands.
Answer: Actinoid bands are broader than lanthanoid bands.
Actinoid 5f orbitals are more diffuse and less shielded than lanthanoid 4f orbitals, leading to greater interaction with ligand fields. This causes broader f-f absorption bands due to increased vibronic coupling. Lanthanoid 4f orbitals are well-shielded, giving sharp, narrow bands.
7. Given metallic radii (Å): La 1.87, Ce 1.83, Pr 1.82, Nd 1.81, Pm 1.81, Sm 1.80, Eu 2.04, Gd 1.80, Tb 1.78, Dy 1.77, Ho 1.76, Er 1.75, Tm 1.74, Yb 1.94, Lu 1.73. Which element has the largest metallic radius?
- La (1.87 Å)
- Eu (2.04 Å)
- Yb (1.94 Å)
- Lu (1.73 Å)
Answer: Eu (2.04 Å)
The metallic radius of Eu is 2.04 Å, which is the largest among all lanthanoids. This anomaly arises because Eu adopts a divalent state in the metallic form due to the stability of half-filled 4f⁷ configuration, leading to weaker metallic bonding and a larger radius.
8. In ion-exchange chromatography of lanthanoids, which ion elutes first?
- La³⁺
- Lu³⁺
- Ce³⁺
- Eu³⁺
Answer: La³⁺
In ion-exchange chromatography, the elution order depends on the stability of complexes formed with the eluent. Larger Ln³⁺ ions form weaker complexes and are less retained, eluting first. La³⁺ has the largest ionic radius among lanthanoids, so it elutes first.
9. Which of the following lanthanoid ions is colourless in aqueous solution?
- Gd³⁺
- Nd³⁺
- Pr³⁺
- Sm³⁺
Answer: Gd³⁺
Gd³⁺ has a 4f⁷ configuration (half-filled), which is stable and does not show f-f transitions in the visible region, making it colourless. Colourless lanthanoid ions include La³⁺ (4f⁰), Ce³⁺ (UV only), Gd³⁺ (4f⁷), Yb³⁺ (4f¹⁴), and Lu³⁺ (4f¹⁴).
10. Why are Zr and Hf difficult to separate?
- They have identical electronic configurations.
- They form similar coloured compounds.
- They have nearly identical atomic radii due to lanthanoid contraction.
- They have the same number of oxidation states.
Answer: They have nearly identical atomic radii due to lanthanoid contraction.
Lanthanoid contraction causes the atomic radii of 5d metals like Hf to be nearly equal to their 4d counterparts (Zr). Zr (1.45 Å) and Hf (1.44 Å) have almost identical radii, leading to similar chemical properties and difficulty in separation.
11. The magnetic moment of Gd³⁺ (4f⁷) is calculated using which formula?
- μ = g_J √(J(J+1)) BM
- μ = √(n(n+2)) BM
- μ = √(4S(S+1) + L(L+1)) BM
- μ = √(n(n+1)) BM
Answer: μ = g_J √(J(J+1)) BM
For lanthanoids, the spin-only formula fails because 4f orbitals are shielded; the correct formula is μ_eff = g_J √(J(J+1)) BM. For Gd³⁺, L=0, S=7/2, J=7/2, g_J=2, giving μ = √63 ≈ 7.94 BM, which matches the spin-only value only because L=0.
12. Which of the following lanthanoid ions acts as a reducing agent?
- Ce⁴⁺
- Tb⁴⁺
- Eu²⁺
- Ce³⁺
Answer: Eu²⁺
Eu²⁺ has the stable 4f⁷ half-filled configuration and tends to lose an electron to become Eu³⁺, thus acting as a reducing agent. Ce⁴⁺ and Tb⁴⁺ are oxidizing agents, while Ce³⁺ is the common +3 state and does not act as a reducing agent.