Inner Transition Elements Lanthanoids - Electronic configuration, oxidation states, and lanthanoid contraction. Actinoids - Electronic configuration and oxidation states.: NEET MCQs with Answers

15 free questions on Inner Transition Elements Lanthanoids - Electronic configuration, oxidation states, and lanthanoid contraction. Actinoids - Electronic configuration and oxidation states. for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
The greater range of oxidation states shown by the actinoids is attributed to [NEET 2017]
  1. actinoid contraction
  2. 5f, 6d and 7s levels having comparable energies
  3. 4f and 5d levels being close in energy
  4. the radioactive nature of the actinoids
Show answer
Answer: (B) 5f, 6d and 7s levels having comparable energies
Explanation: Because the 5f, 6d and 7s energy levels are very close, electrons from all three can be used in bonding, giving oxidation states from +3 up to +7. This is not possible in lanthanoids where the 4f level is much lower in energy.

Question 2

Previous-year question
The electronic configurations of Eu (Z = 63), Gd (Z = 64) and Tb (Z = 65) are respectively [NEET 2016]
  1. [Xe]4f⁷ 6s², [Xe]4f⁸ 6s² and [Xe]4f⁸ 5d¹ 6s²
  2. [Xe]4f⁷ 5d¹ 6s², [Xe]4f⁷ 5d¹ 6s² and [Xe]4f⁹ 6s²
  3. [Xe]4f⁶ 5d¹ 6s², [Xe]4f⁷ 5d¹ 6s² and [Xe]4f⁸ 5d¹ 6s²
  4. [Xe]4f⁷ 6s², [Xe]4f⁷ 5d¹ 6s² and [Xe]4f⁹ 6s²
Show answer
Answer: (D) [Xe]4f⁷ 6s², [Xe]4f⁷ 5d¹ 6s² and [Xe]4f⁹ 6s²
Explanation: Eu stops at the stable half-filled 4f⁷ 6s². Gd puts the extra electron into 5d to preserve the half-filled f⁷ set, giving 4f⁷ 5d¹ 6s². Tb then goes to 4f⁹ 6s².

Question 3

Previous-year question
The reason for lanthanoid contraction is [AIPMT 2014]
  1. the negligible screening effect of the f-orbitals
  2. increasing nuclear charge
  3. decreasing nuclear charge
  4. decreasing screening effect
Show answer
Answer: (A) the negligible screening effect of the f-orbitals
Explanation: The 4f orbitals are diffuse and non-penetrating, so an added 4f electron screens the outer electrons very poorly. The steadily rising nuclear charge is therefore felt almost fully, and the size contracts across the series.

Question 4

Previous-year question
Actinides [AIPMT 1994]
  1. are all synthetic elements
  2. include element 104
  3. have only short-lived isotopes
  4. have variable valency
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Answer: (D) have variable valency
Explanation: Actinoids show a wide range of oxidation states, from +2 up to +7, because the 5f, 6d and 7s levels have comparable energies. Not all are synthetic (Th, Pa and U occur naturally), element 104 lies beyond lawrencium (103), and ²³²Th and ²³⁸U are very long-lived.

Question 5

Previous-year question
Among the lanthanides, the one obtained only by synthetic methods is [AIPMT 1994]
  1. Lu
  2. Pm
  3. Pr
  4. Gd
Show answer
Answer: (B) Pm
Explanation: Promethium (Z = 61) is the only lanthanoid with no stable isotope — every one of its isotopes is radioactive and comparatively short-lived, the longest being ¹⁴⁵Pm with a half-life of about 17.7 years. Any promethium present when the Earth formed has long since decayed, so it is not found in appreciable quantities in nature. It is obtained artificially, either from the fission products of uranium or by neutron bombardment of neodymium. The other three — Lu, Pr and Gd — all occur naturally in monazite and other lanthanoid-bearing ores.

Question 6

Previous-year question
The lanthanide contraction is responsible for the fact that [AIPMT 1997]
  1. Zr and Y have about the same radius
  2. Zr and Nb have similar oxidation states
  3. Zr and Hf have about the same radius
  4. Zr and Zn have the same oxidation state
Show answer
Answer: (C) Zr and Hf have about the same radius
Explanation: Zr (period 5) and Hf (period 6) are in the same group but separated by the lanthanoids. The shrinkage across that series offsets the size gain from a new shell, so their radii are almost equal (about 160 pm and 159 pm). Oxidation states depend on electronic configuration, not on this contraction.

Question 7

Previous-year question
Which one of the following elements shows the maximum number of different oxidation states in its compounds? [AIPMT 1998]
  1. Eu
  2. La
  3. Gd
  4. Am
Show answer
Answer: (D) Am
Explanation: Americium is an actinoid, and in the actinoids the 5f, 6d and 7s levels lie close in energy, so electrons from all three can be used in bonding. Americium accordingly shows +2, +3, +4, +5 and +6 — a much wider range than any lanthanoid. Among the lanthanoids listed, La shows only +3, Gd only +3, and Eu shows +2 and +3. The larger size and lower ionisation enthalpy of the actinoid also help make the higher states accessible.
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Question 8

Previous-year question
Which of the following statements is not correct? [AIPMT 2001]
  1. La(OH)₃ is less basic than Lu(OH)₃.
  2. In the lanthanide series the ionic radius of the Ln³⁺ ion decreases.
  3. La is actually an element of the transition series rather than a lanthanide.
  4. The atomic radii of Zr and Hf are the same because of lanthanide contraction.
Show answer
Answer: (A) La(OH)₃ is less basic than Lu(OH)₃.
Explanation: La³⁺ is the largest Ln³⁺ ion, so La–OH has the greatest ionic character and La(OH)₃ is the most basic; Lu(OH)₃ is the least basic, so this claim is reversed. Ln³⁺ radii do decrease across the series, lanthanum has no 4f electrons, and Zr and Hf have nearly equal radii.

Question 9

Previous-year question
The general electronic configuration of lanthanides is [AIPMT 2002]
  1. (n−2)f¹⁻¹⁴ (n−1)s²p⁶d⁰⁻¹ ns²
  2. (n−2)f¹⁰⁻¹⁴ (n−1)d⁰⁻¹ ns²
  3. (n−2)f⁰⁻¹⁴ (n−1)d¹⁰ ns²
  4. (n−2)d⁰⁻¹ (n−1)f¹⁻¹⁴ ns²
Show answer
Answer: (A) (n−2)f¹⁻¹⁴ (n−1)s²p⁶d⁰⁻¹ ns²
Explanation: For the lanthanoids the outermost shell is the sixth, so n = 6 and the electrons occupy 4f, 5s, 5p, 5d and 6s. Written generally with respect to n: (n−2)f¹⁻¹⁴ (n−1)s²p⁶d⁰⁻¹ ns² The (n−2)f level is the 4f subshell being filled, the (n−1) shell is the completed 5s²5p⁶ set together with an occasionally occupied 5d orbital, and ns² is the 6s pair. The 5d occupancy is 0 for most members but 1 for La, Ce, Gd and Lu.

Question 10

Previous-year question
Lanthanides are [AIPMT 2004]
  1. 14 elements in the sixth period (at. no. 90 to 103) that are filling the 4f sub-level
  2. 14 elements in the seventh period (at. no. 90 to 103) that are filling the 5f sub-level
  3. 14 elements in the sixth period (at. no. 58 to 71) that are filling the 4f sub-level
  4. 14 elements in the seventh period (at. no. 58 to 71) that are filling the 4f sub-level
Show answer
Answer: (C) 14 elements in the sixth period (at. no. 58 to 71) that are filling the 4f sub-level
Explanation: The lanthanoids are the fourteen elements from cerium (Z = 58) to lutetium (Z = 71). They lie in the sixth period and the 4f sub-level fills across the series. The elements from Z = 90 to 103 in the seventh period, filling 5f, are the actinoids.

Question 11

Previous-year question
A larger number of oxidation states are exhibited by the actinides than by the lanthanides. The main reason for this is [AIPMT 2006, 2005]
  1. a greater energy difference between 5f and 6d orbitals than between 4f and 5d orbitals
  2. a lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals
  3. the greater metallic character of the lanthanides than of the corresponding actinides
  4. the more active nature of the actinides
Show answer
Answer: (B) a lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals
Explanation: In the actinoids the 5f and 6d energy levels lie very close together, so electrons can be drawn from both subshells for bonding. This gives access to a wide spread of oxidation states — +3, +4, +5, +6 and even +7 in the case of neptunium. In the lanthanoids the gap between the 4f and 5d levels is comparatively large, and the 4f electrons are buried deep inside the atom where they are shielded by the 5s and 5p shells. They therefore contribute little to bonding, and the +3 state dominates almost exclusively.

Question 12

Previous-year question
Identify the incorrect statement among the following. [AIPMT 2007]
  1. There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.
  2. Lanthanide contraction is the accumulation of successive shrinkages.
  3. As a result of lanthanoid contraction, the properties of the 4d series of transition elements have no similarities with the 5d series of elements.
  4. The shielding power of 4f electrons is quite weak.
Show answer
Answer: (C) As a result of lanthanoid contraction, the properties of the 4d series of transition elements have no similarities with the 5d series of elements.
Explanation: Lanthanoid contraction makes the 5d elements nearly the same size as their 4d counterparts, so pairs such as Zr/Hf, Nb/Ta and Mo/W have very similar properties; claiming no similarity is false. Radii do fall from La³⁺ (103 pm) to Lu³⁺ (86 pm), the contraction is cumulative, and 4f shielding is weak.

Question 13

Previous-year question
Which of the following ions will exhibit colour in aqueous solution? [AIPMT 2010]
  1. La³⁺ (Z = 57)
  2. Ti³⁺ (Z = 22)
  3. Lu³⁺ (Z = 71)
  4. Sc³⁺ (Z = 21)
Show answer
Answer: (B) Ti³⁺ (Z = 22)
Explanation: Colour requires a partly filled d or f subshell so that an electronic transition of suitable energy is possible. Ti³⁺ = [Ar]3d¹ → 1 unpaired d electron, gives purple solutions ✔ La³⁺ = [Xe]4f⁰ → empty f subshell, colourless Lu³⁺ = [Xe]4f¹⁴ → completely filled f subshell, colourless Sc³⁺ = [Ar]3d⁰ → empty d subshell, colourless Only Ti³⁺ can undergo a d–d transition, so only it is coloured.

Question 14

Previous-year question
Which of the following pairs has nearly the same size? [AIPMT 2010]
  1. Fe²⁺, Ni²⁺
  2. Zr⁴⁺, Ti⁴⁺
  3. Zr⁴⁺, Hf⁴⁺
  4. Zn²⁺, Hf⁴⁺
Show answer
Answer: (C) Zr⁴⁺, Hf⁴⁺
Explanation: Zr⁴⁺ (≈ 80 pm) and Hf⁴⁺ (≈ 79 pm) are almost identical in size. Hafnium comes immediately after the fourteen lanthanoids, and the contraction across that series exactly offsets the size increase expected on moving from period 5 to period 6. In the other pairs the sizes differ appreciably: Fe²⁺ is larger than Ni²⁺ because ionic radius decreases across the 3d series; Ti⁴⁺ is distinctly smaller than Zr⁴⁺ since it lies one period higher; and Zn²⁺ and Hf⁴⁺ have neither the same charge nor comparable positions.

Question 15

Previous-year question
Which of the following oxidation states is the most common among the lanthanoids? [AIPMT 2010]
  1. 4
  2. 2
  3. 5
  4. 3
Show answer
Answer: (D) 3
Explanation: The characteristic and most common oxidation state throughout the lanthanoid series is +3. It is reached by losing the two 6s electrons together with one 4f (or 5d) electron, and it is shown by every member of the series. The +2 and +4 states occur only occasionally, and almost always where the resulting ion has a specially stable f⁰, f⁷ or f¹⁴ configuration — for example Ce⁴⁺ (4f⁰), Eu²⁺ (4f⁷), Tb⁴⁺ (4f⁷) and Yb²⁺ (4f¹⁴). No lanthanoid shows a common +5 state.

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