All the actinoids are believed to have the electronic configuration of 7s2 and variable occupancy of the 5f and 6d subshells. The irregularities in the electronic configurations of the actinoids, like those in the lanthanoids are related to the stabilities of the f 0, f 7 and f 14 occupancies of the 5f orbitals. Thus, the configurations of Am and Cm are [Rn] 5f 77s2 and [Rn] 5f 76d17s2. Although the 5f orbitals resemble the 4f orbitals in their angular part of the wave-function, they are not as buried as 4f orbitals and hence 5f electrons can participate in bonding to a far greater extent.
-- NCERT Class 12 Chemistry, Ch. 4, p. 112Actinoids
8 MCQs9-step worked example
Source: NCERT The d and f Block ElementsOfficial key: NTA-verifiedLast updated: 7 Oct 2026
Try this first
Which outer configuration are all the actinoids believed to share?
- A.6s² with a filled 4f subshell
- B.5f¹⁴6d¹⁰ with no s electrons
- C.5d¹6s² only
- D.7s² with variable occupancy of the 5f and 6d subshells
Tap to see the answer
Answer: D. NCERT Class 12 Chemistry, Chapter 4, page 112: all the actinoids are believed to have 7s² and variable occupancy of the 5f and 6d subshells.
A is wrong: A is wrong because 6s² with 4f belongs to the lanthanoid shell pattern; the actinoids are 7s² with 5f and 6d (page 112).
B is wrong: B is wrong because the page gives 7s² as common to all actinoids, with variable 5f and 6d (page 112).
C is wrong: C is wrong because it is a 5d-row pattern; page 112 gives 7s² with variable 5f and 6d occupancy for the actinoids.
Actinoids, explained for NEET
The trap: Students copy the lanthanoid pattern across to the actinoids. Three things differ, and NEET statement questions pick on them: the contraction is greater from element to element, the range of oxidation states is greater, and the 5f electrons take part in bonding far more than 4f electrons do.
Configuration. All the actinoids are believed to have 7s² with variable occupancy of the 5f and 6d subshells. The irregularities, as in the lanthanoids, are related to the stability of f⁰, f⁷ and f¹⁴ occupancies. Am is [Rn] 5f⁷7s² and Cm is [Rn] 5f⁷6d¹7s² (NCERT Class 12 Chemistry, Chapter 4, page 112). The 5f orbitals resemble 4f in their angular part, but they are not as buried, so "5f electrons can participate in bonding to a far greater extent" (page 112).
Size. There is a gradual decrease in the size of atoms or M³⁺ ions across the series, the actinoid contraction. It is "greater from element to element" than in the lanthanoids, from poor shielding by 5f electrons (page 112).
Oxidation states. The range is greater, in part because the 5f, 6d and 7s levels are of comparable energies. In general the actinoids show +3. In the first half of the series higher states are frequent: the maximum rises from +4 in Th to +5, +6 and +7 in Pa, U and Np, then decreases. The actinoids resemble the lanthanoids in having more compounds in +3 than in +4, but +3 and +4 ions tend to hydrolyse (page 112).
Other characteristics. The elements are radioactive, and the latter members have half-lives from a day to 3 minutes for lawrencium (Z = 103) (page 112). The metals are highly reactive, especially when finely divided. Their magnetic properties are more complex than the lanthanoids', and the early actinoids have lower ionisation enthalpies than the early lanthanoids (page 113).
Watch-out: a statement that flips "greater" to "smaller", or +3 to a different common state, is the usual wrong option.
How do you solve a Actinoids question? A worked example
- 1
Given
NCERT lists the maximum oxidation states of Th, Pa, U and Np as +4, +5, +6 and +7 (page 112).
- 2
Required
The step in maximum oxidation state from each element to the next, and the maximum for the series trend in the first half.
- 3
Concept
The first half of the actinoid series frequently shows higher oxidation states, because the 5f, 6d and 7s levels are of comparable energies (NCERT Class 12 Chemistry, Chapter 4, page 112).
- 4
Formula
Step = (maximum oxidation state of the next element) − (maximum oxidation state of this element).
- 5
Substitution
Th to Pa: 5 − 4. Pa to U: 6 − 5. U to Np: 7 − 6.
- 6
Calculation
5 − 4 = 1, 6 − 5 = 1, 7 − 6 = 1. Total rise from Th to Np = 7 − 4 = 3, which is 1 + 1 + 1. The values are counting integers, not measured quantities.
- 7
Final answer
The maximum rises by one unit per element, from +4 (Th) to +7 (Np), and then decreases in succeeding elements (page 112).
- 8
Common trap
Assuming the maximum keeps rising to the end of the series, or that +3 is the highest state because it is the general one. Page 112 says +3 is the general state and that the maximum decreases after Np.
- 9
Similar NEET-style question
Which of Th, Pa, U and Np has a maximum oxidation state of +6? (Answer: U. Starting at Th = +4 and adding one per element: Pa = 4 + 1 = +5, U = 5 + 1 = +6, Np = 6 + 1 = +7, as printed on page 112.)
---
Can you answer these Actinoids MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which outer configuration are all the actinoids believed to share?
Show answer and why every option is right or wrong
Answer: D. NCERT Class 12 Chemistry, Chapter 4, page 112: all the actinoids are believed to have 7s² and variable occupancy of the 5f and 6d subshells.
Why A is wrong: A is wrong because 6s² with 4f belongs to the lanthanoid shell pattern; the actinoids are 7s² with 5f and 6d (page 112).
Why B is wrong: B is wrong because the page gives 7s² as common to all actinoids, with variable 5f and 6d (page 112).
Why C is wrong: C is wrong because it is a 5d-row pattern; page 112 gives 7s² with variable 5f and 6d occupancy for the actinoids.
MCQ 2Easy RecallPractice
Among Th, Pa, U and Np, which shows the highest maximum oxidation state, as given by NCERT?
Show answer and why every option is right or wrong
Answer: A. NCERT Class 12 Chemistry, Chapter 4, page 112: the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np.
Why B is wrong: B is wrong because +6 is the maximum for U, one unit below Np (page 112).
Why C is wrong: C is wrong because +5 is the maximum for Pa (page 112).
Why D is wrong: D is wrong because +4 is the maximum for Th, the lowest of the four (page 112).
MCQ 3Easy RecallPractice
Which oxidation state do the actinoids show in general?
Show answer and why every option is right or wrong
Answer: C. NCERT Class 12 Chemistry, Chapter 4, page 112: the actinoids show in general the +3 oxidation state.
Why A is wrong: A is wrong because page 112 gives +3 as the general state, not +2.
Why B is wrong: B is wrong because the actinoids have more compounds in +3 than in +4 (page 112).
Why D is wrong: D is wrong because +7 is the maximum, reached by Np only, not the general state (page 112).
MCQ 4Direct ApplicationPractice
Why can 5f electrons take part in bonding to a far greater extent than 4f electrons?
Show answer and why every option is right or wrong
Answer: B. NCERT Class 12 Chemistry, Chapter 4, page 112: 5f orbitals resemble 4f in angular part but are not as buried as 4f, so 5f electrons can participate in bonding to a far greater extent.
Why A is wrong: A is wrong because that describes 4f, which is more buried; NCERT says 5f is less buried (page 112).
Why C is wrong: C is wrong because NCERT says the angular part of 5f resembles that of 4f (page 112).
Why D is wrong: D is wrong because 5f orbitals are filled from Pa onwards and are complete at element 103 (page 112).
MCQ 5Direct ApplicationPractice
How does the actinoid contraction compare with the lanthanoid contraction?
Show answer and why every option is right or wrong
Answer: A. NCERT Class 12 Chemistry, Chapter 4, page 112: the contraction is greater from element to element in this series, resulting from poor shielding by 5f electrons. (This is the statement tested in NEET 2021 Q63, where it is true.)
Why B is wrong: B is wrong because the cause named is poor shielding by 5f electrons, so the contraction is greater (page 112).
Why C is wrong: C is wrong because NCERT says there is a gradual decrease in size of atoms or M³⁺ ions across the series, the actinoid contraction (page 112).
Why D is wrong: D is wrong because NCERT says the contraction is greater element to element, not the same (page 112).
MCQ 6Direct ApplicationPractice
The irregularities in actinoid electronic configurations are related to the stability of which occupancies of the 5f orbitals?
Show answer and why every option is right or wrong
Answer: D. NCERT Class 12 Chemistry, Chapter 4, page 112: the irregularities are related to the stabilities of f⁰, f⁷ and f¹⁴ occupancies of the 5f orbitals, like those in the lanthanoids.
Why A is wrong: A is wrong because page 112 names f⁰, f⁷ and f¹⁴, not f¹ and f¹³.
Why B is wrong: B is wrong because the half-filled case is f⁷, not f⁶ (page 112).
Why C is wrong: C is wrong because the empty case f⁰ is named, not f² (page 112).
MCQ 7CalculationPractice
NCERT gives the configurations of Am and Cm. Which pair is correct?
Show answer and why every option is right or wrong
Answer: B. NCERT Class 12 Chemistry, Chapter 4, page 112 gives Am as [Rn] 5f⁷7s² and Cm as [Rn] 5f⁷6d¹7s². Counting along the series from Th (Z = 90), Am is Z = 95 and Cm is Z = 96, which is one more electron, and that extra electron goes into 6d, keeping the half-filled 5f⁷ set.
Why A is wrong: A is wrong because Cm has one more electron than Am, and the page puts it in 6d, not 5f or 7s (page 112).
Why C is wrong: C is wrong because it swaps the two; Am has no 6d electron (page 112).
Why D is wrong: D is wrong because Am is [Rn] 5f⁷7s² with no 6d electron (page 112).
MCQ 8Concept TrapPractice
What does NCERT give as part of the reason for the greater range of oxidation states in the actinoids?
Show answer and why every option is right or wrong
Answer: C. NCERT Class 12 Chemistry, Chapter 4, page 112: the greater range of oxidation states is in part attributed to the 5f, 6d and 7s levels being of comparable energies.
Why A is wrong: A is wrong because page 112 attributes the contraction to poor shielding by 5f electrons, and gives comparable 5f, 6d and 7s energies, not complete shielding, as the oxidation-state reason.
Why B is wrong: B is wrong because 6d occupancy varies, as in Cm [Rn] 5f⁷6d¹7s² (page 112).
Why D is wrong: D is wrong because radioactivity is mentioned as making their study more difficult, not as the cause of the oxidation-state range (page 112).
Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
What to remember before solving Actinoids questions
5 NCERT lines
There is a gradual decrease in the size of atoms or M3+ ions across the series. This may be referred to as the actinoid contraction (like lanthanoid contraction). The contraction is, however, greater from element to element in this series resulting from poor shielding by 5f electrons.
-- NCERT Class 12 Chemistry, Ch. 4, p. 112There is a greater range of oxidation states, which is in part attributed to the fact that the 5f, 6d and 7s levels are of comparable energies. The actinoids show in general +3 oxidation state. The elements, in the first half of the series frequently exhibit higher oxidation states. For example, the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np but decreases in succeeding elements (Table 4.11).
-- NCERT Class 12 Chemistry, Ch. 4, p. 112Key Fact
Early actinoids: lower ionisation enthalpy than early lanthanoids; magnetic properties more complex
The magnetic properties of the actinoids are more complex than those of the lanthanoids. It is evident from the behaviour of the actinoids that the ionisation enthalpies of the early actinoids, though not accurately known, but are lower than for the early lanthanoids. The 5f electrons, will therefore, be more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids.
-- NCERT Class 12 Chemistry, Ch. 4, p. 113The actinoids are highly reactive metals, especially when finely divided. Intext Question 4.10 asks why actinoid contraction is greater from element to element than lanthanoid contraction.
-- NCERT Class 12 Chemistry, Ch. 4, p. 113Actinoids: NEET previous year questions (PYQs) with answers
15 questions in The d and f Block Elements
No question in our NEET 2020–2025 set targets this topic directly.
All 15 past-paper questions from The d and f Block Elements →
More in The d and f Block Elements: 3 exam traps and mistakes · 3 formulas · 3 question patterns from its other lessons.
Sources
NCERT refs: Class 12 Chemistry Chapter 4, p.112
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Every past-paper question from this chapter:
Practice the whole chapter →