Which of the following actinoids show oxidation states upto +7? (i) Am (ii) Pu (iii) U (iv) Np

NCERT Exemplar · Q53

Q. Which of the following actinoids show oxidation states upto +7?

(i) Am

(ii) Pu

(iii) U

(iv) Np

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Concept understanding — Actinoid Contraction Greater

The Intuition First

Imagine you're holding a rope that runs through a series of rings. As you pull the rope tighter, the rings get squeezed closer together. Something similar happens inside the atoms of the actinoid elements — but instead of a rope, it's the pull of the nucleus, and instead of rings, it's the electron cloud.

The actinoid series runs from thorium ( $Z=90$ ) to lawrencium ( $Z=103$ ). As you go from one element to the next, you add one proton to the nucleus and one electron to the 5f subshell. That extra proton yanks harder on all the electrons. But here's the key: the 5f orbitals are shaped like little dumbbells that poke inside the atom, very close to the nucleus. They don't shield the outer electrons well from the growing nuclear charge.

So the outer electrons feel a stronger and stronger pull as you move right across the series. The entire electron cloud shrinks. The atomic radius decreases — steadily, noticeably. That's the actinoid contraction.

The Precise Statement

Actinoid contraction is the steady decrease in atomic (and ionic) radii across the actinoid series (from Th to Lr), caused by the poor shielding ability of 5f electrons.

The contraction is greater than what you see in the lanthanoid series (the 4f elements). Why? Two reasons:

The 5f orbitals are more diffuse and penetrate less than 4f orbitals. They sit further from the nucleus on average, so they shield even more poorly. Each added 5f electron does almost nothing to cancel the pull of the new proton.
Relativistic effects become significant for heavy nuclei (Z > 90). The 5f and 6d electrons move at speeds close to the speed of light, which contracts their orbitals further. This is a subtle but real effect that amplifies the contraction.

The result: the total decrease in ionic radius across the actinoid series is about 0.03–0.04 Å — roughly 10–15% larger than the lanthanoid contraction.

Why It Matters …

Why this formula?

Actinoid Contraction: Why It Is Greater Than Lanthanoid Contraction

Let’s build this from the ground up — understanding the why before the what.

1. What Is Actinoid Contraction?

Actinoid contraction is the steady decrease in atomic and ionic radii across the actinoid series (from Th to Lr, atomic numbers 90–103).

It is greater in magnitude than the analogous lanthanoid contraction (across the lanthanoids, Ce–Lu).

Key result: The contraction per element is larger in actinoids (~2–3 pm per element) than in lanthanoids (~1–2 pm per element).

2. The Core Reason: Poorer Shielding by 5f Electrons

The Shielding Effect

Electrons in inner shells shield the outer electrons from the full nuclear charge.
The effective nuclear charge ( $Z_{\text{eff}}$ ) felt by an electron is:

Z_{\text{eff}} = Z - S

where $Z$ = atomic number, $S$ = shielding constant.

Why 5f Shielding Is Weaker Than 4f

4f orbitals (lanthanoids) are more penetrating — they have a small but significant probability near the nucleus. This gives them better shielding ability.
5f orbitals (actinoids) are more diffuse and less penetrating. They are spread farther from the nucleus, so they shield the outer electrons less effectively.

Result: For the same increase in $Z$ , the $Z_{\text{eff}}$ increases more in actinoids than in lanthanoids.

A larger $Z_{\text{eff}}$ pulls the electron cloud inward more strongly → greater contraction.

3. Mathematical Expression of the Trend

The contraction is described by the change in radius per added proton:

For lanthanoids:

\Delta r_{\text{4f}} \approx -1 \text{ to } -2 \text{ pm per element}

For actinoids:

\Delta r_{\text{5f}} \approx -2 \text{ to } -3 \text{ pm per element}

The ratio of contraction is roughly:

\frac{\Delta r_{\text{5f}}}{\Delta r_{\text{4f}}} \approx 1.5 \text{ to } 2

4. Why the Difference in Shielding? — Orbital Shape Matters

Property	4f orbital (lanthanoids)	5f orbital (actinoids)
Principal quantum number ( $n$ )	4	5
Radial extent	More compact	More diffuse
Penetration near nucleus	Higher (has a small lobe near nucleus)	Lower (less probability near nucleus)
Shielding efficiency	Better	Poorer

The radial distribution function shows that 4f electrons have a secondary maximum close to the nucleus, while 5f electrons lack this — they are more "spread out."

5. The Chain of Reasoning (Step-by-Step)

Add a proton → nuclear charge $Z$ increases by 1.
Add a 5f electron → it shields poorly because of its diffuse shape.
Net effect: $Z_{\text{eff}}$ increases more than it would for a 4f electron. …

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The maximum oxidation state of an actinoid is limited by how many of its 5f, 6d, and 7s electrons it can lose while still reaching a reasonably stable f-configuration. Among the given options, Am, Pu, and Np are all documented to reach +7 (in oxo-species such as $\text{AmO}_4^{3-}$ , $\text{PuO}_5^{3-}$ / $\text{PuO}_4^{3-}$ , and $\text{NpO}_4^{3-}$ ), while U never exceeds +6. The correct options are (i), (ii) and (iv).

The question asks which actinoids can show an oxidation state as high as +7. This is a classic pattern-recognition problem in f-block chemistry, but it is easy to over-apply a simple electron-counting rule — the real answer comes from the actual documented chemistry of each element, not from configuration-counting alone.

Recall the electronic configuration pattern for actinoids.

The general configuration is $[\text{Rn}] \, 5f^{1-14} \, 6d^{0-1} \, 7s^2$ .
Write the configurations for each element and check its documented maximum oxidation state.
- Uranium (U, Z=92): $[\text{Rn}] \, 5f^3 \, 6d^1 \, 7s^2$ . Its highest well-established oxidation state is +6 (as in $\text{UO}_2^{2+}$ , $\text{UF}_6$ ) — U does not reach +7.
- Neptunium (Np, Z=93): $[\text{Rn}] \, 5f^4 \, 6d^1 \, 7s^2$ . Np reaches +7 in strongly alkaline, oxidising conditions (e.g., $\text{NpO}_4^{3-}$ / $\text{NpO}_2(\text{OH})_4^{-}$ ).
- Plutonium (Pu, Z=94): $[\text{Rn}] \, 5f^6 \, 7s^2$ . Although +6 (as $\text{PuO}_2^{2+}$ ) is the most common high state, Pu is also documented to reach +7 (as $\text{PuO}_5^{3-}$ / $\text{PuO}_4^{3-}$ ) under strongly alkaline oxidising conditions — this was in fact one of the earliest-confirmed +7 actinoid states.
- Americium (Am, Z=95): $[\text{Rn}] \, 5f^7 \, 7s^2$ . Am also reaches +7 (as $\text{AmO}_4^{3-}$ ) under similar strongly alkaline oxidising conditions. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2020Set A-11 markMCQ
$p -$
›Reveal solutionSolution
The last $p$ -block element of a period is its noble gas; for period 6 that is Rn ( $Z=86$ ) with a filled $6p^6$ .

Step 1 — Which element is being asked for?

Every period ends with the $p$ -block (group 18). So the last $p$ -block element of the 6th period is the noble gas of period 6, i.e. radon, $Z = 86$ .

Step 2 — Build its configuration by the Aufbau order.

After xenon ( $[Xe]$ , $Z=54$ ) the filling order is
$6s \rightarrow 4f \rightarrow 5d \rightarrow 6p$
so
$\text{Rn} = [Xe]\;4f^{14}\,5d^{10}\,6s^{2}\,6p^{6}, \qquad 54 + 14 + 10 + 2 + 6 = 86 \;\checkmark$
The outermost/valence-shell region is therefore $4f^{14}5d^{10}6s^2 6p^6$ .

Step 3 — Eliminate the distractors.

(B) $7s^2 7p^6$ — principal quantum number $7$ ⇒ this is the 7th period (oganesson), not the 6th.

(C) $5f^{14}6d^{10}7s^2 7p^6$ — again $n=7$ ; this is Og ( $Z=118$ ), the last $p$ -block element of period 7. …
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KCET 2019Set A-11 markMCQ
$Q. The elements in which electrons are progressively filled in 4f orbital are called (A) Actinoids (B) Lanthanoids (C) Transition elements (D) Halogens$
›Reveal solutionSolution
The f-block is the inner-transition series: filling 4f $\Rightarrow$ lanthanoids; filling 5f $\Rightarrow$ actinoids.

Step 1 — Classify by the orbital being filled.

The periodic table is blocked by the sub-shell that receives the last electron:

s-block $\to$ groups 1–2,

p-block $\to$ groups 13–18,

d-block $\to$ transition elements (filling $(n-1)d$ ),

f-block $\to$ inner-transition elements (filling $(n-2)f$ ).

Step 2 — Identify the 4f series.

The lanthanoids — cerium ( $Z=58$ ) through lutetium ( $Z=71$ ) — follow lanthanum and have the general configuration
$[\text{Xe}]\,4f^{\,1-14}\,5d^{\,0-1}\,6s^{2}.$
Here the electrons are progressively added to the 4f orbital. So this is the answer.

Step 3 — Reject the others.

(A) Actinoids (Th, $Z=90$ → Lr, $Z=103$ ): $[\text{Rn}]\,5f^{\,1-14}\,6d^{\,0-1}\,7s^{2}$ — they fill 5f, not 4f. …
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KCET 2018Set A-11 markMCQ
$+2$
›Reveal solutionSolution
The most stable and common oxidation state across all lanthanide elements is $+3$ , due to the favourable energy balance of losing two 6s electrons and one 4f electron.

The lanthanides (elements 57–71, from lanthanum to lutetium) are characterised by the gradual filling of the 4f subshell. Their chemistry is dominated by a single, recurring pattern: the +3 oxidation state. Why is this so universal?

The key lies in the electronic configuration. A typical lanthanide atom has the outer configuration $[\text{Xe}]\,4f^{n}\,6s^{2}$ (with minor exceptions for lanthanum, gadolinium, and lutetium). The 6s electrons are the highest in energy and are always lost first when forming ions. Removing both 6s electrons gives a $+2$ ion. However, the 4f electrons are also relatively loosely held — they are not as deeply buried as inner core electrons. For most lanthanides, losing one additional 4f electron to reach the $+3$ state is energetically favourable enough to make this the dominant oxidation state in aqueous solution and in most solid compounds.

The $+3$ state is so stable that it is the only oxidation state common to all lanthanides. Some lanthanides do show $+2$ or $+4$ states, but these are exceptions tied to special stability of half-filled or empty 4f subshells (e.g., $\text{Eu}^{2+}$ has $4f^{7}$ , $\text{Ce}^{4+}$ has $4f^{0}$ ). No lanthanide shows a $+5$ state under normal conditions.

Let’s check each option:

Option (A): $+2$ — This is not common for all. Only europium ( $\text{Eu}^{2+}$ ), ytterbium ( $\text{Yb}^{2+}$ ), and occasionally samarium ( $\text{Sm}^{2+}$ ) show stable $+2$ states. Most lanthanides do not form $+2$ ions readily. …
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KCET 2018Set A-11 markMCQ
$+3$
›Reveal solutionSolution

Count the electrons in the $+3$ ion and add back the three electrons that were removed.

Step 1 — Read the configuration carefully.

The configuration $[\mathrm{Ar}]3d^5$ belongs to the ion $X^{3+}$ , not to the neutral atom. This is the trap in the question.

Step 2 — Electrons in the ion.

$[\mathrm{Ar}]$ = argon core = $18$ electrons.
$e^-(X^{3+}) = 18 + 5 = 23$
Step 3 — Electrons in the neutral atom.

A $+3$ charge means three electrons were removed, so add them back:
$e^-(X) = 23 + 3 = 26 \quad\Longrightarrow\quad Z = 26$
Step 4 — Identify and cross-check.

$Z = 26$ is iron. Its ground-state configuration is

$\mathrm{Fe}: [\mathrm{Ar}]3d^{6}4s^{2}$ …
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Q.Which of the following actinoids show oxidation states upto +7?

Concept understanding — Actinoid Contraction Greater

The Intuition First

The Precise Statement

Why It Matters …

Why this formula?

Actinoid Contraction: Why It Is Greater Than Lanthanoid Contraction

1. What Is Actinoid Contraction?

2. The Core Reason: Poorer Shielding by 5f Electrons

The Shielding Effect

Why 5f Shielding Is Weaker Than 4f

3. Mathematical Expression of the Trend

4. Why the Difference in Shielding? — Orbital Shape Matters

5. The Chain of Reasoning (Step-by-Step)

Method: Electronic Configuration & Oxidation State Analysis

Steps

Common Mistakes: Actinoid Contraction & Oxidation States up to +7

✗ Mistake 1: Confusing Actinoid Contraction with Lanthanoid Contraction

✗ Mistake 2: Forgetting the +7 Oxidation State Trend

✗ Mistake 3: Misidentifying the Elements in the Options

Q.Which of the following actinoids show oxidation states upto +7?

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