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d- and f-Block Elements: Chemistry | JEE Main

Q: Why is an oxidizing agent even though is a noble gas-like stable configuration?

Although has the stable configuration of Xenon, the +3 oxidation state is thermodynamically favored in solution ( ) due to high hydration and lattice enthalpy stabilization of ions.

Strong reducing and oxidizing agents among the following, respectively, are

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\text{Ce}^{4+}

and

\text{Eu}^{2+}

\text{Ce}^{4+}

and

\text{Tb}^{4+}

\text{Ce}^{3+}

and

\text{Ce}^{4+}

\text{Eu}^{2+}

and

\text{Ce}^{4+}

Step-by-step solutionView

Correct answer

Eu2+ readily oxidizes to the stable +3 state making it a strong reducing agent, whereas Ce4+ readily reduces to the stable +3 state making it a strong oxidizing agent.

Option analysis

Why each option works or fails

A ·

\text{Ce}^{4+}

and

\text{Eu}^{2+}

Identifying the species correctly but reversing the requested order of reducing and oxidizing agents. Ensure the order of species strictly matches 'reducing and oxidizing agents, respectively' as specified in the stem.

B ·

\text{Ce}^{4+}

and

\text{Tb}^{4+}

Believing both Ce4+ and Tb4+ are on opposite sides of redox behavior despite both being in the +4 state. Recognize that both Ce4+ and Tb4+ are prone to reduction back to the common +3 state, meaning both act as oxidizing agents rather than a reducing/oxidizing pair.

C ·

\text{Ce}^{3+}

and

\text{Ce}^{4+}

Assuming that Ce3+ acts as a strong reducing agent because it belongs to a multivalent lanthanoid system. Note that +3 is the most stable and ubiquitous oxidation state for lanthanoids, so Ce3+ does not act as a strong reducing agent; Eu2+ is the classic reducing lanthanoid ion.

D ·

\text{Eu}^{2+}

and

\text{Ce}^{4+}

None. This option correctly identifies Eu2+ as the reducing agent and Ce4+ as the oxidizing agent in the requested sequence. Eu2+ has a 4f7 configuration and readily loses an electron to reach the stable +3 state (reducing agent), while Ce4+ easily gains an electron to reach the stable Ce3+ state (oxidizing agent).

Reviewed route

Solution

StepWorking

01concept

The most common and stable oxidation state of lanthanoids is +3. Species in the +2 oxidation state tend to lose an electron to achieve +3. Because of this, they act as strong reducing agents. Species in the +4 oxidation state tend to gain an electron to reach +3. Therefore, they act as strong oxidizing agents. For example, $\text{Eu}^{2+}$ has the configuration $[\text{Xe}] 4f^7$ . It readily oxidizes to the highly stable $+3$ state, so it is a strong reducing agent. On the other hand, $\text{Ce}^{4+}$ ( $4f^0$ ) easily reduces to $\text{Ce}^{3+}$ ( $E^\circ = +1.74\text{ V}$ ). Thus, it is a strong oxidizing agent.

02option_verdict

Reverses the order requested in the question stem: $\text{Ce}^{4+}$ is an oxidizing agent and $\text{Eu}^{2+}$ is a reducing agent, but the question asks for 'reducing and oxidizing agents respectively'.

03option_verdict

Both $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$ are in the +4 oxidation state and act as oxidizing agents because they both tend to get reduced to the +3 state.

04option_verdict

$\text{Ce}^{3+}$ is already in the common stable +3 oxidation state and does not act as a strong reducing agent under standard conditions.

05option_verdict

$\text{Eu}^{2+}$ is a strong reducing agent (oxidizes to $+3$ ) and $\text{Ce}^{4+}$ is a strong oxidizing agent (reduces to $+3$ ), matching the requested 'reducing and oxidizing agents respectively'.

✓discriminator

Recall: +2 lanthanoid ions ( $\text{Eu}^{2+}, \text{Sm}^{2+}, \text{Yb}^{2+}$ ) act as reducing agents, while +4 lanthanoid ions ( $\text{Ce}^{4+}, \text{Tb}^{4+}$ ) act as oxidizing agents to achieve the stable +3 state. Checking the order 'reducing then oxidizing' uniquely pinpoints Option (3).

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why is

\text{Ce}^{4+}

an oxidizing agent even though

4f^0

is a noble gas-like stable configuration?

Although $\text{Ce}^{4+}$ has the stable $4f^0$ configuration of Xenon, the +3 oxidation state is thermodynamically favored in solution ( $E^\circ_{\text{Ce}^{4+}/\text{Ce}^{3+}} = +1.74\text{ V}$ ) due to high hydration and lattice enthalpy stabilization of $+3$ ions.

Answer

Eu2+ readily oxidizes to the stable +3 state making it a strong reducing agent, whereas Ce4+ readily reduces to the stable +3 state making it a strong oxidizing agent.

Why each option works or fails

A: $\text{Ce}^{4+}$ and $\text{Eu}^{2+}$ - Identifying the species correctly but reversing the requested order of reducing and oxidizing agents. Ensure the order of species strictly matches 'reducing and oxidizing agents, respectively' as specified in the stem.
B: $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$ - Believing both Ce4+ and Tb4+ are on opposite sides of redox behavior despite both being in the +4 state. Recognize that both Ce4+ and Tb4+ are prone to reduction back to the common +3 state, meaning both act as oxidizing agents rather than a reducing/oxidizing pair.
C: $\text{Ce}^{3+}$ and $\text{Ce}^{4+}$ - Assuming that Ce3+ acts as a strong reducing agent because it belongs to a multivalent lanthanoid system. Note that +3 is the most stable and ubiquitous oxidation state for lanthanoids, so Ce3+ does not act as a strong reducing agent; Eu2+ is the classic reducing lanthanoid ion.
D · correct: $\text{Eu}^{2+}$ and $\text{Ce}^{4+}$ - None. This option correctly identifies Eu2+ as the reducing agent and Ce4+ as the oxidizing agent in the requested sequence. Eu2+ has a 4f7 configuration and readily loses an electron to reach the stable +3 state (reducing agent), while Ce4+ easily gains an electron to reach the stable Ce3+ state (oxidizing agent).

Step-by-step solution

concept: The most common and stable oxidation state of lanthanoids is +3. Species in the +2 oxidation state tend to lose an electron to achieve +3. Because of this, they act as strong reducing agents. Species in the +4 oxidation state tend to gain an electron to reach +3. Therefore, they act as strong oxidizing agents. For example, $\text{Eu}^{2+}$ has the configuration $[\text{Xe}] 4f^7$ . It readily oxidizes to the highly stable $+3$ state, so it is a strong reducing agent. On the other hand, $\text{Ce}^{4+}$ ( $4f^0$ ) easily reduces to $\text{Ce}^{3+}$ ( $E^\circ = +1.74\text{ V}$ ). Thus, it is a strong oxidizing agent.
option_verdict: Reverses the order requested in the question stem: $\text{Ce}^{4+}$ is an oxidizing agent and $\text{Eu}^{2+}$ is a reducing agent, but the question asks for 'reducing and oxidizing agents respectively'.
option_verdict: Both $\text{Ce}^{4+}$ and $\text{Tb}^{4+}$ are in the +4 oxidation state and act as oxidizing agents because they both tend to get reduced to the +3 state.
option_verdict: $\text{Ce}^{3+}$ is already in the common stable +3 oxidation state and does not act as a strong reducing agent under standard conditions.
option_verdict: $\text{Eu}^{2+}$ is a strong reducing agent (oxidizes to $+3$ ) and $\text{Ce}^{4+}$ is a strong oxidizing agent (reduces to $+3$ ), matching the requested 'reducing and oxidizing agents respectively'.
discriminator: Recall: +2 lanthanoid ions ( $\text{Eu}^{2+}, \text{Sm}^{2+}, \text{Yb}^{2+}$ ) act as reducing agents, while +4 lanthanoid ions ( $\text{Ce}^{4+}, \text{Tb}^{4+}$ ) act as oxidizing agents to achieve the stable +3 state. Checking the order 'reducing then oxidizing' uniquely pinpoints Option (3).