Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:

Q: Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:

Cerium (Ce, Atomic Number = 58) has the electronic configuration [Xe] , 4f1 5d1 6s2. In the +3 oxidation state, Ce3+ has the configuration [Xe] , 4f1. Although +3 is the most common state for lanthanoids, Cerium readily loses one more electron from its 4f subshell to form Ce4+. The electronic configuration of Ce4+ becomes [Xe] , 4f0, which is the highly stable configuration of the nearest noble gas, Xenon. Cerium shows a +4 oxidation state because it attains a stable empty f-subshell (4f0) confi

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Solution & Explanation

Core Logic

Cerium ( $Ce$ $Ce$, Atomic Number = 58) has the electronic configuration $[Xe] 4f¹ 5d¹ 6s²$ $[Xe] \, 4f^1 5d^1 6s^2$. In the +3 oxidation state, $Ce³⁺$ $Ce^{3+}$ has the configuration $[Xe] 4f¹$ $[Xe] \, 4f^1$. Although +3 is the most common state for lanthanoids, Cerium readily loses one more electron from its $4f$ $4f$ subshell to form $Ce⁴⁺$ $Ce^{4+}$. The electronic configuration of $Ce⁴⁺$ $Ce^{4+}$ becomes $[Xe] 4f⁰$ $[Xe] \, 4f^0$, which is the highly stable configuration of the nearest noble gas, Xenon.

Step 1: Final Conclusion

Cerium shows a +4 oxidation state because it attains a stable empty f-subshell ( $4f⁰$ $4f^0$) configuration.

Pattern Recognition

Anomalous oxidation states in f-block elements (+2, +4) usually occur when the ion can achieve a stable $f⁰$ $f^0$ (empty), $f⁷$ $f^7$ (half-filled), or $f¹⁴$ $f^{14}$ (fully filled) electronic configuration. For Ce (+4), it's $f⁰$ $f^0$.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements

More The d- and f-Block Elements Previous-Year Questions

Q53 neet_2026_03_may_morning Catalytic Properties of Transition Metals

Match List I with List II :

List-I (Transition metal/compound complex)	List-II (Catalytic Role)
A. $V₂O₅$ $V_{2}O_{5}$	(I) Preparation of ammonia from $N₂/H₂$ $N_{2}/H_{2}$ mixture
B. Fe	(II) Polymerisation of alkynes
C. $PdCl₂$ $PdCl_{2}$	(III) Preparation of $H₂SO₄$ $H_{2}SO_{4}$ and $SO₂$ $SO_{2}$
D. Ni complex	(IV) Oxidation of ethyne to ethanal

Choose the correct answer from the options given below.

A. A-III, B-IV, C-I, D-II
B. A-II, B-I, C-IV, D-III
C. A-IV, B-I, C-III, D-II
D. A-III, B-I, C-IV, D-II

Solution

Core Logic

(A) $V₂O₅$ $V_2O_5$ $arrow$ $\rightarrow$ Catalyses the oxidation of $SO₂$ $SO_2$ into $SO₃$ $SO_3$ in the Contact process for the manufacture of $H₂SO₄$ $H_2SO_4$. (Matches III) (B) Fe $arrow$ $\rightarrow$ Acts as a catalyst in Haber's process for the preparation of ammonia from $N₂ / H₂$ $N_2 / H_2$ mixture. (Matches I) (C) $PdCl₂$ $PdCl_2$ $arrow$ $\rightarrow$ Wacker process for the oxidation of ethyne to ethanal. (Matches IV) (D) Ni complex $arrow$ $\rightarrow$ Catalyses the polymerisation of alkynes. (Matches II)

Step 1: Final Match

A $arrow$ $\rightarrow$ III B $arrow$ $\rightarrow$ I C $arrow$ $\rightarrow$ IV D $arrow$ $\rightarrow$ II

Pattern Recognition

Classic industrial catalysts match: Iron = Haber (Ammonia). Vanadium Pentoxide = Contact (Sulfuric Acid). Palladium Chloride = Wacker (Acetaldehyde).

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements

Q57 neet_2026_03_may_morning Magnetic Properties

The calculated 'spin-only' magnetic moment of

Ti²⁺(3d²)

is :

A. 2.84 BM
B. 5.92 BM
C. 4.90 BM
D. 3.87 BM

Solution

Related Formula

μ = \sqrt(n(n + 2)) B.M.

where $n$ $n$ is the number of unpaired electrons.

Core Logic

Electronic configuration of Titanium ( $Ti, Z=22$ $Ti, Z=22$): $[Ar] 4s² 3d²$ $[Ar] \, 4s^2 3d^2$. For $Ti²⁺$ $Ti^{2+}$, two electrons are removed from the 4s orbital first. Electronic configuration of $Ti²⁺ [Ar] 4s⁰ 3d²$ $Ti^{2+} \implies [Ar] \, 4s^0 3d^2$. Number of unpaired electrons ( $n$ $n$) = 2.

Step 1: Calculate Magnetic Moment

μ = \sqrt(2(2 + 2))

μ = \sqrt(2 \times 4) = \sqrt(8)

μ = 2.828 B.M. \approx 2.84 B.M.

Pattern Recognition

The spin-only magnetic moment value is always slightly greater than the number of unpaired electrons ( $n$ $n$). For $n=2$ $n=2$, $μ$ $\mu$ is approx $2.8$ $2.8$. For $n=3$ $n=3$, $μ \approx 3.87$ $\mu \approx 3.87$.

Chapter Mix

Class 12 Chemistry: The d- and f-Block Elements

Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More The d- and f-Block Elements Previous-Year Questions

Solution

Core Logic

Step 1: Final Match

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Magnetic Moment

Pattern Recognition

Chapter Mix

More The d- and f-Block Elements Questions — neet_2026_03_may_morning

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Although +3 oxidation state is most common in lanthanoids, cerium still shows +4 oxidation state because:

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More The d- and f-Block Elements Previous-Year Questions

Solution

Core Logic

Step 1: Final Match

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Magnetic Moment

Pattern Recognition

Chapter Mix

More The d- and f-Block Elements Questions — neet_2026_03_may_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

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