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.

Related Formula

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

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 ≈ 3.87$\mu \approx 3.87$.

Chapter Mix

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

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