The d- and f-Block Elements, on the exam screen.

$\mathrm{V}_{2}\mathrm{O}_{5}$ is the contact-process catalyst linked with $\mathrm{H}_{2}\mathrm{SO}_{4}$/$\mathrm{SO}_{2}$ chemistry (III); Fe is the Haber-process catalyst for ammonia from $\mathrm{N}_{2}$/$\mathrm{H}_{2}$ (I); $\mathrm{PdCl}_{2}$ catalyses Wacker oxidation of ethyne/ethene to ethanal (IV); and Ni complexes catalyse polymerisation of alkynes (II). Hence P-III, Q-I, R-IV, S-II.

Cerium (Z = 58) has the configuration [Xe]4f1 5d1 6s2; on going to Ce4+ it loses its outer electrons to reach the especially stable, empty 4f0 (noble-gas-like) configuration, which is why it readily shows the +4 state. The options citing 4f14, radon, or atomic number 61 are factually wrong for cerium.

The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ Bohr magneton (BM), where n is the number of unpaired electrons. Ti$^{2+}$ is [Ar]3d$^{2}$ with 2 unpaired electrons, so $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84$ BM.

Melting point tracks the strength of the metallic bond, which depends on how many electrons each atom contributes to the delocalised sea. In s-block metals only the ns electrons participate, but in the 3d series the unpaired 3d electrons join the 4s electrons in bonding, producing much stronger interatomic attraction and correspondingly high melting points. The trend peaks near the middle of the series, where the number of unpaired d electrons is greatest. R is therefore the correct explanation of A. NCERT Class 12, The d- and f-Block Elements.

Lanthanoid tripositive ions have the configuration $[\mathrm{Xe}]4\mathrm{f}^{\mathrm{n}}$, where n is the atomic number minus 57. That gives $\mathrm{Ce}^{3+} = 4f^1$, $\mathrm{Nd}^{3+} = 4f^3$, $\mathrm{Tb}^{3+} = 4f^8$ and $\mathrm{Ho}^{3+} = 4f^{10}$. Filling seven f orbitals by Hund's rule, $4f^{10}$ places three pairs and leaves FOUR unpaired electrons, so $\mathrm{Ho}^{3+}$ is the answer. NCERT Class 12, The d- and f-Block Elements (lanthanoids).

Heating permanganate near 513 K decomposes it as 2KMnO$_{4}$ → K$_{2}$MnO$_{4}$ + MnO$_{2}$ + O$_{2}$ at 513 K. The manganate ion $\mathrm{MnO}_{4}^{2-}$ has manganese in the +6 state with a $d^1$ configuration, so it is both intensely green and paramagnetic - permanganate itself, with $d^0$ manganese(VII), is purple and diamagnetic. NCERT Class 12, The d- and f-Block Elements (potassium permanganate).

Chromium is +6 in both CrO$_{4}^{2-}$ and Cr$_{2}$O$_{7}^{2-}$, so the statement that their oxidation states differ is incorrect. The other three statements are valid. Chapter: The d- and f-Block Elements.