Q.Match the catalysts given in Column I with the processes given in Column II.
Column I (Catalyst):

The key to this question is knowing the difference between transition elements (d-block) and inner transition elements (f-block). Inner transition elements are those where the last electron enters the f-orbital — specifically the 4f series (lanthanoids) and the 5f series (actinoids). Elements from the d-block, even if they are transition metals, are not inner transition elements.

Let’s check each set one by one.

1. Set I: Nb, W, Sg

   • Nb (Niobium, atomic number 41) is in the 4d series — a d-block element.
   • W (Tungsten, atomic number 74) is in the 5d series — a d-block element.
   • Sg (Seaborgium, atomic number 106) is in the 6d series — a d-block element. All three are d-block transition metals, not f-block. So Set I is incorrect.

2. Set II: Hs, Mt, Ds

   • Hs (Hassium, atomic number 108) is in the 6d series.
   • Mt (Meitnerium, atomic number 109) is in the 6d series.
   • Ds (Darmstadtium, atomic number 110) is in the 6d series. These are all d-block elements (group 8, 9, 10 respectively). None are f-block. So Set II is incorrect.

3. Set III: Tb, Dy, Cf

   • Tb (Terbium, atomic number 65) is a lanthanoid — 4f-block.
   • Dy (Dysprosium, atomic number 66) is a lanthanoid — 4f-block.
   • Cf (Californium, atomic number 98) is an actinoid — 5f-block. …

Concept & Intuition

Superoxides contain the $O_2^-$ ion, which is relatively large and has a negative charge spread over two oxygen atoms. To form a stable ionic solid, the metal cation must be large enough to accommodate the big $O_2^-$ anion without destabilizing the lattice. Small cations (like Li⁺, Be²⁺, Mg²⁺) have high charge density and polarize the $O_2^-$ ion, causing it to decompose into oxide ($O^{2-}$) and peroxide ($O_2^{2-}$) instead. Only the largest, most electropositive alkali metals—K, Rb, Cs—have cations large enough to stabilize the superoxide ion in the solid state.

Step-by-step reasoning

1. Recall the trend in alkali metal reactivity with oxygen

   • Lithium forms only the normal oxide, Li₂O.
   • Sodium forms mainly the peroxide, Na₂O₂.
   • Potassium, rubidium, and cesium form stable superoxides: KO₂, RbO₂, CsO₂. This trend is directly tied to cation size: as you go down Group 1, the cation radius increases, lowering the lattice energy difference between oxide, peroxide, and superoxide.

2. Why small cations fail

   • Li⁺ is very small and has high charge density. It strongly polarizes the $O_2^-$ ion, pulling electron density toward itself and breaking the O–O bond, leading to oxide formation.
   • Be²⁺ and Mg²⁺ (Group 2) are even smaller and have +2 charge, so they never form superoxides; they only form normal oxides (BeO, MgO) or, under special conditions, peroxides (MgO₂).

3. Why large cations succeed

   • K⁺, Rb⁺, Cs⁺ are large and have low charge density. They cannot polarize the $O_2^-$ ion significantly, so the superoxide ion remains intact. …

The question asks for two elements, X and Y, such that both their dioxides (XO₂, YO₂) and their monoxides (XO, YO) are amphoteric. “Amphoteric” means the oxide can react with both acids and bases. This property is common among elements near the “staircase” line between metals and nonmetals in the periodic table, especially in the p-block.

Why this approach works:

We need to check each pair’s known oxide chemistry. Carbon and silicon are in Group 14, but their monoxides (CO, SiO) are not amphoteric — CO is neutral, SiO is acidic. Germanium’s monoxide (GeO) is amphoteric, but its dioxide (GeO₂) is predominantly acidic. Lead’s monoxide (PbO) is amphoteric, but its dioxide (PbO₂) is strongly oxidizing and not typically amphoteric in the same sense. Only tin and lead have both oxides that show clear amphoteric behavior.

Step-by-step reasoning:

1. Recall the trend in Group 14 (C, Si, Ge, Sn, Pb):

   As we go down the group, metallic character increases. Oxides of nonmetals tend to be acidic; oxides of metals tend to be basic. Amphoteric oxides occur in the middle — where the element is a metalloid or a weak metal.

2. Check each option:

   • (A) C, Si:

     • CO (carbon monoxide) is neutral — not amphoteric.
     • CO₂ is acidic.
     • SiO (silicon monoxide) is not stable; SiO₂ is acidic. → Neither element has both oxides amphoteric. ✗

   • (B) Si, Ge:

     • SiO₂ is acidic; GeO₂ is predominantly acidic (though weakly amphoteric under some conditions).
     • SiO is not amphoteric; GeO is amphoteric. → Not both oxides for both elements. ✗

   • (C) Sn, Pb:

     • SnO₂ (tin dioxide) is amphoteric — reacts with acids to give Sn⁴⁺ salts and with bases to give stannates.
     • SnO (tin monoxide) is also amphoteric — reacts with acids to give Sn²⁺ and with bases to give stannites. …

The outer configuration $ns^2np^1$ places the element squarely in group 13 of the periodic table. All four options — boron, aluminium, gallium, indium — belong here. The clue is the sharp contrast between the crystalline and amorphous forms.

Crystalline boron has a very stable, covalently bonded network structure (icosahedral $B_{12}$ units). This makes it extremely hard and chemically resistant — it does not react with air even at moderate heat. In contrast, amorphous boron is a disordered, more reactive form. When heated in air, it burns readily to form boron trioxide, $B_2O_3$.

Now, the nature of the oxide: $B_2O_3$ is distinctly acidic. It dissolves in water to give boric acid, $H_3BO_3$, a weak acid. This fits the description perfectly.

Aluminium, gallium, and indium behave differently. Their crystalline forms are metallic and react with air — aluminium forms a protective oxide layer, but it does react; gallium and indium also oxidise. More importantly, their oxides ($Al_2O_3$, $Ga_2O_3$, $In_2O_3$) are amphoteric, not purely acidic. Only boron gives a strictly acidic oxide.

1. Identify the group: $ns^2np^1$ → group 13 (boron family).
2. Check crystalline reactivity: Only boron’s crystalline allotrope is inert to air. Al, Ga, In all react with oxygen. …