Match List-I with List-II
| List-I | List-II |
| Process | Metal/catalyst Used |
| (A) Contact process | (I) Iron |
| (B) Haber’s process | (II) Vanadium(V) oxide |
| (C) Wacker process | (III) Nickel |
| (D) Catalytic hydrogenation | (IV) Palladium |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
Transition metals and their compounds are good catalysts because they can show variable oxidation states and can adsorb reactant molecules on their surface, which brings the reactants together and weakens their bonds. Each industrial process uses its own catalyst:
(A) Contact process — This matches (II) Vanadium(V) oxide. \( \ce{V2O5} \) catalyses the oxidation of sulphur dioxide to sulphur trioxide in the manufacture of sulphuric acid.
\( \displaystyle \ce{2SO2(g) + O2(g)} \xrightarrow{\ce{V2O5}} \ce{2SO3(g)} \)
(B) Haber’s process — This matches (I) Iron. Finely divided iron (with molybdenum as promoter) catalyses the synthesis of ammonia.
\( \displaystyle \ce{N2(g) + 3H2(g)} \xrightarrow{\ce{Fe}} \ce{2NH3(g)} \)
(C) Wacker process — This matches (IV) Palladium. \( \ce{PdCl2} \) catalyses the oxidation of ethene to ethanal.
\( \displaystyle \ce{2CH2=CH2 + O2} \xrightarrow{\ce{PdCl2}} \ce{2CH3CHO} \)
(D) Catalytic hydrogenation — This matches (III) Nickel. Finely divided nickel catalyses the addition of hydrogen to C=C bonds, as in the hardening of vegetable oils.
\( \displaystyle \ce{RCH=CH2 + H2} \xrightarrow{\ce{Ni}} \ce{RCH2CH3} \)
Shortcut: The pair (C)–(IV) appears in Options (1) and (2), and (A)–(II) then leaves only Option (1).