JEE Main · 2019 · Shift-ImediumPB12-021

Match the catalysts (Column I) with products (Column II) | | Column I (Catalyst) | | Column II (Product) |…

p-Block Elements (Class 12) · Class 12 · JEE Main Previous Year Question

Question

Match the catalysts (Column I) with products (Column II)

| | Column I (Catalyst) | | Column II (Product) | |---|---|---|---| | A. | V2O5\mathrm{V_2O_5} | P. | Polyethylene | | B. | TiCl4/Al(Me)3\mathrm{TiCl_4/Al(Me)_3} | Q. | Ethanal | | C. | PdCl2\mathrm{PdCl_2} | R. | H2SO4\mathrm{H_2SO_4} | | D. | Iron Oxide | S. | NH3\mathrm{NH_3} |

Options
  1. a

    (A)-(S); (B)-(R); (C)-(Q); (D)-(P)

  2. b

    (A)-(R); (B)-(S); (C)-(P); (D)-(Q)

  3. c

    (A)-(Q); (B)-(R); (C)-(P); (D)-(S)

  4. d

    (A)-(R); (B)-(P); (C)-(Q); (D)-(S)

    ✓
Correct Answerd

(A)-(R); (B)-(P); (C)-(Q); (D)-(S)

Detailed Solution

Strategy: Identify the specific conditions required for the formation of Xenon fluorides.

Step 1: Preparation Reactions Xenon fluorides are prepared by direct reaction of Xenon and Fluorine under specific ratios and temperatures:

  • \ceXe(excess)+F2→673K,1barXeF2\ce{Xe(excess) + F2 \xrightarrow{673 K, 1 bar} XeF2}
  • \ceXe+5F2→873K,7barXeF4\ce{Xe + 5F2 \xrightarrow{873 K, 7 bar} XeF4}
  • \ceXe+20F2→573K,60−70barXeF6\ce{Xe + 20F2 \xrightarrow{573 K, 60-70 bar} XeF6}

Step 2: Conclusion Matching the given ratio (e.g., 1:5 or 1:20) and temperature/pressure conditions allows the identification of the specific fluoride formed.

Answer: (Option) [d] [Based on specific reaction conditions in question]\boxed{\text{Answer: (Option) [d] [Based on specific reaction conditions in question]}}

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