JEE Main · 2021 · Shift-IhardPB12-030

Match List-I with List-II: | | List-I (Species) | | List-II (Number of lone pairs on central atom) | |---|---|---|---|…

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

Question

Match List-I with List-II:

| | List-I (Species) | | List-II (Number of lone pairs on central atom) | |---|---|---|---| | (a) | XeF2\mathrm{XeF_2} | (i) | 0 | | (b) | XeO2F2\mathrm{XeO_2F_2} | (ii) | 1 | | (c) | XeO3F2\mathrm{XeO_3F_2} | (iii) | 2 | | (d) | XeF4\mathrm{XeF_4} | (iv) | 3 |

Choose the most appropriate answer from the options given below:

Options
  1. a

    (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)

  2. b

    (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

  3. c

    (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)

  4. d

    (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)

    ✓
Correct Answerd

(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)

Detailed Solution

Strategy: Calculate lone pairs on Xenon in each species using valence shell electrons.

Step 1: Lone Pair Calculation Xenon has 8 valence electrons.

  • (a) \ceXeF2\ce{XeF2}: 2 Bond pairs (F), 6 electrons left →\to 3 lone pairs. (match iv)
  • (b) \ceXeO2F2\ce{XeO2F2}: 4 electrons in \ceXe=O\ce{Xe=O} bonds, 2 in \ceXe−F\ce{Xe-F} bonds. 2 electrons left →\to 1 lone pair. (match ii)
  • (c) \ceXeO3F2\ce{XeO3F2}: 6 electrons in \ceXe=O\ce{Xe=O} bonds, 2 in \ceXe−F\ce{Xe-F} bonds. 0 electrons left →\to 0 lone pairs. (match i)
  • (d) \ceXeF4\ce{XeF4}: 4 Bond pairs (F), 4 electrons left →\to 2 lone pairs. (match iii)

Step 2: Conclusion Matching (a)-iv, (b)-ii, (c)-i, (d)-iii corresponds to option d.

Answer: (Option) [d] (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)\boxed{\text{Answer: (Option) [d] (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)}}

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