JEE Main · 2025 · Shift-ImediumBOND-185

Match the LIST-I with LIST-II. | LIST-I (Molecule/ion) | LIST-II (Bond pair : lone pair on central atom) | |---|---| |…

Chemical Bonding · Class 11 · JEE Main Previous Year Question

Question

Match the LIST-I with LIST-II. | LIST-I (Molecule/ion) | LIST-II (Bond pair : lone pair on central atom) | |---|---| | A. \ceICl2\ce{ICl2^-} | I. 4 : 2 | | B. \ceH2O\ce{H2O} | II. 4 : 1 | | C. \ceSO2\ce{SO2} | III. 2 : 3 | | D. \ceXeF4\ce{XeF4} | IV. 2 : 2 |

Choose the correct answer from the options given below:

Options
  1. a

    A-IV, B-III, C-II, D-I

  2. b

    A-III, B-IV, C-II, D-I

  3. c

    A-III, B-IV, C-I, D-II

  4. d

    A-II, B-I, C-IV, D-III

Correct Answerb

A-III, B-IV, C-II, D-I

Detailed Solution

🧠 Count bond pairs and lone pairs on the central atom for each species (using total bond pairs, including π-bond contributions).

🗺️ A. \ceICl2\ce{ICl2^-}: I has 2 bonds + 3 LP → 2 : 3(III)

B. \ceH2O\ce{H2O}: O has 2 bonds + 2 LP → 2 : 2(IV)

C. \ceSO2\ce{SO2}: S forms 2 S–O bonds (one double, one single) counting all bond pairs:

  • 1 S=O: contributes 2 bond pairs (σ + π)
  • 1 S–O: contributes 1 bond pair
  • 1 LP on S
  • Total bond pairs = 3 + 1? Using full count: S effectively has 4 bond pairs (double bond = 2 pairs) + 1 LP4 : 1(II)

D. \ceXeF4\ce{XeF4}: Xe has 4 bonds + 2 LP → 4 : 2(I)

Matching: A–III, B–IV, C–II, D–I

Answer: A-III, B-IV, C-II, D-I\boxed{\text{Answer: A-III, B-IV, C-II, D-I}}

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