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JEE Chemistry 2025 Question with Solution

Match the LIST-I with LIST-II.

Table matching question with LIST-I molecules or ions and LIST-II bond pair to lone pair ratios on the central atom.

Choose the correct answer from the options given below :

  • A

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

  • B

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

  • C

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

  • D

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

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A matching question asks for the bond pair : lone pair ratio on the central atom for the species in LIST-I.

Find: The correct matching between the molecules/ions and the ratios in LIST-II using VSEPR theory.

Worked solution table showing LIST-I as ICl2-, H2O, SO2, XeF4 and LIST-II as ratios 4:2, 4:1, 2:3, 2:2.

For each species, count the bond pairs and lone pairs on the central atom.

  1. ICl2−\mathrm{ICl_2^-}: the central iodine has 2 bond pairs and 3 lone pairs. So it matches III.
  2. H2O\mathrm{H_2O}: the central oxygen has 2 bond pairs and 2 lone pairs. So it matches IV.
  3. SO2\mathrm{SO_2}: the central sulfur has 4 bond pairs and 1 lone pair. So it matches II.
  4. XeF4\mathrm{XeF_4}: the central xenon has 4 bond pairs and 2 lone pairs. So it matches I.

Hence the correct matching is:

  • A-III
  • B-IV
  • C-II
  • D-I

Therefore, the correct option is B.

Detailed VSEPR Matching

Given:

  • A = ICl2−\mathrm{ICl_2^-}
  • B = H2O\mathrm{H_2O}
  • C = SO2\mathrm{SO_2}
  • D = XeF4\mathrm{XeF_4}

Find: Match each species with the correct bond pair : lone pair ratio on the central atom.

Using VSEPR theory:

  • In ICl2−\mathrm{ICl_2^-}, iodine is the central atom. It forms 2 bonds with chlorine atoms and retains 3 lone pairs. So the ratio is 2:32:3, which is III.
  • In H2O\mathrm{H_2O}, oxygen forms 2 bonds with hydrogen atoms and has 2 lone pairs. So the ratio is 2:22:2, which is IV.
  • In SO2\mathrm{SO_2}, sulfur has two double bonds with oxygen and one lone pair. Counting bonding pairs on the central atom gives 4 bond pairs and 1 lone pair. So the ratio is 4:14:1, which is II.
  • In XeF4\mathrm{XeF_4}, xenon forms 4 bonds with fluorine atoms and has 2 lone pairs. So the ratio is 4:24:2, which is I.

Thus,

A→IIIB→IVC→IID→I\begin{aligned} A &\rightarrow III \\ B &\rightarrow IV \\ C &\rightarrow II \\ D &\rightarrow I \end{aligned}

Therefore, the correct matching is A-III, B-IV, C-II, D-I, so the correct option is B.

Common mistakes

  • Students often count only the number of atoms attached to the central atom and ignore lone pairs. This is wrong because the question asks for bond pair : lone pair on the central atom. Always determine both bonded electron pairs and non-bonding pairs using VSEPR theory.

  • A common mistake in SO2\mathrm{SO_2} is to count each double bond as one bond pair for this question. That gives an incorrect ratio. Here the solution counts the bonding electron pairs around sulfur as 4 bond pairs and 1 lone pair, so match it accordingly.

  • Students may forget the extra electron in ICl2−\mathrm{ICl_2^-} because of the negative charge. This is wrong because the charge changes the electron count on the central atom. Include the charge before deciding the number of lone pairs.

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