Chemistry Q58 – JEE-MAIN 2025

Home Chemistry Chemical BondingQuestion #250

Match the LIST-I with LIST-I

\begin{array}{lcl} LIST-I & LIST-II \\ Molecule / ion & Bond pair : lone pair \\ (on the central atom) \\ A) I C l_{2}^{-} & I) 4 : 2 \\ B) H_{2} O & II) 4 : 1 \\ C) S O_{2} & III) 2 : 3 \\ D) X e F_{4} & IV) 2 : 2 \end{array}

Choose the correct answer from the options given below:

Clue 1 of 3: VSEPR Theory

Determine the central atom and its valence electrons for each species.

🥷

Ninja Strategy•Common Molecule Recognition

Quickly determine the bond pair and lone pair ratios for common molecules like $H_{2} O$ and $X e F_{4}$ to eliminate multiple options rapidly.

🧠 Full Solution Path

Step 1: Determine Steric Number, Bond Pairs (BP), and Lone Pairs (LP) for each species.✦ Active

We will use the VSEPR theory to determine the number of bond pairs and lone pairs on the central atom for each given molecule/ion. The steric number (SN) is the sum of bond pairs and lone pairs. For species with only single bonds, $S N = (V + M + C) / 2$ , where $V$ is valence electrons of central atom, $M$ is number of monovalent atoms, and $C$ is charge. For species with multiple bonds, each multiple bond counts as one electron domain.

I C l_{2}^{-}

: Central atom is I. Valence electrons of I = 7. Monovalent atoms (Cl) = 2. Charge = -1.

S N = (7 + 2 + 1) / 2 = 5

. BP = 2 (from 2 Cl atoms). LP =

S N - B P = 5 - 2 = 3

. Ratio = 2:3. B)

H_{2} O

: Central atom is O. Valence electrons of O = 6. Monovalent atoms (H) = 2. Charge = 0.

S N = (6 + 2 + 0) / 2 = 4

. BP = 2 (from 2 H atoms). LP =

S N - B P = 4 - 2 = 2

. Ratio = 2:2. C)

S O_{2}

: Central atom is S. Valence electrons of S = 6. Lewis structure shows S forming two bonds with O atoms and having one lone pair. Thus, 2 bonding domains and 1 lone pair.

S N = 2 + 1 = 3

. BP = 2 (from 2 O atoms). LP =

S N - B P = 3 - 2 = 1

. Ratio = 2:1. D)

X e F_{4}

: Central atom is Xe. Valence electrons of Xe = 8. Monovalent atoms (F) = 4. Charge = 0.

S N = (8 + 4 + 0) / 2 = 6

. BP = 4 (from 4 F atoms). LP =

S N - B P = 6 - 4 = 2

. Ratio = 4:2.

Step 2: Match the calculated ratios with LIST-II.○ Expand

Based on the calculated bond pair to lone pair ratios:

I C l_{2}^{-}

(2:3) matches LIST-II (III). B)

H_{2} O

(2:2) matches LIST-II (IV). C)

S O_{2}

(2:1) matches LIST-II (II). D)

X e F_{4}

(4:2) matches LIST-II (I).

Step 3: Identify the correct option.○ Expand

The correct matching sequence is A-III, B-IV, C-II, D-I. This corresponds to option 3.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Chemistry Question 58 – JEE-MAIN 2025

Clue 1 of 3: VSEPR Theory

🧠 Full Solution Path