Chemistry Q57 – JEE-MAIN 2025

Home Chemistry Chemical BondingQuestion #248

Which among the following molecules is (a) involved in

s p^{3} d

hybridization, (b) has different bond lengths and (c) has lone pair of electrons on the central atom ?

Clue 1 of 3: VSEPR Theory and Hybridization

Determine the hybridization and electron geometry for the central atom in each molecule by counting the total number of electron pairs (bond pairs + lone pairs).

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Ninja Strategy•Sequential Condition Check

Quickly eliminate options by checking each condition (hybridization, lone pairs, bond lengths) in sequence, starting with the easiest to verify.

🧠 Full Solution Path

Step 1: Analyze Hybridization and Lone Pairs for Each Molecule✦ Active

For each molecule, determine the number of valence electrons on the central atom, the number of bond pairs, and the number of lone pairs. Summing bond pairs and lone pairs gives the steric number, which dictates hybridization and electron geometry.

\begin{array}{ccccccc} Molecule & Central Atom & {Valence e}^{-} & Bond Pairs & Lone Pairs & Steric Number & Hybridization \\ P F_{5} & P & 5 & 5 & 0 & 5 & s p^{3} d \\ X e F_{2} & X e & 8 & 2 & 3 & 5 & s p^{3} d \\ S F_{4} & S & 6 & 4 & 1 & 5 & s p^{3} d \\ X e F_{4} & X e & 8 & 4 & 2 & 6 & s p^{3} d^{2} \end{array}

Step 2: Evaluate Conditions (a) and (c)○ Expand

From Step 1, we can check conditions (a) $s p^{3} d$ hybridization and (c) presence of lone pairs on the central atom.

\begin{array}{ccc} Molecule & (a) s p^{3} d Hybridization? & (c) Lone Pair on Central Atom? \\ P F_{5} & Yes & No (0 LP) \\ X e F_{2} & Yes & Yes (3 LP) \\ S F_{4} & Yes & Yes (1 LP) \\ X e F_{4} & No (s p^{3} d^{2}) & Yes (2 LP) \end{array}

Only $X e F_{4}$ is eliminated by condition (a). $P F_{5}$ is eliminated by condition (c). $X e F_{2}$ and $S F_{4}$ remain.

💡 Teacher's Secret Hint

Remember that $s p^{3} d$ hybridization corresponds to a steric number of 5.

Step 3: Evaluate Condition (b) for Remaining Molecules○ Expand

Now, consider condition (b) different bond lengths for $X e F_{2}$ and $S F_{4}$ . Both have $s p^{3} d$ hybridization and a trigonal bipyramidal electron geometry.

For $X e F_{2}$ : It has 2 bond pairs and 3 lone pairs. The 3 lone pairs occupy the equatorial positions, and the 2 bond pairs occupy the axial positions, resulting in a linear molecular geometry. All Xe-F bonds are axial and are equivalent in length. Thus, $X e F_{2}$ does not have different bond lengths.

For $S F_{4}$ : It has 4 bond pairs and 1 lone pair. The lone pair occupies an equatorial position, leading to a seesaw molecular geometry. In this geometry, there are two axial S-F bonds and two equatorial S-F bonds. Axial bonds are generally longer than equatorial bonds due to greater repulsion from equatorial bond pairs. Therefore, $S F_{4}$ has different bond lengths.

Since $S F_{4}$ satisfies all three conditions ((a) $s p^{3} d$ hybridization, (b) different bond lengths, and (c) one lone pair on the central atom), it is the correct answer.

💡 Teacher's Secret Hint

Recall the VSEPR rules for placing lone pairs in trigonal bipyramidal geometry (equatorial positions are preferred) and how this affects molecular geometry and bond equivalency.

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Chemistry Question 57 – JEE-MAIN 2025

Clue 1 of 3: VSEPR Theory and Hybridization

🧠 Full Solution Path