Chemistry Q58 – JEE-MAIN 2025

Home Chemistry Chemical BondingQuestion #251

S O_{2}

N O_{2}^{-}

and

N_{3}^{-}

the hybridizations at the central atom are respectively:

Clue 1 of 3: VSEPR Theory

Hybridization of a central atom is determined by its steric number, which is the sum of the number of sigma bonds and lone pairs around it.

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Ninja Strategy•Systematic VSEPR Application

Determine the steric number for each central atom by counting sigma bonds and lone pairs, then assign hybridization to quickly eliminate options.

🧠 Full Solution Path

Step 1: Determine Lewis Structures and Steric Numbers✦ Active

For $S O_{2}$ : The central S atom forms two sigma bonds (one in each S=O double bond) and has one lone pair. Steric number = 2 (sigma bonds) + 1 (lone pair) = 3.

For $N O_{2}^{-}$ : The central N atom forms two sigma bonds (one in N=O, one in N-O) and has one lone pair. Steric number = 2 (sigma bonds) + 1 (lone pair) = 3.

For $N_{3}^{-}$ : The central N atom forms two sigma bonds (one in each N=N double bond in the major resonance structure) and has zero lone pairs. Steric number = 2 (sigma bonds) + 0 (lone pairs) = 2.

Step 2: Assign Hybridization based on Steric Number○ Expand

For $S O_{2}$ : Steric number 3 corresponds to $s p^{2}$ hybridization.

For $N O_{2}^{-}$ : Steric number 3 corresponds to $s p^{2}$ hybridization.

For $N_{3}^{-}$ : Steric number 2 corresponds to $s p$ hybridization.

Step 3: Conclude the Hybridizations○ Expand

The hybridizations for $S O_{2}$ , $N O_{2}^{-}$ , and $N_{3}^{-}$ are $s p^{2}$ , $s p^{2}$ , and $s p$ respectively.

💡 Teacher's Secret Hint

Remember to consider all resonance structures when determining the number of sigma bonds and lone pairs for the central atom.

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Chemistry Question 58 – JEE-MAIN 2025

Clue 1 of 3: VSEPR Theory

🧠 Full Solution Path