Medium MCQ +4 / -1 PYQ · JEE Mains 2024

Number of molecules/ions from the following in which the central atom is involved in $\mathrm{sp}^3$ hybridization is ________.

$\mathrm{NO}_3^{-}, \mathrm{BCl}_3, \mathrm{ClO}_2^{-}, \mathrm{ClO}_3^{-}$

  1. A 2 Correct answer
  2. B 4
  3. C 1
  4. D 3

Solution

<p>To determine the number of molecules/ions in which the central atom is involved in $\mathrm{sp}^3$ hybridization, we must analyze the hybridization state for each central atom. Hybridization is typically determined by the number of sigma bonds and lone pairs on the central atom.</p> <p>Let's evaluate each molecule/ion:</p> <p><b>1. $\mathrm{NO}_3^{-}$</b>:</p> <p>The central atom is nitrogen (N). In $\mathrm{NO}_3^{-}$, nitrogen forms three sigma bonds with oxygen atoms and has no lone pairs. Using the formula for hybridization:</p> <p> Number of hybrid orbitals = number of sigma bonds + number of lone pairs</p> <p>For $\mathrm{NO}_3^{-}$: $3 + 0 = 3$ hybrid orbitals. Therefore, nitrogen is $\mathrm{sp}^2$ hybridized.</p> <p><b>2. $\mathrm{BCl}_3$</b>:</p> <p>The central atom is boron (B). In $\mathrm{BCl}_3$, boron forms three sigma bonds with chlorine atoms and has no lone pairs. Using the same formula:</p> <p> For $\mathrm{BCl}_3$: $3 + 0 = 3$ hybrid orbitals. Therefore, boron is $\mathrm{sp}^2$ hybridized.</p> <p><b>3. $\mathrm{ClO}_2^{-}$</b>:</p> <p>The central atom is chlorine (Cl). In $\mathrm{ClO}_2^{-}$, chlorine forms two sigma bonds with oxygen atoms and has two lone pairs. Therefore,:</p> <p> For $\mathrm{ClO}_2^{-}$: $2 + 2 = 4$ hybrid orbitals. Thus, chlorine is $\mathrm{sp}^3$ hybridized.</p> <p><b>4. $\mathrm{ClO}_3^{-}$</b>:</p> <p>The central atom is chlorine (Cl). In $\mathrm{ClO}_3^{-}$, chlorine forms three sigma bonds with oxygen atoms and has one lone pair. Therefore,:</p> <p> For $\mathrm{ClO}_3^{-}$: $3 + 1 = 4$ hybrid orbitals. Thus, chlorine is $\mathrm{sp}^3$ hybridized.</p> <p>Based on the analysis, the molecules/ions $\mathrm{ClO}_2^{-}$ and $\mathrm{ClO}_3^{-}$ have their central atoms involved in $\mathrm{sp}^3$ hybridization.</p> <p>Therefore, the total number is <strong>2</strong>.</p> <p>Thus, the correct option is <strong>Option A</strong>.</p>

About this question

Subject: Chemistry · Chapter: Chemical Bonding and Molecular Structure · Topic: Ionic and Covalent Bonding

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