Problem 210 - A Complete Resource Book in Chemistry for JEE Main

Question

The hybridization of orbitals of \(\mathrm{N}\) atom in \(\mathrm{NO}_{3}^{-}, \mathrm{NO}_{2}^{+}\) and \(\mathrm{NH}_{4}^{+}\)are respectively: (a) \(s p^{2}, s p^{3}, s p\) (b) \(\mathrm{sp}, \mathrm{sp}^{2}, \mathrm{sp}^{3}\) (c) sp, sp \(^{3}, \mathrm{sp}^{2}\) (d) \(\mathrm{sp}^{2}, \mathrm{sp}, \mathrm{sp}^{3}\)

Step-by-Step Solution

Answer

The answer is (d): \(\mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^3\).

Step 1: Determine Hybridization of \(\mathrm{N}\) in \(\mathrm{NO}_{3}^{-}\)

To find the hybridization of \(\mathrm{N}\) in \(\mathrm{NO}_3^−\), consider the structure: \(\mathrm{N}\) has 3 \(\sigma\) bonds and there is no lone pair on \(\mathrm{N}\) after completing the octet with resonance. Thus, the hybridization is \(\mathrm{sp}^2\) due to the presence of 3 electron groups around it (3 \(\sigma\) bonds).

Step 2: Determine Hybridization of \(\mathrm{N}\) in \(\mathrm{NO}_{2}^{+}\)

In \(\mathrm{NO}_2^+\), \(\mathrm{N}\) forms 2 \(\sigma\) bonds with no lone pairs because the cationic nature reduces electron count. Therefore, it has 2 electron domains, leading to an \(\mathrm{sp}\) hybridization.

Step 3: Determine Hybridization of \(\mathrm{N}\) in \(\mathrm{NH}_{4}^{+}\)

In \(\mathrm{NH}_4^+\), the \(\mathrm{N}\) forms 4 \(\sigma\) bonds with \(\mathrm{H}\) atoms, without any lone pairs. The hybridization is \(\mathrm{sp}^3\) due to 4 electron domains (4 \(\sigma\) bonds).

Step 4: Compare Options with Results

Now that we have determined the hybridizations, for \(\mathrm{NO}_3^-\) it is \(\mathrm{sp}^2\), for \(\mathrm{NO}_2^+\) it is \(\mathrm{sp}\), and for \(\mathrm{NH}_4^+\) it is \(\mathrm{sp}^3\). The correct order is option (d): \(\mathrm{sp}^2, \mathrm{sp}, \mathrm{sp}^3\).

Key Concepts

NO3- Hybridization

In chemistry, hybridization provides insight into how atoms within a molecule bond with each other. Specifically, in \(\mathrm{NO}_3^-\), or the nitrate ion, nitrogen exhibits \(\mathrm{sp}^2\) hybridization. To understand why, let's delve into its molecular structure.

The nitrogen atom in \(\mathrm{NO}_3^-\) forms three \(\sigma\) bonds with the oxygen atoms, resulting from a planar trigonal arrangement. There are no lone pairs on nitrogen, and the presence of 3 electron groups suggests the use of \(\mathrm{sp}^2\) hybrid orbitals.

- **Structure and Bonding:** - Bond arrangement: **Planar trigonal** - Bonds: 3 \(\sigma\) bonds to oxygen - Type: \(\mathrm{sp}^2\) hybridizationThus, the nitrogen atom is involved in forming a double bond with one oxygen atom and single bonds with the other two, necessitating this particular hybridization for bonding.

NO2+ Hybridization

In the nitronium ion (\(\mathrm{NO}_2^+\)), nitrogen exhibits a different kind of hybridization due to its unique electronic environment. In this case, the formation involves \(\mathrm{sp}\) hybridization. Let's examine how:

- **Electron Configuration**: - The \(\mathrm{NO}_2^+\) ion has a formal positive charge, indicating a cationic nature. - Nitrogen forms two \(\sigma\) bonds with oxygen, and there are no lone pairs on the nitrogen due to the electron loss.
Thus, nitrogen in \(\mathrm{NO}_2^+\) is bonded through a linear molecular geometry, aligning with the \(\mathrm{sp}\) hybridization theory.

- **Structure and Bonding:** - Bond arrangement: **Linear** - Bonds: 2 \(\sigma\) bonds to oxygen - Type: \(\mathrm{sp}\) hybridizationThis setup allows the nitrogen to efficiently bond with two oxygen atoms, resulting in a linear geometry characteristic of \(\mathrm{sp}\) hybridization.

NH4+ Hybridization

For the ammonium ion (\(\mathrm{NH}_4^+\)), nitrogen exhibits \(\mathrm{sp}^3\) hybridization because of its structural and bonding needs. Here's how this happens:

- **Electronic Structure**: - Nitrogen forms four \(\sigma\) bonds with hydrogen atoms. - No lone pairs on nitrogen further imply that all valence electrons are used for \(\sigma\) bonding.Consequently, the molecular geometry is tetrahedral, common for atoms with \(\mathrm{sp}^3\) hybridization.

- **Structure and Bonding:** - Bond arrangement: **Tetrahedral** - Bonds: 4 \(\sigma\) bonds to hydrogen - Type: \(\mathrm{sp}^3\) hybridizationThis configuration makes \(\mathrm{NH}_4^+\) a perfectly symmetrical ion with no lone pairs, allowing optimal overlap of orbitals to form stable bonds with hydrogen.