Problem 82 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

(b) \( \mathrm{sp}, \mathrm{sp}^{2}, \mathrm{sp}^{3} \) respectively.

Step 1: Determine Hybridization for \( \mathrm{NO}_{2}^{+} \)

The Lewis structure of \( \mathrm{NO}_{2}^{+} \) shows nitrogen bonded to two oxygen atoms and carrying a positive charge. There are a total of 2 sigma bonds and no lone pairs on the nitrogen. The hybridization of the central atom can be found using the formula \(\text{Number of sigma bonds} + \text{Number of lone pairs} = \text{number of hybrid orbitals}\). For \( \mathrm{NO}_{2}^{+} \), this equals 2 (2 sigma bonds and 0 lone pair). With 2 hybrid orbitals, the hybridization is \( \mathrm{sp} \).

Step 2: Determine Hybridization for \( \mathrm{NO}_{3}^{-} \)

In \( \mathrm{NO}_{3}^{-} \), nitrogen is bonded to three oxygen atoms with one extra electron, forming a resonance structure. Since the central nitrogen makes 3 sigma bonds and has no lone pairs, use the same formula as above. So \(\text{Number of sigma bonds} + \text{Number of lone pairs} = 3 + 0 = 3\) hybrid orbitals. This results in \( \mathrm{sp}^{2} \) hybridization.

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

In \( \mathrm{NH}_{4}^{+} \), nitrogen forms 4 sigma bonds by bonding with four hydrogen atoms. With 4 sigma bonds and no lone pairs, the hybridization is calculated as \(4\) (3 from bonding and 1 from the electron pair), resulting in \( \mathrm{sp}^{3} \) hybridization.

Step 4: Match Choices with Calculated Hybridizations

Now, compare the calculated hybridizations: \( \mathrm{sp} \) for \( \mathrm{NO}_{2}^{+} \), \( \mathrm{sp}^{2} \) for \( \mathrm{NO}_{3}^{-} \), and \( \mathrm{sp}^{3} \) for \( \mathrm{NH}_{4}^{+} \) with the given choices:- (a) \( \mathrm{sp}^{2}, \mathrm{sp}^{3}, \mathrm{sp}^{2} \)- (b) \( \mathrm{sp}, \mathrm{sp}^{2}, \mathrm{sp}^{3} \)- (c) \( \mathrm{sp}^{2}, \mathrm{sp}, \mathrm{sp}^{3} \)- (d) \( \mathrm{sp}^{2}, \mathrm{sp}^{3}, \mathrm{sp} \)The correct sequence based on our determinations is option (b).

Key Concepts

Hybridization of Nitrogen in NO2+ Ion

In the nitric oxide ion, known as the \( \mathrm{NO}_{2}^{+} \) ion, nitrogen forms a key structural role. Here, nitrogen is bonded to two oxygen atoms with a positive charge affecting the bonding structure. The hybridization concept helps us understand this bonding better.
Nitrogen in \( \mathrm{NO}_{2}^{+} \) forms two sigma bonds and has no lone pairs. The lone pairs and sigma bonds often dictate what type of hybridization we observe. Using the formula \( \text{Number of sigma bonds} + \text{Number of lone pairs} = \text{number of hybrid orbitals} \), we find that this nitrogen has 2 hybrid orbitals, suggesting \( \mathrm{sp} \) hybridization.
In simpler terms, nitrogen uses one s orbital and one p orbital to bond with the oxygens, efficiently using its available orbitals to achieve the stable structure observed in the \( \mathrm{NO}_{2}^{+} \) ion.

Hybridization of Nitrogen in NO3- Ion

Just as in other nitrate compounds, nitrogen in the nitrate ion \( \mathrm{NO}_{3}^{-} \) bonds in a fascinating way. The central nitrogen atom is bonded to three oxygen atoms. Additionally, the -1 charge means there's an extra electron involved in the overall structure.
The result is something called a resonance structure: a way of representing molecules where electron density can shift among multiple valid molecular structures. This phenomenon is captured using hybridization. For \( \mathrm{NO}_{3}^{-} \), nitrogen bonds through 3 sigma bonds, lacking lone pairs altogether.
Utilizing the hybrid orbital equation \( \text{Number of sigma bonds} + \text{Number of lone pairs} = 3 + 0 = 3 \), the type of hybridization present in nitrogen is \( \mathrm{sp}^{2} \). The three bonds are consistent with using an s orbital and two p orbitals. Such hybridization allows for a planarity and shared electron distribution, characteristic of resonance in nitrate.

Hybridization of Nitrogen in NH4+ Ion

When examining the ammonium ion \( \mathrm{NH}_{4}^{+} \), nitrogen forms a quite different structure compared to nitrates. In this ion, nitrogen binds with four hydrogen atoms, resulting in a perfectly symmetrical shape.
In terms of hybridization, nitrogen undertakes \( \mathrm{sp}^{3} \), meaning it uses one s and three p orbitals to bond.This is consistent with nitrogen’s formation of 4 sigma bonds, completing its octet with no lone pairs.
The tetrahedral geometry of \( \mathrm{NH}_{4}^{+} \) is a key character arising from \( \mathrm{sp}^{3} \) hybridization.Every hydrogen atom receives an equivalent share in the bonding, resulting in the stability and symmetrical nature seen in ammonium ions.