Problem 12 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The species in which the \(\mathrm{N}\) atom is in a state of \(s p\) hybridisation is : (a) \(\mathrm{NO}_{3}^{-}\) (b) \(\mathrm{NO}_{2}\) (c) \(\mathrm{NO}_{2}^{+}\) (d) \(\mathrm{NO}_{2}^{-}\)

Step-by-Step Solution

Answer

The species is \(\text{NO}_{2}^{+}\) (Option c).

Step 1: Determining the Hybridization

To determine the hybridization of the nitrogen atom in each species, we consider the steric number of the nitrogen atom, which is the number of atoms bonded to it plus the number of lone pairs. Hybridization is determined by the formula: \( ext{Steric number} = ext{number of hybrid orbitals} \). For \(sp\) hybridization, the steric number is 2 as it involves two hybrid orbitals (one \(s\) and one \(p\)).

Step 2: Analyze \(\text{NO}_{3}^{-}\)

In \(\text{NO}_{3}^{-}\), the nitrogen atom is central and bonded to three oxygen atoms and has no lone pair. Steric number is 3 (three bonds), indicating \(sp^2\) hybridization.

Step 3: Analyze \(\text{NO}_{2}\)

In \(\text{NO}_{2}\), the nitrogen atom is bonded to two oxygen atoms and has one electron (equivalent to half a lone pair in terms of hybridization calculation). The steric number is 2.5, which after approximation suggests a mixed hybridization case rather than simple \(sp\), \(sp^2\), or \(sp^3\).

Step 4: Analyze \(\text{NO}_{2}^{+}\)

In \(\text{NO}_{2}^{+}\), nitrogen is bonded to two oxygen atoms and has no lone pair. The steric number is 2, as there are only two bonds, matching the requirements for \(sp\) hybridization.

Step 5: Analyze \(\text{NO}_{2}^{-}\)

In \(\text{NO}_{2}^{-}\), nitrogen is bonded to two oxygen atoms and has one lone pair. The steric number is 3, indicating \(sp^2\) hybridization.

Step 6: Conclusion

Based on the analysis, only in \(\text{NO}_{2}^{+}\) does the nitrogen atom have a steric number of 2, which corresponds to \(sp\) hybridization.

Key Concepts

Steric Number

The steric number is an important concept in chemistry that helps us understand the hybridization state of atoms in molecules. Essentially, the steric number is the sum of the number of atoms bonded to a central atom and the number of lone pairs on that atom. This count gives us insights into the possible hybridization state.
For example, if a nitrogen atom in a molecule is bonded to two other atoms and has no lone pairs, the steric number would be 2. This helps to indicate the possibility of an \( sp \) hybridization. If the steric number is 3, the nitrogen could likely involve \( sp^2 \) hybridization, which involves forming three hybrid orbitals.
Here is a simplified breakdown of steric numbers and corresponding hybridizations:

• Steric number 1 : \( s \) hybrid - though more theoretical, often leads to unhybridized state
• Steric number 2 : \( sp \) hybridization
• Steric number 3 : \( sp^2 \) hybridization
• Steric number 4 : \( sp^3 \) hybridization

Understanding the steric number helps predict the shape and bonding characteristics of a molecule by providing a straightforward method to identify the hybridization state.

Nitrogen Atom Hybridization

Hybridization is a key notion when discussing the bonding and molecular structure of nitrogen atoms in different compounds. The hybridization of a nitrogen atom affects its geometry and chemical properties. Nitrogen usually has five valence electrons, allowing it to form several types of hybrid orbitals.
In general, the types of hybridization that nitrogen can undergo are:

• \( sp \) hybridization: In this state, the nitrogen atom adopts a linear geometry and forms two hybrid orbitals, combining one \( s \) and one \( p \) orbital. It's commonly noted when nitrogen bonds to two atoms without lone pairs.
• \( sp^2 \) hybridization: This involves three hybrid orbitals, from one \( s \) and two \( p \) orbitals, resulting in trigonal planar geometry. It occurs typically in molecules where nitrogen forms three bonds or has one lone pair.
• \( sp^3 \) hybridization: Here, four hybrid orbitals form from one \( s \) and all three \( p \) orbitals. This is associated with tetrahedral geometry and is often seen in situations where nitrogen forms three bonds plus one lone pair or four bonds.

By analyzing nitrogen's bonding scenario and the number of lone pairs, we can deduce the hybridization state, which aids in understanding its role within a molecule.

sp Hybridization

The \( sp \) hybridization refers to a hybridization state where one \( s \) orbital and one \( p \) orbital mix to form two equivalent \( sp \) orbitals. This hybridization results in a linear geometry with a bond angle of \( 180^\circ \).
\( sp \) hybridization is often associated with situations where a nitrogen atom forms two bonds without having lone pairs. For instance, in \( \mathrm{NO}_2^+ \), the nitrogen atom is bonded linearly to two oxygen atoms, with no lone pairs, ideal for \( sp \) hybridization as shown by its steric number of 2.
Key identifiers for \( sp \) hybridization include:

• Bonding atoms on opposite sides of the central atom
• Absence of lone pairs in the hybridized orbitals conceptually simplifying the electronic structure
• A linear molecular shape, ideal for many diatomic molecules and ions like acetylene and certain diazonium salts.

Because of its straightforward linear nature, \( sp \) hybridization is a helpful concept in understanding some simple structures and correlations of geometry to function.