Problem 63 - A Complete Resource Book in Chemistry for JEE Main

Question

The bond between carbon atom (1) and carbon atom (2) in compound \(\mathrm{N} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\) involves the hybridization (a) \(\mathrm{sp}^{2}\) and \(\mathrm{sp}^{2}\) (b) \(\mathrm{sp}^{3}\) and sp (c) sp and sp \(^{2}\) (d) sp and sp

Step-by-Step Solution

Answer

The bond involves \(\mathrm{sp}\) and \(\mathrm{sp}^{2}\) hybridization, so the correct answer is (c).

Step 1: Identify the Bond Type

Look at the compound structure: \(\mathrm{N}\equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\). The bond we are considering is between carbon atom (1) and carbon atom (2). This is a \(\mathrm{C}-\mathrm{C}\) single bond immediately following the \(\mathrm{C}\equiv \mathrm{N}\) triple bond.

Step 2: Assess the Hybridization of Each Carbon Atom

Carbon atom (1) is part of the \(\mathrm{N}\equiv \mathrm{C}\) triple bond. In a triple bond, a carbon atom is usually \(\mathrm{sp}\) hybridized. Carbon atom (2) is bonded through a double bond as seen in \(-\mathrm{CH}=\). This typically indicates \(\mathrm{sp}^{2}\) hybridization.

Step 3: Match Hybridization to Options

The hybridization of carbon atom (1) is \(\mathrm{sp}\), and the hybridization of carbon atom (2) is \(\mathrm{sp}^{2}\). Comparing these to the given options:- (a) \(\mathrm{sp}^{2}\) and \(\mathrm{sp}^{2}\)- (b) \(\mathrm{sp}^{3}\) and \(\mathrm{sp}\)- (c) \(\mathrm{sp}\) and \(\mathrm{sp}^{2}\)- (d) \(\mathrm{sp}\) and \(\mathrm{sp}\)The correct match is option (c) \(\mathrm{sp}\) and \(\mathrm{sp}^{2}\).

Key Concepts

Triple Bond Hybridization

Triple bonds are a fascinating aspect of organic chemistry. In these bonds, carbon tends to use an \( sp \) hybridization. But what does that mean? When a carbon atom forms a triple bond, like the one in \( \, ext{C} \equiv ext{N} \, \), it needs three pairs of electrons to form the bonds. This is typical for molecules such as alkynes or nitriles. To minimize energy, one of the \( s \) orbitals mixes (or hybridizes) with one \( p \) orbital to form two \( sp \) hybrid orbitals.This configuration allows the carbon atom to:

• Form a sigma bond using one of the \( sp \) orbitals.
• Make two pi bonds using the remaining unhybridized \( p \) orbitals.

These three bonds create a linear arrangement, with the participating atoms spread out in a straight line, offering the most stable and lowest energy configuration for the triple bond.

Double Bond Hybridization

Double bonds, such as those in alkenes, play a crucial role in the chemistry of organic molecules. In a double bond like \( -\text{CH}= \ \text{CH}_{2} \), the first bond is a sigma bond, while the second is a pi bond. The carbon involved in the double bond features \( sp^2 \) hybridization. Let's break down how this works:

• Three \( sp^2 \) hybrid orbitals form when one \( s \) orbital mixes with two \( p \) orbitals.
• These orbitals arrange themselves into a planar shape, spaced 120-degree angles apart.
• The unhybridized \( p \) orbital forms the pi bond, giving the molecule its rigidity and planar nature.

This hybridization results in increased bond energy and decreased bond length. The planar structure allows for resonance and other electronic interactions, enhancing the reactivity and versatility of molecules containing double bonds.

Carbon-Carbon Bond Hybridization

Carbon-carbon bonds are the backbone of organic chemistry. Their hybridization sets the structure and reactivity of the organic molecules. Each type of hybridization affects the bond's length, angle, and strength:- **Single Bond (\( sp^3 \) Hybridization):** Here, all four \( sp^3 \) hybrid orbitals form sigma bonds, creating a tetrahedral structure. This arrangement allows for rotation around the bond axis, giving alkanes their flexibility.- **Double Bond (\( sp^2 \) Hybridization):** This involves one pi and one sigma bond, leading to a planar geometry and fixed position that prevents rotation, affecting molecules like ethene.- **Triple Bond (\( sp \) Hybridization):** This structure forms one sigma and two pi bonds. The linear geometry results in a much stronger, shorter bond than the previous two, common in acetylene.Understanding the hybridization of carbon-carbon bonds is key to predicting and explaining organic compounds' chemical behavior.