Problem 85 - A Complete Resource Book in Chemistry for JEE Main

Question

In the compound, \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{CH}\), the \(\mathrm{C}_{2}-\mathrm{C}_{3}\) bond is of the type (a) \(\mathrm{sp}-\mathrm{sp}^{2}\) (b) \(s p^{3}-s p^{3}\) (c) \(\mathrm{sp}-\mathrm{sp}^{3}\) (d) \(\mathrm{sp}^{2}-\mathrm{sp}^{3}\)

Step-by-Step Solution

Answer

(d) \(\mathrm{sp}^2-\mathrm{sp}^3\)

Step 1: Identify the Structure of Compound

The given compound is: \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{CH}\). Let's label the carbon atoms: \(\text{C}_1 =\mathrm{CH}_2=\), \(\text{C}_2 = \mathrm{CH}-, \text{C}_3 = \mathrm{CH}_2-, \text{C}_4 = \mathrm{CH}_2-, \text{C}_5 \equiv \mathrm{CH}\). We are focusing on the bond between C2 and C3.

Step 2: Determine Hybridization of C2

\(\text{C}_2\) has a double bond involving one \(\pi\) and one \(\sigma\) bond. The hybridization involved here is \(\mathrm{sp}^2\) as it forms three \(\sigma\)-bonds: one with \(\text{C}_1\), one with \(\text{C}_3\), and one with hydrogen.

Step 3: Determine Hybridization of C3

\(\text{C}_3\) forms single bonds with \(\text{C}_2, \text{C}_4,\) and two hydrogens. This total of four \(\sigma\)-bonds means it uses \(\mathrm{sp}^3\) hybridization.

Step 4: Conclude the Bond Type

The bond between \(\text{C}_2\) and \(\text{C}_3\) is therefore formed by \(\mathrm{sp}^2\) orbitals on \(\text{C}_2\) and \(\mathrm{sp}^3\) orbitals on \(\text{C}_3\). Thus, the bond is categorized as \(\mathrm{sp}^2-\mathrm{sp}^3\).

Key Concepts

Hybridization

Hybridization is a concept that helps us understand how different atomic orbitals mix to form new hybrid orbitals, which are used to form chemical bonds in molecules.
When carbon forms compounds, it often changes the distribution of its electrons across orbitals by hybridizing.
Let's break this down with our example compound.

• The first carbon (\(\text{C}_2\)) is involved in a double bond. It uses \(\text{sp}^2\) hybridization. From a total of four valence electrons, three occupy the hybrid orbitals while one remains unhybridized, forming a \(\pi\) bond with a neighboring carbon.
• In contrast, the second carbon (\(\text{C}_3\)) forms single \(\sigma\) bonds with neighboring atoms, using \(\text{sp}^3\) hybridization. This hybridization is a result of mixing one \(s\) and three \(p\) orbitals.

In summary, recognition of hybridization types (\(\text{sp}, \text{sp}^2, \text{sp}^3\)) is essential to predicting and explaining molecular structure and properties.

Bonding

Bonding in molecules explains how atoms connect to form chemical structures. The type of bonds depends on the hybridization of the involved atoms.
In our compound example, \(\text{C}_2\) and \(\text{C}_3\) are bonded through a \(\text{sp}^2\)-\(\text{sp}^3\) overlap.

• The \(\text{sp}^2\) hybridized \(\text{C}_2\) overlaps with the \(\text{sp}^3\) hybridized \(\text{C}_3\), forming a \(\sigma\) bond.
• This \(\sigma\) bond means a direct overlap of orbitals, essential for the structure's stability.
• The \(\pi\) bond at \(\text{C}_2\) does not affect this connection but contributes to the rigidity and planarity in the molecule's structure.

Understanding the nature and formation of bonds helps predict the geometry and reactivity of a molecule. Without bonds, the atoms can't hold a chemical structure together.

Chemical Structure Analysis

Chemical structure analysis helps to identify the organization and connectivity of atoms within a compound. It's crucial to understand this in organic molecules for predicting behavior and properties.
In the molecule we examined, such analysis involves:

• Recognizing different parts of the molecule, such as single, double, and triple bonds.
• Labeling carbon atoms helps visualize the entire structure. By doing this, we identify how atoms interact and their role in the molecule.
• Determining each atom's hybridization aids in mapping out the compound's geometry and bond angles.

By analyzing these characteristics, chemists can hypothesize a molecule's reactivity, function, or how it might interact with other chemicals. The focus on the bond between \(\text{C}_2\) and \(\text{C}_3\) shows the type of atomic interactions in play, specifically recognizing the \(\text{sp}^2\) and \(\text{sp}^3\) roles in forming chemical bonds. Through this method, complete understanding is achieved for complex organic molecules.