Question

Consider the given statements about the molecule: \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{2} \mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\) 1\. Three carbons are sp \(^{3}\) hybridized 2\. Three carbons are sp \(^{2}\) hybridized 3\. Two carbons are sp hybridized Of these statements (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2 and 3 are correct (d) All are correct

Short answer

(d) All are correct.

Step-by-step solution

Identify Hybridization of Each Carbon

Examine the structure of the molecule to determine the hybridization of each carbon atom. Carbons in single bonds are typically sp \(^{3}\), those in double bonds are typically sp \(^{2}\), and those in triple bonds are usually sp hybridized.

Analyze Single Bond Carbons

The first two methyl groups, \(\mathrm{H}_{3} \mathrm{C}-\), have carbons that are each attached to three hydrogen atoms and one other carbon. This configuration indicates sp \(^{3}\) hybridization because they have four single bonds.

Analyze Double Bond Carbons

The carbon in \(\mathrm{-CH} = \mathrm{CH}-\) that forms part of the double bond uses sp \(^{2}\) hybridization. This is because it is bonded to three other atoms (one hydrogen and two carbons), forming one double bond.

Analyze Triple Bond Carbons

The carbon atoms in the section \(\mathrm{-C} \equiv \mathrm{C}-\) are each sp hybridized, as they are forming one triple bond with each other and one single bond with a neighboring atom.

Count Carbons by Hybridization Type

Based on the analysis: There are three sp \(^{3}\) hybridized carbons (all in single bonds within the \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{2}\)), three sp \(^{2}\) hybridized carbons (the ones involved in double bonds, \(\mathrm{CH}=\mathrm{CH}-\)), and two sp hybridized carbons (the ones in \(\mathrm{-C} \equiv \mathrm{C}-\)).

Determine Correct Statements

According to our count, statements 1, 2, and 3 are all correct as we have three carbons if each hybridization: sp \(^{3}\), sp \(^{2}\), and sp.

Key concepts

sp3 hybridization

In the world of organic chemistry, the concept of hybridization explains the bonding and shape of molecules. For sp3 hybridization, carbon's 2s and three 2p orbitals mix to form four equivalent sp3 hybrid orbitals. These orbitals arrange themselves in a tetrahedral geometry to minimize electron repulsion.

The angle between these orbitals is about 109.5 degrees, providing a stable structure. In the given exercise, the carbons in the two methyl groups \[(\text{H}_3\text{C})_2 \text{CH}-\]have sp3 hybridization. Each carbon in these groups forms four sigma bonds — one with another carbon and three with hydrogen atoms. This hybridization gives these segments of the molecule their three-dimensional tetrahedral shape.

• Occurs mainly with single bonds.
• Tetrahedral geometry.
• Bond angle approximately 109.5°.

sp2 hybridization

Hybridization doesn't stop at sp3; sp2 hybridization also plays a crucial role, especially in molecules with double bonds. In sp2 hybridization, the 2s orbital mixes with two 2p orbitals, leaving one unhybridized p orbital. The result is three sp2 hybrid orbitals forming a triangular planar geometry.

The unhybridized p orbital lies perpendicular to the plane of the sp2 orbitals. It overlaps with another p orbital to form the pi bond component of a double bond. In the molecule under study, the carbons involved in the\[-\text{CH}=\text{CH}-\]segment exhibit sp2 hybridization. Each of these bonds involves three \(sp^2\)hybridized orbitals for sigma bonds and one p orbital for the pi bond, giving these atoms a planar arrangement with bond angles of about 120 degrees.

• Occurs with one sigma and one pi bond (usually in double bonds).
• Planar geometry with a bond angle of 120°.

sp hybridization

Sp hybridization, while involving fewer bonds, offers stability in the presence of triple bonds. In this type of hybridization, the carbon's 2s orbital combines with only one 2p orbital. The remaining two p orbitals do not participate in hybridization and remain unhybridized.

The two sp orbitals arrange themselves linearly, resulting in a bond angle of 180 degrees. This linear arrangement allows the two unhybridized p orbitals to overlap with each other, creating two pi bonds. It forms a triple bond consisting of one sigma and two pi bonds. In the molecule we are examining, the atoms in the\[\text{-C} \equiv \text{C}-\]portion are sp hybridized. This section is characterized by its strong, linear triple bond.

• Typically involves a triple bond (one sigma and two pi bonds).
• Linear geometry with a bond angle of 180°.