Problem 81 - Advanced Problems in Organic Chemistry for Competitive Examinations

Question

Which of the following statements is NOT true for ethylene \(\left(\mathrm{CH}_{2}=\mathrm{CH}_{2}\right)\) molecule? (a) Both carbons are sp \(^{2}\) hybridised. (b) \(\mathrm{C}=\mathrm{C}\) bond length is shorter than the \(\mathrm{C}-\mathrm{C}\) bond length in ethane. (c) The two \(C=C\) bonds are equally strong. (one is pi and the other is sigma.) (d) The entire molecule has a planar geometry.

Step-by-Step Solution

Answer

Statement (c) is NOT true.

Step 1: Analyze Carbon Hybridization

In the ethylene molecule, each carbon atom forms three sigma bonds: two with hydrogen atoms and one with the other carbon atom. To accommodate this bonding configuration, each carbon atom undergoes \( sp^2 \) hybridization. Thus, statement (a) is true.

Step 2: Compare Bond Lengths

The \( C=C \) (double bond) consists of one sigma and one pi bond which is shorter and stronger than a single \( C-C \) bond. Ethane has a single \( C-C \) bond, so the \( C=C \) bond in ethylene is indeed shorter than the \( C-C \) bond in ethane. Statement (b) is true.

Step 3: Evaluate Bond Strengths

In a \( C=C \) double bond, there is one sigma bond and one pi bond connecting the two carbon atoms. These two bonds are not equally strong; the sigma bond is stronger than the pi bond. Therefore, statement (c) is NOT true.

Step 4: Assess Molecular Geometry

Due to \( sp^2 \) hybridization, the geometry around each carbon in ethylene makes the entire molecule planar with bond angles of approximately 120°. Thus, statement (d) is true.

Key Concepts

Carbon Hybridization in Ethylene

Ethylene, also known as ethene (\(\mathrm{C}_2\mathrm{H}_4\)), features a rich and insightful example of carbon hybridization. Each carbon atom in the ethylene molecule forms three sigma bonds: two with hydrogen atoms and one with the other carbon atom. To facilitate this, carbon undergoes a process called \(sp^2\) hybridization. \(sp^2\) hybridization involves the mixing of one "s" orbital and two "p" orbitals to form three equivalent hybrid orbitals.
These hybrid orbitals are used to form sigma bonds, which are characterized by their end-to-end overlap leading to a single line of electron density between the bonded atoms. The third "p" orbital, which remains unhybridized, is responsible for forming the pi bond.
Thus, the \(sp^2\) hybridization in ethylene allows for the creation of a double bond configuration that is essential for its structure, with a combination of stronger pi bond and sigma bonds between carbon atoms.

Bond Lengths in Ethylene

In the ethylene molecule, the carbon-carbon double bond (\(C=C\)) is a combination of one sigma bond and one pi bond. This double bond structure results in a bond length that is significantly shorter than the carbon-carbon single bond (\(C-C\)) found in molecules such as ethane.
The shorter bond length can be attributed to the additional electron density present in the pi bond, which pulls the two carbon atoms closer together. Compare this to a single bond, where less electron density exists between the nucleus of the bonded atoms.

• The typical \(C=C\) bond length is about 1.34 Å.
• On the other hand, a typical \(C-C\) bond length in ethane is approximately 1.54 Å.

These values emphasize the compact nature of the double bond in ethylene compared to the relatively longer single bond in ethane.

Bond Strength Comparison in Ethylene

When discussing the bond strength in ethylene, it's crucial to compare the sigma and pi components of the carbon-carbon double bond. Although both these bonds contribute to the overall strength of the double bond, they do not share the same strength individually.
The sigma bond, with its direct end-to-end overlap, is much stronger than the pi bond, which results from the lateral overlap of orbitals. The sigma bond is the primary contributor to the bond strength of the double bond in ethylene.
In summary:

• Sigma bonds provide major stability and strength due to maximum overlap.
• Pi bonds, while providing additional bonding, are comparatively weaker and introduce additional structural rigidity but lesser strength than sigma bonds.

Understanding this comparison helps us know why not all statements about equal bond strength in ethylene hold true.

Planar Geometry of Ethylene

The geometry of the ethylene molecule is predominantly influenced by the \(sp^2\) hybridization of the carbon atoms. This configuration leads to a planar geometry due to the arrangement of the hybridized orbitals and the pi bond structure.
Each carbon atom in ethylene is bonded in a trigonal planar arrangement, causing the molecule to sit on a flat plane. The bond angles created by this arrangement are approximately 120°, which helps facilitate the \(C=C\) double bond and the attached hydrogen atoms.

• This planar geometry allows for pi bond interactions which are essential for the double bond strength.
• It provides the molecule with stability and minimizes electron pair repulsion.

The planar nature of ethylene is a fundamental aspect that influences its reactivity and interaction with other molecules.