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Hybridization of 1 and 2 carbon atom in \({ }^{1} \mathrm{CH}_{2}={ }^{2} \mathrm{C}=\mathrm{CH}_{2}\) are: (a) sp, sp (b) \(\mathrm{sp}^{2}, \mathrm{sp}^{2}\) (c) \(\mathrm{sp}^{2}\), sp (d) \(\mathrm{sp}^{3}, \mathrm{sp}^{2}\)

Short Answer

Expert verified
The hybridization of carbon 1 is \(\mathrm{sp}^{2}\) and carbon 2 is \(\mathrm{sp}\).

Step by step solution

01

Identify the structure of the molecule

The given molecule is \(^{1}\mathrm{CH}_{2}=^{2}\mathrm{C}=\mathrm{CH}_{2}\). It has a carbon-carbon double bond in the middle with two other carbons bonded on each side.
02

Assign hybridization to carbon atoms

First decide on the hybridization of each carbon based on the bonds. A carbon with a double bond is usually \(sp^2\) hybridized, as it involves one \(\sigma\) bond and one \(\pi\) bond, and a total of three sigma bond equivalents (including the \(\pi\) system as one equivalent count for hybridization purposes).
03

Determine hybridization of first carbon (C1)

The first carbon \(^{1}\mathrm{CH}_{2}\) has two \(\sigma\) bonds (one with a hydrogen and one with \(^{2}\mathrm{C}\)) and one \(\pi\) bond (part of the \(C=C\) bond). Therefore, it has an \(sp^2\) hybridization.
04

Determine hybridization of second carbon (C2)

The second carbon \(^{2}\mathrm{C}\) is also involved in two \(\pi\) bonds—one with \(^{1}\mathrm{C}\) and another with the third carbon \(\mathrm{CH}_{2}\). It only forms two \(\sigma\) bonds (one on either side), suggesting \(sp\) hybridization for \(^{2}\mathrm{C}\).
05

Conclude the correct option

From the analysis, carbon \(^{1}\) is \(sp^2\) hybridized, and carbon \(^{2}\) is \(sp\) hybridized. This matches with option (c) \(\mathrm{sp}^{2}\), sp.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Bonding
Chemical bonding is the force that holds atoms together in a molecule. These bonds form when atomic orbitals overlap, allowing for the sharing or transfer of electrons. There are several types of chemical bonds, but in organic chemistry, we mainly deal with covalent bonds. Covalent bonds occur when two atoms share electron pairs, which can include single (13 14), double (12), or triple bonds (3), each differing in the number of electrons shared. The kind of bonds in any molecule greatly affects its stability, reactivity, and physical properties. When atoms form a chemical bond, they may also undergo hybridization, which involves the mixing of atomic orbitals to form new, equivalent hybrid orbitals. This process helps in achieving optimal geometry that minimizes electron repulsion between bonded atoms.
  • Single Bonds: Generally involve sigma (13 14) bonds, formed by the head-on overlap of atomic orbitals.
  • Double Bonds: Contain one sigma and one pi (12) bond, formed by parallel overlapping of p orbitals.
  • Triple Bonds: Consist of one sigma and two pi bonds, giving the molecule a linear shape.
Understanding chemical bonding is fundamental in exploring how different atoms create vast and diverse structures in organic chemistry.
Sigma and Pi Bonds
Sigma (13 14) and pi (12) bonds are essential components of chemical bonding, especially in organic molecules. Sigma bonds are the strongest type of covalent chemical bond and serve as the building blocks for molecules. Sigma bonds are formed by the head-on overlap of orbitals, which allows for free rotation of the bonded atoms around the bond axis. This characteristic makes sigma bonds very stable. Pi bonds, on the other hand, originate from the side-to-side overlap of p orbitals. They are present in addition to a sigma bond in double and triple bonds. Because of their electron cloud location above and below the bond axis, pi bonds restrict the rotation, giving fixed geometrical shapes to molecules.
  • The presence of pi bonds usually indicates a higher level of unsaturation in a molecule.
  • Pi bonds are generally weaker than sigma bonds but very important as they affect the molecule's reactivity and stability.
  • Double bonds are always constructed by one sigma and one pi bond, while triple bonds include one sigma and two pi bonds.
Knowing how sigma and pi bonds interact can help understand how molecules function and how they can be manipulated to form new structures drastically."
Organic Chemistry
Organic chemistry is the study of carbon-containing compounds, which are the basis of life. Atoms in organic molecules are primarily carbon and hydrogen, but they can also include oxygen, nitrogen, sulfur, and other elements. This branch of chemistry delves into the structure, properties, composition, reactions, and synthesis of these compounds. Why is the focus on carbon? Carbon is unique because it can form stable covalent bonds with many elements, including itself, leading to a variety of structures such as chains, rings, and complex three-dimensional shapes. This ability is crucial for creating the diverse array of organic molecules found in nature, from simple hydrocarbons to complex biomolecules like proteins and DNA. In organic chemistry:
  • Hybridization is a concept used to explain the geometry and type of bonds in a molecule. It helps us understand how carbon is able to form four equivalent bonds in methane (14b).
  • Reactions often involve the breaking and forming of both sigma and pi bonds.
  • Understanding intermediate structures like carbocations, radicals, and carbanions is essential for predicting and explaining reactions.
The depth and variety of organic chemistry provide insight into many biological processes and synthetic materials, making it a pivotal science in pharmaceuticals, biotechnology, and materials science.

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Most popular questions from this chapter

\(\mathrm{H}_{2} \mathrm{O}\) is dipolar, whereas \(\mathrm{BeF}_{2}\) is not. It is because: (a) The electronegativity of \(\mathrm{F}\) is greater than that of \(\mathrm{O}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) involves hydrogen bonding whereas \(\mathrm{BeF}_{2}\) is a discrete molecule (c) \(\mathrm{H}_{2} \mathrm{O}\) is linear and \(\mathrm{BeF}_{2}\) is angular (d) \(\mathrm{H}_{2} \mathrm{O}\) is angular and \(\mathrm{BeF}_{2}\) is linear

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