2. Bonding Capacity of C element

​Carbon (₆C) has configuration 1s²2s²2p² with 4 valence electrons (2 unpaired). In bonding, one 2s electron is promoted to 2p, giving 4 unpaired electrons. The 2s and 3×2p orbitals then hybridize to form four identical sp³ orbitals.

​Example : Explain the bond structure of the CF₄ molecule by using electron dot structure.

​Carbon forms 4 sp³ orbitals, each bonding with 1 unpaired electron from 4 F atoms → CF₄.
C–F bonds are polar, but tetrahedral symmetry cancels dipoles → CF₄ is nonpolar.

​Atoms like carbon, oxygen, and nitrogen can form single, double, or triple bonds. When orbitals overlap, two bond types arise: sigma (σ) and pi (π). A single bond is always a σ bond. π bonds form only after a σ bond. Thus, a double bond has 1 σ + 1 π, and a triple bond has 1 σ + 2 π. Let’s look at how σ and π bonds form.

​SIGMA (σ) BONDS

​Sigma (σ) bonds form by end-to-end overlap of orbitals, which can be s, p, or hybrid orbitals. In CH₄, the carbon atom’s four sp³ orbitals overlap with the 1s orbitals of four hydrogen atoms, forming four C–H σ bonds.

​In C₂H₆, each C–H σ bond forms by end-to-end overlap of a carbon sp³ orbital with a hydrogen 1s orbital. The C–C σ bond forms from the end-to-end overlap of the two carbon sp³ orbitals. Thus, C₂H₆ has six C–H σ bonds and one C–C σ bond, totaling seven σ bonds.

​Pi (π) BONDS

​Pi (π) bonds form by the side-by-side overlap of two parallel p orbitals. The electron cloud of a π bond lies above and below the plane of the σ bonds. π bonds are generally weaker than σ bonds.

​A π bond cannot exist by itself; it forms only after a σ bond if unhybridized p orbitals are available. In other words, a σ bond must form first before a π bond can be created.

​Formation of The Pi (π) Bond in The Ethylene Molecule

​In C₂H₄, each carbon is sp² hybridized, forming five σ bonds (four C–H and one C–C) and one π bond between the carbons, giving a total of six bonds.

​Formation of Pi (π) Bonds in The Acetylene Molecule

​In C₂H₂, both carbons are sp hybridized. Each forms a C–H σ bond, and their sp orbitals form a C–C σ bond. The unhybridized p orbitals overlap to form two C–C π bonds, giving one σ and two π bonds between the carbons.

​Example : The chemical formula of methyl methacrylate, a key monomer used in making plexiglass, is given below: How many σ (sigma) and π (pi) bonds are present in this molecule?

​Example : The chemical formula of styrene, an important monomer in the production of polystyrene plastics, is given below: How many σ (sigma) and π (pi) bonds are present in this molecule?

Multiple Choice

What type of bond is formed by the head-to-head overlap of orbitals?

• Sigma (σ) bond

• Pi (π) bond

• Hydrogen bond

• Ionic bond

Multiple Choice

Greater the overlap of the orbitals, _________________________

• lesser the bond strength

• greater the bond strength

• bond strength remains the same

• no bond formation

Multiple Choice

A π (pi) bond is the result of the

• overlap of two s orbitals

• overlap of an s orbital and a p orbital

• overlap of two p orbitals along their axes

• sidewise overlap of two parallel p orbitals

• sidewise overlap of two s orbitals

Multiple Choice

What is the angle between the sp hybrid orbitals?

• 90°

• 120°

• 180°

• 109.5°

Multiple Choice

The hybridisation of the carbon atom is?

• sp hybridization

• sp² hybridization

• sp³ hybridization

• sp³d hybridization

Multiple Choice

What is the hybridization of each Carbon atom in benzene (C₆H₆)?

• sp

• sp³

• sp²

• sp³ d

Multiple Choice

What is the hybridization of the Carbon atom indicated by the arrow?

• sp hybridization

• sp² hybridization

• sp³ hybridization

• dsp³ hybridization

Multiple Choice

Type of hybridization in C number 3

• sp4

• sp3

• sp2

• sp

Multiple Choice

How many sigma bond in this molecule?

• 1

• 2

• 3

• 4

Multiple Choice

Type of hybridization in C number 2

• sp4

• sp3

• sp2

• sp

Multiple Choice

Type of hybridization in C number 1

• sp4

• sp3

• sp2

• sp

​In some molecules, the theoretical structure may differ from reality. For ozone (O₃), rules suggest one single and one double bond so that each oxygen completes its octet, giving the predicted structure.

​An O–O single bond is normally longer than an O=O double bond, but in the O₃ molecule, both oxygen–oxygen bonds are equal in length. This bond length is intermediate—shorter than a single bond but longer than a double bond.