Question

What are the characteristic hybrid orbitals employed by (a) carbon in an alkane, (b) carbon in a double bond in an alkene, (c) carbon in the benzene ring, (d) carbon in a triple bond in an alkyne?

Short answer

The characteristic hybrid orbitals employed by carbon are (a) sp3 in an alkane, (b) sp2 in a double bond in an alkene, (c) sp2 in a benzene ring, and (d) sp in a triple bond in an alkyne.

Step-by-step solution

Determine the bonding in an alkane

In an alkane, each carbon atom forms single bonds with other carbon atoms and hydrogen atoms. Each carbon is tetrahedral in shape and has a bond angle of 109.5°.

Identify the hybridization in an alkane

As the carbon atom in an alkane is forming 4 sigma bonds using its four valence electrons (one 2s and three 2p electrons), the hybridization is sp3, resulting from the combination of one s and three p atomic orbitals.

The characteristic hybrid orbital in an alkane

The answer for (a) is sp3 hybridization. ####Step 2: Hybrid orbitals in carbon in a double bond in an alkene

Determine the bonding in an alkene

In an alkene, each carbon atom forms a double bond with another carbon atom and single bonds with hydrogen atoms. Each carbon has a trigonal planar geometry and a bond angle of 120°.

Identify the hybridization in an alkene

Since the carbon atom in an alkene forms 3 sigma bonds using its four valence electrons, the hybridization is sp2, which is a combination of one s and two p atomic orbitals.

The characteristic hybrid orbital in an alkene

The answer for (b) is sp2 hybridization. ####Step 3: Hybrid orbitals in carbon in a benzene ring

Determine the bonding in a benzene ring

In a benzene ring, each carbon atom forms single bonds with its neighboring carbon atoms and one hydrogen atom, and a delocalized π bond formed by the overlapping of p orbitals along the ring. The geometry around each carbon atom is trigonal planar with a bond angle of 120°.

Identify the hybridization in a benzene ring

As each carbon atom in a benzene ring forms 3 sigma bonds using its four valence electrons, the hybridization is also sp2, just like in an alkene, which is a combination of one s and two p atomic orbitals.

The characteristic hybrid orbital in a benzene ring

The answer for (c) is sp2 hybridization. ####Step 4: Hybrid orbitals in carbon in a triple bond in an alkyne

Determine the bonding in an alkyne

In an alkyne, each carbon atom forms a triple bond with another carbon atom and a single bond with a hydrogen atom. The geometry around each carbon is linear with a bond angle of 180°.

Identify the hybridization in an alkyne

Since the carbon atom in an alkyne forms 2 sigma bonds using its four valence electrons, the hybridization is sp, which is a combination of one s and one p atomic orbitals.

The characteristic hybrid orbital in an alkyne

The answer for (d) is sp hybridization.

Key concepts

sp3 Hybridization

Carbon atoms exhibit sp3 hybridization when they form single bonds, as in alkanes. This hybridization involves the mixing of one s orbital with three p orbitals, producing four equivalent sp3 orbitals. These orbitals arrange themselves in a tetrahedral shape, with bond angles of approximately 109.5°.
This geometry allows each carbon to form four sigma (σ) bonds.

• Example: Methane (CH₄) is a perfect example of an alkane where carbon displays sp3 hybridization.
• In such structures, the carbon atom is bonded to four hydrogen atoms forming single covalent bonds.

sp2 Hybridization

In compounds like alkenes, where a carbon-carbon double bond is present, carbon atoms show sp2 hybridization. This type of hybridization results from the mixing of one s orbital and two p orbitals, forming three sp2 hybrid orbitals and leaving one unhybridized p orbital.
These three sp2 hybrid orbitals lie in a plane, forming a trigonal planar shape with 120° bond angles.

• Example: Ethene (C₂H₄) illustrates sp2 hybridization where each carbon atom is bonded to two hydrogens and one other carbon atom using sp2 orbitals.
• The p orbitals overlap to form the π-bond of the double bond.

sp Hybridization

When a carbon atom forms a triple bond, as seen in alkynes, it undergoes sp hybridization. This involves the mixing of one s orbital and one p orbital, resulting in two equivalent sp hybrid orbitals and two unhybridized p orbitals.
The sp hybrid orbitals align linearly, creating a bond angle of 180°, which allows for the linear shape of molecules.

• Example: In ethyne (acetylene, C₂H₂), each carbon atom forms a triple bond with another carbon and a single bond with a hydrogen, using sp orbitals.
• The two remaining p orbitals form two π-bonds, completing the triple bond.

Alkane

Alkanes are hydrocarbons consisting solely of carbon and hydrogen atoms connected by single bonds. They are also known as saturated hydrocarbons because each carbon atom is bonded to the maximum number of hydrogen atoms.
The molecular formula for alkanes is CₙH₂ₙ₊₂, where n is the number of carbon atoms.

• Example: Propane (C₃H₈) is a simple alkane with three carbon atoms.
• Alkanes have a tetrahedral shape around each carbon due to sp3 hybridization.

Alkene

Alkenes are hydrocarbons that contain at least one carbon-carbon double bond, which introduces unsaturation to the molecule. The presence of this double bond means that alkenes have fewer hydrogen atoms compared to alkanes of the same size.
The general formula for alkenes is CₙH₂ₙ, reflecting this unsaturation.

• Example: Propene (C₃H₆) is an example of an alkene with one double bond.
• Alkenes have a trigonal planar geometry due to sp2 hybridization around the carbon atoms involved in the double bond.

Alkyne

Alkynes are characterized by at least one carbon-carbon triple bond, making them even less saturated than alkenes. The presence of a triple bond leaves two fewer hydrogens compared to the analogous alkane.
The general formula for alkynes is CₙH₂ₙ₋₂, representing the additional degree of unsaturation compared to both alkanes and alkenes.

• Example: Butyne (C₄H₆) includes one triple bond.
• The linear geometry arises from sp hybridization at the carbon atoms involved in the triple bond.

Benzene Ring

The benzene ring represents a unique case among hydrocarbons due to its cyclic structure and delocalized electrons. Each carbon atom in benzene is sp2 hybridized, forming a hexagonal ring with all carbon-carbon bonds being equivalent.
This equivalence is due to the electron delocalization across the ring, often visualized as resonance structures.

• Example: Benzene (C₆H₆) is the simplest aromatic compound, noted for its stability and special characteristic.
• The geometry around each carbon atom is trigonal planar, with a bond angle of 120°.