Question

Underlined carbon is sp hybridized in (a) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}\)

Short answer

The underlined carbon is sp hybridized in option (d): \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN} \).

Step-by-step solution

Analyze the Hybridization Concept

Hybridization is the process of mixing atomic orbitals into new hybrid orbitals suitable for the pairing of electrons in chemical bonds. For a carbon atom, **sp hybridization** occurs when it forms two bonds and involves the mixing of one s orbital and one p orbital, resulting in two sp orbitals arranged linearly at 180°.

Evaluate Option (a)

In the compound \( \mathrm{CH}_{3} \mathrm{CH} = \mathrm{CH}_{2} \), the underlined carbon is part of a double bond. Double bonds typically involve \( sp^2 \) hybridization, meaning one s orbital and two p orbitals mix. Thus, this carbon is \( sp^2 \) hybridized, not \( sp \).

Evaluate Option (b)

In \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} \), the underlined carbon is single-bonded to other atoms, indicating \( sp^3 \) hybridization due to the involvement of four sigma bonds. Therefore, this carbon is \( sp^3 \) and not \( sp \).

Evaluate Option (c)

In the compound \( \mathrm{CH}_{3} \mathrm{CONH}_{2} \), the underlined carbon is attached via a double bond to oxygen ( a carbonyl group). Such a carbon is \( sp^2 \) hybridized due to the combination of three groups (a double bond, a carbon, and a nitrogen). Thus, it is not \( sp \) hybridized.

Evaluate Option (d)

The compound is \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{\underline{C}N} \). The underlined carbon in \( \mathrm{C}N \) is part of a triple bond. Triple bonds are characteristic of \( sp \) hybridization, involving one s and one p orbital, leaving two p orbitals for pi bond formation. Therefore, this underlined carbon is indeed \( sp \) hybridized.

Key concepts

sp hybridization

In organic chemistry, hybridization is a crucial concept for understanding how atoms form bonds. **Sp hybridization** in particular arises when a carbon atom is involved in forming two sigma bonds. During this process, one s orbital and one p orbital from the carbon atom mix to create two sp hybrid orbitals. These new equivalent orbitals are oriented linearly, 180 degrees apart. This geometric arrangement minimizes electron pair repulsion, favoring a shape that is stable and energetically favorable.

When we analyze molecules for sp hybridization, it is typically associated with the presence of a triple bond. This is because the remaining two p orbitals do not partake in hybridization and instead form part of the pi bonds in a triple bond.

triple bond

A triple bond forms when two atoms share three pairs of electrons. This type of bond is stronger and shorter than a single or double bond due to the greater electron sharing between the atoms.

Triple bonds are made up of

• One sigma bond: This bond is formed from the overlap of sp hybrid orbitals, providing the primary axis of attachment.
• Two pi bonds: These arise from the side-to-side overlap of the two remaining p orbitals that do not undergo hybridization.

The presence of a triple bond leads to linear molecular geometry, which is characteristic of molecules such as acetylene (C₂H₂). In organic structures, the involvement of triple bonds directly leads to the sp hybridization of the participating carbon atoms.

atomic orbitals

Atomic orbitals describe regions around a nucleus where the probability of finding an electron is highest. An understanding of these orbitals is essential for comprehending hybridization in chemistry.

The basic types of atomic orbitals include:

• s orbital: Spherically shaped and can hold two electrons.
• p orbital: Dumbbell-shaped and can hold up to six electrons when three p orbitals are filled.

During hybridization, these atomic orbitals combine to form new hybrid orbitals that are sterically suitable for forming bonds in molecules. For sp hybridization, one s and one p orbital combine, leaving two p orbitals unhybridized. This process smartly aligns with molecular geometry to stabilize a molecule as it participates in chemical bonding.