Question

Which of the following represents the given mode of hybridization \(\mathrm{sp}^{2}-\mathrm{sp}^{2}-\mathrm{sp}-\mathrm{sp}\) from left to right? (a) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{NCH}\) (b) \(\mathrm{CH} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{N}\) (c) \(\mathrm{CH}_{2}=\mathrm{C}=\mathrm{C}=\mathrm{CH}_{2}\) (d) \(\mathrm{CH}_{2}\)

Short answer

The correct answer is (a) \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{NCH} \).

Step-by-step solution

Understand the hybridization notation

The notation \( \mathrm{sp}^{2}-\mathrm{sp}^{2}-\mathrm{sp}-\mathrm{sp} \) implies that we are looking for a molecule where the first two atoms have \( \mathrm{sp}^2 \) hybridization and the next two atoms have \( \mathrm{sp} \) hybridization in the order listed.

Analyze option (a)

Option (a) is \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{NCH} \). The first carbon \( \mathrm{CH}_{2} \) forms a double bond with another carbon \( \mathrm{CH} \), indicating \( \mathrm{sp}^2 \) hybridization. The second carbon \( \mathrm{CH} \) also forms a double bond, indicating \( \mathrm{sp}^2 \) hybridization. The third carbon \( \mathrm{C} \) is part of a triple bond, which leads to \( \mathrm{sp} \) hybridization. The fourth atom in the series, \( \mathrm{N} \), has a triple bond to carbon, indicating \( \mathrm{sp} \) hybridization.

Analyze option (b)

Option (b) is \( \mathrm{CH} \equiv \mathrm{C}-\mathrm{C} \equiv \mathrm{N} \). Both the \( \mathrm{CH} \) and \( \mathrm{C} \), as well as the \( \mathrm{C} \) in the triple bond, have \( \mathrm{sp} \) hybridizations, which does not match the required order \( \mathrm{sp}^{2}-\mathrm{sp}^{2}-\mathrm{sp}-\mathrm{sp} \).

Analyze option (c)

Option (c) is \( \mathrm{CH}_{2}=\mathrm{C}=\mathrm{C}=\mathrm{CH}_{2} \). Each of these carbons forms double bonds, so they all have \( \mathrm{sp}^2 \) hybridization, which does not match the required hybridization pattern.

Analyze option (d)

Option (d) is \( \mathrm{CH}_{2} \). This is a single group and cannot complete the sequence of \( \mathrm{sp}^{2}-\mathrm{sp}^{2}-\mathrm{sp}-\mathrm{sp} \) hybridizations needed.

Key concepts

sp hybridization

Understanding hybridization in chemistry is crucial for determining the shape and bonding of molecules. One key type is \( sp \) hybridization, which occurs when one \( s \) orbital mixes with one \( p \) orbital. This combination results in two equivalent \( sp \) hybrid orbitals.
These orbitals are linear, meaning they form a 180-degree angle from each other. It's commonly seen in molecules where carbon forms a triple bond, as in acetylene \( (\mathrm{HC} \equiv \mathrm{CH}) \). Each carbon atom in this molecule uses one \( sp \) hybrid orbital to bond with the other carbon and a hydrogen atom.
You don't need to worry about double bonds here. Triple bonds mean the bonds are shorter and stronger.

• Linear geometry.
• Commonly seen with triple bonds.
• Two equivalent hybrid orbitals.

Recognizing \( sp \) hybridization can help predict molecule shapes and reactivity.

sp2 hybridization

Another essential hybridization form is \( sp^2 \) hybridization, involving one \( s \) and two \( p \) orbitals. This mingling produces three equivalent \( sp^2 \) hybrid orbitals.
These orbitals spread out to create a trigonal planar shape, approximately 120 degrees apart. In a molecule like ethylene \( (\mathrm{CH}_2=\mathrm{CH}_2) \), each carbon atom is \( sp^2 \) hybridized and forms a double bond.
The additional un-hybridized \( p \) orbital is responsible for forming the \( \pi \) bond, characteristic of double bonds. This double-bond feature reduces rotation around the bond, giving rise to the molecule's planar structure.

• 120-degree trigonal planar shape.
• Involved in double-bond formations.
• Three equivalent hybrid orbitals.

Knowing the concept of \( sp^2 \) hybridization allows us to predict how certain molecules will interact and behave.

Molecular structure

The molecular structure of a compound is deeply influenced by its atoms' hybridization.
An excellent example is the compound \( \mathrm{CH}_2=\mathrm{CH}-\mathrm{C} \equiv \mathrm{NCH} \), which integrates both \( sp^2 \) and \( sp \) hybridized atoms.

• Two terminal carbon atoms are \( sp^2 \) hybridized, engaging in double bonds.
• The central carbon and nitrogen atoms are \( sp \) hybridized, forming a triple bond.

This diverse hybridization explains the compound's linear and planar segments, affecting its reactivity and stability.
Understanding how hybridization influences molecular structure helps make accurate predictions about a molecule's properties, such as boiling points and solubility.
Remember that each type of hybridization contributes differently, from shape and angle to strength of bonds. By mastering these concepts, you'll gain a deeper insight into molecular chemistry.