Chapter 8, Problem 63A

(a)

Interpretation Introduction

Interpretation:

Hybrid orbitals in methane are to be determined.

Concept Interpretation:

An orbital that is created by fusing two or more atomic orbitals is known as a hybrid orbital.

Answer to Problem 63A

  $\text{CH}$ is ${\text{sp}}^{\text{3}}$hybridized.

Explanation of Solution

Methane $\left({\text{CH}}_{\text{4}}\right)$ has $\text{'C'}$ as a central atom that is single-bonded with four $\text{H}$ atoms. Methane has a tetrahedral structure. It needed one $\text{2s}$ and three $\text{2p}$ orbitals to form ${\text{sp}}^{\text{3}}$ .

(b)

Interpretation Introduction

Interpretation:

Hybridization in ethyl is to be determined.

Concept Interpretation:

The atomic orbitals with equivalent energies are mixed during the hybridization process.

Answer to Problem 63A

  ${\text{H}}_{\text{2}}{\text{C=CH}}_{\text{2}}$ (Ethylene) have 3 sigma and 1 pie bond. So, ${\text{H}}_{\text{2}}{\text{C=CH}}_{\text{2}}$ is ${\text{sp}}^2$ hybridized.

Explanation of Solution

Due to the double bond in this ethene molecule, ${\text{sp}}^{\text{2}}$ hybridization occurs. Carbon uses one $\text{2s}$ and two $\text{2p}$ orbitals to form ${\text{sp}}^2$ hybrid orbitals.

(c)

Interpretation Introduction

Interpretation:

Bonding in $\text{HC}\equiv \text{CH}$ is to be determined.

Concept Interpretation:

The concept of hybridization is the joining of two atomic orbitals to form a new class of hybridized orbitals.

Answer to Problem 63A

  $\text{HC}\equiv \text{CH}$ (ethyne) have 2sigma and 2pie bonds. So, $\text{HC}\equiv \text{CH}$ is $\text{sp}$ hybridized.

Explanation of Solution

  $\text{sp}$ hybridization occurs because the molecule has a triple bond. Carbon only uses one $\text{2s}$ and one $\text{2p}$ orbitals to form $\text{sp}$ hybrid orbitals.

(d)

Interpretation Introduction

Interpretation:

Hybridization in $\text{N}\equiv \text{C-C}\equiv \text{N}$ is to be determined.

Concept Interpretation:

To redistribute energy and create a new set of orbitals with a comparable shape and energy, orbitals with slightly different energies are mixed in a process known as hybridization.

Answer to Problem 63A

  $\text{N}\equiv \text{C-C}\equiv \text{N}$ (Cyanogen) have 2 sigma and 2 pie bonds. So $\text{N}\equiv \text{C-C}\equiv \text{N}$ is $\text{sp}$ hybridized.

Explanation of Solution

  $\text{sp}$ hybridization occurs because the molecule has a triple bond. Treat triple bonds as single bonds when analyzing molecules using hybrid orbitals and VSEPR. Carbon only uses one $\text{2s}$ and one $\text{2p}$ orbitals to form $\text{sp}$ hybrid orbitals.