Chapter 7, Problem 38QRT

(a)

Interpretation Introduction

Interpretation:

The hybridization and approximate bond angle for each carbon atoms in ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{CCH}$ has to be identified.

Concept Introduction:

Hybridization is the mixing of valence atomic orbitals to get equivalent hybridized orbitals that having similar characteristics and energy.

Geometry of different types of molecule with respect to the hybridizations are mentioned are mentioned below,

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& \begin{array}{l}Atomicorbitals\\ usedforhybridaization\end{array}& Geometry\\ A{X}_2& sp& 1s+1p& Linear\\ A{X}_3,A{X}_{2}B& s{p}^2& 1s+2p& Trigonalplanar\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& 1s+3p& Tetrahedral\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& 1s+3p+1d& Trigonalbipyramidal\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& 1s+3p+2d& Octahedral\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Bond angle is the angle between two bonds of a molecule and it is determined based on the electron-domain geometry.

Explanation of Solution

Given molecule is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{CCH}$.

The Lewis electron dot structure for alanine is given below:

  .

Consider the first carbon. There will be four electron regions in the molecule and hence the electron-region geometry will be tetrahedral. For a molecule having tetrahedral geometry, the hybridization will be ${\text{sp}}^{\text{3}}$. For a tetrahedral molecule, the bond angle will be $\text{109}{\text{.5}}^{\text{o}}$.

Consider the second carbon. There will be four electron regions in the molecule and hence the electron-region geometry will be tetrahedral. For a molecule having tetrahedral geometry, the hybridization will be ${\text{sp}}^{\text{3}}$. For a tetrahedral molecule, the bond angle will be $\text{109}{\text{.5}}^{\text{o}}$.

Consider the third carbon. There will be two electron regions in the molecule and hence the electron-region geometry will be linear. For a molecule having linear geometry, the hybridization will be $\text{sp }$. For a linear molecule, the bond angle will be ${\text{180}}^{\text{o}}$.

Consider the fourth carbon atom. There will be two electron regions in the molecule and hence the electron-region geometry will be linear. For a molecule having linear geometry, the hybridization will be $\text{sp }$. For a linear molecule, the bond angle will be ${\text{180}}^{\text{o}}$.

(b)

Interpretation Introduction

Interpretation:

The shortest carbon-to-carbon bond length in molecule has to be identified.

Concept Introduction:

Bond length is the distance between the nuclei in a bond and it is related to the sum of the covalent radii at the bonded atoms.

Higher the bond order, shorter will be the bond length.

Explanation of Solution

Given molecule is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{CCH}$.

The Lewis electron dot structure for alanine is given below:

  .

In the given molecule, the triple bond is present between $\text{C-3}\text{to}\text{C-4}$. Thus, the shortest carbon-to-carbon bond is between $\text{C-3}\text{to}\text{C-4}$.

(c)

Interpretation Introduction

Interpretation:

The strongest carbon-to-carbon bond length in molecule has to be identified.

Concept Introduction:

Comparing triple bond, double bond and single bond, the bond with higher energy is the triple bond.

Explanation of Solution

Given molecule is ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{CCH}$.

The Lewis electron dot structure for alanine is given below:

  .

In the given molecule, the triple bond is present between $\text{C-3}\text{to}\text{C-4}$.

The strength of a triple bond is more than the single bond.

Thus, the strongest carbon-to-carbon bond is between $\text{C-3}\text{to}\text{C-4}$.