Chapter 2, Problem 2E.17E

(a)

Interpretation Introduction

Interpretation:

Bond angle at central atom for ${\text{O}}_{\text{3}}$ molecule has to be predicted.

Concept Introduction:

Valence Shell Electron Pair Repulsion model predicts shape by inclusion of bond angles and most distant arrangement of atoms that leads to minimum repulsion. For the molecules that have no lone pairs around the central atom the bonded-atom unshared -pair arrangement is decided by the table as follows:

$\begin{array}{cc}\begin{array}{l}\text{Number of}\\ \text{electron pairs}\end{array}& \text{Molecular shape}\\ 2& \text{Linear}\\ 3& \text{Trigonal Planar}\\ 4& \text{Tetrahedral}\\ 5& \text{Trigonal Bipyramidal}\\ 6& \text{Octahedral}\\ 7& \text{Pentagonal Bipyramidal}\end{array}$

In order to determine the shape the steps to be followed are indicated as follows:

1. 1. Lewis structure of molecule should be written.
2. 2. The type electron arrangement around the central atom should be identified around the central atom. This essentially refers to determination of bond pairs and unshared or lone pairs around central atoms.
3. 3. Then bonded-atom unshared -pair arrangement that can maximize the distance of electron pairs about central atom determines the shape.

For molecules that have lone pairs around central atom, lone pairs influence shape, because there are no atoms at the positions occupied by these lone pairs. The key rule that governs the molecular shape, in this case, is the extent of lone –lone pair repulsions are far greater than lone bond pair or bond pair-bond pair repulsions. The table that summarized the molecular shapes possible for various combinations of bonded and lone pairs are given as follows:

$\begin{array}{cccc}\text{Steric number}& \text{Number of lone pairs}& \text{Molecular geometry}& \text{Bond angles}\\ 2& 0& \text{Linear}& 180°\\ 3& \begin{array}{c}0\\ 1\end{array}& \begin{array}{c}\text{Trigonal planar}\\ \text{Bent}\end{array}& 120°\\ 4& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{Tetrahedral}\\ \text{Trigonal pyramidal}\\ \text{Bent}\end{array}& 109.5°\\ 5& \begin{array}{c}0\\ 1\\ 2\\ 3\end{array}& \begin{array}{c}\text{Trigonal Bi-pyramidal}\\ \text{See-Saw}\\ \text{T-shaped}\\ \text{Linear}\end{array}& 90°\text{,120 }°,180°\\ 6& \begin{array}{c}0\\ 1\\ 2\end{array}& \begin{array}{c}\text{Octahedral}\\ \text{Square pyramidal}\\ \text{Square planar}\end{array}& 90°\text{,}180°\end{array}$

Answer to Problem 2E.17E

In ${\text{O}}_{\text{3}}$ molecule bond angle is less than $\text{120 }°$.

Explanation of Solution

${\text{O}}_{\text{3}}$ has $\text{O}$ as central atom. $\text{O}$ possesses 6 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in ${\text{O}}_{\text{3}}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=6\left(3\right)\\ =18\end{array}$

The skeleton structure in ${\text{O}}_{\text{3}}$ has two bond pairs that comprise 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=18-4\\ =14\end{array}$

These 7 electron pairs are allotted as lone pairs to satisfy respective octets. Hence, the Lewis structure in ${\text{O}}_{\text{3}}$ is illustrated below:

It is evident that ${\text{O}}_{\text{3}}$ has trigonal planar arrangement due to two bond pairs and one lone pair. The lone pair is localized into one of the trigonal planar corners and it tends to occupy more space thus the bond angle reduces from usual $\text{120 }°$ to about $\text{119}\text{.5 }°$ so as to minimize lone pair–bond pair repulsions in accordance with VSPER model.

(b)

Interpretation Introduction

Interpretation:

Bond angle at central atom for ${\text{N}}_{\text{3}}{}^{-}$ has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.17E

In azide ion, bond angle is $\text{180 }°$.

Explanation of Solution

${\text{N}}_{\text{3}}{}^{-}$ has $\text{N}$ as central atom. $\text{N}$ possesses 6 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in ${\text{N}}_{\text{3}}{}^{-}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=5\left(3\right)+1\\ =16\end{array}$

The skeleton structure in ${\text{N}}_{\text{3}}{}^{-}$ has two bond pairs that comprise 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=16-4\\ =12\end{array}$

These 6 electron pairs are allotted multiple bonds to satisfy respective octets. Hence, the Lewis structure in ${\text{N}}_{\text{3}}{}^{-}$ is illustrated below:

It is evident that ${\text{N}}_{\text{3}}{}^{-}$ has linear arrangement due to two bond pairs and zero lone pair. These bond pairs form a linear pair so as to minimize bond pair–bond pair repulsions in accordance with VSPER model and thus bond angle is $\text{180 }°$.

(c)

Interpretation Introduction

Interpretation:

Bond angle at central atom for ${\text{CNO}}^{-}$ has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.17E

In ${\text{CNO}}^{-}$ bond angle is $\text{180 }°$.

Explanation of Solution

${\text{CNO}}^{-}$ has $\text{C}$ as central atom. $\text{N}$ possesses 5 valence electrons and $\text{O}$ possesses 6 valence electrons.

Thus total valence electrons is sum of the valence electrons on each atom along with charge in ${\text{CNO}}^{-}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=4+5+6+1\\ =16\end{array}$

The skeleton structure in ${\text{CNO}}^{-}$ has two bonds that comprises 4 electrons. The electrons left to be allocated are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=16-4\\ =12\end{array}$

These 6 electron pairs are allotted multiple bonds to satisfy respective octets. Hence, the Lewis structure in ${\text{N}}_{\text{3}}{}^{-}$ is illustrated below:

It is evident that ${\text{CNO}}^{-}$ has linear arrangement due to two bond pairs and zero lone pair. These bond pairs form a linear pair so as to minimize bond pair–bond pair repulsions in accordance with VSPER model and thus bond angle is $\text{180 }°$.

(d)

Interpretation Introduction

Interpretation:

Bond angle at central atom for ${\text{H}}_{\text{3}}{\text{O}}^{+}$ has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.17E

Bond angle in ${\text{H}}_{\text{3}}{\text{O}}^{+}$ is $\text{107 }°$.

Explanation of Solution

${\text{H}}_{\text{3}}{\text{O}}^{+}$ has $\text{O}$ as central atom. $\text{O}$ has 6 valence electrons, hydrogen possesses 1 valence electron.

Total valence electrons are sum of the valence electrons on atom in ${\text{H}}_{\text{3}}{\text{O}}^{+}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=6+3\left(1\right)-1\\ =8\end{array}$

The skeleton structure in ${\text{H}}_{\text{3}}{\text{O}}^{+}$ has three bond pairs that comprise 6 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=8-6\\ 2\end{array}$

This 1 electron pair is allotted as lone pair. Hence, the Lewis structure in ${\text{H}}_{\text{3}}{\text{O}}^{+}$ along with bond angle is illustrated below:

It is evident that in ${\text{H}}_{\text{3}}{\text{O}}^{+}$ the central oxygen atom has three bond pairs and one lone pairs. Such three electron pairs and one lone pair correspond to tetrahedral arrangement that has lone pair localized at apex of the tetrahedron so as to have minimum repulsions in accordance with VSPER model. This results in trigonal pyramidal shape for ${\text{H}}_{\text{3}}{\text{O}}^{+}$ each oriented at angle close to $\text{107 }°$.