Chapter 2, Problem 2E.13E

(a)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for ${\text{I}}_3{}^{-}$ 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.13E

The shape for ${\text{I}}_3{}^{-}$ molecule is linear, corresponding VSEPR formula is ${\text{X}}_2{\text{E}}_3$ and bond angle is $180°$.

Explanation of Solution

${\text{I}}_3{}^{-}$ has $\text{I}$ as central atom. $\text{I}$ possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in ${\text{I}}_3{}^{-}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=7\left(3\right)+1\\ =22\left(11\text{ pairs}\right)\end{array}$

The skeleton structure in ${\text{I}}_3{}^{-}$ has two bond pairs that comprise 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=22-4\\ =18\end{array}$

These 9 electron pairs are allotted as lone pairs of each of the iodine atoms to satisfy their octet. Hence, the Lewis structure in ${\text{I}}_3{}^{-}$ is illustrated below:

It is evident that in ${\text{I}}_3{}^{-}$, the central iodine atom has two bond pairs and three lone pairs. Lone pairs tend to occupy the equatorial locations of trigonal plane in trigonal bi-pyramidal arrangement so as to have minimum repulsions in accordance with VSPER model. This results in linear ${\text{I}}_3{}^{-}$ molecule. The shape is linear so bond angle is $180°$.

If lone pairs are represented by E, and other attached bond pairs by X, then for any linear species the VSEPR formula is predicted as ${\text{X}}_2{\text{E}}_3$.

(b)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for molecule has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.13E

The shape for ${\text{POCl}}_{\text{3}}$ is tetrahedral, corresponding VSEPR formula is ${\text{AX}}_4$ and bond angle is $109.5°$.

Explanation of Solution

${\text{POCl}}_{\text{3}}$ has $\text{P}$ as central atom. $\text{P}$ possesses five valence electrons and $\text{O}$ and $\text{Cl}$ have 6 and 7 electrons respectively.

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

  $\begin{array}{c}\text{Total valence electrons}=5\left(1\right)+6\left(1\right)+7\left(3\right)\\ =32\left(16\text{ pairs}\right)\end{array}$

The skeleton structure in ${\text{POCl}}_{\text{3}}$ has four bond pairs that comprise 8 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=32-8\\ =24\end{array}$

These 12 electron pairs are allotted as lone pairs or multiple bonds to satisfy respective octets. Hence, the Lewis structure in ${\text{POCl}}_{\text{3}}$ and corresponding VSEPR geometry and bond angle is illustrated below:

It is evident that in ${\text{POCl}}_{\text{3}}$ the central phosphorus atom has four bond pairs because the double bond is also regarded as single unit. To minimize repulsion all bond pairs can easily be accommodated in a tetrahedral shape in accordance with VSPER model.

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any tetrahedral species the VSEPR formula is predicted to be ${\text{AX}}_4$.

(c)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for ${\text{IO}}_3{}^{-}$ molecule has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.13E

The shape for ${\text{IO}}_3{}^{-}$ is trigonal pyramidal, corresponding VSEPR formula is ${\text{AX}}_{\text{4}}{\text{E}}_{\text{2}}$ and bond angle is less than $109.5°$.

Explanation of Solution

${\text{IO}}_3{}^{-}$ has $\text{I}$ as central atom. $\text{I}$ has seven valence electrons and oxygen possesses 6 valence electrons.

Total valence electrons are sum of the valence electrons on atom in ${\text{IO}}_3{}^{-}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=7\left(1\right)+6\left(3\right)+1\\ =26\left(13\text{ pairs}\right)\end{array}$

The skeleton structure in ${\text{IO}}_3{}^{-}$ has three bond pairs that comprise 6 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=26-6\\ =20\end{array}$

These 10 electron pairs are allotted as lone pairs to each of the oxygen to satisfy its octet. Hence, the Lewis structure in ${\text{IO}}_3{}^{-}$ along with corresponding VSEPR geometry is illustrated below:

It is evident that in ${\text{IO}}_3{}^{-}$ the central iodine atom has three bond pairs and one lone pairs. Such four electron pairs correspond to tetrahedral arrangement.

Lone pairs tend to be localized on apical position while $109.5°$ bond angle reduces slightly from usual tetrahedral bond angle so as to have minimum repulsions in accordance with VSPER model. This results in trigonal pyramidal shape for ${\text{IO}}_3{}^{-}$.

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any trigonal pyramidal species the VSEPR formula is predicted as ${\text{AX}}_3\text{E}$.

(d)

Interpretation Introduction

Interpretation:

The VSEPR formula, shape for the Lewis structure of ${\text{N}}_{\text{2}}\text{O}$ molecule along with bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.13E

The shape for ${\text{N}}_{\text{2}}\text{O}$ molecule is linear with bond angle as $180°$ and corresponding VSEPR formula is ${\text{X}}_2{\text{E}}_3$.

Explanation of Solution

${\text{N}}_{\text{2}}\text{O}$ has $\text{N}$ as central atom that possesses 5 valence electrons. Oxygen possesses 6 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in ${\text{N}}_{\text{2}}\text{O}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=5\left(2\right)+6\\ =16\end{array}$

The skeleton structure in ${\text{N}}_{\text{2}}\text{O}$ 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 as lone pairs or double bonds on each of the atom in ${\text{N}}_{\text{2}}\text{O}$ to satisfy the octet. Hence, the Lewis structure in ${\text{N}}_{\text{2}}\text{O}$ and corresponding VSPER geometry is illustrated below:

It is evident that in ${\text{N}}_{\text{2}}\text{O}$, the central iodine atom has two bond pairs (since each triple or double bond is treated as one super pair). This results in linear ${\text{N}}_{\text{2}}\text{O}$ in accordance with VSPER model and consequently, bond angle must be $180°$.

If central atom is represented by A, and other attached bond pairs by X, then for any linear species the VSEPR formula is predicted as ${\text{AX}}_2$.