Chapter 2, Problem 2E.16E

(a)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle, and molecular shape for ${\text{PCl}}_3{\text{F}}_2$ molecule have 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.16E

The shape for ${\text{PCl}}_3{\text{F}}_2$ molecule is trigonal bi-pyramidal, bond angles are $\text{90 }°$ and $\text{120 }°$ respectively and corresponding VSEPR formula is ${\text{AX}}_3{\text{X}}_2\text{'}$.

Explanation of Solution

${\text{PCl}}_3{\text{F}}_2$ has $\text{P}$ as central atom. $\text{I}$ possesses 7 valence electrons, $\text{P}$ has 5 and $\text{F}$ , $\text{Cl}$ have 7 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in ${\text{PCl}}_3{\text{F}}_2$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=5+7\left(3\right)+7\left(2\right)\\ =40\end{array}$

The skeleton structure in ${\text{PCl}}_3{\text{F}}_2$ has five bond pairs that comprise 10 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=40-10\\ =30\end{array}$

These 15 electron pairs are allotted as lone pairs to satisfy respective octets. Hence, the Lewis structure in ${\text{PCl}}_3{\text{F}}_2$ is illustrated below:

It is evident that ${\text{PCl}}_3{\text{F}}_2$ has four bond pairs and one lone pairs. Thus with total 5 pairs around central sulphur atom, it is predicted to correspond to trigonal bi-pyramidal arrangement.

 One lone pair is localized on equatorial positions so as to minimize lone pair–bond pair repulsions in accordance with VSPER model. This leads see-saw shape for ${\text{PCl}}_3{\text{F}}_2$. Clearly the bond angles for such a molecular shape is  $\text{90 }°$ and $\text{120 }°$ respectively since the equivalent fluorine atom, each atom oriented at $\text{120 }°$, in trigonal plane will be at right angles to the axial chlorine atoms in ${\text{PCl}}_3{\text{F}}_2$.

If lone pairs are represented by E, central atom with A and each unique atom attached by X and $\text{X'}$ respectively then for such trigonal bi-pyramidal the VSEPR formula is predicted to be ${\text{AX}}_3{\text{X}}_2\text{'}$ .

(b)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle, and molecular shape for ${\text{SnF}}_4$ molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.16E

The shape for ${\text{SnF}}_4$ molecule is see-saw, bond angles are $\text{90 }°$ and $\text{120 }°$ corresponding VSEPR formula is ${\text{AX}}_4$.

Explanation of Solution

${\text{SnF}}_4$ has $\text{Sn}$ as central atom. $\text{Sn}$ possesses 4 valence electrons and $\text{F}$ have 7 electrons respectively.

Total valence electrons are sum of the valence electrons on atom in ${\text{SnF}}_4$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=4+7\left(4\right)\\ =32\end{array}$

The skeleton structure in ${\text{SnF}}_4$ 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 to satisfy respective octets. Hence, the Lewis structure in ${\text{SnF}}_4$ along with corresponding VSEPR geometry and bond angle is illustrated below:

It is evident that ${\text{SnF}}_4$ has four bond pairs and zero lone pairs. Thus with total 5 pairs around central $\text{Sn}$ atom, it is predicted to correspond to tetrahedral arrangement So as to minimize lone pair–bond pair repulsions in accordance with VSPER model. This leads to bond angles as $\text{109}\text{.5 }°$.

If lone pairs are represented by E, central atom with A and other attached bon pairs by X, then for any tetrahedral species with no one pairs the VSEPR formula is predicted to be ${\text{AX}}_4$.

(c)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle, and molecular shape for ${\text{SnF}}_6{}^{2-}$ molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.16E

The shape for ${\text{SnF}}_6{}^{2-}$ molecule is trigonal planar, bond angle is $\text{120 }°$ and corresponding VSEPR formula is ${\text{AX}}_3$.

Explanation of Solution

${\text{SnF}}_6{}^{2-}$ has $\text{Sn}$ as central atom. $\text{Sn}$ possesses 4 valence electrons and $\text{F}$ have 7 electrons respectively.

Total valence electrons are sum of the valence electrons on atom along with two negative charges  in ${\text{SnF}}_6{}^{2-}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=4+7\left(6\right)+2\\ =48\end{array}$

The skeleton structure in ${\text{SnF}}_6{}^{2-}$ has six bond pairs that comprise 12 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=48-12\\ =36\end{array}$

These 18 electron pairs are allotted as lone pairs on each fluorine atom to satisfy respective octets. Hence, the Lewis structure in ${\text{SnF}}_6{}^{2-}$ along with corresponding VSEPR geometry and bond angle is illustrated below:

It is evident that in ${\text{SnF}}_6{}^{2-}$ the central tin atom has 6 bond pairs and zero lone pairs. Such six electron pairs correspond to octahedral arrangement with VSEPR formula as ${\text{AX}}_6$. so as to have minimum repulsions as per VSPER model.

(d)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle and molecular shape for ${\text{IF}}_5$ molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.16E

The shape for ${\text{IF}}_5$ molecule is square pyramidal, bond angle is $\text{90 }°$ and $\text{180 }°$ while corresponding VSEPR formula is ${\text{AX}}_5\text{E}$.

Explanation of Solution

${\text{IF}}_5$ has $\text{I}$ as central atom. $\text{I}$ has seven valence electrons and fluorine also possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each fluorine and central iodine in ${\text{IF}}_5$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=7\left(1\right)+7\left(5\right)\\ =42\end{array}$

The skeleton structure in ${\text{IF}}_5$ has five bond pairs that comprises 10 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=42-10\\ =32\end{array}$

These 16 electron pairs are allotted as lone pairs of each of the fluorine atoms and one on central iodine to satisfy respective octet. Hence, the Lewis structure ${\text{IF}}_5$ along with predicted VSEPR shape and corresponding bond angle is illustrated below:

It is evident that in ${\text{IF}}_5$ the central iodine atom has five bond pairs and one lone pair. Lone pairs tend to occupy one of the apical position in octahedral arrangement so as to have minimum repulsions in accordance with VSPER model. This results in square pyramidal ${\text{IF}}_5$ molecule.

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any square planar species the VSEPR formula is predicted as ${\text{AX}}_{\text{4}}{\text{E}}_{\text{2}}$.

(e)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle and molecular shape for ${\text{XeO}}_4$ molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.16E

The shape for ${\text{XeO}}_4$ is trigonal pyramidal, bond angle is approximately less than $109.5°$ and corresponding VSEPR formula is ${\text{AX}}_3\text{E}$.

Explanation of Solution

${\text{XeO}}_4$ has $\text{Xe}$ as central atom that has 8 valence electrons and $\text{O}$ has 6 valence electrons.

Total valence electrons are sum of the valence electrons on atom in ${\text{XeO}}_4$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=8+6\left(4\right)\\ =32\end{array}$

Thus, Lewis structure in ${\text{XeO}}_4$ has four bond pairs to each of the four $\text{O}$. 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 either lone pairs or multiple bonds with $\text{O}$ atoms. Hence, the Lewis structure in ${\text{XeO}}_4$ along with VSEPR geometry and corresponding bond angle is illustrated below:

It is evident that in ${\text{XeO}}_4$ the central xenon atom has four bond pairs (since each double bond is regarded as single super pair). Such four electron pairs correspond to tetrahedral arrangement.

So $109.5°$ occurs for tetrahedral ${\text{XeO}}_4$ so as to have minimum repulsions 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 as ${\text{AX}}_4$ .