Chapter 2, Problem 2E.14E

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{SiCl}}_{\text{4}}$ has to be drawn and VSEPR formula, molecular shape, and bond angle 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 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 pair–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.14E

The shape for ${\text{SiCl}}_{\text{4}}$ molecule is tetrahedral, VSEPR formula is ${\text{AX}}_{\text{4}}$ and corresponding bond angle is $109.5°$.

Explanation of Solution

${\text{SiCl}}_{\text{4}}$ has $\text{Si}$ as central atom. $\text{Si}$ has four valence electrons while $\text{Cl}$ possesses seven valence electrons.

Total valence electrons are sum of the valence electrons on each atom in ${\text{SiCl}}_{\text{4}}$ calculated as follows:

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

The skeleton structure in ${\text{SiCl}}_{\text{4}}$ has four bond pairs that comprise of 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 assigned as lone pairs of each of the $\text{Cl}$ atoms to satisfy its octet. Hence, the Lewis structure of ${\text{SiCl}}_{\text{4}}$ along with VSEPR molecular geometry and corresponding bond angle is illustrated below.

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

It is evident that in ${\text{SiCl}}_{\text{4}}$,the central silicon atom has four bond pairs and no lone pairs. Thus four bonds are separated at an angle of $109.5°$ so as to have minimum bond pair- bond pair repulsions in accordance with VSPER model. This results in tetrahedral shape in ${\text{SiCl}}_{\text{4}}$ molecule.

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{PF}}_5$ has to be drawn and VSEPR formula, molecular shape, and bond angle have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.14E

The shape for ${\text{PF}}_5$ is trigonal pyramidal, corresponding VSEPR formula is ${\text{AX}}_5$ and the bond angles are $\text{120 }°$ , $\text{180 }°$ and $\text{90 }°$.

Explanation of Solution

${\text{PF}}_5$ has $\text{P}$ as central atom. $\text{P}$ has five valence electrons while $\text{F}$ possesses seven valence electrons.

Total valence electrons are sum of the valence electrons on each atom in ${\text{PF}}_5$ calculated as follows:

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

The skeleton structure in ${\text{PF}}_5$ 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 assigned as lone pairs of each of the $\text{F}$ atoms to satisfy respective octets. Hence, the Lewis structure ${\text{PF}}_5$ that has shape corresponding trigonal bi-pyramidal arrangement is illustrated below.

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any trigonal pyramidal geometry the VSEPR formula is predicted as ${\text{AX}}_5$.

It is evident that in ${\text{PF}}_5$, the central phosphorus atom has five bond pairs and no lone pairs. to have minimum repulsions three fluorine form bonds in trigonal equatorial plane while two other fluorine atoms form bonds in axial plane to have minimum repulsions in accordance with VSPER model. This results in trigonal pyramidal shape in ${\text{PF}}_5$ molecule.

The bond angles are $\text{120 }°$ in the trigonal equatorial plane and $\text{180 }°$ for axial fluorine. The equatorial plane is at $\text{90 }°$ to axial plane.

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{SBr}}_2$ has to be drawn and VSEPR formula, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.14E

The shape for ${\text{SBr}}_2$ molecule is angular or bent, VSEPR formula is ${\text{AX}}_2\text{E}$ and corresponding bond angle is less than $105°$.

Explanation of Solution

${\text{SBr}}_2$ has $\text{S}$ as central atom. $\text{S}$ has six valence electrons and $\text{Br}$ also possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each chlorine and central iodine in ${\text{SBr}}_2$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=6+7\left(2\right)\\ =20\end{array}$

The skeleton structure in ${\text{SBr}}_2$ has two bond pairs that comprises 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=20-4\\ =16\end{array}$

These 8 electron pairs are allotted as lone pairs to each of the bromine atoms and central sulfur to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in ${\text{SBr}}_2$ is illustrated below:

It is evident that in ${\text{SBr}}_2$, the central sulfur atom has two bond pairs and two lone pairs.  This corresponds to tetrahedral arrangement. However, as the two localized lone pairs tend to occupy maximum space the bond angle reduces slightly from usual tetrahedral bond angle to nearly $105°$ so as to have minimum repulsions in accordance with VSPER model. This results in angular or bent shaped for ${\text{SBr}}_2$.

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any see-saw species the VSEPR formula is predicted as ${\text{AX}}_2\text{E}$.

(d)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{ICl}}_2{}^{+}$ has to be drawn and VSEPR formula, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.14E

The shape for ${\text{ICl}}_2{}^{+}$ molecule is linear, VSEPR formula is ${\text{AX}}_2{\text{E}}_3$ and corresponding bond angles are $\text{120 }°$ , $\text{180 }°$ and $\text{90 }°$.

Explanation of Solution

${\text{ICl}}_2{}^{+}$ has $\text{I}$ as central atom. $\text{I}$ has seven valence electrons and chlorine also possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each chlorine and central iodine in ${\text{ICl}}_2{}^{+}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=7+7\left(2\right)-1\\ =20\end{array}$

The skeleton structure in ${\text{ICl}}_2{}^{+}$ has two bond pairs that comprises 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=20-4\\ =16\end{array}$

These 8 electron pairs are allotted as lone pairs of each of the chlorine atoms and central iodine to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in ${\text{ICl}}_2{}^{+}$ is illustrated below:

It is evident that in ${\text{ICl}}_2{}^{+}$ the central iodine atom has two bond pairs to two chlorine atoms and three lone pairs. If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any see-saw species the VSEPR formula is predicted as ${\text{AX}}_2{\text{E}}_3$.

Lone pairs tend to occupy the equatorial locations of trigonal plane so that they are $120°$ apart to have minimum repulsions in accordance with VSPER model while the two chlorine atoms take up axial position so that they are $180°$ apart from each other and are at right angles to equatorial lone pairs. This results in linear shape for ${\text{ICl}}_2{}^{+}$.