Chapter 2, Problem 2E.11E

(a)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for sulfur tetrachloride 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.11E

The shape for sulfur tetrachloride molecule is square planar and VSEPR formula is ${\text{AX}}_{\text{4}}\text{E}$.

Explanation of Solution

Sulfur tetrachloride has sulphur as central atom. Sulfur has six valence electrons while chlorine possesses seven valence electrons.

Total valence electrons are sum of the valence electrons on each chlorine and central sulfur in ${\text{SCl}}_{\text{4}}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=6\left(1\right)+7\left(4\right)\\ =34\left(17\text{ pairs}\right)\end{array}$

The skeleton structure in ${\text{SCl}}_{\text{4}}$ has four bond pairs that comprise 8 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=34-8\\ =26\end{array}$

These 13 electron pairs are assigned as lone pairs of each of the chlorine atoms to satisfy its octet.

Hence, the Lewis structure of ${\text{SCl}}_{\text{4}}$ that has shape corresponding to square planar arrangement is illustrated below.

It is evident that in ${\text{SCl}}_{\text{4}}$ the central sulfur atom has four bond pairs and two lone pairs. Lone pair on central sulfur tends to be localized in the equatorial triangular plane so as to have minimum repulsions in accordance with VSPER model. This results in see-saw ${\text{SCl}}_{\text{4}}$ 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 ${\text{AX}}_{\text{4}}\text{E}$.

(b)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for iodine trichloride molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.11E

The shape for iodine trichloride is T-shape and VSEPR formula is ${\text{AX}}_3{\text{E}}_{\text{2}}$.

Explanation of Solution

Iodine trichloride 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 ${\mathrm{ICl}}_3$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=7\left(1\right)+7\left(3\right)\\ =28\left(14\text{ pairs}\right)\end{array}$

The skeleton structure in ${\mathrm{ICl}}_3$ has four bond pairs that comprise 8 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=28-6\\ =22\end{array}$

These 11 electron pairs are allotted as lone pairs of each of the chlorine atoms to satisfy its octet. Hence, the Lewis structure and corresponding T-shape in ${\mathrm{ICl}}_3$ is illustrated below:

It is evident that in ${\text{ICl}}_{\text{3}}$ the central iodine atom has three bond pairs and two 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 T-shaped ${\text{ICl}}_{\text{3}}$ molecule.

If lone pairs are represented by E, central atom with A and other attached bond pairs by X, then for any bent or T-shaped species the VSEPR formula is predicted to be ${\text{AX}}_3{\text{E}}_{\text{2}}$.

(c)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for ${\text{IF}}_4{}^{-}$ molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.11E

The shape for ${\text{IF}}_4{}^{-}$ is T-shaped and corresponding VSEPR formula is ${\text{AX}}_{\text{4}}{\text{E}}_{\text{2}}$.

Explanation of Solution

${\text{IF}}_4{}^{-}$ 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 fluorine and central iodine in ${\text{IF}}_4{}^{-}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=7\left(1\right)+7\left(4\right)+1\\ =36\left(18\text{ pairs}\right)\end{array}$

The skeleton structure in ${\text{IF}}_4{}^{-}$ has four bond pairs that comprise 8 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=36-8\\ =28\end{array}$

These 14 electron pairs are allotted as lone pairs of each of the fluorine atoms to satisfy its octet. Hence, the Lewis structure and corresponding T-shape in ${\text{IF}}_4{}^{-}$ is illustrated below:

In ${\text{IF}}_4{}^{-}$, the central iodine atom has four bond pairs and two lone pairs. Lone pairs tend to occupy the apical locations of octahedral arrangement so as to have minimum repulsions in accordance with VSPER model. This results in square planar ${\text{IF}}_4{}^{-}$.

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 ${\text{AX}}_{\text{4}}{\text{E}}_{\text{2}}$.

(d)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for xenon trioxide molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.11E

The shape for xenon trioxide molecule is trigonal pyramidal and corresponding VSEPR formula is ${\text{AX}}_3\text{E}$.

Explanation of Solution

Xenon trioxide has $\text{Xe}$ as central atom. $\text{Xe}$ has eight valence electrons and oxygen possesses 6 valence electrons.

Total valence electrons are sum of the valence electrons on each oxygen atom and central $\text{Xe}$ is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=8\left(1\right)+6\left(3\right)\\ =26\left(13\text{ pairs}\right)\end{array}$

The skeleton structure in ${\text{XeO}}_3$ has three bond pairs that include three doubly bonded oxygen atom. This comprises of 12 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=26-12\\ =14\end{array}$

These 7 electron pairs are allotted as lone pairs of each of the oxygen atoms to satisfy its octet. Thus, the Lewis structure and shape of ${\text{XeO}}_3$ is illustrated below:

It is evident that in ${\text{XeO}}_3$ the central xenon atom has three bond pairs and one lone pair. Lone pairs tend to occupy one of the apex positions of tetrahedron so as to have minimum repulsions in accordance with VSPER model. This results in trigonal pyramidal ${\text{XeO}}_3$ molecule.

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