Chapter 2, Problem 2E.12E

(a)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for ${\text{PF}}_4{}^{-}$ 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.12E

The shape for ${\text{PF}}_4{}^{-}$ molecule is see-saw and corresponding VSEPR formula is ${\text{AX}}_{\text{4}}\text{E}$.

Explanation of Solution

${\text{PF}}_4{}^{-}$ 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 along with uni-negative charge in ${\text{PF}}_4{}^{-}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=5+7\left(4\right)+1\\ =34\end{array}$

The skeleton structure in ${\text{PF}}_4{}^{-}$ 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 $\text{F}$ atoms to satisfy its octet.

Hence, the Lewis structure ${\text{PF}}_4{}^{-}$ that has shape corresponding trigonal bi-pyramidal arrangement is illustrated as follows:

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 ${\text{PF}}_4{}^{-}$ the central phosphorus atom has four bond pairs and one lone pair. Lone pair tends to be localized in the equatorial trigonal plane so as to have minimum repulsions in accordance with VSPER model. This results in see-saw shape in ${\text{PF}}_4{}^{-}$ molecule.

(b)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for ${\text{ICl}}_4{}^{+}$ ion have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.12E

The shape for ${\text{ICl}}_4{}^{+}$ is see-saw and corresponding VSEPR formula is ${\text{AX}}_{\text{4}}\text{E}$.

Explanation of Solution

${\text{ICl}}_4{}^{+}$ 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 chlorine and central iodine in ${\text{ICl}}_4{}^{+}$ calculated as follows:

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

The skeleton structure in ${\text{ICl}}_4{}^{+}$ 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 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 ${\text{ICl}}_4{}^{+}$ is illustrated below.

It is evident that in ${\text{ICl}}_4{}^{+}$ the central iodine atom has four bond pairs to four chlorine atoms and one lone pair. 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 ${\text{AX}}_{\text{4}}\text{E}$.

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 see-saw shaped ${\text{ICl}}_4{}^{+}$.

(c)

Interpretation Introduction

Interpretation:

The VSEPR formula and shape for ${\text{PF}}_5$ molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.12E

The shape for ${\text{PF}}_5$ is trigonal pyramidal and corresponding VSEPR formula is ${\text{AX}}_5$.

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 along with uni-negative charge 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 comprises 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, Lewis structure of ${\text{PF}}_5$ 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.

(d)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.12E

The shape for xenon terafluoride molecule is square planar and corresponding VSEPR formula is ${\text{AX}}_{\text{4}}{\text{E}}_{\text{2}}$.

Explanation of Solution

Xenon terafluoride has $\text{Xe}$ as central atom. $\text{Xe}$ has eight valence electrons and fluorine possesses 7 valence electrons.

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

  $\begin{array}{c}\text{Total valence electrons}=8+7\left(4\right)\\ =36\end{array}$

The skeleton structure in ${\text{XeF}}_4$ has four bond pairs. This comprises of 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 and central xenon to satisfy respective octets. Thus, the Lewis structure and corresponding VSEPR geometry ${\text{XeF}}_4$ is illustrated below:

It is evident that in ${\text{XeF}}_4$ the central xenon atom has four bond pairs and two lone pairs. Lone pairs tend to occupy the two opposite apex positions of octahedron so as to have minimum repulsions in accordance with VSPER model. This results in square planar ${\text{XeF}}_4$ molecule.

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