Chapter 2, Problem 2E.20E

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{PBr}}_5$ has to be drawn, number of lone pairs on central atom, molecular shape, and bond angle 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 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.20E

The shape for ${\text{PBr}}_5$ 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

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

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

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

The skeleton structure in ${\text{PBr}}_5$ 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 $\text{Br}$ atoms to satisfy respective octets.

Hence, the Lewis structure ${\text{PBr}}_5$ 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 ${\text{PBr}}_5$, the central phosphorus atom has five bond pairs and no lone pairs. To have minimum repulsions three bromine form bond in trigonal equatorial plane while two other bromine atoms form bonds in axial plane to have minimum repulsions in accordance with VSPER model. This results in trigonal pyramidal shape in ${\text{PBr}}_5$ 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.

(b)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{XeOF}}_2$ has to be drawn, number of lone pairs on central atom, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.20E

The shape for ${\text{XeOF}}_2$ molecule T-shaped, number of lone pairs on central xenon is two and bond angles are $\text{120 }°$ , $\text{180 }°$ and $\text{90 }°$.

Explanation of Solution

${\text{XeOF}}_2$ has $\text{Xe}$ as central atom. $\text{Xe}$ has eight valence electrons and ${\text{XeOF}}_2$ fluorine possesses 7 valence electrons.

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

  $\begin{array}{c}\text{Total valence electrons}=8+6+7\left(2\right)\\ =28\end{array}$

The skeleton structure in ${\text{XeOF}}_2$ has three bond pairs. This comprises of 6 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 fluorine, oxygen atoms and central xenon to satisfy respective octets. Thus, the Lewis structure and corresponding VSEPR geometry ${\text{XeOF}}_2$ is illustrated below:

It is evident that in ${\text{XeOF}}_2$, the central xenon atom has three bond pairs and two lone pair. Lone pairs tend to occupy the two equatorial positions so as to stay at $\text{120 }°$ apart from each other and at right angles to axial fluorine are $180°$ apart from each other in accordance with VSPER model.  Thus bond angles that correspond to T-shaped ${\text{XeOF}}_2$ are $\text{120 }°$ , $\text{180 }°$ and $\text{90 }°$.

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{SF}}_5{}^{+}$ has to be drawn and number of lone pairs on central atom, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.20E

The shape for ${\text{SF}}_5{}^{+}$ molecule trigonal pyramidal, number of lone pairs on central sulphur is zero and bond angles are $\text{120 }°$ , $\text{180 }°$ and $\text{90 }°$.

Explanation of Solution

${\text{SF}}_5{}^{+}$ has $\text{S}$ as central atom. $\text{S}$ has six valence electrons and $\text{F}$ possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each chlorine and central iodine in ${\text{SF}}_5{}^{+}$ calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=6+7\left(5\right)-1\\ =40\end{array}$

The skeleton structure in ${\text{SF}}_5{}^{+}$ has two bond pairs that comprises 4 electrons. The electrons left to be allocated as lone pairs are determined as follows:

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

These 15 electron pairs are allotted as lone pairs to each of the $\text{F}$ atoms to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in ${\text{SF}}_5{}^{+}$ is illustrated below:

It is evident that in ${\text{SF}}_5{}^{+}$, the central sulfur atom has five bond pairs and zero lone pairs.  This corresponds to trigonal bi-pyramidal arrangement. The equatorial bonds locations in trigonal plane are $120°$ apart so as to have minimum repulsions in accordance with VSPER model while the axial are $180°$ apart from each other and are at right angles to equatorial bonds.

(d)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{IF}}_3$ has to be drawn and number of lone pairs on central atom, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.20E

The shape for ${\text{IF}}_3$ molecule T-shape, number of lone pairs on central iodine is two pairs and corresponding bond angles are $\text{120 }°$ , $\text{180 }°$ and $\text{90 }°$.

Explanation of Solution

${\text{IF}}_3$ has $\text{I}$ as central atom. $\text{I}$ has seven valence electrons and $\text{F}$ also possesses 7 valence electrons.

Total valence electrons are sum of the valence electrons on each $\text{F}$ and central iodine in ${\text{IF}}_3$ calculated as follows:

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

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

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

These 11 electron pairs are allotted as lone pairs of each of the fluorine atoms and central iodine to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in ${\text{IF}}_3$ is illustrated below:

It is evident that in ${\text{IF}}_3$ the central iodine atom has three bond pairs to three chlorine atoms and two 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 ${\text{AX}}_3{\text{E}}_2$.

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 three chlorine atoms take up two axial positions and one equatorial position so that they are $180°$ apart from each other and are at right angles to equatorial lone pairs. This result in linear shape for ${\text{IF}}_3$.

(e)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{BrO}}_3{}^{-}$ has to be drawn and number of lone pairs on central atom, molecular shape, and bond angle has to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.20E

The shape for ${\text{BrO}}_3{}^{-}$ molecule trigonal pyramidal, number of lone pairs on central bromine is one and corresponding bond angles are $\text{107 }°$.

Explanation of Solution

${\text{BrO}}_3{}^{-}$ has $\text{Br}$ as central atom. $\text{Br}$ has seven valence electrons and $\text{O}$ possesses 6 valence electrons.

Total valence electrons are sum of the valence electrons on each atom in ${\text{BrO}}_3{}^{-}$ along with uni-negative charge calculated as follows:

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

The skeleton structure in ${\text{BrO}}_3{}^{-}$ has three bond pairs that comprises 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 or multiple bonds to satisfy respective octets. Hence, the Lewis structure and corresponding VSPER geometry in ${\text{BrO}}_3{}^{-}$ is illustrated below:

It is evident that in ${\text{BrO}}_3{}^{-}$ the central bromine atom has three bond pairs to three oxygen atoms as the double bond is also regarded as single bond unit. No of lone pair on central bromine is one.

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 ${\text{AX}}_3\text{E}$.

The bond pairs in ${\text{BrO}}_3{}^{-}$ are less than $109.5°$ apart to have minimum repulsions in accordance with VSPER model. The three oxygen atoms take up triangular plane in base. This results in trigonal pyramidal for ${\text{BrO}}_3{}^{-}$.