Chapter 2, Problem 2E.15E

(a)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle, and molecular shape for ${\text{CF}}_3\text{Cl}$ 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.15E

The shape for ${\text{CF}}_3\text{Cl}$ molecule is tetrahedral, bond angle is $109.5°$ and corresponding VSEPR formula is ${\text{AX}}_4$.

Explanation of Solution

${\text{CF}}_3\text{Cl}$ has $\text{C}$ as central atom. $\text{I}$ possesses 7 valence electrons, carbon has four and $\text{F}$ has 7 valence electrons.

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

  $\begin{array}{c}\text{Total valence electrons}=4+7\left(3\right)+7\\ =32\end{array}$

The skeleton structure in ${\text{CF}}_3\text{Cl}$ 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 ${\text{CF}}_3\text{Cl}$ is illustrated below:

It is evident that in ${\text{CF}}_3\text{Cl}$ the central carbon atom has four bond pairs and zero lone pair. to have minimum repulsions ${\text{CF}}_3\text{Cl}$ adopts tetrahedral accordance with VSPER model and thus the bond angle is $109.5°$.

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

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.15E

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

Explanation of Solution

${\text{TeCl}}_4$ has $\text{Te}$ as central atom. $\text{Te}$ possesses 6 valence electrons and $\text{Cl}$ have 7 electrons respectively.

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

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

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

It is evident that ${\text{TeCl}}_4$ has four bond pairs and one lone pairs. Thus with total 5 pairs around central $\text{Te}$ atom, it is predicted to correspond to trigonal bipyramidal 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 ${\text{TeCl}}_4$. Clearly, the bond angles for such a molecular shape is $\text{90 }°$ and $\text{120 }°$ respectively since the trigonal plane will be at right angles to the equatorial plane that has each atom oriented at $\text{120 }°$.

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 to be ${\text{AX}}_4\text{E}$ .

(c)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle and molecular shape for ${\text{COF}}_2$ molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.15E

The shape for ${\text{COF}}_2$ molecule is trigonal planar, bond angle is $\text{120 }°$ and corresponding VSEPR formula is ${\text{AX}}_3$.

Explanation of Solution

${\text{COF}}_2$ has $\text{C}$ as central atom. $\text{C}$ has 4 valence electrons, oxygen possesses 6 valence electrons and $\text{F}$ has 7 valence electrons.

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

  $\begin{array}{c}\text{Total valence electrons}=4+6+7\left(2\right)\\ =24\left(12\text{ pairs}\right)\end{array}$

The skeleton structure in ${\text{COF}}_2$ has three bond pairs that comprise 6 electrons. The electrons left to be allocated as lone pairs are determined as follows:

  $\begin{array}{c}\text{Remaining electrons}=24-6\\ =18\end{array}$

These 9 electron pairs are allotted as lone pairs or as multiple bonds to oxygen so as to satisfy octets. Hence, the Lewis structure in ${\text{COF}}_2$ along with corresponding VSEPR geometry, bond angle is illustrated below:

It is evident that in ${\text{COF}}_2$ the central carbon atom has three bond pairs and zero lone pairs. Such four electron pairs correspond to trigonal planar arrangement so as to have minimum repulsions in accordance with VSPER model. This results in trigonal planar shape for ${\text{COF}}_2$ each oriented at $\text{120 }°$.

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

(d)

Interpretation Introduction

Interpretation:

Lewis structure, VSEPR formula, bond angle and molecular shape for ${\text{CH}}_3{}^{-}$ molecule have to be predicted.

Concept Introduction:

Refer to part (a).

Answer to Problem 2E.15E

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

Explanation of Solution

${\text{CH}}_3{}^{-}$ has $\text{C}$ as central atom that has 4 valence electrons.

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

  $\begin{array}{c}\text{Total valence electrons}=4+1\left(3\right)+1\\ =8\end{array}$

Thus, Lewis structure in ${\text{CH}}_3{}^{-}$ has three bond pairs to each of the three $\text{H}$ that leaves one pair to be assigned as lone pair. Hence, the Lewis structure in ${\text{CH}}_3{}^{-}$ along with VSEPR geometry and corresponding bond angle is illustrated below:

It is evident that in ${\text{CH}}_3{}^{-}$ the central carbon atom has three bond pairs and one lone pair. Such four electron pairs correspond to tetrahedral arrangement.

Lone pair tend to be localized on apical position and so $109.5°$ bond angle reduces slightly from usual tetrahedral bond angle so as to have minimum repulsions in accordance with VSPER model. This results in trigonal pyramidal shape for ${\text{CH}}_3{}^{-}$.

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