Chapter 8, Problem 24PS

Give approximate values for the indicated bond angles:

1. (a) Cl—S—Cl in SCl₂
2. (b) N—N—O in N₂O
3. (c) Bond angles 1, 2, and 3 in vinyl alcohol (a component of polymers and a molecule found in outer space)

Interpretation Introduction

Interpretation:

The approximate value for $Cl-S-Cl$ bond angle  in $SC{l}_2$ molecule should be determined.

Concept Introduction:

VSEPR Theory:

As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then tend to be in position in order to minimize the repulsions.

According to VSEPR theory, the geometry is predicted by the minimizing the repulsions between electron-pairs in the bonds and lone-pairs of electrons. The VSEPR theory is summarized in the given table as,

$\begin{array}{cccc}\text{Electron-pair}& \text{lone-pair}& \text{Electron-pair}\text{geometry}& \text{Molecular}\text{shape}\\ 2& 0& \text{Linear}& \text{Linear}\\ 3& 0& \text{Trigonal}\text{planar}& \text{Trigonal}\text{planar}\\ 2& 1& \text{Trigonal}\text{planar}& \text{Bent}\\ 4& 0& \text{Tetrahedral}& \text{Tetrahedral}\\ 3& 1& \text{Tetrahedral}& \text{Pyramidal}\\ 2& 2& \text{Tetrahedral}& \text{V}-\text{shape}\\ 5& 0& \text{Trigonal}\text{bipyramidal}& \text{Trigonal}\text{bipyramidal}\\ 4& 1& \text{Trigonal}\text{bipyramidal}& \text{See}-\text{saw}\\ 3& 2& \text{Trigonal}\text{bipyramidal}& \text{T}-\text{shape}\\ 2& 3& \text{Trigonal}\text{bipyramidal}& \text{Linear}\\ 6& 0& \text{Octahedral}& \text{Octahedral}\\ 5& 1& \text{Octahedral}& \text{Square}\text{pyramidal}\\ 4& 2& \text{Octahedral}& \text{Square}\text{planar}\end{array}$

Bond angle: The bond angle is the angle formed between three atoms at least two bonds.

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& Geometry& Bondangle\\ A{X}_2& sp& Linear& {180}^{°}\\ A{X}_3,A{X}_{2}B& s{p}^2& Trigonalplanar& {120}^{°}\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& Tetrahedral& {109.5}^{°}\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& Trigonalbipyramidal& {120}^{°},{90}^{°}\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& Octahedral& {90}^{°}\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

Structure of $SC{l}_2$ molecule is,

Sulfur is the central atom in $Cl-S-Cl$ bond in the $SC{l}_2$ molecule. There are four electron pairs (2 lone pair and 2 bond pair) around the central atom and it is $s{p}^3$ hybridized. Therefore $Cl-S-Cl$ bond angle value in the $SC{l}_2$ molecule is ${109.5}^o$

Interpretation Introduction

Interpretation:

The approximate value for $N-N-O$ bond angle in $N_{2}O$ molecule should be determined.

Concept Introduction:

VSEPR Theory:

As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then tend to be in position in order to minimize the repulsions.

According to VSEPR theory, the geometry is predicted by the minimizing the repulsions between electron-pairs in the bonds and lone-pairs of electrons. The VSEPR theory is summarized in the given table as,

$\begin{array}{cccc}\text{Electron-pair}& \text{lone-pair}& \text{Electron-pair}\text{geometry}& \text{Molecular}\text{shape}\\ 2& 0& \text{Linear}& \text{Linear}\\ 3& 0& \text{Trigonal}\text{planar}& \text{Trigonal}\text{planar}\\ 2& 1& \text{Trigonal}\text{planar}& \text{Bent}\\ 4& 0& \text{Tetrahedral}& \text{Tetrahedral}\\ 3& 1& \text{Tetrahedral}& \text{Pyramidal}\\ 2& 2& \text{Tetrahedral}& \text{V}-\text{shape}\\ 5& 0& \text{Trigonal}\text{bipyramidal}& \text{Trigonal}\text{bipyramidal}\\ 4& 1& \text{Trigonal}\text{bipyramidal}& \text{See}-\text{saw}\\ 3& 2& \text{Trigonal}\text{bipyramidal}& \text{T}-\text{shape}\\ 2& 3& \text{Trigonal}\text{bipyramidal}& \text{Linear}\\ 6& 0& \text{Octahedral}& \text{Octahedral}\\ 5& 1& \text{Octahedral}& \text{Square}\text{pyramidal}\\ 4& 2& \text{Octahedral}& \text{Square}\text{planar}\end{array}$

Bond angle: The bond angle is the angle formed between three atoms at least two bonds.

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& Geometry& Bondangle\\ A{X}_2& sp& Linear& {180}^{°}\\ A{X}_3,A{X}_{2}B& s{p}^2& Trigonalplanar& {120}^{°}\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& Tetrahedral& {109.5}^{°}\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& Trigonalbipyramidal& {120}^{°},{90}^{°}\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& Octahedral& {90}^{°}\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

Structure of $N_{2}O$ molecule is,

In $N_{2}O$ molecule the central nitrogen atom is $sp$ hybridized. Therefore the molecule has a linear geometry, and so the $N-N-O$ bond angle in $N_{2}O$ molecule is ${180}^0$

Interpretation Introduction

Interpretation:

The approximate values of the angles marked 1, 2, and 3 in the given molecule should be determined.

Concept Introduction:

VSEPR Theory:

As the name itself indicates that the basis for this theory is the electron pair that is bonded electron present in either single or double bonds or lone pair electrons, present in the valence shell tends to repel each other which then tend to be in position in order to minimize the repulsions.

According to VSEPR theory, the geometry is predicted by the minimizing the repulsions between electron-pairs in the bonds and lone-pairs of electrons. The VSEPR theory is summarized in the given table as,

$\begin{array}{cccc}\text{Electron-pair}& \text{lone-pair}& \text{Electron-pair}\text{geometry}& \text{Molecular}\text{shape}\\ 2& 0& \text{Linear}& \text{Linear}\\ 3& 0& \text{Trigonal}\text{planar}& \text{Trigonal}\text{planar}\\ 2& 1& \text{Trigonal}\text{planar}& \text{Bent}\\ 4& 0& \text{Tetrahedral}& \text{Tetrahedral}\\ 3& 1& \text{Tetrahedral}& \text{Pyramidal}\\ 2& 2& \text{Tetrahedral}& \text{V}-\text{shape}\\ 5& 0& \text{Trigonal}\text{bipyramidal}& \text{Trigonal}\text{bipyramidal}\\ 4& 1& \text{Trigonal}\text{bipyramidal}& \text{See}-\text{saw}\\ 3& 2& \text{Trigonal}\text{bipyramidal}& \text{T}-\text{shape}\\ 2& 3& \text{Trigonal}\text{bipyramidal}& \text{Linear}\\ 6& 0& \text{Octahedral}& \text{Octahedral}\\ 5& 1& \text{Octahedral}& \text{Square}\text{pyramidal}\\ 4& 2& \text{Octahedral}& \text{Square}\text{planar}\end{array}$

Bond angle: The bond angle is the angle formed between three atoms at least two bonds.

$\begin{array}{l}\begin{array}{cccc}Typeofmolecule& Hybridaization& Geometry& Bondangle\\ A{X}_2& sp& Linear& {180}^{°}\\ A{X}_3,A{X}_{2}B& s{p}^2& Trigonalplanar& {120}^{°}\\ A{X}_4,A{X}_{3}B,A{X}_2{B}_2& s{p}^3& Tetrahedral& {109.5}^{°}\\ A{X}_5,A{X}_{4}B,A{X}_3{B}_2,A{X}_2{B}_3& s{p}^{3}d& Trigonalbipyramidal& {120}^{°},{90}^{°}\\ A{X}_6,A{X}_{5}B,A{X}_4{B}_2& s{p}^3{d}^2& Octahedral& {90}^{°}\end{array}\\ A\to Centralatom\\ X\to AtomsbondedtoA\\ B\to NonbondingelectronpairsonA\end{array}$

Explanation of Solution

Given structure of vinyl alcohol is,

Value of the angles marked as 1, 2, and 3 in the molecule can be determined as follows,

1 and 2 are the bond angles between $H-C-H$ and $C=C-H$ bonds respectively. carbon atom in both of these bonds are $s{p}^2$ hybridized and so the bond angle value will be ${120}^o$

3 is the bond angle between $C-O-H$ bonds. The central oxygen atom in the bond is $s{p}^3$ hybridized and so the bond angle value will be ${109.5}^o$