Chapter 8.3, Problem 26LC

(a)

Interpretation Introduction

Interpretation: Using VSEPR theory to predict bond angles in given molecules.

Concept introduction: The Valence Shell Electron Pair Repulsion Theory abbreviated as VSEPR theory is based on the premise that there is repulsion between the pairs of electrons in all atoms.

Answer to Problem 26LC

Bond angles for the covalently bonded molecules are:

Methane = $\text{109}{\text{.5}}^{\text{o}}$

Explanation of Solution

We will use the formula, that is

Number of electrons $\text{=}\frac{\text{1}}{\text{2}}\text{[V+N-C+A]}$

Here, V= number of valence electrons present in the central atoms.

N= number of monovalent atoms bound to central atoms

C= charge of the cation

A= charge of the anion

Methane: in the given molecule we have carbon as the central atom and there are four hydrogens as monovalent atoms,

By formula we get

  $\begin{array}{l}\text{=}\frac{\text{1}}{\text{2}}\text{[4+4+-0+0]}\\ \text{=4}\end{array}$

As now we know the hybridization will be ${\text{sp}}^{\text{3}}$ and the geometry of the molecule will be tetrahedral. Therefore the number of electrons is four.

So, the bond angle for the tetrahedral geometry is $\text{109}{\text{.5}}^{\text{o}}$ .

(b)

Interpretation Introduction

Interpretation: Using VSEPR theory to predict bond angles in given molecules.

Concept introduction: The Valence Shell Electron Pair Repulsion Theory abbreviated as VSEPR theory is based on the premise that there is repulsion between the pairs of electrons in all atoms.

Answer to Problem 26LC

Bond angles for the covalently bonded molecules are:

Ammonia = ${\text{107}}^{\text{o}}$

Explanation of Solution

Ammonia: Now in this nitrogen is the central atom and there are three hydrogens as monovalent atoms.

Now, from the formula we get $\text{=}\frac{\text{1}}{\text{2}}\text{[5+3-0+0]=4}$

So, the hybridization will be $s{p}^3$ and the geometry of the molecule will be pyramidal.

And the Bond angle for this is ${\text{107}}^{\text{o}}$ .

(c)

Interpretation Introduction

Interpretation: Using VSEPR theory to predict bond angles in given molecules.

Concept introduction: The Valence Shell Electron Pair Repulsion Theory abbreviated as VSEPR theory is based on the premise that there is repulsion between the pairs of electrons in all atoms.

Answer to Problem 26LC

Bond angles for the covalently bonded molecules are:

Water = $\text{104}{\text{.5}}^{\text{o}}$

Explanation of Solution

Water: Now in this water oxygen is the central atom and there are two hydrogens as monovalent atoms.

Now, from the formula we get $\text{=}\frac{\text{1}}{\text{2}}\text{[6+2-0+0]=4}$

The hybridization will be ${\text{sp}}^{\text{3}}$ and the geometry of the molecule will be bent as there will be two bond pairs and two lone pairs.

The bond angle for the pyramidal geometry is $\text{104}{\text{.5}}^{\text{o}}$ .

The bond angle reduces further due to greater lone pair repulsions.