Chapter 12, Problem 58QAP

Give the total number of valence electrons in each of the following molecules.

msp;  $\begin{array}{l}\text{a}.{\text{ B}}_{\text{2}}{\text{O}}_{\text{3}} \text{c}.{\text{ C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}\\ \text{b}.{\text{ CO}}_{\text{2}} \text{}\text{}\text{}\text{} \text{d}.{\text{ NO}}_{\text{2}}\end{array}$

Interpretation Introduction

(a)

Interpretation:

The number of valence electrons of the given compound is to be stated.

Concept Introduction:

The distribution of the electrons that exists in the atomic orbital of an atom is collectively known as electron configuration. The description of every electron that is moving freely in an orbital is given by the electron configuration of that atom.

The electrons that are present in the outermost shell of an atom are known as valence electrons. The outermost shell of an atom is known as valence shell. The valence electrons are more loosely held than the electrons which are closer to the nucleus. The formula for the calculation of number of valence electrons in a molecule is:

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_{\text{1}}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_2\right)+\\ \mathrm{.......}+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_{\text{m}}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Answer to Problem 58QAP

The number of valence electrons in ${\text{B}}_{\text{2}}{\text{O}}_3$ is $24$.

Explanation of Solution

Total valence electrons in ${\text{B}}_{\text{2}}{\text{O}}_3$ is calculated as shown below.

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{B}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{O}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Substitute the values of $\text{n}$, valence electrons of $\text{B}$ and $\text{O}$ in the above expression.

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{B}\right)+\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{O}\right)\\ =2\left(3\right)+3\left(6\right)\\ =24\text{valence}\text{electrons}\end{array}$

Thus the number of valence electrons in ${\text{B}}_{\text{2}}{\text{O}}_3$ is $24$.

Interpretation Introduction

(b)

Interpretation:

The number of valence electrons of the given compound is to be stated.

Concept Introduction:

The distribution of the electrons that exists in the atomic orbital of an atom is collectively known as electron configuration. The description of every electron that is moving freely in an orbital is given by the electron configuration of that atom.

The electrons that are present in the outermost shell of an atom are known as valence electrons. The outermost shell of an atom is known as valence shell. The valence electrons are more loosely held than the electrons which are closer to the nucleus. The formula for the calculation of number of valence electrons in a molecule is:

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_{\text{1}}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_2\right)+\\ \mathrm{.......}+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_{\text{m}}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Answer to Problem 58QAP

The number of valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$ is $20$.

Explanation of Solution

Total valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$ is calculated as shown below.

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{C}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{H}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{O}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Substitute the values of $\text{n}$, valence electrons of $\text{C,}\text{H}$ and $\text{O}$ in the above expression.

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{C}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{H}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{O}\right)\end{array}\right)\\ =2\left(4\right)+6\left(1\right)+1\left(6\right)\\ =20\text{valence}\text{electrons}\end{array}$

Thus the number of valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}$ is $20$.

Interpretation Introduction

(c)

Interpretation:

The number of valence electrons of the given compound is to be stated.

Concept Introduction:

The distribution of the electrons that exists in the atomic orbital of an atom is collectively known as electron configuration. The description of every electron that is moving freely in an orbital is given by the electron configuration of that atom.

The electrons that are present in the outermost shell of an atom are known as valence electrons. The outermost shell of an atom is known as valence shell. The valence electrons are more loosely held than the electrons which are closer to the nucleus. The formula for the calculation of number of valence electrons in a molecule is:

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_{\text{1}}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_2\right)+\\ \mathrm{.......}+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_{\text{m}}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Answer to Problem 58QAP

The number of valence electrons in ${\text{CO}}_2$ is $16$.

Explanation of Solution

Total valence electrons in ${\text{CO}}_2$ is calculated as shown below.

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{C}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{O}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Substitute the values of $\text{n}$, valence electrons of $\text{C}$ and $\text{O}$ in the above expression.

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{C}\right)+\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{O}\right)\\ =1\left(4\right)+2\left(6\right)\\ =16\text{valence}\text{electrons}\end{array}$

Thus the number of valence electrons in ${\text{CO}}_2$ is $16$.

Interpretation Introduction

(d)

Interpretation:

The number of valence electrons of the given compound is to be stated.

Concept Introduction:

The distribution of the electrons that exists in the atomic orbital of an atom is collectively known as electron configuration. The description of every electron that is moving freely in an orbital is given by the electron configuration of that atom.

The electrons that are present in the outermost shell of an atom are known as valence electrons. The outermost shell of an atom is known as valence shell. The valence electrons are more loosely held than the electrons which are closer to the nucleus. The formula for the calculation of number of valence electrons in a molecule is:

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_{\text{1}}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_2\right)+\\ \mathrm{.......}+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}{\text{Atom}}_{\text{m}}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Answer to Problem 58QAP

The number of valence electrons in ${\text{NO}}_2$ is $17$.

Explanation of Solution

Total valence electrons in ${\text{NO}}_2$ is calculated as shown below.

$\text{Total}\text{valence}\text{electrons}=\left(\begin{array}{l}\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{N}\right)+\\ \text{n}\left(\text{Valence}\text{electrons}\text{in}\text{O}\right)\end{array}\right)$

Where,

• $\text{n}$ is the number of atoms.

Substitute the values of $\text{n}$, valence electrons of $\text{N}$ and $\text{O}$ in the above expression.

$\begin{array}{c}\text{Total}\text{valence}\text{electrons}=\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{N}\right)+\text{n}\left(\text{Valence}\text{electrons}\text{in}\text{O}\right)\\ =1\left(5\right)+2\left(6\right)\\ =17\text{valence}\text{electrons}\end{array}$

Thus the number of valence electrons in ${\text{NO}}_2$ is $17$.