Chapter 17, Problem 1CE

Interpretation Introduction

(a)

Interpretation:

The oxidation number for nonmetal X in ${\text{X}}_{\text{2}}$ is to be stated.

Concept introduction:

The oxidation of an element is the tendency of an element to lose electrons. Oxidation number is the charge carried by an atom on formation of compound. A neutral molecule with identical elements have zero oxidation state.

Answer to Problem 1CE

The oxidation number for non metal X in ${\text{X}}_{\text{2}}$ is zero.

Explanation of Solution

In ${\text{X}}_{\text{2}}$, the compound formed is between two identical elements with same electronegativity. In this, no donation of electrons takes place between them and a covalent bond will be formed by sharing of electrons. This leads to the development of zero charge on both the elements.

Conclusion

Zero is the oxidation number for non metal X in ${\text{X}}_{\text{2}}$.

Interpretation Introduction

(b)

Interpretation:

The oxidation number for non metal X in ${\text{X}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

The oxidation of an element is the tendency of an element to lose electrons. Oxidation number is the charge carried by an atom on formation of compound. A neutral molecule with different elements have elements present in different oxidation state.

Answer to Problem 1CE

The oxidation number for non metal X in ${\text{X}}_{\text{2}}\text{O}$ is $+1$.

Explanation of Solution

In ${\text{X}}_{\text{2}}\text{O}$, the compound formed is between two identical elements with same electronegativity and one oxygen atom with different electronegativity. In this, donation of electrons takes place between them but the overall charge on the molecule is zero. In this compound, both $\text{X}$ donate electron to oxygen and therefore, oxygen attains $-2$ charge.

Assume the oxidation number of $\text{X}$ to be y then, oxidation number on 2 $\text{X}$ will be 2y.

Calculate the oxidation number of $\text{X}$ in ${\text{X}}_{\text{2}}\text{O}$ as shown below.

$\begin{array}{rll}2y-2& =& 0\\ 2y& =& 2\\ y& =& +1\end{array}$

Therefore, the oxidation number of X in ${\text{X}}_{\text{2}}\text{O}$ is $+1$.

Conclusion

The oxidation number for non metal X in ${\text{X}}_{\text{2}}\text{O}$ is found to be $+1$.

Interpretation Introduction

(c)

Interpretation:

The oxidation number for nonmetal X in ${\text{CaX}}_{\text{2}}$ is to be stated.

Concept introduction:

The oxidation of an element is the tendency of an element to lose electrons. Oxidation number is the charge carried by an atom on formation of compound. A neutral molecule with different elements have elements present in different oxidation state.

Answer to Problem 1CE

The oxidation number for nonmetal X in ${\text{CaX}}_{\text{2}}$ is $-1$.

Explanation of Solution

In ${\text{CaX}}_{\text{2}}$, the compound formed is between two identical elements with same electronegativity and one calcium atom with different electronegativity. In this, donation of electrons takes place between them but the overall charge on the molecule is zero. In this compound, both $\text{X}$ accept electron from calcium and therefore, calcium attains $+2$ charge.

Assume the oxidation number of $\text{X}$ to be y then, oxidation number on 2 $\text{X}$ will be 2y.

Calculate the oxidation number of $\text{X}$ in ${\text{CaX}}_{\text{2}}$ as shown below.

$\begin{array}{rll}2+2y& =& 0\\ 2y& =& -2\\ y& =& -1\end{array}$

Therefore, the oxidation number of X in ${\text{CaX}}_{\text{2}}$ is $-1$.

Conclusion

The oxidation number for nonmetal X in ${\text{CaX}}_{\text{2}}$ is found to be $-1$.<!--

Interpretation Introduction

(d)

Interpretation:

The oxidation number for nonmetal X in ${\text{HXO}}_{\text{4}}$ is to be stated.

Concept introduction:

The oxidation of an element is the tendency of an element to lose electrons. Oxidation number is the charge carried by an atom on formation of compound. A neutral molecule with different elements have elements present in different oxidation state.

Answer to Problem 1CE

The oxidation number for non metal X in ${\text{HXO}}_{\text{4}}$ is $+7$.

Explanation of Solution

In ${\text{HXO}}_{\text{4}}$, the compound formed is between four identical oxygen elements with same electronegativity and two atom with different electronegativity. In this, donation of electrons takes place between them but the overall charge on the molecule is zero. In this compound, oxygen accepts electron from hydrogen and $\text{X}$ and therefore, oxygen attains $-2$ charge. There are $4$ oxygen atoms, so, the total charge on oxygen atoms will be $-8$.

Assume the oxidation number of $\text{X}$ in ${\text{HXO}}_{\text{4}}$ to be y.

Calculate the oxidation number of $\text{X}$ in ${\text{HXO}}_{\text{4}}$ as shown below.

$\begin{array}{rll}+1+y-8& =& 0\\ y& =& +7\end{array}$

Therefore, the oxidation number of X in ${\text{HXO}}_{\text{4}}$ is $+7$.

Conclusion

The oxidation number for non metal X in ${\text{HXO}}_{\text{4}}$ is found to be $+7$.