Chapter 8, Problem 8.91P

(a)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare hypochlorous acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is dichlorine oxide and its chemical formula is ${\text{Cl}}_2\text{O}$.

Explanation of Solution

The formula of hypochlorous acid is $\text{HClO}$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of chlorine in $\text{HClO}$ is as follows:

  $\left[\begin{array}{l}\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of chlorine }\right)+\\ \left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (1)

Rearrange equation (1) for the oxidation number of chlorine as follows:

  $\text{Oxidation number of chlorine}=\left[\begin{array}{l}-\left(\text{oxidation number of hydrogen}\right)\\ -\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (2)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (2).

  $\begin{array}{c}\text{Oxidation number of chlorine}=\left[-1-\left(-2\right)\right]\\ =\left[-1+2\right]\\ =+1.\end{array}$

The oxidation number of chlorine is $+1$ so the oxide formed is dichlorine oxide with the chemical formula ${\text{Cl}}_2\text{O}$.

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.

(b)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare chlorous acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is dichlorine trioxide and its chemical formula is ${\text{Cl}}_2{\text{O}}_3$.

Explanation of Solution

The formula of chlorous acid is ${\text{HClO}}_2$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of chlorine in ${\text{HClO}}_2$ is as follows:

  $\left[\begin{array}{l}\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of chlorine }\right)+\\ 2\left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (3)

Rearrange equation (3) for the oxidation number of chlorine as follows:

  $\text{Oxidation number of chlorine}=\left[\begin{array}{l}-\left(\text{oxidation number of hydrogen}\right)\\ -2\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (4)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (4).

  $\begin{array}{c}\text{Oxidation number of chlorine}=\left[-1-2\left(-2\right)\right]\\ =\left[-1+4\right]\\ =+3.\end{array}$

The oxidation number of chlorine is $+3$ so the oxide formed is dichlorine trioxide with the chemical formula ${\text{Cl}}_2{\text{O}}_3$.

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.

(c)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare chloric acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is dichlorine pentaoxide and its chemical formula is ${\text{Cl}}_2{\text{O}}_5$.

Explanation of Solution

The formula of chloric acid is ${\text{HClO}}_3$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of chlorine in ${\text{HClO}}_3$ is as follows:

  $\left[\begin{array}{l}\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of chlorine }\right)+\\ 3\left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (5)

Rearrange equation (5) for the oxidation number of chlorine as follows:

  $\text{Oxidation number of chlorine}=\left[\begin{array}{l}-\left(\text{oxidation number of hydrogen}\right)\\ -3\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (6)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (6).

  $\begin{array}{c}\text{Oxidation number of chlorine}=\left[-1-3\left(-2\right)\right]\\ =\left[-1+6\right]\\ =+5.\end{array}$

The oxidation number of chlorine is $+5$ so the oxide formed is dichlorine pentaoxide with the chemical formula ${\text{Cl}}_2{\text{O}}_5$

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.

(d)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare perchloric acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is dichlorine heptaoxide and its chemical formula is ${\text{Cl}}_2{\text{O}}_7$.

Explanation of Solution

The formula of perchloric acid is ${\text{HClO}}_4$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of chlorine in ${\text{HClO}}_4$ is as follows:

  $\left[\begin{array}{l}\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of chlorine }\right)+\\ 4\left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (7)

Rearrange equation (7) for the oxidation number of chlorine as follows:

  $\text{Oxidation number of chlorine}=\left[\begin{array}{l}-\left(\text{oxidation number of hydrogen}\right)\\ -4\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (8)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (8).

  $\begin{array}{c}\text{Oxidation number of chlorine}=\left[-1-4\left(-2\right)\right]\\ =\left[-1+8\right]\\ =+7.\end{array}$

The oxidation number of chlorine is $+7$ so the oxide formed is dichlorine heptaoxide with the chemical formula ${\text{Cl}}_2{\text{O}}_7$

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.

(e)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare sulphuric acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is sulfur trioxide and its chemical formula is ${\text{SO}}_3$.

Explanation of Solution

The formula of sulfuric acid is ${\text{H}}_2{\text{SO}}_4$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of sulfur in ${\text{H}}_2{\text{SO}}_4$ is as follows:

  $\left[\begin{array}{l}2\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of sulphur }\right)+\\ 4\left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (9)

Rearrange equation (9) for the oxidation number of sulfur as follows:

  $\text{Oxidation number of sulphur}=\left[\begin{array}{l}-2\left(\text{oxidation number of hydrogen}\right)\\ -4\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (10)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (10).

  $\begin{array}{c}\text{Oxidation number of sulphur}=\left[2\left(-1\right)-4\left(-2\right)\right]\\ =\left[-2+8\right]\\ =+6.\end{array}$

The oxidation number of sulfur is $+6$ so the oxide formed is sulfur trioxide with the chemical formula ${\text{SO}}_3$

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.

(f)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare sulfurous acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is sulfur dioxide and its chemical formula is ${\text{SO}}_2$.

Explanation of Solution

The formula of sulfurous acid is ${\text{H}}_2{\text{SO}}_3$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of sulfur in ${\text{H}}_2{\text{SO}}_3$ is as follows:

  $\left[\begin{array}{l}2\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of sulphur }\right)+\\ 3\left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (11)

Rearrange equation (11) for the oxidation number of sulfur as follows:

  $\text{Oxidation number of sulphur}=\left[\begin{array}{l}-2\left(\text{oxidation number of hydrogen}\right)\\ -3\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (12)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (12).

  $\begin{array}{c}\text{Oxidation number of sulphur}=\left[2\left(-1\right)-3\left(-2\right)\right]\\ =\left[-2+6\right]\\ =+4.\end{array}$

The oxidation number of sulfur is $+4$ so the oxide formed is sulfur dioxide with the chemical formula ${\text{SO}}_2$

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.

(g)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare nitric acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is dinitrogen pentaoxide and its chemical formula is ${\text{N}}_2{\text{O}}_5$.

Explanation of Solution

The formula of nitric acid is ${\text{HNO}}_3$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of nitrogen in ${\text{HNO}}_3$ is as follows:

  $\left[\begin{array}{l}2\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of nitrogen }\right)+\\ 3\left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (13)

Rearrange equation (13) for the oxidation number of nitrogen as follows:

  $\text{Oxidation number of nitrogen}=\left[\begin{array}{l}-\left(\text{oxidation number of hydrogen}\right)\\ -3\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (14)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (14).

  $\begin{array}{c}\text{Oxidation number of nitrogen}=\left[-1-3\left(-2\right)\right]\\ =\left[-1+6\right]\\ =+5.\end{array}$

The oxidation number of nitrogen is $+5$ so the oxide formed is dinitrogen pentaoxide with the chemical formula ${\text{N}}_2{\text{O}}_5$

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.

(h)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare nitrous acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is dinitrogen trioxide and its chemical formula is ${\text{N}}_2{\text{O}}_3$.

Explanation of Solution

The formula of nitrous acid is ${\text{HNO}}_2$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of nitrogen in ${\text{HNO}}_2$ is as follows:

  $\left[\begin{array}{l}\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of nitrogen }\right)+\\ 2\left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (15)

Rearrange equation (15) for the oxidation number of nitrogen as follows:

  $\text{Oxidation number of nitrogen}=\left[\begin{array}{l}-\left(\text{oxidation number of hydrogen}\right)\\ -2\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (16)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (14).

  $\begin{array}{c}\text{Oxidation number of nitrogen}=\left[-1-2\left(-2\right)\right]\\ =\left[-1+4\right]\\ =+3.\end{array}$

The oxidation number of nitrogen is $+3$ so the oxide formed is dinitrogen trioxide with the chemical formula ${\text{N}}_2{\text{O}}_3$

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.

(i)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare carbonic acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is carbon dioxide and its chemical formula is ${\text{CO}}_2$.

Explanation of Solution

The formula of carbonic acid is ${\text{H}}_2{\text{CO}}_3$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of carbon in ${\text{H}}_2{\text{CO}}_3$ is as follows:

  $\left[\begin{array}{l}2\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of carbon }\right)+\\ 3\left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (17)

Rearrange equation (17) for the oxidation number of carbon is as follows:

  $\text{Oxidation number of carbon}=\left[\begin{array}{l}-2\left(\text{oxidation number of hydrogen}\right)\\ -3\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (18)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (18).

  $\begin{array}{c}\text{Oxidation number of carbon}=\left[2\left(-1\right)-3\left(-2\right)\right]\\ =\left[-2+6\right]\\ =+4.\end{array}$

The oxidation number of carbon is $+4$ so the oxide formed is carbon dioxide with the chemical formula ${\text{CO}}_2$

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.

(j)

Interpretation Introduction

Interpretation:

The name and the formula of the oxide that is used to prepare phosphoric acid are to be determined.

Concept introduction:

Oxidation of a species involves the loss of electrons by that species and reduction of a species involves the gain of electrons by that species.

The oxidation number is defined as the formal charge an atom would gain if all the bonds attached to it in a compound are heterolytically cleaved. Oxidation number can be a positive or negative number but cannot be fractional.

Answer to Problem 8.91P

The name of the oxide is diphosphorus pentaoxide and its chemical formula is ${\text{P}}_2{\text{O}}_5$.

Explanation of Solution

The formula of phosphoric acid is ${\text{H}}_3{\text{PO}}_4$.

The oxidation number of hydrogen is $+1$ and the oxidation state of oxygen is $-2$.

The expression to calculate the oxidation number of phosphorous in ${\text{H}}_3{\text{PO}}_4$ is as follows:

  $\left[\begin{array}{l}3\left(\text{oxidation number of hydrogen}\right)+\\ \left(\text{oxidation number of phosphorous }\right)+\\ 4\left(\text{oxidation number of oxygen}\right)\end{array}\right]=0$        (19)

Rearrange equation (19) for the oxidation number of phosphorous is as follows:

  $\text{Oxidation number of P}=\left[\begin{array}{l}-3\left(\text{oxidation number of hydrogen}\right)\\ -4\left(\text{oxidation number of oxygen}\right)\end{array}\right]$        (20)

Substitute $+1$ for oxidation number of hydrogen and $-2$ for oxidation number of oxygen in equation (20).

  $\begin{array}{c}\text{Oxidation number of phosphorous}=\left[3\left(-1\right)-4\left(-2\right)\right]\\ =\left[-3+8\right]\\ =+5.\end{array}$

The oxidation number of phosphorous is $+5$ so the oxide formed is diphosphorus pentaoxide with the chemical formula ${\text{P}}_2{\text{O}}_5$

Conclusion

The sum of the oxidation numbers of various elements in a neutral compound is zero whereas that in an ionic compound is equal to the charge on the ion.