Chapter 17, Problem 1PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in ${\text{CuCO}}_{\text{3}}$ has to be determined.

Concept Introduction:

Oxidation number is integer value allotted to every element. It is formal charge occupied by atom if all of its bonds are dissociated heterolytically. Below mentioned are rules to assign oxidation numbers to various elements.

1. Elements present in their free state have zero oxidation number.

2. Oxidation number of hydrogen is generally $+1$, except for metal hydrides.

3. Oxidation number of oxygen is $-2$, except for peroxides.

4. Metals have positive oxidation numbers.

5. Negative oxidation numbers are assigned to most electronegative element in covalent compounds.

6. Sum of oxidation numbers of different elements in neutral atom is zero.

7. Sum of oxidation numbers of various elements in polyatomic ion is equal to charge present on ion.

Explanation of Solution

Since ${\text{CuCO}}_{\text{3}}$ is not peroxide compound, oxidation state of $\text{O}$ is $-2$. Since copper belongs to $d\text{-block}$ and most common oxidation state of these elements is $+2$, oxidation state of $\text{Cu}$ is $+2$.

Expression for oxidation number in ${\text{CuCO}}_{\text{3}}$ is as follows:

  $\left[\begin{array}{l}\text{Oxidation number of Cu}+\\ \text{Oxidation number of C}+\\ 3\left(\text{Oxidation number of O}\right)\end{array}\right]=0$        (1)

Rearrange equation (1) for oxidation number of $\text{C}$.

  $\text{Oxidation number of C}=\left[\begin{array}{c}-\text{Oxidation number of Cu-}\\ 3\left(\text{Oxidation number of O}\right)\end{array}\right]$        (2)

Substitute $-2$ for oxidation number of $\text{O}$ and $+2$ for oxidation number of $\text{Cu}$ in equation (2).

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

Hence, oxidation number of $\text{C}$ is $+4$, that of $\text{O}$ is $-2$ and that of $\text{Cu}$ is $+2$.

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in ${\text{CH}}_4$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since ${\text{CH}}_4$ is not hydride compound, oxidation number of $\text{H}$ is $+1$.

Expression for oxidation number in ${\text{CH}}_4$ is as follows:

  $\left[\begin{array}{l}\text{Oxidation number of C}+\\ 4\left(\text{Oxidation number of H}\right)\end{array}\right]=0$        (3)

Rearrange equation (3) for oxidation number of $\text{C}$.

  $\text{Oxidation number of C}=-4\left(\text{Oxidation number of H}\right)$        (4)

Substitute $+1$ for oxidation number of $\text{H}$ in equation (4).

  $\begin{array}{c}\text{Oxidation number of C}=-4\left(\text{+1}\right)\\ =-4\end{array}$

Hence, oxidation number of $\text{C}$ is $-4$ and that of $\text{H}$ is $+1$.

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in $\text{IF}$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since fluorine is more electronegative than iodine and both are members of halogen family, oxidation number of $\text{F}$ is $-1$.

Expression for oxidation number in $\text{IF}$ is as follows:

  $\left[\begin{array}{l}\text{Oxidation number of I}+\\ \text{Oxidation number of F}\end{array}\right]=0$        (5)

Rearrange equation (5) for oxidation number of $\text{I}$.

  $\text{Oxidation number of I}=-\text{Oxidation number of F}$        (6)

Substitute $-1$ for oxidation number of $\text{F}$ in equation (6).

  $\begin{array}{c}\text{Oxidation number of I}=-\left(-1\right)\\ =+1\end{array}$

Hence, oxidation number of $\text{F}$ is $-1$ and that of $\text{I}$ is $+1$.

(d)

Interpretation Introduction

Interpretation:

Oxidation number of each element in ${\text{CH}}_2{\text{Cl}}_2$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since ${\text{CH}}_2{\text{Cl}}_2$ is not hydride compound, oxidation state of $\text{H}$ is $+1$. Since chlorine is member of halogen group, its oxidation number is $-1$.

Expression for oxidation number in ${\text{CH}}_2{\text{Cl}}_2$ is as follows:

  $\left[\begin{array}{l}\text{Oxidation number of C}+\\ 2\left(\text{Oxidation number of H}\right)+\\ 2\left(\text{Oxidation number of Cl}\right)\end{array}\right]=0$        (7)

Rearrange equation (7) for oxidation number of $\text{C}$.

  $\text{Oxidation number of C}=\left[\begin{array}{c}-2\left(\text{Oxidation number of H}\right)-\\ 2\left(\text{Oxidation number of Cl}\right)\end{array}\right]$        (8)

Substitute $-1$ for oxidation number of $\text{Cl}$ and $+1$ for oxidation number of $\text{H}$ in equation (8).

  $\begin{array}{c}\text{Oxidation number of C}=\left[-2\left(+1\right)-2\left(-1\right)\right]\\ =-2+2\\ =0\end{array}$

Hence, oxidation number of $\text{C}$ is 0, that of $\text{H}$ is $+1$ and that of $\text{Cl}$ is $-1$.

(e)

Interpretation Introduction

Interpretation:

Oxidation number of each element in ${\text{SO}}_2$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since ${\text{SO}}_2$ is not peroxide, oxidation number of $\text{O}$ is $-2$.

Expression for oxidation number in ${\text{SO}}_2$ is as follows:

  $\left[\begin{array}{l}\text{Oxidation number of S}+\\ 2\left(\text{Oxidation number of O}\right)\end{array}\right]=0$        (9)

Rearrange equation (9) for oxidation number of $\text{S}$.

  $\text{Oxidation number of S}=-2\left(\text{Oxidation number of O}\right)$        (10)

Substitute $-2$ for oxidation number of $\text{O}$ in equation (10).

  $\begin{array}{c}\text{Oxidation number of S}=-2\left(-2\right)\\ =+4\end{array}$

Hence, oxidation state of $\text{O}$ is $-2$ and that of $\text{S}$ is $+4$.

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CrO}}_{\text{4}}$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CrO}}_{\text{4}}$ is composed of ${\text{NH}}_4^{+}$ and ${\text{CrO}}_4^{2-}$ ions. Since it is not hydride compound, oxidation number of hydrogen is generally $+1$.

Expression for oxidation number in ${\text{NH}}_4^{+}$ is as follows:

  $\left[\begin{array}{l}\text{Oxidation number of N}+\\ 4\left(\text{Oxidation number of H}\right)\end{array}\right]=+1$        (11)

Rearrange equation (11) for oxidation number of $\text{N}$.

  $\text{Oxidation number of N}=+1-4\left(\text{Oxidation number of H}\right)$        (12)

Substitute $+1$ for oxidation number of $\text{H}$ in equation (12).

  $\begin{array}{c}\text{Oxidation number of N}=+1-4\left(+1\right)\\ =+1-4\\ =-3\end{array}$

Since ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{CrO}}_{\text{4}}$ is not peroxide, oxidation number of oxygen is $-2$.

Expression for oxidation number in ${\text{CrO}}_4^{2-}$ is as follows:

  $\left[\begin{array}{l}\text{Oxidation number of Cr}+\\ 4\left(\text{Oxidation number of O}\right)\end{array}\right]=-2$        (13)

Rearrange equation (13) for oxidation number of $\text{Cr}$.

  $\text{Oxidation number of Cr}=-2-4\left(\text{Oxidation number of O}\right)$        (14)

Substitute $-2$ for oxidation number of $\text{O}$ in equation (14).

  $\begin{array}{c}\text{Oxidation number of Cr}=-2-4\left(-2\right)\\ =-2+8\\ =+6\end{array}$

Hence, oxidation number of $\text{N}$ is $-3$, that of $\text{H}$ is $+1$, that of $\text{Cr}$ is $+6$ and that of $\text{O}$ is $-2$.