Chapter 17, Problem 2PE

(a)

Interpretation Introduction

Interpretation:

Oxidation number of each element in ${\text{CHF}}_{\text{3}}$ 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 fluorine is member of halogen group, its oxidation number is $-1$. Since ${\text{CHF}}_{\text{3}}$ is not hydride compound, oxidation state of $\text{H}$ is $+1$.

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

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

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

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

Substitute $-1$ for oxidation number of $\text{F}$ and $+1$ for oxidation number of $\text{H}$ in equation (2).

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

Hence, oxidation number of $\text{C}$ is $+2$, that of $\text{H}$ is $+1$ and that of $\text{F}$ is $-1$.

(b)

Interpretation Introduction

Interpretation:

Oxidation number of each element in ${\text{P}}_2{\text{O}}_5$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since ${\text{P}}_2{\text{O}}_5$ is not peroxide, oxidation number of $\text{O}$ is $-2$.

Expression for oxidation number in ${\text{P}}_2{\text{O}}_5$ is as follows:

  $\left[\begin{array}{l}2\left(\text{Oxidation number of P}\right)+\\ 5\left(\text{Oxidation number of O}\right)\end{array}\right]=0$        (3)

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

  $\text{Oxidation number of P}=\left[\frac{-5\left(\text{Oxidation number of O}\right)}{2}\right]$        (4)

Substitute $-2$ for oxidation number of $\text{O}$ in equation (4).

  $\begin{array}{c}\text{Oxidation number of P}=\left[\frac{-5\left(-2\right)}{2}\right]\\ =\frac{+10}{2}\\ =+5\end{array}$

Hence, oxidation number of $\text{P}$ is $+5$ and that of $\text{O}$ is $-2$.

(c)

Interpretation Introduction

Interpretation:

Oxidation number of each element in ${\text{SF}}_6$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since fluorine is member of halogen family, oxidation number of $\text{F}$ is $-1$.

Expression for oxidation number in ${\text{SF}}_6$ is as follows:

  $\left[\begin{array}{l}\text{Oxidation number of S}+\\ 6\left(\text{Oxidation number of F}\right)\end{array}\right]=0$        (5)

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

  $\text{Oxidation number of S}=-6\left(\text{Oxidation number of F}\right)$        (6)

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

  $\begin{array}{c}\text{Oxidation number of S}=-6\left(-1\right)\\ =+6\end{array}$

Hence, oxidation number of $\text{F}$ is $-1$ and that of $\text{S}$ is $+6$.

(d)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since ${\text{SnSO}}_4$ is not peroxide compound, oxidation state of $\text{O}$ is $-2$. Since chlorine is member of halogen group, its oxidation number is $-1$. $\text{Sn}$ is member of carbon family but due to inert pair effect, its oxidation number is $+2$.

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

  $\left[\begin{array}{l}\text{Oxidation number of Sn}+\\ \text{Oxidation number of S}+\\ 4\left(\text{Oxidation number of O}\right)\end{array}\right]=0$        (7)

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

  $\text{Oxidation number of S}=\left[\begin{array}{c}-\text{Oxidation number of Sn}-\\ 4\left(\text{Oxidation number of O}\right)\end{array}\right]$        (8)

Substitute $-2$ for oxidation number of $\text{O}$ and $+2$ for oxidation number of $\text{Sn}$ in equation (8).

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

Hence, oxidation number of $\text{Sn}$ is $+2$, that of $\text{S}$ is $+6$ and that of $\text{O}$ is $-2$.

(e)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since ${\text{CH}}_{\text{3}}\text{OH}$ is not peroxide, oxidation number of $\text{O}$ is $-2$. It is not hydride compound so oxidation number of $\text{H}$ is $+1$.

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

  $\left[\begin{array}{l}\text{Oxidation number of C}+\\ 4\left(\text{Oxidation number of H}\right)+\\ \text{Oxidation number of O}\end{array}\right]=0$        (9)

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

  $\text{Oxidation number of C}=\left[\begin{array}{c}-4\left(\text{Oxidation number of H}\right)-\\ \text{Oxidation number of O}\end{array}\right]$        (10)

Substitute $-2$ for oxidation number of $\text{O}$ and $+1$ for oxidation number of $\text{H}$ in equation (10).

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

Hence, oxidation state of $\text{O}$ is $-2$, that of $\text{H}$ is $+1$ and that of $\text{C}$ is $-2$.

(f)

Interpretation Introduction

Interpretation:

Oxidation number of each element in ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Since ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is neither hydride nor peroxide compound, oxidation number of $\text{H}$ is $+1$ and that of $\text{O}$ is $-2$.

Expression for oxidation number in ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is as follows:

  $\left[\begin{array}{l}3\left(\text{Oxidation number of H}\right)+\\ \text{Oxidation number of P}+\\ 4\left(\text{Oxidation number of O}\right)\end{array}\right]=0$        (11)

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

  $\text{Oxidation number of P}=\left[\begin{array}{c}-3\left(\text{Oxidation number of H}\right)-\\ 4\left(\text{Oxidation number of O}\right)\end{array}\right]$        (12)

Substitute $+1$ for oxidation number of $\text{H}$ and $-2$ for oxidation number of $\text{O}$ in equation (12).

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

Hence, oxidation number of $\text{H}$ is $+1$, that of $\text{P}$ is $+5$ and that of $\text{O}$ is $-2$.