Chapter 14, Problem 14.109P

(a)

Interpretation Introduction

Interpretation:

The equation for the reaction of interhalogen $\text{IF}$ has to be written from the given depiction.

Concept Introduction:

Interhalogen compounds:

The compounds which are only composed of two different halogens are known as Interhalogen.

Disproportionation redox reaction:

The formation of products of oxidized and reduced substances from a simultaneous redox reaction of single reactant is known as disproportionation redox reaction.

Explanation of Solution

Given depiction is,

    

Figure 1

From the given depiction, brown colour = iodine and green coloured are Fluorine atoms.

Therefore, the depiction refers the formation of ${\text{IF}}_5$ and ${\text{2I}}_2$ molecules from five $\text{IF}$ molecules by disproportionation redox reaction.

Hence, the balanced equation for the give reaction is,

    $\text{5IF}\stackrel{}{\to }{\text{IF}}_{\text{5}}{\text{+2I}}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation:

Name of the product of given redox reaction of $\text{IF}$ has to be given.

Concept Introduction:

Disproportionation redox reaction:

The formation of products of oxidized and reduced substances from a simultaneous redox reaction of single reactant is known as disproportionation redox reaction.

Interhalogen compounds:

The compounds which are only composed of two different halogens are known as Interhalogen.

Explanation of Solution

Given depiction is,

    

Figure 1

From the given depiction, brown colour = iodine and green coloured are Fluorine atoms.

Therefore, the depiction refers the formation of ${\text{IF}}_5$ and ${\text{2I}}_2$ molecules from five $\text{IF}$ molecules by disproportionation redox reaction.

The balanced equation for the give reaction is,

    $\text{5IF}\stackrel{}{\to }{\text{IF}}_{\text{5}}{\text{+2I}}_{\text{2}}$

Product of above reaction is Iodine penta-fluoride.

(c)

Interpretation Introduction

Interpretation:

Name of given redox reaction of $\text{IF}$ has to be given.

Concept Introduction:

Disproportionation redox reaction:

The formation of products of oxidized and reduced substances from a simultaneous redox reaction of single reactant is known as disproportionation redox reaction.

Interhalogen compounds:

The compounds which are only composed of two different halogens are known as Interhalogen.

Explanation of Solution

Given depiction is,

    

Figure 1

From the given depiction, brown colour = iodine and green coloured are Fluorine atoms.

Therefore, the depiction refers the formation of ${\text{IF}}_5$ and ${\text{2I}}_2$ molecules from five $\text{IF}$ molecules by disproportionation redox reaction.

The balanced equation for the give reaction is,

    $\text{5IF}\stackrel{}{\to }{\text{IF}}_{\text{5}}{\text{+2I}}_{\text{2}}$

Therefore, the above reaction shows given reaction is disproportionation reaction.

(d)

Interpretation Introduction

Interpretation:

Mass of product that formed from $\text{2}\text{.50}\times {10}^{-3}\text{mol}\text{IF}$ has to be calculated.

Concept Introduction:

Disproportionation redox reaction:

The formation of products of oxidized and reduced substances from a simultaneous redox reaction of single reactant is known as disproportionation redox reaction.

Interhalogen compounds:

The compounds which are only composed of two different halogens are known as Interhalogen.

Mole:

Mole of substance that present in taken mole of sample is calculated by mass of substance is divided by molar mass of the substance.

    $\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Explanation of Solution

Given depiction is,

    

Figure 1

From the given depiction, brown = iodine and green = Fluorine atoms.

Therefore, the depiction refers the formation of ${\text{IF}}_5$ and ${\text{2I}}_2$ molecules from five $\text{IF}$ molecules by disproportionation redox reaction.

The balanced equation for the give reaction is,

    $\text{5IF}\stackrel{}{\to }{\text{IF}}_{\text{5}}{\text{+2I}}_{\text{2}}$

Therefore, the above reaction shows given reaction is disproportionation reaction.

Mass of ${\text{IF}}_5$ formed from the reaction of $\text{2}\text{.50}\times {10}^{-3}\text{mol}\text{IF}$ is,

Molar mass of ${\text{IF}}_{\text{5}}$ is $\text{221}\text{.9}\text{g}$

    $\begin{array}{l}\text{Mass}{\text{IF}}_{\text{5}}\text{=}\left(\text{7}\text{IF}\text{molecules}\right)\left(\frac{\text{2}{\text{.50×10}}^{\text{-3}}\text{mol}\text{IF}}{\text{1}\text{mol}\text{IF}}\right)\left(\frac{\text{1}\text{mol}{\text{IF}}_{\text{5}}}{\text{5mol}\text{IF}}\right)\left(\frac{\text{221}\text{.9}\text{g}}{\text{1}\text{mol}\text{IF}}\right)\\ \text{=}\text{0}\text{.777}\text{g}{\text{IF}}_5\end{array}$

Mass of ${\text{I}}_2$ formed from the reaction of $\text{2}\text{.50}\times {10}^{-3}\text{mol}\text{IF}$ is,

Molar mass of ${\text{I}}_2$ is $\text{253}\text{.8 g}$

$\begin{array}{l}\text{Mass}{\text{I}}_{\text{2}}\text{=}\left(\text{7}\text{IF}\text{molecules}\right)\left(\frac{\text{2}{\text{.50×10}}^{\text{-3}}\text{mol}\text{IF}}{\text{1}\text{mol}\text{IF}}\right)\left(\frac{\text{2}\text{mol}{\text{I}}_{\text{2}}}{\text{5mol}\text{IF}}\right)\left(\frac{\text{253}\text{.8}\text{g}{\text{I}}_{\text{2}}}{\text{1}\text{mol}\text{IF}}\right)\\ \text{=}\text{1}\text{.78}\text{g}{\text{I}}_{\text{2}}\end{array}$

Hence, $\text{0}\text{.777}\text{g}{\text{IF}}_5$ and $\text{1}\text{.78}\text{g}{\text{I}}_{\text{2}}$ are formed from $\text{2}\text{.50}\times {10}^{-3}\text{mol}\text{IF}$.