Q14.143CP - Chemistry: Molecular Nature Of Matter And Change

Question

The interhalogen IF undergoes the reaction depicted below (I is purple and F is green):

(a) Write the balanced equation. 

(b) Name the interhalogen product. 

(c) What type of reaction is shown? 

(d) If each molecule of IF represents $\mathbf{2}.\mathbf{50}\times {\mathbf{10}}^{-\mathbf{3}}$ mol, what mass of each product forms?

Step-by-Step Solution

Answer

1. The balanced equation is: $5\mathrm{IF}\to {\mathrm{IF}}_5+2{\mathrm{I}}_2$ .
2. The interhalogen product's name is Iodine pentafluoride.
3. It is a disproportionation redox reaction
4. 0.77 g IF₅  and 1.78 g  I₂ formed from 7 molecules, as one molecule of IF contains data-custom-editor="chemistry" $2.50\times {10}^{-3}$ mole.

Step 1: What are the terms "oxidation," "reduction," "oxidising agent," "reducing agent," and "balanced chemical equation"?

Oxidation:

Oxidation is a chemical reaction in which one or more of the following changes occur:

1. Accumulation of oxygen atoms

2. Electrons are lost.

3. The atom of hydrogen is lost.

4. Increasing the number of oxidations

 

Reduction:

Reduction is a procedure that involves one or more of the following changes:

1. Oxygen atoms are lost.

2. Electron acquisition

3. The addition of a hydrogen atom.

4. Lowering the oxidation number.

The processes of oxidation and reduction are in reverse order.

 

Agent of oxidation:

An oxidizing agent is a chemical reactant that produces oxidation by receiving electrons.

 

Agent of reduction:

A reducing agent is a reaction's reactant that causes a reduction through donating an electron.

 

A balanced chemical equation :

A balanced chemical equation contains the same number of atoms and elements on both sides of the reaction.

Step 2: (a) The balanced equation

The chemical reaction of inter-halogen, iodine fluoride IF compound has the following balancing equation:

$5\mathrm{IF}\to {\mathrm{IF}}_5+2{\mathrm{I}}_2$

Step 3: (b) The interhalogen product's name

The representation reaction of inter-halogen, iodine fluoride compound has the following balancing equation:

$5\mathrm{IF}\to {\mathrm{IF}}_5+2{\mathrm{I}}_2$

         Iodine 

       pentafluoride

Iodine pentafluoride is generated in this process, along with iodine.

Step 4: (c) What kind of reaction is displayed?

The oxidation numbers of both halogens will change in this reaction as follows:

Here, IF serves as both an oxidizing and a reducing agent. As a result, it is an example of a disproportionation redox reaction, which is reduced and oxidized concurrently to produce two different products,  IF₅ and I₂  .

This redox reaction is a disproportionation redox reaction because the oxidation number of fluorine changes from (1-) to (1-) and (0), while the oxidation number of iodine changes from (1+) to (5+).

Step 5: (d) What is the mass of each product if each molecule of IF represents 2 . 50 × 10 - 3 mol?

To begin, determine the total number of moles in 7 molecules of IF, as one molecule of IF has $2.50\times {10}^{-3}$  mole.

$\begin{array}{rcl}\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{moles} \mathrm{IF}& =& 7 \mathrm{IF} \mathrm{molecules}\times \frac{2.50\times {10}^{-3}}{1 \mathrm{IF} \mathrm{molecule}}\\ & =& 17.5\times {10}^{-3} \mathrm{IF} \mathrm{mole}\end{array}$

Calculate the number of products using the following chemical reaction:

$5\mathrm{IF}\to {\mathrm{IF}}_5+2{\mathrm{I}}_2$

IF₅ mole numbers:

$\begin{array}{rcl}\mathrm{Number} \mathrm{of} \mathrm{IF} \mathrm{moles}& =& \frac{1 \mathrm{mol} {\mathrm{IF}}_5}{5 \mathrm{mol} \mathrm{IF}}\times 17.5\times {10}^{-3} \mathrm{mole} \mathrm{IF}\\ & =& 3.50\times {10}^{-3} {\mathrm{IF}}_5 \mathrm{mole}\end{array}$

The number of  I₂ moles is:

$\begin{array}{rcl}\mathrm{Number} \mathrm{of} {\mathrm{I}}_2 \mathrm{moles}& =& \frac{2 \mathrm{mole} {\mathrm{I}}_2}{5 \mathrm{mole} \mathrm{IF}}\times 17.5\times {10}^{-3} \mathrm{mole} \mathrm{IF}\\ & =& 7.00\times {10}^{-3} {\mathrm{I}}_{2 } \mathrm{mole}\end{array}$

 

Now multiply the mass of the products by their molar mass:

Mole of IF₅ mass:

$\begin{array}{rcl}\mathrm{Mass} \mathrm{of} {\mathrm{IF}}_5& =& \frac{221.9 \mathrm{g} {\mathrm{IF}}_5}{1 \mathrm{mole} {\mathrm{IF}}_5}\times 3.50\times {10}^{-3} {\mathrm{IF}}_5 \mathrm{mole}\\ & =& 0.777 \mathrm{g} {\mathrm{IF}}_5\end{array}$ 

Mole of   I₂   mass is:

 $\begin{array}{rcl}\mathrm{Mass} \mathrm{of} {\mathrm{I}}_2& =& \frac{253.8 \mathrm{g} {\mathrm{I}}_2}{1 \mathrm{mole} {\mathrm{I}}_2}\times 7.00\times {10}^{-3} {\mathrm{I}}_{2 } \mathrm{mole}\\ & =& 1.78 \mathrm{g} {\mathrm{I}}_2\end{array}$

Hence 0.77 g  IF₅   and 1.78 g    I₂  formed from 7 molecules of, as one molecule of IF contains  $2.50\times {10}^{-3}$mole.