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?

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.

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

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

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.

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+).

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:

localid="1663357513868" $\begin{array}{rcl}\mathrm{Number} \mathrm{of} \mathrm{IF} \mathrm{moles}& =& \frac{1 \mathrm{mol} {\mathrm{IF}}_5}{5\mathrm{mol}\mathrm{IF}}\times 17.5\times {10}^{-3}\mathrm{mole}\mathrm{IF}\\ & =& 3.50\times {10}^{-3} {\mathrm{IF}}_5\mathrm{mole}\end{array}$

The number of I₂ moles is:

localid="1663357536092" $\begin{array}{rcl}\mathrm{Number} \mathrm{of} {\mathrm{I}}_2 \mathrm{moles}& =& \frac{2 \mathrm{mole} {\mathrm{I}}_2}{5 \mathrm{mole} \mathrm{IF}}\times 17.5\times {10}^{-3} \mathrm{mole} \mathrm{IF}\\ & =& 7.00\times {10}^{-3} {\mathrm{I}}_2 \mathrm{mole}\end{array}$

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

Mole ofIF₅ mass:

localid="1663357409949" $\begin{array}{rcl}\mathrm{Mass} \mathrm{of}{\mathrm{IF}}_5& =& \frac{221.9 \mathrm{g} {\mathrm{IF}}_5}{1 \mathrm{mole} {\mathrm{IF}}_5}\times 3.50\times {10}^{-3} {\mathrm{IF}}_5 \mathrm{mole}\\ & =& 0.777 \mathrm{g} {\mathrm{IF}}_5\end{array}$

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.