Question

Write the following equation in ionic form.
\[Mn{{O}_{2}}+4HCl\to MnC{{l}_{2}}+2{{H}_{2}}O+C{{l}_{2}}\]

Answer 1

Hint: By ionic form, it means that you have to write this equation in a form that will establish the type of dissolved ions present in the solution. For doing that, you have to split the molecules into their ionic forms keeping in mind their respective oxidation numbers.

Complete step by step solution: Let’s describe the steps involved one-by-one so that we can grasp all that we have to do for answering the given question.
First let us establish the states of all the compounds present in the chemical equation. It is as follows:

\[Mn{{O}_{2}}(s)+4HCl(aq)\to MnC{{l}_{2}}(aq)+2{{H}_{2}}O(l)+C{{l}_{2}}(g)\]

The oxides are generally solids and insoluble in solutions. They settle down at the bottom of the containers as they precipitate. The hydrochloric acid is a strong electrolyte and therefore in a solution it will always exist in its dissociated ionic from. Manganese chloride will dissociate into its ionic form when dissolved in a solution. As the solvent in this solution is water, it won’t change its form and therefore remain as a liquid. The chlorine gas will escape from the solution.
So keeping the above facts in mind, we can now write the complete ionic equation as:

\[Mn{{O}_{2}}(s)+4{{H}^{+}}(aq)+4C{{l}^{-}}(aq)\to M{{n}^{2+}}(aq)+2C{{l}^{-}}(aq)+2{{H}_{2}}O(l)+C{{l}_{2}}(g)\]

As you can see, there are common ions on both the side of the reactions. We have four chlorine ions on the reactant side and two of them on the product side. These ions are known as spectator ions because they can be cancelled out, just as we cancel the same values across a mathematical equation. This will generate the net ionic reaction, which is as follows:

\[Mn{{O}_{2}}(s)+4{{H}^{+}}(aq)+2C{{l}^{-}}(aq)\to M{{n}^{2+}}(aq)+2{{H}_{2}}O(l)+C{{l}_{2}}(g)\]

So the above equation is the ionic form of the equation we were provided with.

Note: Any reactant or product that is present in bulk will not be included in its ionic form. It is because its concentration and therefore its state of matter will not undergo a drastic change when the reaction proceeds.
To find the charges on the ions of each element the knowledge of finding out oxidation number is very important.
Also you might have marked that, the equation is equivalently charged on both the sides of the reaction. Failing to do so means you have a wrong reaction.