Question

Which of the following is not an example of a disproportionation reaction
a) $4ClO_{3}^{-}\to C{{l}^{-}}+3ClO_{4}^{-}$
b) $2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}$
c) $2N{{O}_{2}}+2O{{H}^{-}}\to NO_{2}^{-}+NO_{3}^{-}+{{H}_{2}}O$
d) $TiC{{l}_{4}}+2Mg\to Ti+2MgC{{l}_{2}}$

Answer 1

Hint: The disproportionation reaction is also called the dismutation reaction. In such a reaction, one element in the reactants forms two compounds in the products, one with higher oxidation state and the other with lower oxidation state. The answer to this question is a reaction which has not formed two compounds in the product but there is some replacement of the element.

Complete step by step answer:
First, we will know about the disproportionation reactions and then we will be able to find the correct answer.
The disproportionation reaction can be defined as the redox reaction in which one compound of the intermediate oxidation state is converted into two compounds one with lower oxidation state and the other one with higher oxidation state.
In general, the disproportion reaction can be written as -
$2X\to {{X}^{n+}}+{{X}^{n-}}$
Now, let us see the examples given to us and find out which is not an example of a disproportionation reaction.
We will see the oxidation state of the central atom in each reaction and then see how it changes in products.
The first reaction is -
$4ClO_{3}^{-}\to C{{l}^{-}}+3ClO_{4}^{-}$
The central atom is Chlorine. The $\text{H}\,\text{=}\,\text{+1,O}\,\text{= -2}$
So, chlorine in reactant has oxidation state,$\text{Cl}\,\text{=}\,\text{+5}$
Thus, the change can be written as-
$4\overset{+5}{\mathop{Cl}}\,O_{3}^{-}\to \overset{-1}{\mathop{Cl}}\,+3\overset{+7}{\mathop{Cl}}\,O_{4}^{-}$
So, it is a disproportionation reaction.
Now, the second equation is-
$2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}$
In this, the central atom is oxygen.
The change in oxidation state can be written as-
$2{{H}_{2}}\overset{-1}{\mathop{{{O}_{2}}}}\,\to 2{{H}_{2}}\overset{-2}{\mathop{{{O}_{2}}}}\,+\overset{0}{\mathop{{{O}_{2}}}}\,$
So, even it is a disproportionation reaction.
The third reaction is-
$2N{{O}_{2}}+2O{{H}^{-}}\to NO_{2}^{-}+NO_{3}^{-}+{{H}_{2}}O$
The change in the oxidation state of central metal atom N is as-
$2\overset{+4}{\mathop{N}}\,{{O}_{2}}+2O{{H}^{-}}\to \overset{+3}{\mathop{N}}\,O_{2}^{-}+\overset{+5}{\mathop{N}}\,O_{3}^{-}+{{H}_{2}}O$
Here, also there is a change in oxidation state in both the compounds. So, it is also a disproportionation reaction.
The fourth reaction is-
$TiC{{l}_{4}}+2Mg\to Ti+2MgC{{l}_{2}}$
In this reaction, we observe that Magnesium has displaced Titanium. There is not any single reactant that formed two compounds with different oxidation states. It is a displacement reaction and not a disproportion reaction.

So, the option (d) is the answer.

Note:
It must be noted that we have to tell the reaction which is not disproportionate. The reaction which is something other than disproportionation. The easiest way to identify the disproportionation is that the central atom in the reactant forms two compounds in the product. Further, even the sum of change of oxidation state in products is equal to the oxidation state of the reactant.