Question

Identify disproportionation reaction:
A. $C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O$
B. $C{H_4} + 4C{l_2} \to CC{l_4} + 4HCl$
C. $2{F_2} + 2O{H^ - } \to {F^ - } + O{F_2} + {H_2}O$
D. $2N{O_2} + 2O{H^ - } \to NO_2^ - + NO_3^ - + {H_2}O$

Answer 1

A disproportionation reaction is when a substance is both oxidised and reduced forming different products. Check for the oxidation number of the elements in compounds and compare them on both sides.

Complete step by step answer:
We are asked to identify disproportionation reactions out of the four reactions. Before that let us be clear about disproportionation reactions. In these reactions, the same substance will undergo both oxidation and reduction. What we will do here is go through each of the above reactions one by one and check the oxidation numbers of the substances. General Thumb rule is to find the most electronegative and electropositive of substances and assign the oxidation number accordingly. Rest of the calculations is to find oxidation numbers of the others to equate it with net charge on the substance.
We have, $C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O$
We will list the oxidation numbers of all the substances before and after reaction as follows:

Elements	Before	After
C	-4	+4
H	+1	+1
O	0	-2

So we see that there is not a single element which has gone through both oxidation and reduction but instead only oxidation or reduction. The second reaction is: $C{H_4} + 4C{l_2} \to CC{l_4} + 4HCl$
As we did for the first reaction, we will list the oxidation numbers

Elements	Before	After
Cl	0	-1,-1
C	-4	+4
H	+1	+1

In the above example, we have two products with chlorine but both are reduced for chlorine.
The third reaction is: $2{F_2} + 2O{H^ - } \to {F^ - } + O{F_2} + {H_2}O$
In the second product above, fluorine is most electronegative and thus oxygen will be oxidised and there will increase in its oxidation number. The table is as follows:

Elements	Before	After
F	0	-1,-1
H	+1	+1
O	-2	+2,-2

In the above example we have two different oxidation states of oxygen but when we compare this with reactant, we see that we have only an example of oxidation of oxygen but no reduction. So this option is not an example of disproportionation reaction.
The fourth reaction is: $2N{O_2} + 2O{H^ - } \to NO_2^ - + NO_3^ - + {H_2}O$
In the above reaction we have two products and a compound of no charge on the product side. Nitrogen is expected to show oxidation and reduction. The table is as follows:

Elements	Before	After
N	+4	+3,+5
H	+1	+1
O	0	-2

In this reaction as expected, nitrogen in $N{O_2}$ oxidises from +4 to +5 in $NO_3^ -$ and reduces from +4 to +3 in $NO_2^ -$. Thus this is an example of a disproportionation reaction. We can say that option D is the correct choice. So, the correct answer is “Option D”.

Note: Generally, we can save some time by selecting the last option as the correct one as we have deduced that other options are incorrect. But it is better to be sure by evaluating it completely. But as mentioned earlier, noting the electronegative and electropositive of all the substances can help to distinguish the reaction and oxidation number.