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The correct statement among the following for disproportional of four moles of $ClO_4^ -$ ion to $C{l^ - }$ and $ClO_4^ - $ ion is
A. Only two moles of chlorine undergo increase in oxidation number
B. Only one mole of chlorine shows increase in oxidation number
C. One mole of chlorine shows decrease in oxidation number
D. Three moles of chlorine undergo decrease in oxidation number

Answer
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Hint: In this question, we need to identify the correct statement for the disproportional of four moles of $ClO_4^ - $ ion to $C{l^ - }$ and $ClO_4^ - $ ion. We will proceed by explaining the concept of disproportionation reaction to check whether the given question.

Complete step by step solution:
For the chlorite ion, the following reaction and exchange of the ions will take place,
\[
  3C\mathop l\limits^{ + 1} O \to 2\mathop C\limits^{ + 5} {l^ - } + C\mathop l\limits^{ + 5} O_3^ - \\
  Cl{O^{2 - }}({\text{chlorite ion}}) \\
 \]
For the chlorate ion, the following reaction and exchange of the ions will take place,
\[
   + {\text{3 - 1 }} + {\text{5}} \\
  3ClO_2^ - \to C{l^ - } + 2ClO_3^ - \\
  ClO_3^ - ({\text{chlorate ion)}} : \\
 \]
For the perchlorate ion, the following reaction and exchange of the ions will take place,
\[
   + {\text{5 - 1 }} + {\text{7}} \\
  4ClO_3^ - \to C{l^ - } + 4ClO_4^{ - 1} \\
 \]
Electronic configuration of Cl is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}$
$ClO_4^ - $ cannot show any disproportionation reaction. The oxidation state of Cl is $ClO_4^ - $ ion is +7.
It is the maximum oxidation state which it can have. It can decrease the same by undergoing reduction and not increase it anymore hence $ClO_4^ - $ ion does not undergo a disproportionation reaction.
In $ClO_4^ - $, chlorine is present in +7 oxidation state so it cannot increase its oxidation state.

So, option (2) only one mole of chlorine shows increase in oxidation number is the correct answer.

Note: One must know that a disproportionate reaction is a special type of redox reaction in which an element simultaneously gives electrons and accepts electrons to form different products.