Total number of species from the following which can undergo disp… — Solution | JEE Chemistry

MCQMediumJEE 2024Oxidation Number & Redox Reactions

Total number of species from the following which can undergo disproportionation reaction: $\mathrm{H_2O_2}$ , $\mathrm{ClO_3^-}$ , $\mathrm{P_4}$ , $\mathrm{Cl_2}$ , $\mathrm{Ag^+}$ , $\mathrm{F_2}$ , $\mathrm{NO_2}$ , $\mathrm{K}$

A
$6$
B
$7$
C
$8$
D
$9$

Answer

Correct answer:A

Step-by-step solution

the solution unavailable

Given: The species are $\mathrm{H_2O_2}$ , $\mathrm{ClO_3^-}$ , $\mathrm{P_4}$ , $\mathrm{Cl_2}$ , $\mathrm{Ag^+}$ , $\mathrm{F_2}$ , $\mathrm{NO_2}$ , and $\mathrm{K}$ .

Find: The total number of species that can undergo disproportionation reaction.

The solution could not be extracted, so the detailed working is unavailable from the source. Using the provided correct answer field, the correct option corresponds to $6$ .

Therefore, the correct option is A.

Common mistakes

Counting $\mathrm{F_2}$ as undergoing disproportionation. This is incorrect because fluorine is the most electronegative element and does not show positive oxidation states. Do not include $\mathrm{F_2}$ .
Assuming every elemental form can disproportionate. This is incorrect because the element must be able to undergo both oxidation and reduction from its initial oxidation state. Check the accessible oxidation states before counting.
Ignoring species in intermediate oxidation states such as $\mathrm{H_2O_2}$ or $\mathrm{NO_2}$ . These are often capable of both oxidation and reduction. Evaluate the oxidation state of the central atom carefully before deciding.

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