Consider four species A, B, C, and D.

A + B \(\stackrel{\rm KI, H^{+}}{\longrightarrow}\) C

Oxidation of A with C in an acidic medium gives D.

A, B, C, and D are, respectively

Concept:-

• The first equation represents a reaction between species A and B in the acidic medium in the presence of I^- ion to form species C.
• The second equation represents the oxidation of S2O\(_3^{2−}\)(species A) by I₂ (species C) to form  S4O\(_6^{2−}\) (species D) and I- ions.

Explanation:-

• The balanced chemical equations are:

KIO₃ + 5I^- + 6H⁺ = 3I₂ + 3H₂O + K⁺

I₂ + S₂O₃^2- = S₄O₆^2- +2NaI

• The Ultimate reaction is

S2O\(_3^{2−}\) (A) + KIO3 (​B) \(\stackrel{\rm KI, H^{+}}{\longrightarrow}\) I₂ (C)

• S2O\(_3^{2−}\) (A) can be titrated against I₂ (C) in an acidic medium to give S4O\(_6^{2−}\) (D).
• The balanced chemical equations are:

I₂ + S2O\(_3^{2−}\) = S4O\(_6^{2−}\) + 2I^-

Comparing the given species with these equations, we see that:

• A is S2O\(_3^{2−}\)
• B is KIO₃
• C is I₂
• D is S₄O\(_6^{2−}\)

Conclusion:-

• Therefore, the correct answer is option 3, 
• Hence, A, B, C, and D are, respectively 

S2O\(_3^{2−}\), KIO3, I2 and S4O\(_6^{2−}\)