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
The question asks us to identify four chemical species A, B, C, and D based on two given reactions. Let's analyze the reactions provided:
We are given options for the identities of A, B, C, and D. Let's examine the options and see which set of species fits the descriptions of the reactions.
The second reaction is more specific: "Oxidation of A with C in an acidic medium gives D". This implies A is the substance being oxidized and C is the oxidizing agent, resulting in D.
Let's look at the third option, as the provided correct answer corresponds to it:
Let's test these species in Reaction 2:
Oxidation of A (\( \text{S}_2\text{O}_3^{2−} \)) with C (\( \text{I}_2 \)) gives D (\( \text{S}_4\text{O}_6^{2−} \)).
The reaction between thiosulfate ions and iodine is a well-known redox reaction:
\( 2\text{S}_2\text{O}_3^{2−} \text{(aq)} + \text{I}_2 \text{(aq)} \longrightarrow \text{S}_4\text{O}_6^{2−} \text{(aq)} + 2\text{I}^− \text{(aq)} \)
In this reaction, the thiosulfate ion (\( \text{S}_2\text{O}_3^{2−} \)) is oxidized (oxidation state of sulfur changes from +2 to +2.5 in \( \text{S}_4\text{O}_6^{2−} \)), and iodine (\( \text{I}_2 \)) is reduced (oxidation state changes from 0 to -1 in \( \text{I}^− \)). The product of the oxidation of thiosulfate by iodine is indeed the tetrathionate ion (\( \text{S}_4\text{O}_6^{2−} \)). This reaction perfectly fits the description of Reaction 2 where A is oxidized by C to give D (along with iodide ions, which are not explicitly mentioned as D, but D is stated as \( \text{S}_4\text{O}_6^{2−} \)).
Now let's consider Reaction 1: \( \text{A} + \text{B} \stackrel{\rm KI, H^{+}}{\longrightarrow} \text{C} \)
Using the species from Option 3:
A = \( \text{S}_2\text{O}_3^{2−} \)
B = \( \text{KIO}_3 \)
C = \( \text{I}_2 \)
The reaction conditions are KI and H\(^+\) (acidic medium). Potassium iodate (\( \text{KIO}_3 \)) contains the iodate ion (\( \text{IO}_3^− \)). In an acidic medium, iodate ions react with iodide ions (from KI) to produce iodine (\( \text{I}_2 \)). The balanced equation is:
\( \text{IO}_3^− \text{(aq)} + 5\text{I}^− \text{(aq)} + 6\text{H}^+ \text{(aq)} \longrightarrow 3\text{I}_2 \text{(aq)} + 3\text{H}_2\text{O} \text{(l)} \)
This reaction shows that B (\( \text{KIO}_3 \)) reacts with KI in acidic medium to produce C (\( \text{I}_2 \)). The statement "A + B \(\stackrel{\rm KI, H^{+}}{\longrightarrow}\) C" implies that mixing A and B under these conditions leads to the formation of C. While A (\( \text{S}_2\text{O}_3^{2−} \)) doesn't participate in the formation of C (\( \text{I}_2 \)) from B (\( \text{KIO}_3 \)) and KI/H\(^+\), it is present in the mixture where C is formed. The question might be interpreted as "When A and B are mixed along with KI and H\(^+\), C is formed from the reactants present". The formation of I\(_{2}\) (C) from KIO\(_{3}\) (B) and KI/H\(^+\) is a standard reaction, consistent with the identification of B as KIO\(_{3}\) and C as I\(_{2}\).
Based on the analysis of both reactions, especially the specific and definitive reaction between thiosulfate and iodine, the species A = \( \text{S}_2\text{O}_3^{2−} \), B = \( \text{KIO}_3 \), C = \( \text{I}_2 \), and D = \( \text{S}_4\text{O}_6^{2−} \) are consistent with the descriptions provided in the question.
Consider the following pairs of compounds.
(i) NH4Cl and FeO
(ii) H3N∙BF3 and BCl3
(iii) HSO3F and HF
The more acidic species in (i), (ii) and (iii) are, respectively
Which of the following has the lowest boiling point?
Choose the equilibria from the following that are NOT favoured to go to right:
(A) As2S5 + 5HgO ⇌ As2O5 + 5 HgS
(B) La2(CO3)3 + Bi2S3 ⇌ La2S3 + Bi2(CO3)3
(C) CdSO4 + CaS ⇌ CdS + CaSO4
(D) BeF2 + HgI2 ⇌ BeI2 + HgF2
The reaction of XeF6 with a limited amount of quartz gives compound A. Then on reaction with an equivalent amount of XeO3, A gives B. The products A and B are, respectively