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Identify the two liquids \((\mathrm{X})\) and \((\mathrm{Y})\) made up of same atoms, both \((\mathrm{X})\) and \((\mathrm{Y})\) are diamagnetic. (X) turns blue litmus to red, but (Y) does not. Also \(\mathrm{X}+\mathrm{KI}+\mathrm{H}^{+} \stackrel{\text { Starch }}{\longrightarrow}\) Blue colour \(\mathrm{Y}+\mathrm{KI}+\mathrm{H}^{+} \stackrel{\text { Starch }}{\longrightarrow}\) No colour (a) \(\mathrm{X}=\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Y}=\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{X}=\mathrm{H}_{2} \mathrm{O}, \mathrm{Y}=\mathrm{H}_{2} \mathrm{O}_{2}\) (c) Both \(\mathrm{X}\) and \(\mathrm{Y}\) are \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{X}=\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Y}=\mathrm{O}_{3}\)

Short Answer

Expert verified
X is H_2O_2, and Y is H_2O.

Step by step solution

01

Understand Properties of X and Y

Both liquids X and Y are composed of the same atoms, are diamagnetic, and X turns blue litmus red while Y does not. X reacts with KI, H+, and starch to give a blue color, whereas Y does not produce any color in the same reaction.
02

Analyze Reaction with Litmus Paper

X turns blue litmus to red, indicating that X is acidic. Y, however, does not affect the litmus, suggesting it is neutral.
03

Understand the Reaction with KI and Starch

The reaction shows that X turns blue with KI, H+, and starch, indicating the presence of a peroxide ion or similar oxidizing agent that forms iodine, which reacts with starch to form a blue color. Y does not cause any color change, indicating it is not oxidizing.
04

Identify Liquid X

Since X turns blue litmus red (acidic) and reacts with the KI and starch to form a blue color, it is likely an oxidizing agent, such as hydrogen peroxide (H_2O_2).
05

Identify Liquid Y

Y does not turn litmus red and does not affect the KI and starch reaction, indicating it is neutral and non-oxidizing, likely water (H_2O).
06

Compare X and Y

X is hydrogen peroxide (H_2O_2), and Y is water (H_2O), fitting all given properties and reaction descriptions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diagnetism
Diamagnetism is a fascinating property displayed by certain substances that makes them repel magnetic fields. Unlike paramagnetic substances, which possess unpaired electrons and are attracted by magnetic fields, diamagnetic substances have all their electrons paired. This means they are weakly repelled by a magnetic field. For example, both water (H₂O) and hydrogen peroxide (H₂O₂) exhibit diamagnetism because they have no unpaired electrons. This intrinsic property is a critical consideration when analyzing chemical substances in experiments or reactions involving magnets or magnetic fields. To better understand, consider water: when brought near a magnetic field, it seeks to avoid the field due to its diamagnetic nature. The same goes for hydrogen peroxide, providing a nice example of how two different liquids can exhibit similar magnetic properties through diamagnetism despite differing chemical behavior.
Litmus Test
The litmus test is a simple and rapid indicator used to determine the acidity or basicity of a solution. It involves using a strip of litmus paper that changes color in response to the pH level of a liquid. - If the paper turns red, the solution is acidic, typically having a pH less than 7. - If the paper turns blue, the solution is basic or alkaline, with a pH greater than 7. - Neutral solutions, close to pH 7, do not change the color of litmus paper. In the context of the exercise, substance X turns blue litmus red, indicating it is acidic. This behavior suggests the presence of a mild acid like hydrogen peroxide (H₂O₂), which can release hydrogen ions into the solution, making it acidic. On the contrary, water (H₂O) does not change the litmus color because it is neutral, with a pH close to 7.
Oxidizing Agents
Oxidizing agents play a crucial role in redox reactions as they gain electrons and induce oxidation in other substances. They are typically characterized by their ability to transfer oxygen atoms or accept electrons. Common examples include oxygen, ozone, and hydrogen peroxide. In the exercise, hydrogen peroxide acts as an oxidizing agent. It reacts with potassium iodide (KI) in the presence of acid (H⁺) to form iodine. The iodine then combines with starch, resulting in a blue coloration, signaling oxidation. This reaction is significant because it demonstrates the oxidative property of hydrogen peroxide, differentiating it from water, which does not exhibit such oxidative behavior. Identifying X as hydrogen peroxide due to its oxidizing nature is critical in the exercise since it accounts for the observable chemical change.
Hydrogen Peroxide Properties
Hydrogen peroxide (H₂O₂) is a well-known compound often used for bleaching and as a disinfectant. Chemically, it comprises two hydrogen atoms and two oxygen atoms, creating a clear liquid that appears similar to water but behaves differently. - It is a mild acid, capable of turning blue litmus paper red, thereby indicating its acidity. - Hydrogen peroxide is a strong oxidizing agent, making it useful in numerous household and industrial applications like cleaning and sterilizing. - Unlike many oxidizing agents, hydrogen peroxide decomposes into water and oxygen easily, ensuring it is environmentally friendly. In the described exercise, hydrogen peroxide is identified as liquid X due to its reactive properties with the potassium iodide and starch, leading to a distinctive blue color change. Such reactions highlight its unique characteristics that offer a clear distinction from water, which is neutral and non-reactive in similar conditions.

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Most popular questions from this chapter

The dissociation constant of water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) is \(1.0 \times 10-{ }^{14}\) and that of heavy water \(\left(\mathrm{D}_{2} \mathrm{O}\right)\) is \(3.0 \times 10^{-15}\). This is because the extent of dissociation of \(\mathrm{H}_{2} \mathrm{O}\) is about (a) \(1 / 3\) times that of \(\mathrm{D}_{2} \mathrm{O}\) (b) 3 times that of \(\mathrm{D}_{2} \mathrm{O}\) (c) 3000 times that of \(\mathrm{D}_{2} \mathrm{O}\) (d) 30 times that of \(\mathrm{D}_{2} \mathrm{O}\)

The absorption of ultraviolet radiation by \(\mathrm{O}_{3}\) (a) makes both the function effective (b) makes no function effective (c) maintains an equilibrium between the concentrations of \(\mathrm{O}_{2}\) and \(\mathrm{O}_{3}\) (d) protects the inhabitants of out planet from injurious radiation

For decolourization of one mole of \(\mathrm{KMnO}_{4}\) the number of moles of \(\mathrm{H}_{2} \mathrm{O}_{2}\) required are (a) \(1 / 2\) (b) \(3 / 2\) (c) \(5 / 2\) (d) \(7 / 2\)

An orange solid (A) on heating gives a green residue (B), a colourless gas (C) and water vapours. The dry gas (C) on passing over heated magnesium gave a white solid (D). (D) on reaction with water gives a gas (E) which gives dense white fumes with HCl. Here, (E) will be (a) \(\mathrm{N}_{2}\) (b) \(\mathrm{NO}_{2}\) (c) \(\mathrm{NH}_{3}\) (d) \(\mathrm{N}_{2} \mathrm{O}\)

The reaction of \(\mathrm{H}_{2} \mathrm{O}\) with \(\mathrm{X}\) does not liberate gaseous product. Which of the following is \(X\) ? (a) \(\mathrm{PbO}_{2}\) (b) \(\mathrm{KMnO}_{4} / \mathrm{H}^{+}\) (c) \(\mathrm{PbS}\) (d) \(\mathrm{Cl}_{2}\)

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