Problem 96 - A Complete Resource Book in Chemistry for JEE Main

Question

The oxidation states of sulphur in the anions \(\mathrm{SO}_{3}^{2-}\), \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) and \(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}\) follow the order (a) \(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}\) (b) \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{SO}_{3}^{2-}\) (c) \(\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}\)

Step-by-Step Solution

Answer

(d) \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-} \)

Step 1: Calculate Oxidation State of Sulphur in \( \mathrm{SO}_{3}^{2-} \)

For \( \mathrm{SO}_{3}^{2-} \), let the oxidation state of sulfur be \( x \). The sum of the oxidation states of one sulfur (S) and three oxygens (O) should equal the charge of the ion, which is -2. Since the oxidation state of oxygen is usually -2, the equation is:\[ x + 3(-2) = -2 \]\[ x - 6 = -2 \]Solving for \( x \), we get:\[ x = +4 \]

Step 2: Calculate Oxidation State of Sulphur in \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} \)

For \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} \), let the oxidation states of the two sulfurs be \( 2x \). The equation becomes:\[ 2x + 4(-2) = -2 \]\[ 2x - 8 = -2 \]Solving for \( 2x \), we add 8 to both sides:\[ 2x = 6 \]Divide by 2 to find the average oxidation state of each sulfur:\[ x = +3 \]

Step 3: Calculate Oxidation State of Sulphur in \( \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \)

For \( \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \), assume each sulfur has the same oxidation state \( x \). The equation follows:\[ 2x + 6(-2) = -2 \]\[ 2x - 12 = -2 \]Adding 12 to both sides, we find:\[ 2x = 10 \]Divide by 2 to find the average oxidation state of each sulfur:\[ x = +5 \]

Step 4: Compare the Oxidation States

The calculated oxidation states of the sulfurs are:1. \( \mathrm{SO}_{3}^{2-} \) : \( +4 \)2. \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} \) : \( +3 \)3. \( \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \) : \( +5 \)Ordering these from lowest to highest gives:\( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} < \mathrm{SO}_{3}^{2-} < \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \).

Key Concepts

Sulphur Compounds

Sulphur is a versatile element present in various compounds. These compounds often include oxygen, creating structures known as oxoanions. The chemistry of sulphur is vast because it can exist in multiple oxidation states, making it an element of interest in both academic and industrial chemistry. Some of the common sulphur compounds include sulphur dioxide

• Sulphur Dioxide \( (\mathrm{SO}_2) \)
• Sulphur Trioxide \( (\mathrm{SO}_3) \)
• Thiosulfate \( (\mathrm{S}_2\mathrm{O}_3^{2-}) \)

These compounds are notable for their sulfur-oxygen bonds and their ability to participate in various chemical reactions.Sulphur oxoanions, like those found in the exercise, are an essential part of studying the oxidation states of sulfur. These compounds can reveal much about the redox behavior of sulfur, displaying how its oxidation state changes across different molecular structures. Understanding the variety in sulfur's behavior helps chemists predict reactions and applications of these compounds.

Anions

In chemistry, anions are negatively charged ions. They are formed when atoms gain extra electrons, resulting in a net negative charge. Anions are crucial in a wide range of chemical reactions and are often found in both inorganic and organic compounds. In the context of sulphur compounds, anions are present as oxoanions, where sulfur is centrally bonded with oxygen atoms. This configuration is typical for sulfur due to its ability to stabilize multiple oxidation states.

• Common examples of sulphur anions include sulfate \( \mathrm{SO}_4^{2-} \), sulfite \( \mathrm{SO}_3^{2-} \), and thiosulfate \( \mathrm{S}_2\mathrm{O}_3^{2-} \).

The exercise specifically mentions anions like \( \mathrm{SO}_3^{2-} \) and variations like \( \mathrm{S}_2\mathrm{O}_4^{2-} \) and \( \mathrm{S}_2\mathrm{O}_6^{2-} \). All of these have a sulfur center bonded with multiple oxygen atoms, each contributing negatively to the structure's overall charge. The presence of these anions plays a critical role in understanding chemical reactivity, electrochemistry, and industrial applications.

Oxidation Reactions

Oxidation is a fundamental concept in chemistry that involves the loss of electrons from an atom, thereby increasing its oxidation state. In contrast, reduction refers to the gain of electrons. Often, oxidation and reduction occur simultaneously in what are called redox reactions. In the exercise, determining the oxidation states of sulfur in various anions requires understanding its role in oxidation reactions. Each change in sulfur's oxidation state corresponds to a transfer of electrons between sulfur and the atoms it bonds with.

• In \( \mathrm{SO}_3^{2-} \), sulfur has an oxidation state of \(+4\).
• In \( \mathrm{S}_2\mathrm{O}_4^{2-} \), sulfur has an oxidation state of \(+3\).
• Lastly, in \( \mathrm{S}_2\mathrm{O}_6^{2-} \), sulfur's oxidation state is \(+5\).

These oxidation states reflect how electrons are distributed within these molecules and how they might react with other substances. The step-by-step calculations provided in the original solution demonstrate these principles, as one must establish the algebraic sum of the charges to match the ion's overall charge. Understanding oxidation states is vital for predicting the chemical behavior of compounds during reactions and is widely applied in everything from biochemistry to environmental science.