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 in SO3^2-

In the sulfate ion \(\mathrm{SO}_3^{2-}\), let the oxidation state of sulfur be \(x\). We know the oxidation state of oxygen is \(-2\). Thus, the equation becomes \[ x + 3(-2) = -2 \] Solving gives \[ x = +4 \].

Step 2: Calculate Oxidation State in S2O4^2-

In the sulfite ion \(\mathrm{S}_2\mathrm{O}_4^{2-}\), let the oxidation state of sulfur be \(x\). Thus, the equation becomes \[ 2x + 4(-2) = -2 \] Solving gives \[ 2x - 8 = -2 \] \[ 2x = 6 \] \[ x = +3 \].

Step 3: Calculate Oxidation State in S2O6^2-

In the dithionate ion \(\mathrm{S}_2\mathrm{O}_6^{2-}\), let the oxidation state of sulfur be \(x\). Thus, the equation becomes \[ 2x + 6(-2) = -2 \] Solving gives \[ 2x - 12 = -2 \] \[ 2x = 10 \] \[ x = +5 \].

Step 4: Compare Oxidation States

We have found the oxidation states: - \( \mathrm{SO}_3^{2-} : +4 \)- \( \mathrm{S}_2\mathrm{O}_4^{2-} : +3 \)- \( \mathrm{S}_2\mathrm{O}_6^{2-} : +5 \)The order is \( \mathrm{S}_2\mathrm{O}_4^{2-} < \mathrm{SO}_3^{2-} < \mathrm{S}_2\mathrm{O}_6^{2-} \).

Key Concepts

Sulphur Compounds

Sulphur compounds are a fascinating area of chemistry because of their diverse chemical properties and applications. These compounds often contain sulphur atoms that can bond with various elements to form different molecular structures. The oxidation state of sulphur within these compounds can vary widely, from

• -2, where sulphur is the least oxidized,
• up to +6, which is the most oxidized state achieved in sulphuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)

This versatility makes sulphur compounds quite significant in both industrial applications and biological systems. In the example of the sulphur anions given, such as \(\mathrm{SO}_{3}^{2-}\), \(\mathrm{S}_{2}\mathrm{O}_{4}^{2-}\), and \(\mathrm{S}_{2}\mathrm{O}_{6}^{2-}\), the understanding of their oxidation states is essential for understanding their redox behavior and their role in chemical reactions. Sulphur can form both simple ions like \(\mathrm{HS}^{-}\) and more complex structures such as those found in thiosulphates and sulfonic acids, providing a rich chemistry landscape for exploration.

Anion Chemistry

Anions are negatively charged ions that form when atoms gain extra electrons. In the realm of chemistry, understanding anion structure and behavior is key to exploring a broad range of chemical reactions. The anions discussed within this exercise, such as \(\mathrm{SO}_{3}^{2-}\), \(\mathrm{S}_{2}\mathrm{O}_{4}^{2-}\), and \(\mathrm{S}_{2}\mathrm{O}_{6}^{2-}\), illustrate the diversity of sulphur-containing anions. Adding electrons gives these ions their charge, influencing both their stability and reactivity. Within ionic compounds, anions often combine with cations (positively charged ions) to produce a neutral molecule.

• For example, combining \(\mathrm{SO}_{3}^{2-}\) with calcium ions (\(\mathrm{Ca}^{2+}\)) results in calcium sulfite.
• Sulfide (\(\mathrm{S}^{2-}\)) and sulfate (\(\mathrm{SO}_{4}^{2-}\)) are other popular sulfur anions showcasing how sulphur can varyingly expand its valence shell.

Anion chemistry is crucial in the formation of substances ranging from simple salts to complex biochemical molecules. As we investigate these anions more, we unravel the complex, yet structured fabric of chemistry at both atomic and molecular levels.

Redox Reactions

Redox reactions involve the transfer of electrons between chemical species. These reactions are fundamental to a wide array of natural and industrial processes such as metabolism, battery operation, and corrosion. In redox reactions, one species undergoes oxidation, losing electrons, while another undergoes reduction, gaining electrons.
The comprehension of oxidation states is crucial in identifying which species are oxidized and reduced. For instance, considering the sulphur anions \(\mathrm{SO}_{3}^{2-}\), \(\mathrm{S}_{2}\mathrm{O}_{4}^{2-}\), and \(\mathrm{S}_{2}\mathrm{O}_{6}^{2-}\), recognizing their differing oxidation states allows us to predict their behavior in redox reactions. A lower oxidation state generally means that the species is more inclined to donate electrons and act as a reducing agent.

• In redox reactions,\(\mathrm{S}_{2}\mathrm{O}_{4}^{2-}\) might act as a reducing agent due to its lower oxidation state compared to \(\mathrm{S}_{2}\mathrm{O}_{6}^{2-}\).
• Conversely,\(\mathrm{S}_{2}\mathrm{O}_{6}^{2-}\), being in a higher oxidation state, is more prone to accepting electrons and can act as an oxidizing agent.

Understanding these dynamics is essential not only for chemical applications but also for leveraging redox reactions in broad technological advancements.