Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
Option (d) is correct: \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} < \mathrm{SO}_{3}^{2-} < \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \).

Step by step solution

01

Determine the Oxidation State of Sulfur in \( \mathrm{SO}_{3}^{2-} \)

First apply the sum of oxidation states rule. Let the oxidation state of sulfur be \( x \). Oxygen typically has an oxidation state of \(-2\). For \( \mathrm{SO}_{3}^{2-} \): \[ x + 3(-2) = -2 \] Solve for \( x \): \[ x - 6 = -2 \] \[ x = +4 \] So, the oxidation state of sulfur in \( \mathrm{SO}_{3}^{2-} \) is +4.
02

Determine the Oxidation State of Sulfur in \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} \)

Consider the compound \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} \). Let the oxidation state of each sulfur be \( y \). Calculate using the total charge:\[ 2y + 4(-2) = -2 \] \[ 2y - 8 = -2 \] \[ 2y = +6 \] \[ y = +3 \] So, the oxidation state of each sulfur in \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} \) is +3.
03

Determine the Oxidation State of Sulfur in \( \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \)

Now consider \( \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \). Let the oxidation state of each sulfur be \( z \). The total charge gives:\[ 2z + 6(-2) = -2 \] \[ 2z - 12 = -2 \] \[ 2z = +10 \] \[ z = +5 \] So, the oxidation state of each sulfur in \( \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \) is +5.
04

Rank the Oxidation States

The oxidation states are now known:- \( \mathrm{SO}_{3}^{2-} \): +4- \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} \): +3- \( \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \): +5Rank the oxidation states from lowest to highest: \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} < \mathrm{SO}_{3}^{2-} < \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \).
05

Select the Correct Answer

Match the order obtained: \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} < \mathrm{SO}_{3}^{2-} < \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \) Option \( (d) \) \( \mathrm{S}_{2} \mathrm{O}_{4}^{2-} < \mathrm{SO}_{3}^{2-} < \mathrm{S}_{2} \mathrm{O}_{6}^{2-} \) is the correct choice.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Sulfur Compounds
Sulfur compounds are fascinating because they exhibit a variety of chemical behaviors due to their diverse oxidation states. In nature and in chemistry, sulfur can form multiple types of compounds, from simple diatomic molecules like sulfur dioxide (SO extsubscript{2}) to large complex structures like proteins in biological systems. Sulfur is found in different states in compounds, typically ranging from ext{-2} to +6.
This versatility is driven by sulfur's ability to form bonds with other sulfur atoms and various elements, including oxygen, resulting in a wide array of compounds. Sulfur compounds can be gaseous, liquid, or solid and are important in both environmental chemistry and industrial processes.
Some notable sulfur-containing compounds include:
  • Sulfides (like H extsubscript{2}S) where sulfur has an oxidation state of ext{-2}.
  • Sulfates (like Na extsubscript{2}SO extsubscript{4}) where sulfur is at the +6 oxidation state.
Understanding these states helps in predicting how sulfur compounds react, especially in redox reactions.
Anion Chemistry
Anions are negatively charged ions formed when atoms gain electrons. In chemistry, the behavior of anions is crucial for understanding various chemical reactions. When sulfur combines with oxygen to form anions such as SO extsubscript{3} extsuperscript{2-} , S extsubscript{2}O extsubscript{4} extsuperscript{2-}, and S extsubscript{2}O extsubscript{6} extsuperscript{2-}, it involves the interaction of their oxidation states and charges.
Anion chemistry often involves the concept of charge balance and stoichiometry. When calculating oxidation states for elements within anions, such calculations ensure that the net charge is consistent with the total of all oxidation states.
Key points about anions include:
  • Anions are frequently formed by non-metals, like sulfur and oxygen, due to their tendency to gain electrons.
  • They play a pivotal role in the chemistry of salts, where they are paired with cations to form neutral compounds.
Mastering the understanding of anion chemistry aids significantly in predicting the behavior and reactivity of complex ions, especially in aqueous solutions.
Redox Reactions
Redox reactions are chemical processes involving the transfer of electrons between two substances. These reactions include oxidation, where a substance loses electrons, and reduction, where a substance gains electrons.
In the context of sulfur compounds, redox reactions can drastically change the oxidation state of sulfur, influencing the properties and reactivity of the compound.
For example, consider sulfur's transition from SO extsubscript{3} extsuperscript{2-} to S extsubscript{2}O extsubscript{6} extsuperscript{2-}:
  • Involves an oxidation of sulfur from +4 to +5.
  • Indicates loss of electrons (oxidation step).
  • Accompanied by a gain of electrons elsewhere in the reaction (reduction step).
Redox reactions are fundamental in energy conversion processes, such as in batteries and metabolic pathways. Understanding these reactions allows chemists to create and manipulate chemical processes for desired outcomes, from industrial manufacturing to biochemistry applications. Studying redox reactions enhances comprehension of broader chemical principles and is critical for anyone looking to advance in chemistry fields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Match the following: List I (Manufacturing process) 1\. Deacon's process for chlorine 2\. Hydrogenation of vegetable oils 3\. Ostwald's process for nitric acid 4\. Haber's process for ammonia The correct matching is List II (Catalyst used) (i) finely divided iron with molybdenum as promoter (ii) copper(II)chloride (iii) finely divided nickel powder (iv) platinum gauze 1 3 4 (a) (iv) (iii) (ii) (b) (i) (iv) (ii) (iii) (c) (ii) (iii) (i) (iv) (d) (ii) (iii) (iv) (i)

Compound \(\mathrm{X}\) on reduction with \(\mathrm{LiAlH}_{4}\) gives a hydride Y containing \(21.72 \%\) hydrogen along with other products. The compound \(Y\) reacts with air explosively resulting in boron trioxide. Identify \(\mathrm{Y}\). (a) \(\mathrm{B}_{4} \mathrm{H}_{10}\) (b) \(\mathrm{B}_{2} \mathrm{H}_{6}\) (c) \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7}\) (d) both (a) and (b)

The increase in boiling points of noble gases from He to \(\mathrm{Xe}\) is due to the (a) increase in atomic volume (b) increase in electron affinity (c) increase in polarizability (d) decrease in ionization energy

Which of the following species involve \(\mathrm{M}-\mathrm{M}\) bond in addition to \(\mathrm{M}-\mathrm{O}\) bonds? (a) Carbon suboxide (b) Pyrosulphate ion (c) Pyrophosphoric acid (d) Dithionate ion

The pair of species having identical shapes for molecules of both species is (a) \(\mathrm{PF}_{\mathrm{s}}, \mathrm{IF}\), (b) \(\mathrm{XeF}_{2}, \mathrm{CO}_{2}\) (c) \(\mathrm{CF}_{4}, \mathrm{SF}\) (d) \(\mathrm{BF}_{3}, \mathrm{PCl}_{\underline{ }}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free