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The oxidation number of sulphur in \(\mathrm{S}_{8}, \mathrm{~S}_{2} \mathrm{~F}_{2}, \mathrm{H}_{2} \mathrm{~S}\) respectively, are (a) \(0,+1\) and \(-2\) (b) \(+2,+1\) and \(-2\) (c) \(0,+1\) and \(+2\) (d) \(-2,+1\) and \(-2\).

Short Answer

Expert verified
The correct answer is (a) 0,+1 and -2.

Step by step solution

01

Determine Oxidation Number of Sulfur in S8

In the molecule \( \mathrm{S}_8 \), sulfur is in its elemental form, so the oxidation number is \( 0 \). This is because elements in their natural, uncombined state have an oxidation number of \( 0 \).
02

Determine Oxidation Number of Sulfur in S2F2

In \( \mathrm{S}_2\mathrm{F}_2 \), fluoride has a known oxidation number of \(-1\). Let \( x \) represent the oxidation number of sulfur. The equation is \( 2x + 2(-1) = 0 \). Solving: \( 2x - 2 = 0 \Rightarrow 2x = 2 \Rightarrow x = +1 \). Therefore, the oxidation number of sulfur in \( \mathrm{S}_2\mathrm{F}_2 \) is \(+1\).
03

Determine Oxidation Number of Sulfur in H2S

In \( \mathrm{H}_2\mathrm{S} \), hydrogen has an oxidation number of \(+1\). Let \( y \) represent the oxidation number of sulfur. The equation is \( 2(+1) + y = 0 \). Solving: \( 2 + y = 0 \Rightarrow y = -2 \). Thus, the oxidation number of sulfur in \( \mathrm{H}_2\mathrm{S} \) is \(-2\).
04

Match Calculated Values with Options

From the previous steps: sulfur in \( \mathrm{S}_8 \) is \( 0 \), in \( \mathrm{S}_2\mathrm{F}_2 \) is \(+1\), and in \( \mathrm{H}_2\mathrm{S} \) is \(-2\). These match with the values in option (a).

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Key Concepts

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

Sulfur Compounds
Understanding sulfur compounds is crucial in chemistry, especially given sulfur's versatility as an element. Sulfur can adopt a variety of oxidation states and can form multiple types of compounds, including elemental sulfur, sulfides, and sulfates. Sulfur compounds are present in many forms in nature.
  • **Elemental Sulfur** - Referring to sulfur in its natural state, like in the molecule \( \mathrm{S}_8 \), where it holds an oxidation number of zero. This form is found in volcanic areas and decaying organic matter.
  • **Sulfides** - These are compounds where sulfur is combined with another element, usually with a more positive oxidation number.
  • **Sulfates and Sulfuric Acids** - In these compounds, sulfur typically exhibits its higher oxidation states like +6, contributing to the importance of sulfuric acid in industry.
Sulfur's ability to form various compounds with different elements and its occurrence in differing oxidation states significantly affects its chemical reactivity and applications.
Chemical Bonding
Chemical bonding is the attractive force that holds atoms together, allowing them to form compounds. Understanding chemical bonding helps explain how sulfur, like many elements, forms various compounds.
Sulfur can participate in different types of bonding:
  • **Covalent Bonds** - Sulfur typically forms covalent bonds, where it shares its valence electrons with other atoms. This sharing results in common compounds such as hydrogen sulfide (\( \mathrm{H}_2\mathrm{S} \)), where sulfur shares electrons with hydrogen.
  • **Ionic Bonds** - While less common for sulfur, it can also form ionic bonds, primarily seen in its more electropositive compounds like sulfides.
When understanding and calculating oxidation numbers, recognizing the type of bond is critical, as this affects how electrons are distributed between atoms. This distribution is essential to determine the oxidation state of an element within the molecule.
Oxidation States in Molecules
Oxidation states, or oxidation numbers, are theoretical charges assigned to atoms to describe electron transfer in molecules. They help chemists understand electron balance in compounds, which is fundamental for chemical reactions.
  • **Assigning Oxidation States** - The rules for assigning oxidation numbers include setting an oxidation state of zero for elements in their standard state and known states for other very common elements like -1 for fluorine.
  • **Calculating Oxidation Numbers** - Using algebraic equations based on the known oxidation numbers and the neutral charge of molecules, as shown in calculating sulfur’s oxidation state in \( \mathrm{S}_2\mathrm{F}_2 \) and \( \mathrm{H}_2\mathrm{S} \).
  • **Chemical Reactions** - Tracking oxidation states is essential in identifying redox reactions, where oxidation numbers change as electrons are transferred between atoms.
Understanding oxidation states not only aids in solving problems related to electron distribution and chemical bonding but also informs the study of chemical properties and interactions of compounds.

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Most popular questions from this chapter

The limiting molar conductivities \(\Lambda^{\circ}\) for \(\mathrm{NaCl}, \mathrm{KBr}\) and \(\mathrm{KCl}\) are 126,152 and \(150 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) respectively. The \(\Lambda^{\circ}\) for \(\mathrm{NaBr}\) is (a) \(278 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) (b) \(178 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) (c) \(128 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) (d) \(306 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)

In the electrolytic cell, flow of electrons is from (a) cathode to anode in solution (b) cathode to anode through external supply (c) cathode to anode through internal supply (d) anode to cathode through internal supply

Consider the following \(E^{\circ}\) values \(E\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=+0.77 \mathrm{~V}\) \(E\left(\mathrm{Sn}^{2+} / \mathrm{Sn}\right)=-0.14 \mathrm{~V}\) Under standard conditions, the potential for the reaction \(\mathrm{Sn}(\mathrm{s})+2 \mathrm{Fe}^{3+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Sn}^{2+}(\mathrm{aq})\) is (a) \(1.68 \mathrm{~V}\) (b) \(0.91 \mathrm{~V}\) (c) \(0.63 \mathrm{~V}\) (d) \(1.46 \mathrm{~V}\)

\(\mathrm{aHNO}_{3}+\mathrm{bH}_{2} \mathrm{~S} \longrightarrow \mathrm{cNO}+\mathrm{d} \mathrm{H}_{2} \mathrm{O}+3 \mathrm{~S} .\) What are the values of \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} ?\) (a) \(2,3,2,2\) (b) \(2,3,2,4\) (c) \(3,3,2,4\) (d) \(4,3,1,2\)

The standard emf of a galvanic cell involving cell reaction with \(\mathrm{n}=2\) is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction would be (Given \(\left.F=96500 \mathrm{C} \mathrm{mol}^{-1} ; \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(2.0 \times 10^{11}\) (b) \(4.0 \times 10^{12}\) (c) \(1.0 \times 10^{2}\) (d) \(1.0 \times 10^{10}\)

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