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Chapter 9: Problem 154

Identify the compounds in which the sulphur atoms are in different oxidation states? (a) \(\mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\) (b) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) (d) \(\mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{\mathrm{y}}\)

Short Answer

Expert verified
The compound with sulfur atoms in different oxidation states is \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\).

Step by step solution

01

Calculate Sulfur Oxidation State in K2S2O7

In the compound \(\mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\), potassium (K) has an oxidation state of +1 and oxygen (O) has an oxidation state of -2. Let sulfur's oxidation state be \(x\). The compound's charge is 0. The equation is:\[2(1) + 2(x) + 7(-2) = 0\]Simplifying, \(2 + 2x - 14 = 0 \) leads to \(2x = 12\) and thus \(x = +6\). The sulfur atoms are in the same oxidation state, +6.
02

Calculate Sulfur Oxidation State in Na2S2O3

For \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\), sodium (Na) has +1, and oxygen (O) has -2 oxidation states. Let sulfur's oxidation state be \(x\). The total compound charge is 0:\[2(1) + 2(x) + 3(-2) = 0\]Solving gives \(2 + 2x - 6 = 0\), so \(2x = 4\) and \(x = +2\). The sulfur atoms are in the same oxidation state, +2.
03

Calculate Sulfur Oxidation State in Na2S4O6

For \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\), sodium (Na) is +1 and oxygen is -2. Let sulfur鈥檚 oxidation states be \(x\) and \(y\) due to potential variances:\[2(1) + 4x + 6(-2) = 0\]Solving gives \(2 + 4x - 12 = 0\), which simplifies to \(4x = 10\) or \(x = +2.5\), suggesting averaging of two different states to 5. Assigning each pair average values gives plausible states like +5 and 0 confirming different oxidation states are possible for \(x\) and \(y\).
04

Conclusion

In comparing oxidation states, - \(\mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\) and \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) have sulfur atoms in the same oxidation state. However, - \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) has unique sulfur oxidation states due to interspersed values, and possibly having representative states, as explained, indicative of differing actual states.

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Key Concepts

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

Oxidation State Calculation
Calculating oxidation states helps us understand how electrons are distributed in compounds. Each element has a typical oxidation state, but it can vary based on the compound it's in. To find the oxidation state of sulfur in a compound:
  • Start by knowing the oxidation states of other elements. For instance, sodium (Na) is usually +1, potassium (K) also +1, and oxygen (O) -2.
  • Set up an equation considering the total charge of the compound, which is often zero for neutral compounds. Assume sulfur's oxidation state as a variable, denoted by \(x\).
  • Solve by balancing the total sum of charges to zero (or the given net charge).
The step-by-step calculation allows you to pinpoint oxidation states critically, such as finding sulfur's state in each compound presented. This systematic approach brightens the puzzle of electron sharing and distribution.
Sulfur Compounds
Sulfur forms a diverse range of compounds, which makes it a fascinating element to study in inorganic chemistry. Its ability to exist in various oxidation states is central to understanding its chemical behavior:
  • Sulfur forms compounds like sulfates and thiosulfates, often paired with oxygen or metals.
  • Compounds such as \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) (sodium thiosulfate) show sulfur with multiple oxidation states in some cases, adding complexity to its chemical properties.
  • In molecules like \(\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}\), sulfur can unusually show differential oxidation states, which influences their reactivity and stability.
Understanding these compounds enhances comprehension of sulfur鈥檚 versatile bonding and its role in various chemical processes.
Inorganic Chemistry
Within the broad field of chemistry, inorganic chemistry focuses on compounds not primarily based on carbon-hydrogen bonds, like those of sulfur. Here is what makes it unique:
  • It involves the study of metals, minerals, and organometallic compounds.
  • Electrons in inorganics follow specific rules of coordination and ionic bonding, differing from those in organic compounds.
  • Sulfur's role in inorganic chemistry includes forming sulfide minerals and participating in redox reactions as an electron donor or accepter.
This field provides vivid insight into the behaviors of sulfur in coordination compounds and ionic lattices, showcasing how sulfur's varied oxidation states are integral to its chemistry.
Differential Oxidation States
Differential oxidation states refer to the presence of the same element in varying oxidation states within a single molecule. This concept is especially important in compounds such as \(\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}\):
  • It signifies that sulfur atoms in the compound do not all have the same oxidation number, implying unique electronic environments.
  • The variation in states can lead to differential chemical properties and reactivity within the same molecule.
  • This results in compounds that show unforeseen or mixed behavior, such as differing solubility, electrical conductivity, and reactivity with other chemicals.
Recognizing differential oxidation states in sulfur compounds is key to predicting and understanding their chemical behavior and applications.

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Most popular questions from this chapter

\(\mathrm{Cu}^{+}(\mathrm{aq})\) is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction, \(2 \mathrm{Cu}^{+}(\mathrm{aq})=\mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) choose correct \(E^{\circ}\) for above reaction if \(E^{\circ}\left(\mathrm{Cu}^{2+} / \mathrm{Cu}\right)=0.34 \mathrm{~V}\) and \(E^{\circ}\left(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\right)=0.15 \mathrm{~V}\) (a) \(-0.38 \mathrm{~V}\) (b) \(+0.49 \mathrm{~V}\) (c) \(+0.38 \mathrm{~V}\) (d) \(-0.19 \mathrm{~V}\)

The brown ring complex compound is formulated as \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{NO}\right] \mathrm{SO}_{4} .\) The oxidation state of iron in the compound is (a) 1 (b) 2 (c) 3 (d) 0

A compound contains atoms \(X, Y, Z\) the oxidation number of \(\mathrm{X}\) is \(+2, \mathrm{Y}\) is \(+5\) and \(\mathrm{Z}\) is \(-2\) The possible form of the compound is (a) \(\mathrm{XY}_{1} \mathrm{Z}_{2}\) (b) \(\mathrm{Y}_{2}\left(\mathrm{XZ}_{3}\right)_{2}\) (c) \(\mathrm{X}_{3}\left(\mathrm{YZ}_{4}\right)_{2}\) (d) \(\mathrm{X}_{3}\left(\mathrm{Y}_{4} \mathrm{Z}\right)_{2}\)

Oxidation numbers of carbon in \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}, \mathrm{CH}_{4}\) and diamond respectively are (a) \(+3,4\) and \(+4\) (b) \(+3,-4\) and zero (c) \(+6,+4\) and zero (d) \(+6,+4\) and \(+4\)

The equivalent conductance at infinite dilution of \(\mathrm{HCl}\) and \(\mathrm{NaCl}\) are \(426.15\) and \(126.15 \mathrm{mho} \mathrm{cm}^{2} \mathrm{~g} \mathrm{eq}^{-1}\) respectively. It can be said that the mobility of (a) \(\mathrm{H}^{+}\)ions is much more than that of \(\mathrm{Cl}\) ions (b) \(\mathrm{Cl}\) - ions is much more than that of \(\mathrm{H}^{+}\)ions (c) \(\mathrm{H}^{+}\)ions is much more than that of \(\mathrm{Na}^{+}\)ions (d) \(\mathrm{Na}^{+}\)ions is much more than that of \(\mathrm{H}^{+}\)ions
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