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Chapter 8: Problem 21

The difference in the oxidation numbers of the two types of sulphur atoms in \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) is

Short Answer

Expert verified
The difference in oxidation numbers is 6.

Step by step solution

01

Determine Oxidation Rules

Identify the general rules to assign oxidation states: 1) The sum of oxidation numbers in a neutral compound must be zero; 2) The oxidation state of sodium (Na) is +1; 3) The oxidation state of oxygen (O) is -2.
02

Assign Known Oxidation States

For \(\mathrm{Na}_{2}\mathrm{~S}_{4}\mathrm{O}_{6} \), each Na has an oxidation number of +1 and each O has an oxidation number of -2. Since there are two Na atoms, their total contribution is \(+2\). For the six oxygen atoms, the total contribution is \(-12\).
03

Calculate Overall Charge Contribution

Combine the contributions of the known oxidation states. The positive contribution from sodium is +2, and the negative contribution from oxygen is -12. Thus, collectively, they contribute \(+2 - 12 = -10\).
04

Calculate Sum of Sulfur Contribution

Since the compound is neutral, the total oxidation number for sulfur atoms must offset the -10 from other atoms, making sulfur's total sum \(+10\).
05

Distribution of Oxidation Numbers Among Sulfur Atoms

We know sulfur can have multiple oxidation states in the same compound. Assign hypothetical oxidation numbers to sulfur atoms: two sulfur atoms in the center are likely to show a higher oxidation number due to more bonds with oxygen; they are often +5 each. The other two sulfur atoms, part of disulfide linkage ( -S-S- ), typically show oxidation states of -1 each.
06

Validate the Distribution

Validate the oxidation states: for sulfur atoms \(2 imes -1 = -2\) and \(2 imes +5 = +10\). The total for sulfur becomes \(-2 + 10 = +8\), this overall sulfur contribution complements \(-10\) from other atoms, confirming the structure balance to zero.
07

Calculate Difference in Oxidation Numbers

Subtract the oxidation number of disulfide sulfur atoms \((-1)\) from those bonded with more oxygens \((+5)\). The difference is \(+5 - (-1) = 6\).

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Key Concepts

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

Sulfur Oxidation State
When dealing with compounds like \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\), it's important to understand that sulfur can exist in multiple oxidation states within a single molecule. In this case, the sulfur atoms are involved in two different environments:
  • The central sulfur atoms typically participate in numerous bonds involving oxygen, which often results in a higher oxidation state. For this compound, these central sulfur atoms are each assigned an oxidation state of +5.
  • The sulfur atoms that form part of a disulfide linkage, represented as \(-\mathrm{S-S}-\), usually have fewer bonds with oxygen. Here, these atoms are typically assigned an oxidation state of -1.
This distribution aligns with known chemical behaviors, whereby sulfur exhibits variable valency, adapting based on its bonding scenario. By assigning oxidation numbers based on structure, chemists can predict the oxidation states and thus understand the behavior and reactivity of different sulfur-containing compounds.
Neutral Compound
A neutral compound is one where the total charge from all atoms is zero. This is a critical concept in chemistry because it helps in determining how oxidation numbers are distributed among the atoms in a molecule.
  • In our example of \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\), this means the sum of the oxidation states of all sodium, sulfur, and oxygen atoms must be zero.
  • For the conjunctive knowledge, sodium has a known oxidation state of +1, while oxygen typically exhibits an oxidation state of -2. Adding their contributions helps us establish the balance needed from sulfur atoms.
Knowing a compound is neutral allows us to use this information as a guiding principle to allocate unknown oxidation states, ensuring the sum balances to zero without any leftover charges.
Oxidation Rules
Oxidation rules provide a systematic way to assign oxidation states to diverse elements in a compound, ensuring chemical equations are balanced and predictions about reactivity can be made. The key rules to consider include:
  • The oxidation number of an element in its standard state is always zero.
  • The sum of the oxidation states in a neutral compound must be zero.
  • Sodium (\(\mathrm{Na}\)) typically takes an oxidation state of +1, owing to its single valence electron, which it donates during bonding.
  • Oxygen almost always takes an oxidation state of -2, given its electronegative nature, causing it to gain electrons.
By applying these rules, chemically valid oxidation numbers are determined, as shown in the \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) compound. This ensures not only the balance of charges but also allows for a deep understanding of how atoms transform through redox reactions.

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Most popular questions from this chapter

Given : \(\mathrm{Co}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Co}^{2+} ; \mathrm{E}^{\circ}=+1.81 \mathrm{~V}\) \(\mathrm{Pb}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Pb}^{2+} ; \mathrm{E}^{0}=+1.67 \mathrm{~V}\) \(\mathrm{Ce}^{4+}+\mathrm{e}^{-} \rightarrow \mathrm{Ce}^{3+} ; \mathrm{E}^{0}=+1.61 \mathrm{~V}\) \(\mathrm{Bi}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Bi} ; \mathrm{E}^{\circ}=+0.20 \mathrm{~V}\) oxidizing power of the species will increase in the order: (a) \(\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Co}^{3+}\) (b) \(\mathrm{Bi}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Co}^{3+}\) (c) \(\mathrm{Co}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Pb}^{4+}\) (d) \(\mathrm{Co}^{3+}<\mathrm{Pb}^{4+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}\)

In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of \(\mathrm{CO}_{2}\) is: (a) 1 (b) 10 (c) 2 (d) 5

Given that \(\mathrm{E}_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}^{\circ}=+1.23 \mathrm{~V} ;\) \(\mathrm{E}_{\mathrm{S}_{2} \mathrm{O}_{8}^{2-} / \mathrm{SO}_{4}^{2-}}=2.05 \mathrm{~V}\) \(\mathrm{E}_{\mathrm{Br}_{2} / \mathrm{Br}^{-}}^{\circ}=+1.09 \mathrm{~V}\) \(\mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}^{\circ}=+1.4 \mathrm{~V}\) The strongest oxidising agent is : (a) \(\mathrm{Au}^{3+}\) (b) \(\mathrm{O}_{2}\) (c) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\) (d) \(\mathrm{Br}_{2}\)

Arrange the following in increasing oxidation number of iodine. [1986-1 Mark] \(\mathrm{I}_{2}, \mathrm{HI}, \mathrm{HIO}_{4}, \mathrm{ICl}\)

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 is : [1987-1 Mark] (a) 1 (b) 2 (c) 3 (d) 0
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