The ion OSCN \(^{-}\) is found in human saliva. Discuss the problems in assigning oxidation numbers to the atoms in this ion. Suggest a reasonable set of oxidation numbers for the atoms in OSCN \(^{-}\).

Short Answer

Expert verified
The oxidation numbers of OSCN- are complicated by multiple electronegative elements and possible bonding structures. A reasonable suggestion for the oxidation numbers is O = -2, S = +2, C = +4, and N = -3, ensuring the sum equates to the overall charge of -1.

Step by step solution

01

Understand the concept of oxidation numbers

Oxidation numbers are a bookkeeping system in chemistry used to keep track of electrons in molecules and ions. They are assigned based on a set of rules, considering electronegativity and the molecule or ion structure.
02

Identify the challenges

Assigning oxidation numbers to each atom in OSCN- is challenging because of the presence of multiple electronegative elements (O, S, and N) and the potential for multiple bonding scenarios. Moreover, the compound is an ion with an overall negative charge, which must be accounted for.
03

Assign oxidation numbers using known rules

Oxygen usually has an oxidation number of -2. Assuming Sulfur, which is less electronegative, to have a positive oxidation state, the Cyanide ion (CN-) is typically assigned a -1 charge as a whole since it behaves as a pseudohalogen. Sulfur (S) is known to have multiple oxidation states; thus, it could have any positive value that balances the charge. The overall charge is -1, so the sum of oxidation numbers should equal -1.
04

Suggest a reasonable set of oxidation numbers

A reasonable set of oxidation numbers for OSCN- considering the above rules could be: oxygen (O) as -2, sulfur (S) as +2, carbon (C) as +4, and nitrogen (N) as -3. This assignment maintains the overall charge of -1 and adheres to common oxidation states for these atoms.

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Key Concepts

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

Oxidation States in Ions
Understanding the oxidation states in ions is pivotal to grasping many chemical concepts involving reactions and bonding. An oxidation state, often referred to as an oxidation number, is a theoretical charge an atom would have if electrons were distributed according to certain rules. In a neutral molecule, the sum of all oxidation states equals zero; in an ion, it equals the ion's charge.

When determining the oxidation states for atoms in an ion, it's essential to begin with known values. Elements from Group 1 and 2 of the periodic table typically have oxidation numbers of +1 and +2, respectively. Oxygen, in most cases, has an oxidation state of -2. Hydrogen is usually +1 except when bonded to metals where it can be -1. For ions, after assigning oxidation states to the known elements, the remaining values are given to other atoms such that the total equals the charge of the ion.

Challenges with Transition Metals and Polyatomic Ions

Transition metals and elements with multiple oxidation states can be more complex to assign. The same element can have different oxidation states in various compounds or ions. For polyatomic ions, like the OSCN- ion found in human saliva, this task becomes even more challenging, requiring a deep understanding of the compound's structure and bonding.
Electronegativity and Oxidation Numbers
Electronegativity is a key factor in determining the oxidation numbers within a compound or an ion. It refers to the tendency of an atom to attract electrons in a chemical bond. The more electronegative an atom is, the more likely it is to attract electrons and have a negative oxidation state.

The electronegativity values follow a trend on the periodic table, increasing from left to right and bottom to top. Accounting for electronegativity helps in predicting which atoms will have positive or negative oxidation states. For instance, in a bond between carbon and oxygen, oxygen is more electronegative and is typically assigned an oxidation number of -2, whereas carbon is assigned a positive value that reflects its loss of electron density to oxygen.

Balancing Electron Exchange

When assigning oxidation numbers based on electronegativity, one must ensure that the sum of the oxidation numbers is equal to the charge of the ion or molecule. This balancing act takes into consideration both the known values of electronegativity and the need to satisfy the overall charge.
OSCNS- Ion Chemistry
The OSCNS- ion is a fascinating subject in the field of ion chemistry due to its presence in human saliva and its unique chemical structure, which presents challenges in assigning oxidation numbers. As outlined in textbook solutions, determining the oxidation numbers for OSCN- involves understanding the individual elements' tendencies and the overall charge of the ion.

For the OSCNS- ion, oxygen typically has an oxidation number of -2; however, sulfur and nitrogen, having a wide range of possible oxidation states, can complicate the situation. Nitrogen usually has a -3 oxidation state in most of its stable compounds. Sulfur's state is often positive and varies greatly depending on the compound.

Cyanide Group Considerations

The cyanide group (CN-) within OSCNS- usually carries a -1 charge and is treated as a unit. Carbon, being less electronegative than nitrogen, tends to have positive oxidation states. By considering these factors and ensuring that the sum of oxidation numbers equals the ion's overall charge, a reasonable convention can be suggested where oxygen is -2, sulfur is +2, carbon is +4, and nitrogen is -3, fulfilling the ion's -1 charge and providing insight into its electron distribution and bonding characteristics.

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Most popular questions from this chapter

If the oxidation number of nitrogen in a certain molecule changes from +3 to -2 during a reaction, is the nitrogen oxidized or reduced? How many electrons are gained or lost by the nitrogen atom?

For the following reactions, identify the substance oxidized, the substance reduced, the oxidizing agent, and the reducing agent. $$ \begin{array}{l} \text { (a) } 2 \mathrm{HNO}_{3}+3 \mathrm{H}_{3} \mathrm{AsO}_{3} \longrightarrow \\ 2 \mathrm{NO}+3 \mathrm{H}_{3} \mathrm{AsO}_{4}+\mathrm{H}_{2} \mathrm{O} \\ \text { (b) } \mathrm{NaI}+3 \mathrm{HOCl} \longrightarrow \mathrm{NaIO}_{3}+3 \mathrm{HCl} \end{array} $$ (c) \(2 \mathrm{KMnO}_{4}+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow\) $$ 10 \mathrm{CO}_{2}+\mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+8 \mathrm{H}_{2} \mathrm{O} $$ (d) \(6 \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{Al} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{SO}_{2}+6 \mathrm{H}_{2} \mathrm{O}\)

Hydroiodic acid reduces chlorine to hydrochloric acid and iodine. Write a balanced net ionic equation for the reaction.

Sulfites are used worldwide in the wine industry as antioxidant and antimicrobial agents. However, sulfites have also been identified as causing certain allergic reactions suffered by asthmatics, and the FDA mandates that sulfites be identified on the label if they are present at levels of 10 ppm (parts per million) or higher. The analysis of sulfites in wine uses the "Ripper method" in which a standard iodine solution, prepared by the reaction of iodate and iodide ions, is used to titrate a sample of the wine. The iodine is formed in the reaction $$ \mathrm{IO}_{3}^{-}+5 \mathrm{I}^{-}+6 \mathrm{H}^{+} \longrightarrow 3 \mathrm{I}_{2}+3 \mathrm{H}_{2} \mathrm{O} $$ The iodine is held in solution by adding an excess of \(\mathrm{I}^{-}\), which combines with \(\mathrm{I}_{2}\) to give \(\mathrm{I}_{3}^{-}\). In the titration, the \(\mathrm{SO}_{3}^{2-}\) is converted to \(\mathrm{SO}_{2}\) by acidification, and the reaction during the titration is $$ \mathrm{SO}_{2}+\mathrm{I}_{3}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{SO}_{4}^{2-}+3 \mathrm{I}^{-}+4 \mathrm{H}^{+} $$ Starch is added to the wine sample to detect the end point, which is signaled by the formation of a dark blue color when excess iodine binds to the starch molecules. In a certain analysis, \(0.0421 \mathrm{~g}\) of \(\mathrm{NaIO}_{3}\) was dissolved in dilute acid and excess NaI was added to the solution, which was then diluted to a total volume of \(100.0 \mathrm{~mL}\) A \(50.0 \mathrm{~mL}\) sample of wine was then acidified and titrated with the iodine- containing solution. The volume of iodine solution required was \(2.47 \mathrm{~mL}\). (a) What was the molarity of the iodine (actually, \(\left.\mathrm{I}_{3}^{-}\right)\) in the standard solution? (b) How many grams of \(\mathrm{SO}_{2}\) were in the wine sample? (c) If the density of the wine was \(0.96 \mathrm{~g} / \mathrm{mL}\), what was the percentage of \(\mathrm{SO}_{2}\) in the wine? (d) Parts per million (ppm) is calculated in a manner similar to percent (which is equivalent to parts per hundred). $$ \mathrm{ppm}=\frac{\text { grams of component }}{\text { grams of sample }} \times 10^{6} \mathrm{ppm} $$ What was the concentration of sulfite in the wine, expressed as parts per million \(\mathrm{SO}_{2} ?\)

Why must both oxidation and reduction occur simultaneously during a redox reaction? What is an oxidizing agent and what happens to it in a redox reaction? What is a reducing agent and what happens to it in a redox reaction?

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