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Assign oxidation numbers to the elements in the and following: (a) \(\mathrm{MnCl}_{2}\), (b) \(\mathrm{MnO}_{4}^{-}\) (c) \(\mathrm{MnO}_{4}^{2-},\) (d) \(\mathrm{MnO}_{2}\)

Short Answer

Expert verified
(a) Mn(+2)Cl2(-1 each), (b) Mn(+7)O4(-2 each), (c) Mn(+6)O4(-2 each), (d) Mn(+4)O2(-2 each)

Step by step solution

01

- Assign oxidation numbers for MnCl2

Chlorine typically has an oxidation number of -1. Since we have two chlorine atoms in MnCl2, the total charge contributed by chlorine is -2. Manganese (Mn) must balance this with a +2 oxidation number, giving us Mn(+2)Cl2(-1 each).
02

- Assign oxidation numbers for MnO4-

In the permanganate ion, MnO4-, oxygen typically has an oxidation number of -2. There are four oxygen atoms, totaling -8. The overall charge of the ion is -1, thus manganese (Mn) must have an oxidation number of +7 to balance the charge (+7 - 8 = -1).
03

- Assign oxidation numbers for MnO42-

In the manganate ion, MnO42-, oxygen again has an oxidation number of -2. Four oxygen atoms total -8 in oxidation number. The overall charge is -2, meaning manganese (Mn) must have an oxidation number of +6 to balance the charge (+6 - 8 = -2).
04

- Assign oxidation numbers for MnO2

In manganese dioxide, MnO2, oxygen has an oxidation number of -2. With two oxygens, the total oxidation number coming from oxygen is -4. Manganese (Mn) must have an oxidation number of +4 to balance the oxygen and make the compound neutral (+4 - 4 = 0).

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Key Concepts

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

Oxidation State Chemistry
Understanding oxidation states, also known as oxidation numbers, is pivotal in the study of chemical reactions, especially when delving into redox processes. Oxidation states are theoretical charges assigned to an atom assuming that electrons in a bond are completely transferred to the more electronegative element. It's a helpful bookkeeping device employed to track electron transfer.

For instance, in the exercise provided, the oxidation state of manganese (Mn) in different compounds is determined according to certain rules. A key concept is that the sum of oxidation states in a neutral molecule must equal zero, and for ions, it should equal the ion's charge. This is why when we assign a -1 oxidation number to each chlorine atom in MnCl2, manganese must be assigned a +2 to balance the equation, reflecting the neutral charge of the compound. Similarly, in ions like MnO4-, we consider the charge of the ion (-1) and the expected oxidation state of oxygen (-2 typically) to calculate the oxidation state of manganese.

Note that these values do not represent true charges on the atoms but rather serve as a way to predict and understand the formation and breakage of bonds during chemical reactions. Regularly referring to the oxidation states can simplify the understanding of the chemical behavior of elements and compounds.
Redox Reactions
Redox reactions are a family of reactions that involve the transfer of electrons between chemical species. The term 'redox' is a combination of reduction (gain of electrons) and oxidation (loss of electrons). In a redox process, one reactant is oxidized and the other is reduced, essentially exchanging electrons and altering oxidation states.

In our textbook exercise, knowing the oxidation states is crucial for identifying redox reactions. The changes in oxidation numbers signal a redox event. For instance, when manganese (Mn) changes from an oxidation state of +7 in MnO4- to +6 in MnO42-, it indicates a reduction has occurred, and Mn has gained an electron. Conversely, if we were to see Mn going from +2 in MnCl2 to +3 in another compound, we would deduce it has been oxidized.

Identifying these redox couples is fundamental when balancing redox equations, predicting reaction outcomes, and even in understanding the fundamentals of electrochemistry where redox reactions are harnessed to produce electrical energy or enact chemical change through the application of an electrical current.
Chemical Nomenclature
Chemical nomenclature is the system for naming chemical compounds, providing a way to communicate the composition and structure of compounds unambiguously. It relies on a set of agreed-upon rules to determine the name of a compound based on its constituents and their quantities. For example, in MnCl2, nomenclature indicates it contains manganese and chlorine with a 1:2 ratio.

In our exercise, understanding how to name compounds and ions, like MnO4- being called permanganate and MnO42- being called manganate, aids in remembering and understanding their chemical behavior, including oxidation states. The nomenclature often gives a hint about the oxidation state and chemistry of the compound. For example, the '-ate' suffix in permanganate and manganate implies that manganese is in a higher oxidation state than it is in oxide forms, such as in MnO2 (manganese dioxide).

Mastering chemical nomenclature not only helps in organizational terms but also in predictive terms, as it is often possible to infer chemical properties and reactivity patterns from a compound's name. Therefore, chemical nomenclature is an essential tool for students to grasp the broad and detailed aspects of chemistry.

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Most popular questions from this chapter

One way to analyze a sample for nitrite ion is to acidify a solution containing \(\mathrm{NO}_{2}^{-}\) and then allow the \(\mathrm{HNO}_{2}\) that is formed to react with iodide ion in the presence of excess \(\mathrm{I}^{-}\). The reaction is $$ 2 \mathrm{HNO}_{2}+2 \mathrm{H}^{+}+3 \mathrm{I}^{-} \longrightarrow 2 \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{3}- $$ Then the \(\mathrm{I}_{3}^{-}\) is titrated with \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution using starch as an indicator. $$ \mathrm{I}_{3}^{-}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow 3 \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-} $$ In a typical analysis, a \(1.104 \mathrm{~g}\) sample that was known to contain \(\mathrm{NaNO}_{2}\) was treated as described above. The titration required \(29.25 \mathrm{~mL}\) of \(0.3000 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solu- tion to reach the end point. (a) How many moles of \(\mathrm{I}_{3}^{-}\) had been produced in the first reaction? (b) How many moles of \(\mathrm{NO}_{2}^{-}\) had been in the original \(1.104 \mathrm{~g}\) sample? (c) What was the percentage by mass of \(\mathrm{NaNO}_{2}\) in the original sample?

The other major air pollutant is \(\mathrm{NO}_{2}\). What are the oxidation numbers of the atoms in \(\mathrm{NO}_{2} ?\)

In each of the following pairs, choose the metal that would most likely react more rapidly with a nonoxidizing acid such as HCl: (a) aluminum or iron, (b) zinc or nickel, and (c) cadmium or magnesium.

Assign oxidation numbers to each atom in the following: (a) \(\mathrm{Na} \mathrm{OBr}\) (b) \(\mathrm{NaBrO}_{2}\), (c) \(\mathrm{NaBrO}_{3},\) and (d) \(\mathrm{NaBrO}_{4}\)

Use oxidation numbers to show that the fermentation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) to carbon dioxide and ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) is a redox reaction.

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