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Order the following molecules from lowest to highest oxidation state of the nitrogen atom: \(\mathrm{HNO}_{3}, \mathrm{NH}_{4} \mathrm{Cl}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\), \(\mathrm{NaNO}_{2}\)

Short Answer

Expert verified
The molecules can be ordered from lowest to highest oxidation state of the nitrogen atom as follows: \( NH_4Cl (-3) < N_2O (+1) < NaNO_2 (+3) < NO_2 (+4) < HNO_3 (+5) \).

Step by step solution

01

Determine the oxidation state of nitrogen in each molecule

For each molecule, find the oxidation state of nitrogen by using the rules of oxidation states. Here are some rules we can use: - The oxidation state of an atom in a pure element is zero, e.g., N2 has nitrogen with an oxidation state of 0. - The oxidation state of hydrogen is usually +1, and the oxidation state of oxygen is usually -2. - The algebraic sum of the oxidation states of all atoms in a molecule must equal the overall charge of the molecule.
02

Calculate the oxidation state of nitrogen in HNO3

In HNO3, we have one hydrogen atom (+1), one nitrogen atom, and three oxygen atoms (-2 each). Applying the rules from step 1: 1 + x - 6 = 0, so the oxidation state of nitrogen in HNO3 is x = +5.
03

Calculate the oxidation state of nitrogen in NH4Cl

In NH4Cl, we can directly look at the oxidation state of nitrogen in NH4+ (the ammonium ion). There are four hydrogen atoms (+1 each) and one nitrogen atom, so 4(1) + x = +1. Thus, the oxidation state of nitrogen in NH4Cl is x = -3.
04

Calculate the oxidation state of nitrogen in N2O

In N2O, there are two nitrogen atoms and one oxygen atom (-2). Applying the rules: 2x - 2 = 0. Thus, the average oxidation state of nitrogen in N2O is x = +1. Note that the individual nitrogen atoms in N2O have different oxidation states, but we're comparing the average oxidation state of nitrogen in these molecules.
05

Calculate the oxidation state of nitrogen in NO2

In NO2, there is one nitrogen atom and two oxygen atoms (-2 each). Applying the rules: x - 4 = 0. Therefore, the oxidation state of nitrogen in NO2 is x = +4.
06

Calculate the oxidation state of nitrogen in NaNO2

NaNO2 is an ionic compound, so we can look at the oxidation state of nitrogen in the NO2- ion. There is one nitrogen atom and two oxygen atoms (-2 each): x - 4 = -1. Thus, the oxidation state of nitrogen in NaNO2 is x = +3.
07

Order the molecules from lowest to highest oxidation state of the nitrogen atom

Now that we have the oxidation state of nitrogen in each molecule, we can arrange them in order from lowest to highest: NH4Cl (-3) < N2O (+1) < NaNO2 (+3) < NO2 (+4) < HNO3 (+5).

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Key Concepts

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

Determining Oxidation Numbers
Understanding how to determine oxidation numbers is essential in chemistry, particularly when dealing with reactions involving electron transfer. Oxidation numbers indicate the degree of oxidation of an atom within a compound, resembling the number of electrons an atom has gained, lost, or shared at a given time.

To find the oxidation number of an atom, one must follow a set of rules: first, the oxidation state of any elemental atom is zero, and for a monoatomic ion, it equals the ion's charge. Molecules and polyatomic ions have a net oxidation state equal to their charge. Hydrogen is generally assigned an oxidation state of +1, oxygen -2, and other elements follow their respective group trends unless they're in a peroxide or have other unusual bonding.

By systematically applying these rules, as shown in the exercise with nitrogen compounds, students can calculate the oxidation state of any element within a molecule. Taking the oxidation states into account helps to balance chemical equations, particularly in the context of redox reactions.
Redox Chemistry
Redox chemistry, also known as oxidation-reduction chemistry, is the study of reactions in which atoms have their oxidation states changed. This can occur through the transfer of electrons between chemical species, or through covalent bonding where electron sharing results in changes to oxidation numbers.

In a redox reaction, one species will lose electrons and become oxidized while the other gains electrons and becomes reduced. Hence the name redox, combining 'reduction' and 'oxidation'. These reactions are fundamental to many processes including metabolism, combustion, and corrosion. Determining the oxidation numbers, as in the given exercise with nitrogen compounds, is the first step to understanding and balancing redox reactions.
Chemical Bonding
Chemical bonding is the force that holds atoms together in molecules and compounds. There are three primary types of chemical bonds: ionic, covalent, and metallic. Ionic bonding occurs when there is a complete transfer of electrons from one atom to another, leading to the formation of positively and negatively charged ions.

Covalent bonding involves the sharing of electrons between atoms, often leading to a more equal distribution of charge. This can affect oxidation states as shared electrons are assigned to the more electronegative atom in the bond. Determining how electrons are shared or transferred in chemical bonds is crucial for establishing the correct oxidation numbers, contributing to an understanding of the compound's properties and behavior.
Inorganic Compounds
Inorganic compounds are substances that do not contain carbon-hydrogen bonds, which are characteristic of organic compounds. Examples include salts, metals, and substances like water and nitrogen oxides. The study of inorganic compounds includes understanding their physical and chemical properties, which are influenced by the types of atoms they contain and the bonds that hold those atoms together.

Determining the oxidation numbers in inorganic compounds, such as the ones with nitrogen in the exercise, helps us predict their reactivity, especially in redox reactions. Many inorganic compounds play key roles in industry and technology, where their redox behavior is often exploited - for example, nitrogen oxides in the production of fertilizers or as indicator substances in various chemical processes.

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Most popular questions from this chapter

Three students were asked to find the identity of the metal in a particular sulfate salt. They dissolved a \(0.1472-\mathrm{g}\) sample of the salt in water and treated it with excess barium chloride, resulting in the precipitation of barium sulfate. After the precipitate had been filtered and dried, it weighed \(0.2327 \mathrm{~g}\). Each student analyzed the data independently and came to different conclusions. Pat decided that the metal was titanium. Chris thought it was sodium. Randy reported that it was gallium. What formula did each student assign to the sulfate salt? Look for information on the sulfates of gallium, sodium, and titanium in this text and reference books such as the \(C R C\) Handbook of Chemistry and Physics. What further tests would you suggest to determine which student is most likely correct?

Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. \(\operatorname{copper}(\) II \()\) sulfate \((a q)+\) iron \((s) \longrightarrow\) \(\operatorname{copper}(s)+\) iron(II) sulfate \((a q)\) copper(II) sulfate \((a q)+\operatorname{iron}(s) \longrightarrow\) \(\operatorname{copper}(s)+\) iron(III) sulfate \((a q)\) You place \(87.7 \mathrm{~mL}\) of a \(0.500-M\) solution of copper(II) sulfate in a beaker. You then add \(2.00 \mathrm{~g}\) of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate \(2.27 \mathrm{~g}\) of copper. Which equation above describes the reaction that occurred? Support your answer.

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid-base reactions. a. \(\mathrm{HNO}_{3}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) b. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{KOH}(a q) \rightarrow\) c. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{HCl}(a q) \rightarrow\)

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of \(0.5894 \mathrm{~g}\) were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of \(\mathrm{BaSO}_{4}\) obtained was \(0.5032 \mathrm{~g}\). What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?

How would you prepare \(1.00 \mathrm{~L}\) of a \(0.50-M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" ( \(18 \mathrm{M}\) ) sulfuric acid b. HCl from "concentrated" (12 \(M\) ) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO \(_{3}\) from "concentrated" ( \(16 M\) ) reagent e. Sodium carbonate from the pure solid

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