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Ozone, \(\mathrm{O}_{3},\) is an allotrope of oxygen and is one of the oxidants responsible for the haze over the Smoky Mountains. What is the oxidation number of the oxygen atoms in ozone?

Short Answer

Expert verified
The oxidation number of each oxygen atom in ozone ((O_3)) is 0.

Step by step solution

01

Understanding the concept of oxidation numbers

The oxidation number of an atom is a number that represents the total number of electrons that an atom either gains or loses when it forms chemical bonds. In a neutral molecule, the sum of the oxidation numbers of all atoms must equal zero.
02

Applying the oxidation number rule for molecules

For ozone ((O_3)), the molecule is neutral, and therefore, the sum of the oxidation numbers for all the oxygen atoms must be zero. Since ozone consists of three identical oxygen atoms that are bonded to each other, each oxygen atom must have the same oxidation number.
03

Determining the oxidation number of oxygen in ozone

To calculate the oxidation number of the oxygen atoms in ozone, we set up the equation: 3x = 0, where x is the oxidation number of each oxygen atom in the ozone molecule. Solving for x gives: x = 0 / 3 = 0. Therefore, the oxidation number of each oxygen atom in ozone is 0.

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Key Concepts

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

Ozone
Ozone, with the molecular formula \( \mathrm{O}_{3} \), is a fascinating molecule that plays a critical role in the Earth's atmosphere. It has a pale blue color and a distinctive sharp smell. Ozone forms a layer in the Earth's stratosphere, which serves as a vital shield, protecting us from the harmful ultraviolet radiation from the sun.

It is also present at ground level, where it can be a component of smog, and the molecule that creates the haze over the Smoky Mountains, as mentioned in the exercise. In the context of the oxidation state, each oxygen atom in the ozone molecule carries an oxidation number of 0. This matches with the fact that ozone is a neutral molecule which means that the total charge is balanced and equal to zero. Despite its simple composition, ozone is highly reactive due to the presence of a strained ring structure where three oxygen atoms form an unstable arrangement, making it a powerful oxidizing agent.
Allotropes of Oxygen
The term 'allotropes' refers to different physical forms in which an element can exist. Oxygen, a diatomic molecule \( \mathrm{O}_{2} \) in its most common form, also exists as ozone (\( \mathrm{O}_{3} \)). These allotropes display unique properties despite being made of the same element - oxygen.

The key difference between them lies in their molecular structure and bonding. While \( \mathrm{O}_{2} \) is composed of two oxygen atoms joined by a stable double bond, ozone is made of three oxygen atoms in a less stable, bent formation, which affects its chemical reactivity and physical properties. Knowledge of oxygen's allotropes is important in understanding environmental phenomena, such as the formation of the ozone layer and its depletion and the processes that lead to air pollution.
Chemical Bonds
Chemical bonds are the attractive forces that hold atoms together in a compound. They form when atoms share or transfer valence electrons to achieve a full outer shell, which is associated with stability in the atomic world. Fundamental types of chemical bonds include ionic, covalent, and metallic bonds.

In the case of ozone (\( \mathrm{O}_{3} \)), the atoms are bonded covalently, meaning they share electrons to stay together. Specifically, ozone has a resonance structure that distributes the single and double bonds among the three oxygen atoms in a dynamic equilibrium. This resonance contributes to the molecule's overall stability despite being a powerful oxidizing agent. The bonds within ozone are a perfect example of how molecular structure influences the properties and reactivity of a substance, demonstrating the core principles that govern chemical interactions.

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Most popular questions from this chapter

Define combustion, rusting, and tarnishing.

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) \(6 \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{Al} \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{SO}_{2}+6 \mathrm{H}_{2} \mathrm{O}\)

Zirconia, which is \(\mathrm{Zr} \mathrm{O}_{2}\), is used to make ceramic knives. What are the oxidation states of zirconium and oxygen in zirconia?

The following equation is not balanced. $$ \mathrm{Ag}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Ag}^{+}+\mathrm{Fe} $$ Why? Use the ion-electron method to balance it. Can you balance this reaction by inspection?

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?

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