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What is the oxidation state of the atoms in an uncombined element? Does it depend on whether the element occurs as a diatomic molecule \(\left(\mathrm{O}_{2}, \mathrm{N}_{2}\right)\) or as a larger molecule \(\left(\mathrm{P}_{4}, \mathrm{S}_{8}\right) ?\)

Short Answer

Expert verified
The oxidation state of the atoms in an uncombined element is zero, and it does not depend on whether the element occurs as a diatomic molecule (O2, N2) or as a larger molecule (P4, S8).

Step by step solution

01

Define Oxidation State

Oxidation state or oxidation number is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. It is used to keep track of the electrons transferred during a chemical reaction.
02

Understand Uncombined Elements

An uncombined element is a pure element that is not bonded to any other element. In this case, we are considering the following uncombined elements: 1. Diatomic molecules: O2 (Oxygen) and N2 (Nitrogen) 2. Larger molecules: P4 (Phosphorus) and S8 (Sulfur)
03

Determine Oxidation State

In an uncombined element, all the atoms are of the same element, and hence there is no transfer of electrons. Therefore, the oxidation state of all the atoms in an uncombined element is zero, regardless of whether it occurs as a diatomic molecule or a larger molecule. So, the oxidation state of the atoms in O2, N2, P4, and S8 is 0.
04

Conclusion

The oxidation state of the atoms in an uncombined element is zero, and it does not depend on whether the element occurs as a diatomic molecule (O2, N2) or as a larger molecule (P4, S8).

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Key Concepts

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

Oxidation Number
Understanding the oxidation number concept is crucial for studying chemistry, particularly when dealing with chemical reactions. Think of the oxidation number as a tag assigned to atoms that helps us track how electrons are distributed during reactions. This 'tag' is a hypothetical charge that an atom would possess if all its bonds to different elements were entirely ionic in nature.

In simpler terms, an oxidation number indicates whether an atom has a positive, negative, or neutral charge assuming that the electrons in its bonds are pulled completely towards the more electronegative element. For instance, in water (H2O), hydrogen has an oxidation number of +1, while oxygen has an oxidation number of -2, which corresponds to how electrons are shared in reality.

To clarify, when dealing with an uncombined element—simply put, an element in its purest form, like nitrogen in N2—each atom technically doesn't gain or lose any electrons to another element. Therefore, each atom's oxidation number is zero. This rule applies no matter the complexity of the molecule, be it diatomic, like oxygen in O2, or a larger molecular structure like phosphorus in P4.
Chemical Reactions
Chemical reactions are the heart of chemistry. They involve the transformation of substances through the breaking and forming of chemical bonds. Throughout these processes, atoms are rearranged, and energy is exchanged, leading to the creation of new substances with different properties from the starting materials.

Each chemical reaction involves a change in oxidation states as atoms lose or gain electrons. Consider the simple synthesis of water: hydrogen reacts with oxygen to form H2O. In this reaction, hydrogen, which starts with an oxidation number of 0, ends up with a +1 oxidation number, and oxygen goes from 0 to -2. This change effortlessly illustrates the concept of electron transfer and highlights how oxidation numbers can help trace such changes.

An important thing to remember is that in any chemical reaction, the number of electrons lost is always equal to the number of electrons gained, maintaining an overall electrical balance. This fundamental principle is the basis for the concept of conservation of mass in chemical equations.
Diatomic Molecules
Diatomic molecules are composed of two atoms, which may be of the same or different chemical elements. The most familiar diatomic molecules are those composed of the same element, such as O2 (oxygen) or N2 (nitrogen)—these are also key players in our atmosphere.

The diatomic nature of these molecules has significant implications for their oxidation states. In O2 or N2, the two atoms share electrons evenly because they are of the same element, hence possessing equal electronegativity. This equal sharing means that no atom gains or loses electrons when we talk about oxidation states in the context of reactions. As a result, each atom in a diatomic molecule like O2 or N2 has an oxidation state of zero. This is a fundamental rule that is true for all elemental molecules—whether diatomic or polyatomic—so an element like chlorine would also follow this rule in its diatomic Cl2 form.
Electron Transfer
Electron transfer is the foundation of many types of chemical reactions, especially oxidation-reduction (redox) reactions. Electrons are the currency of chemical reactions, and the transfer of electrons from one atom to another is what drives the formation of new substances. Oxidation involves the loss of electrons, while reduction entails the gain of electrons.

But how is the concept of electron transfer relevant to uncombined elements? Actually, in such cases, there's a little twist—since the atoms in an uncombined element, such as S8 or P4, are the same, there is no net transfer of electrons among them. The electrons are perfectly balanced, and no oxidation or reduction is occurring, leaving each atom with an oxidation number of zero.

In the broader context of redox reactions, keeping track of electron transfer using oxidation numbers becomes a powerful tool. It allows us to systematically understand the intricacies of reactions and predict the behavior of chemicals during those reactions. This also ties back to the conservation of charge, as the total number of electrons lost in oxidation must equal the number gained during reduction.

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Most popular questions from this chapter

At which electrode (anode/cathode) do species gain electrons in a galvanic cell?

In each of the following reactions, identify which element is oxidized and which is reduced by assigning oxidation numbers. a. \(2 \mathrm{HNO}_{3}(a q)+3 \mathrm{H}_{2} \mathrm{S}(g) \rightarrow\) \(2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+3 \mathrm{S}(s)\) b. \(2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) c. \(2 \mathrm{ZnS}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{SO}_{2}(g)\) d. \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{HCl}(g)\)

What is an oxidizing agent? Is an oxidizing agent itself oxidized or reduced when it acts on another species?

Why is fluorine always assigned an oxidation state of \(-1 ?\) What oxidation number is \(u\) sually assigned to the other halogen elements when they occur in compounds? In an interhalogen compound involving fluorine (such as ClF), which atom has a negative oxidation state?

For each of the following oxidation-reduction reactions of metals with nonmetals, identify which element is oxidized and which is reduced. a. \(4 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Na}_{2} \mathrm{O}(s)\) b. \(\operatorname{Fe}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{FeSO}_{4}(a q)+\mathrm{H}_{2}(g)\) c. \(2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \rightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g)\) d. \(3 \mathrm{Mg}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Mg}_{3} \mathrm{N}_{2}(s)\)

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