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Chapter 4: Problem 48

Give the oxidation number for the following species: \(\mathrm{H}_{2}, \mathrm{Se}_{8}, \mathrm{P}_{4}, \mathrm{O}, \mathrm{U}, \mathrm{As}_{4}, \mathrm{~B}_{12}\)

Short Answer

Expert verified
The oxidation numbers of \(\mathrm{H}_{2}, \mathrm{Se}_{8}, \mathrm{P}_{4}, \mathrm{O}, \mathrm{U}, \mathrm{As}_{4}, \mathrm{B}_{12}\) are all zero.

Step by step solution

01

Identifying the forms of elements

Recognize that all the given species are in their elemental form, which means they are not combined with other elements.
02

Determining the oxidation number

In chemistry, the rule is that the oxidation state of a free element, in its uncombined state, is always zero. This is the case for all the given species: \(\mathrm{H}_{2}, \mathrm{Se}_{8}, \mathrm{P}_{4}, \mathrm{O}, \mathrm{U}, \mathrm{As}_{4}, \mathrm{B}_{12}\).
03

Writing the results

We therefore determine that each of these species has an oxidation number of zero.

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Key Concepts

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

Elemental Forms
Elements in their purest form are simply known as "elemental forms." This means they exist as single elements and are not bonded with any other different elements. For example, when we see substances like \(\mathrm{H}_{2}\), \(\mathrm{Se}_{8}\), or \(\mathrm{P}_{4}\), these are all examples of elements in their elemental forms. They are composed entirely of one type of element.
In chemistry, knowing that a substance is in its elemental form is crucial for understanding its chemical properties and reactions, since they act quite differently from when they are part of compounds.
For instance:
  • Hydrogen gas (\(\mathrm{H}_{2}\)) is just hydrogen in its natural diatomic form.
  • Selenium can exist as \(\mathrm{Se}_{8}\), forming an octatomic structure common to selenium in its uncombined form.
  • Phosphorus regularly forms tetratomic molecules represented by \(\mathrm{P}_{4}\) in its natural state.
Free Elements
"Free elements" refer to elements that exist in their pure, uncombined state, often as single atoms or molecules made up solely of the one type of element. They have not reacted with other elements to form compounds. For example, the noble gases like neon and argon are frequently found as free elements because they are stable and generally do not bond with other elements.
To identify free elements in practice:
  • Take iron (\(\mathrm{Fe}\)) found naturally as pure metallic iron or platinum found as a pure elemental metal.
  • Carbon is often encountered in its free elemental forms as diamond or graphite.
Free elements are crucial in chemistry because they are often starting materials for chemical reactions. Understanding their free state is essential for adjusting chemical reactants and predicting reaction outcomes.
Uncombined State
When we talk about elements in their "uncombined state," it is very similar to discussing their "elemental forms" or them being "free." An element in its uncombined state is not chemically bonded with any other kinds of atoms or compounds. This often means the substance is in a stable and natural form, ready to participate in chemical reactions.
In this state, you can interpret these elements as being just themselves without any external influence altering their basic structure or behavior:
  • Uranium (\(\mathrm{U}\)) often exists in nature as a metal in its uncombined, stable state before processing or usage.
  • Uncombined boron can be seen in nature as a crystalline solid, found as \(\mathrm{B}_{12}\).
Oxidation State Rules
Understanding oxidation state rules is foundational in chemistry. One key rule is that any free element, when it is in its uncombined state, always has an oxidation number of zero. This helps chemists easily understand and balance chemical equations.
Whenever you encounter an element, like \(\mathrm{O}\) or \(\mathrm{As}_{4}\), which is in its pure elemental form, you automatically recognize its oxidation number as zero, aiding in various calculations and chemical analysis.
  • For instance, the oxidation number for oxygen (as a pure element, \(\mathrm{O}_{2}\)) is zero.
  • Similarly, for semimetals like arsenic, found in a tetrahedral cluster as \(\mathrm{As}_{4}\), their oxidation state is zero.
This simple rule not only simplifies the process of determining oxidation numbers but also provides the basis for further chemical understanding and application.

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Most popular questions from this chapter

Describe how to prepare \(1.00 \mathrm{~L}\) of \(0.646 \mathrm{M} \mathrm{HCl}\) solution, starting with a \(2.00 M \mathrm{HCl}\) solution.

Barium hydroxide, often used to titrate weak organic acids, is obtained as the octahydrate, \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\). What mass of \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) would be required to make \(500.0 \mathrm{~mL}\) of a solution that is \(0.1500 \mathrm{M}\) in hydroxide ions?

Hydrogen halides (HF, HCl, HBr, HI) are highly reactive compounds that have many industrial and laboratory uses. (a) In the laboratory, HF and \(\mathrm{HCl}\) can be generated by reacting \(\mathrm{CaF}_{2}\) and \(\mathrm{NaCl}\) with concentrated sulfuric acid. Write appropriate equations for the reactions. (Hint: These are not redox reactions.) (b) Why is it that HBr and HI cannot be prepared similarly-that is, by reacting NaBr and NaI with concentrated sulfuric acid? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger oxidizing agent than both \(\mathrm{Br}_{2}\) and \(\mathrm{I}_{2} .\) ) (c) HBr can be prepared by reacting phosphorus tribromide \(\left(\mathrm{PBr}_{3}\right)\) with water. Write an equation for this reaction.

Explain how you would prepare potassium iodide (KI) by means of (a) an acid- base reaction and (b) a reaction between an acid and a carbonate compound.

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. If \(24.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{KMnO}_{4}\) solution is needed to titrate \(1.00 \mathrm{~g}\) of a sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample? The net ionic equation is \(2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow\) \(2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\)
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