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

Short Answer

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The oxidation number for all species \(\mathrm{H}_{2}, \mathrm{Se}_{g}, \mathrm{P}_{4}, \mathrm{O}, \mathrm{U}, \mathrm{As}_{4}, \mathrm{B}_{12}\) is 0.

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01

Understanding Oxidation Numbers

The oxidation number (or oxidation state) indicates the degree of oxidation of an atom in a chemical compound. For free elements, whether they are monoatomic or polyatomic, the oxidation number is always zero.
02

Identifying the Species as Free Elements

The species given are \(\mathrm{H}_2\), \(\mathrm{Se}_g\), \(\mathrm{P}_4\), \(\mathrm{O}\), \(\mathrm{U}\), \(\mathrm{As}_4\), and \(\mathrm{B}_{12}\). Each of these species is a pure element in its elemental form, not combined with other elements. This means they are all free elements.
03

Assigning Oxidation Numbers to Free Elements

Since each of these species is a free element, the oxidation number for each of them is zero. This rule applies universally to all free elements, regardless of the number of atoms in the molecule for diatomic, polyatomic or single atoms.

Key Concepts

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

Free Elements
In chemistry, the term "free elements" refers to elements that are found in their pure, uncombined form. Importantly, free elements can be either single atoms or consist of multiple atoms bonded together, like \(\mathrm{H}_2\) or \(\mathrm{P}_4\). Despite their molecular complexity, the defining characteristic of free elements is that they do not combine with atoms of a different element. For example:
  • \(\mathrm{H}_2\) represents hydrogen gas made entirely of hydrogen atoms.
  • \(\mathrm{U}\) is simply pure uranium.
  • \(\mathrm{As}_4\) is elemental arsenic in its natural form.
Free elements play a foundational role in chemical reactions and provide a basis for understanding how elements combine in compounds. One of the unchanging rules in chemistry is that the oxidation number of any free element is zero. This is because there is no gain or loss of electrons as the atoms are not in a compound. Remember, this rule holds true no matter how many atoms are in a free element molecule.
Chemical Compounds
Chemical compounds are substances formed when two or more elements are chemically bonded together. These bonds can be covalent or ionic, depending on the nature of the atoms involved and how they share or transfer electrons. When elements bond to create a compound, the oxidation number comes into play, as it helps us understand how electrons are distributed among the atoms. Unlike free elements, the oxidation numbers in a compound can vary depending on the type and number of atoms involved.To differentiate, consider these points:
  • A compound like water, \(\mathrm{H}_2\mathrm{O}\), is not a free element, but a chemical compound with hydrogen and oxygen atoms bonded together.
  • The oxidation number for hydrogen in \(\mathrm{H}_2\mathrm{O}\) is +1, while for oxygen it is -2, reflecting their electron-sharing behavior in the compound.
Understanding these differences is crucial, as compounds exhibit a richness of behaviors and properties that are distinct from their constituent elements. This makes them fundamental to the study of chemistry and its applications.
Oxidation State
The oxidation state, also known as the oxidation number, is a concept that describes the degree of oxidation of an atom within a molecule. It indicates how many electrons an atom gains, loses, or appears to use when forming compounds. This helps chemists determine how atoms interact within a compound and predict possible reactions.Some key points include:
  • The oxidation state of a free element, as discussed earlier, is always zero due to no electron transfer occurring.
  • In a chemical compound, elements have oxidation states that reflect their bonding and electron interaction. For instance, in \(\mathrm{NaCl}\), sodium has an oxidation state of +1 and chlorine -1, illustrating their electron transfer relationship.
  • Oxidation states can also guide the balancing of chemical equations, aiding in the understanding of how atoms combine and rearrange during reactions.
Overall, understanding oxidation states is essential for any study of redox reactions, where electrons are transferred between species. Through mastering the concept of oxidation states, you can gain a deeper understanding of a broad range of chemical processes.

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Most popular questions from this chapter

Arrange the following species in order of increasing oxidation number of the sulfur atom: (a) \(\mathrm{H}_{2} \mathrm{~S},(\mathrm{~b}) \mathrm{S}_{8}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{S}^{2-}\) (e) HS \(^{-}\), (f) \(\mathrm{SO}_{2},(\mathrm{~g}) \mathrm{SO}_{3}\)

Determine which of the following metals can react with acid: (a) \(\mathrm{Au},(\mathrm{b}) \mathrm{Ni},(\mathrm{c}) \mathrm{Zn},(\mathrm{d}) \mathrm{Ag},(\mathrm{e}) \mathrm{Pt}\).

Describe in each case how you would separate the cations or anions in the following aqueous solutions: (a) \(\mathrm{NaNO}_{3}\) and \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{K} \mathrm{NO}_{3},\) (c) \(\mathrm{KBr}\) and \(\mathrm{KNO}_{3},\) (d) \(\mathrm{K}_{3} \mathrm{PO}_{4}\) and \(\mathrm{KNO}_{3},\) (e) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaNO}_{3}\)

Determine how many grams of each of the following solutes would be needed to make \(2.50 \times 10^{2} \mathrm{~mL}\) of a \(0.100-M\) solution: (a) cesium iodide (CsI), (b) sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right),(\mathrm{c})\) sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right),\) (d) potassium dichromate \(\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right),\) (e) potassium permanganate \(\left(\mathrm{KMnO}_{A}\right)\)

Water is added to \(25.0 \mathrm{~mL}\) of a \(0.866 \mathrm{M} \mathrm{KNO}_{3}\) solution until the volume of the solution is exactly \(500 \mathrm{~mL}\). What is the concentration of the final solution?

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