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Assign oxidation numbers to each element in (a) \(\mathrm{P}_{2} \mathrm{O}_{5}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{CO}_{3}{ }^{2-}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (e) \(\mathrm{N}_{2} \mathrm{H}_{4}\)

Short Answer

Expert verified
Answer: The oxidation numbers are as follows: \(P\) in \(\mathrm{P}_{2} \mathrm{O}_{5}\) is +5, \(N\) in \(\mathrm{NH}_{3}\) is -3, \(H\) in \(\mathrm{NH}_{3}\) is +1, \(C\) in \(\mathrm{CO}_{3}^{2-}\) is +4, \(O\) in \(\mathrm{CO}_{3}^{2-}\) is -2, \(S\) in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is +2, \(O\) in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is -2, \(N\) in \(\mathrm{N}_{2} \mathrm{H}_{4}\) is -2, and \(H\) in \(\mathrm{N}_{2} \mathrm{H}_{4}\) is +1.

Step by step solution

01

Assign oxidation numbers to P in \(\mathrm{P}_{2} \mathrm{O}_{5}\)

The oxidation number of oxygen is usually -2. In \(\mathrm{P}_{2} \mathrm{O}_{5}\), there are 5 oxygen atoms, giving a total oxidation number of -10 for all oxygens. Since the compound is neutral, the 2 phosphorus atoms must have a combined oxidation number of +10. Therefore, the oxidation number of P in \(\mathrm{P}_{2} \mathrm{O}_{5}\) is +5.
02

Assign oxidation numbers to N and H in \(\mathrm{NH}_{3}\)

The oxidation number of hydrogen is usually +1. In \(\mathrm{NH}_{3}\), there are 3 hydrogen atoms, giving a total oxidation number of +3 for all hydrogens. Since the molecule is neutral, the nitrogen atom must have an oxidation number of -3 to balance out the total charge. So, the oxidation number for N is -3 in \(\mathrm{NH}_{3}\).
03

Assign oxidation numbers to C and O in \(\mathrm{CO}_{3}^{2-}\)

The oxidation number of oxygen is -2. In \(\mathrm{CO}_{3}^{2-}\), there are 3 oxygen atoms, giving a total oxidation number of -6 for all oxygens. Since the ion has a 2- charge, the carbon atom must have an oxidation number of +4 to balance the charges. Hence, the oxidation number for C in \(\mathrm{CO}_{3}^{2-}\) is +4.
04

Assign oxidation numbers to S and O in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\)

The oxidation number of oxygen is -2. In \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), there are 3 oxygen atoms, giving a total oxidation number of -6 for all oxygens. Since the ion has a 2- charge, the 2 sulfur atoms must have a combined oxidation number of +4 to balance the charges. Therefore, the oxidation number for S in \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is +2.
05

Assign oxidation numbers to N and H in \(\mathrm{N}_{2} \mathrm{H}_{4}\)

The oxidation number of hydrogen is usually +1. In \(\mathrm{N}_{2} \mathrm{H}_{4}\), there are 4 hydrogen atoms, giving a total oxidation number of +4 for all hydrogens. Since the molecule is neutral, the 2 nitrogen atoms must have a combined oxidation number of -4. Thus, the oxidation number of N in \(\mathrm{N}_{2} \mathrm{H}_{4}\) is -2.

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Key Concepts

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

Chemical Bonding
Chemical bonding is the force that holds atoms together in molecules and compounds, forming the basis of all substances. There are three primary types of chemical bonds: ionic, covalent, and metallic.

Ionic bonds form when electrons are transferred from one atom to another, leading to the creation of ions with opposite charges that attract each other. For example, sodium and chlorine atoms form an ionic bond in table salt, with sodium losing an electron and chlorine gaining one.

Covalent bonds occur when two atoms share pairs of electrons, generating molecules with shared electron clouds. Oxygen molecules (O2), for instance, consist of two oxygen atoms sharing a pair of electrons through a covalent bond.

Metallic bonds are found in metals, where electrons are free to move around many positively charged metal ions, thus providing metals with their characteristic properties like conductivity and malleability.

Understanding chemical bonding is crucial when assigning oxidation numbers because the type of bond influences the distribution of electrons among atoms, consequently affecting their oxidation states.
Redox Reactions
Redox reactions, or oxidation-reduction reactions, are processes where electrons are transferred between atoms, leading to changes in oxidation states. These reactions encompass two half-reactions: oxidation, where an atom, ion, or molecule loses electrons, and reduction, where it gains electrons.

The concept of redox reactions is vital in chemistry because it underpins various chemical processes from metabolic paths in the biological domain to industrial applications like battery operation. In the context of practicing problems or analyzing substances, oxidation states help determine which atom is oxidized and which is reduced.

For example, in a reaction where iron rust forms (Fe2O3), iron (Fe) is oxidized as it loses electrons, while oxygen gains them and is reduced. By identifying the changes in oxidation states, students can comprehend the flow of electrons and the transformation of matter in a reaction.
Oxidation States
Oxidation states, also known as oxidation numbers, are values assigned to atoms in molecules and ions to describe the degree of oxidation or reduction they undergo. These numerical designations provide insight into the transfer of electrons in chemical reactions, especially redox reactions.

Oxidation states are determined based on a set of rules: elements in their elemental form have an oxidation number of 0, and in most cases, the sum of oxidation numbers in a neutral compound or ion must equate to its overall charge.

For instance, in water (H2O), oxygen has an oxidation state of -2, while hydrogen has +1. Hence, oxidation numbers play a pivotal role in balancing redox reactions and understanding chemical bonding, as they reflect how electrons are distributed among atoms—crucial for predicting compound behavior and reaction outcomes.

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Most popular questions from this chapter

A solution contains both iron(II) and iron(III) ions. A \(50.00-\mathrm{mL}\) sample of the solution is titrated with \(35.0 \mathrm{~mL}\) of \(0.0280 \mathrm{M} \mathrm{KMnO}_{4}\), which oxidizes \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+} .\) The permanganate ion is reduced to manganese(II) ion. Another \(50.00-\mathrm{mL}\) sample of the solution is treated with zinc, which reduces all the \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\). The resulting solution is again titrated with \(0.0280 \mathrm{M}\) \(\mathrm{KMnO}_{4} ;\) this time \(48.0 \mathrm{~mL}\) is required. What are the concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the solution?

Write a balanced net ionic equation for each of the following acid-base reactions in water. (a) acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) with strontium hydroxide (b) diethylamine, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\), with sulfuric acid (c) aqueous hydrogen cyanide (HCN) with sodium hydroxide

How would you prepare from the solid and pure water (a) \(0.400 \mathrm{~L}\) of \(0.155 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2} ?\) (b) \(1.75 \mathrm{~L}\) of \(0.333 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} ?\)

A capsule of vitamin \(\mathrm{C}\), a weak acid, is analyzed by titrating it with \(0.425 M\) sodium hydroxide. It is found that \(6.20 \mathrm{~mL}\) of base is required to react with a capsule weighing \(0.628 \mathrm{~g}\). What is the percentage of vitamin \(\mathrm{C}\) \(\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\) in the capsule? (One mole of vitamin \(\mathrm{C}\) reacts with one mole of hydroxide ion.)

Using circles to represent cations and squares to represent anions, show pictorially the reactions that occur between aqueous solutions of (a) \(\mathrm{Ba}^{2+}\) and \(\mathrm{OH}^{-}\) (b) \(\mathrm{Co}^{3+}\) and \(\mathrm{PO}_{4}^{3-}\)

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