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Name the following compounds and assign oxidation states to the halogens in them: (a) \(\mathrm{Fe}\left(\mathrm{ClO}_{3}\right)_{3},(\mathbf{b}) \mathrm{HClO}_{2}\) (c) \(\mathrm{XeF}_{6},(\mathbf{d}) \mathrm{BrF}_{5},(\mathbf{e}) \mathrm{XeOF}_{4},(\mathbf{f}) \mathrm{HIO}_{3}\)

Short Answer

Expert verified
The names of the compounds and the oxidation states of the halogens are as follows: a) Iron(III) chlorate, Cl oxidation state: +5 b) Chlorous acid, Cl oxidation state: +3 c) Xenon hexafluoride, F oxidation state: -1 d) Bromine pentafluoride, F oxidation state: -1 e) Xenon oxytetrafluoride, F oxidation state: -1 f) Iodic acid, I oxidation state: +5

Step by step solution

01

a) Naming Fe(ClO3)3 and finding the oxidation state of Cl

The compound \(\mathrm{Fe}\left(\mathrm{ClO}_{3}\right)_{3}\) is an ionic compound consisting of an iron(III) cation (Fe) and chlorate anions (ClO3). The name of this compound is iron(III) chlorate. To find the oxidation state of chlorine in this compound, we can use the rule that the sum of the oxidation states of all the atoms in the compound must be zero. Since oxygen has an oxidation state of -2 and iron has an oxidation state of +3, the equation for the sum of the oxidation states is: \( +3 + 3x + 9(-2) = 0 \) Solving for x, which is the oxidation state of Cl, we get \( x= +5 \) So, the oxidation state of chlorine in \(\mathrm{Fe}\left(\mathrm{ClO}_{3}\right)_{3}\) is +5.
02

b) Naming HClO2 and finding the oxidation state of Cl

The compound \(\mathrm{HClO}_{2}\) is a molecular compound consisting of hydrogen, chlorine, and oxygen. It is called chlorous acid. To find the oxidation state of Cl in this compound, we use the rule that the sum of the oxidation states of all the atoms in the compound must be zero. Hydrogen has an oxidation state of +1, and oxygen has an oxidation state of -2. The equation for the sum of the oxidation states is: \( +1 + x + 2(-2) = 0 \) Solving for x, we get \( x= +3 \) So, the oxidation state of chlorine in \(\mathrm{HClO}_{2}\) is +3.
03

c) Naming XeF6 and finding the oxidation state of F

The compound \(\mathrm{XeF}_{6}\) is a molecular compound consisting of xenon and fluorine. It is called xenon hexafluoride. The oxidation state of F in any compound is always -1. Therefore, the oxidation state of F in \(\mathrm{XeF}_{6}\) is -1.
04

d) Naming BrF5 and finding the oxidation state of F

The compound \(\mathrm{BrF}_{5}\) is a molecular compound consisting of bromine and fluorine. It is called bromine pentafluoride. As mentioned earlier, the oxidation state of F in any compound is always -1. Therefore, the oxidation state of F in \(\mathrm{BrF}_{5}\) is -1.
05

e) Naming XeOF4 and finding the oxidation state of F

The compound \(\mathrm{XeOF}_{4}\) is a molecular compound consisting of xenon, oxygen, and fluorine. It is called xenon oxytetrafluoride. The oxidation state of F in any compound is always -1. Therefore, the oxidation state of F in \(\mathrm{XeOF}_{4}\) is -1.
06

f) Naming HIO3 and finding the oxidation state of I

The compound \(\mathrm{HIO}_{3}\) is a molecular compound consisting of hydrogen, iodine, and oxygen. It is called iodic acid. To find the oxidation state of I in this compound, we use the rule that the sum of the oxidation states of all the atoms in the compound must be zero. Hydrogen has an oxidation state of +1, and oxygen has an oxidation state of -2. The equation for the sum of the oxidation states is: \( +1 + x + 3(-2) = 0 \) Solving for x, we get \( x= +5 \) So, the oxidation state of iodine in \(\mathrm{HIO}_{3}\) is +5.

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Key Concepts

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

Naming Compounds
Naming chemical compounds is essential to communicate their structure and composition clearly. For ionic compounds, the name typically reflects the cation followed by the anion. For example, in Fe(ClO3)3, the cation is iron(III) reflecting its +3 oxidation state, and the anion is chlorate, thus, the compound is named iron(III) chlorate.
Molecular compounds, composed of non-metals, use prefixes like mono-, di-, tri- to indicate the number of atoms. An example is HClO2, known as chlorous acid, where the suffix 'ous' signifies the lower oxidation state of chlorine, compared to chloric acid (HClO3).
Many compounds also derive their names from systematic IUPAC naming conventions, which provides consistency across international contexts. Understanding these naming conventions is vital when referencing various chemical literature.
Halogen Chemistry
Halogens are highly reactive non-metals found in Group 17 of the periodic table. They include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Each has unique properties but generally forms salts (halides) with metals and covalent bonds with non-metals.
Their high reactivity is due to having seven valence electrons, so they often gain an electron to achieve a stable octet configuration, resulting in a -1 oxidation state. However, in oxyanions like ClO3^- or ClO2^-, they can exhibit positive oxidation states.
It's crucial to grasp the versatility of halogen oxidation states, which contributes significantly to their reactivity and compounds' diversity, such as in bromine pentafluoride (BrF5) or iodic acid (HIO3).
Chemical Compounds
Chemical compounds form when two or more different elements are chemically bonded together. They are classified as ionic or molecular based on the nature of the bond.
Chemical compounds exhibit distinct physical and chemical properties different from the individual elements that comprise them. Their stability, solubility, and reactivity are determined by their molecular structure and types of bonds present.
For instance, sulfuric acid (H2SO4) behaves differently compared to its individual elements (hydrogen, sulfur, and oxygen) due to its strong acid properties and ability to donate hydrogen ions in solution. Understanding how compounds interact at the molecular level is central to mastering chemistry principles.
Ionic and Molecular Compounds
Ionic and molecular compounds differ primarily in the type of bonds that hold their atoms together. Ionic compounds are formed through the transfer of electrons from metal to non-metal, resulting in oppositely charged ions that attract each other. A classic example is sodium chloride (NaCl).
Molecular compounds, on the other hand, involve the sharing of electrons between non-metals, leading to stable configurations that hold atoms together using covalent bonds. Water (H2O) and carbon dioxide (CO2) are common examples.
The properties of these compounds such as melting points, solubility, and electrical conductivity vary significantly. Ionic compounds generally have high melting points and conduce electricity when dissolved in water. Molecular compounds tend to have lower melting points and do not conduct electricity. Understanding these distinctions helps predict and explain compound behaviors in various chemical reactions.

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Most popular questions from this chapter

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: (a) \(\mathrm{NH}_{4}^{+},(\mathbf{b}) \mathrm{NO}_{2}^{-},(\mathbf{c}) \mathrm{N}_{2} \mathrm{O},(\mathbf{d}) \mathrm{NO}_{2}\)

In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+},(\mathbf{b}) \mathrm{Br}_{2}\) to \(\mathrm{Br}^{-},(\mathbf{c}) \mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+},(\mathbf{d}) \mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\) . In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.

(a) How does the structure of diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) differ from that of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) ?(\mathbf{b})\) Explain why diborane adopts the geometry that it does. (c) What is the significance of the statement that the hydrogen atoms in diborane are described as "hydridic"?

When aluminum replaces up to half of the silicon atoms in \(\mathrm{SiO}_{2},\) a mineral class called feldspars result. The feldspars are the most abundant rock-forming minerals, comprising about 50\(\%\) of the minerals in Earth's crust. Orthoclase is a feldspar in which Al replaces one-fourth of the Si atoms of \(\mathrm{SiO}_{2},\) and charge balance is completed by \(\mathrm{K}^{+}\) ions. Determine the chemical formula for orthoclase.

Write a balanced equation for each of the following reactions. (You may have to guess at one or more of the reaction products, but you should be able to make a reasonable guess, based on your study of this chapter.) (a) Hydrogen selenide can be prepared by reaction of an aqueous acid solution on aluminum selenide. (b) Sodium thiosulfate is used to remove excess \(\mathrm{Cl}_{2}\) from chlorine-bleached fabrics. The thiosulfate ion forms \(\mathrm{SO}_{4}^{2-}\) and elemental sulfur, while \(\mathrm{Cl}_{2}\) is reduced to \(\mathrm{Cl}^{-}\) .

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