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Rank the halogens \(\mathrm{Cl}_{2}, \mathrm{Br}_{2},\) and \(\mathrm{I}_{2}\) in order of increasing oxidizing strength based on their products with the metal rhenium (Re): \(\operatorname{ReCl}_{6}\), ReBr \(_{5}\), ReI \(_{4}\). Explain your ranking.

Short Answer

Expert verified
The order of increasing oxidizing strength is \(\text{I}_2 < \text{Br}_2 < \text{Cl}_2\).

Step by step solution

01

- Understand the Products

The products formed when rhenium combines with halogens are \(\text{ReCl}_{6}\), \(\text{ReBr}_{5}\), and \(\text{ReI}_{4}\). The subscript indicates the number of halogen atoms bonded to the rhenium.
02

- Correlate Oxidizing Strength

Oxidizing strength is related to the ability of a substance to gain electrons. The higher the number of halogen atoms bonded to rhenium, the stronger the oxidizing agent.
03

- Rank by Oxidizing Strength

Comparing the products: \(\text{ReCl}_{6}\) has 6 chlorine atoms, \(\text{ReBr}_{5}\) has 5 bromine atoms, and \(\text{ReI}_{4}\) has 4 iodine atoms. Therefore, \(\text{ReCl}_{6}\) indicates chlorine (Cl) is the strongest oxidizing agent, followed by bromine (Br), and iodine (I) is the weakest.
04

- Final Ranking

Based on the number of halogen atoms bonded to rhenium, the order of increasing oxidizing strength is: \(\text{I}_2 < \text{Br}_2 < \text{Cl}_2\).

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Key Concepts

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

Halogens
Halogens are a group of elements found in Group 17 of the periodic table. They include fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). These elements are known for their high reactivity and are commonly found in nature in the form of salts.
Halogens have seven electrons in their outermost shell, which makes them highly eager to gain one more electron to achieve a stable octet configuration. This characteristic explains their vigorous reactivity, especially with metals, to form ionic compounds.
Key properties of halogens include:
  • High electronegativity
  • High ionization energy
  • Presence as diatomic molecules (e.g., Cl2, Br2, I2) in their elemental form
  • Forming salts when reacting with metals
In chemical reactions, halogens often act as oxidizing agents because of their strong tendency to gain electrons and form negative ions (anions).
Oxidizing Agents
An oxidizing agent is a substance that has the ability to oxidize other substances by gaining electrons. In other words, it removes electrons from another substance and, in the process, gets reduced itself.
When ranking the oxidizing strength of halogens, as seen in the provided exercise, we look at their ability to attract and gain electrons. The more readily a halogen gains electrons, the stronger its oxidizing ability.
Here's how to understand the strength of oxidizing agents among halogens:
  • Chlorine (Cl2) is the strongest oxidizing agent among the three mentioned in the exercise.
  • Bromine (Br2) is a moderate oxidizing agent, stronger than iodine but weaker than chlorine.
  • Iodine (I2) is the weakest oxidizing agent among the three.
This ranking is evident based on the number of halogen atoms bonded to rhenium (Re) in the products of their reactions.
Chemical Bonding
Chemical bonding involves the interaction between atoms to form molecules and compounds. The key types of bonds include ionic, covalent, and metallic bonds.
Halogens typically participate in ionic and covalent bonds:
  • Ionic Bonds: Formed when a halogen reacts with a metal. The halogen gains an electron (reduction) becoming an anion, while the metal loses an electron (oxidation) becoming a cation. An example is the formation of sodium chloride (NaCl).
  • Covalent Bonds: Formed when halogens share electrons with other nonmetals. For example, in a molecule like Cl2, each chlorine atom shares one electron to form a covalent bond.
In the context of the given exercise, the products like ReCl6, ReBr5, and ReI4 illustrate interactions where halogens bond with rhenium to form complex compounds.
Understanding chemical bonding helps clarify why halogens act as strong oxidizing agents due to their propensity to gain electrons and engage in stable bond formations.

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Most popular questions from this chapter

Rank the following species in order of decreasing acid strength: \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{H}_{2} \mathrm{SO}_{3}, \mathrm{HSO}_{3}^{-}\)

Disulfur decafluoride is intermediate in reactivity between \(\mathrm{SF}_{4}\) and \(\mathrm{SF}_{6}\). It disproportionates at \(150^{\circ} \mathrm{C}\) to these monosulfur fluorides. Write a balanced equation for this reaction, and give the oxidation state of \(\mathrm{S}\) in each compound.

Given the following information, $$\begin{array}{rlrr}\mathrm{H}^{+}(g)+\mathrm{H}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(g) & \Delta H=-720 \mathrm{~kJ} \\\ \mathrm{H}^{+}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q) & & \Delta H=-1090 \mathrm{~kJ} \\ \mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H= & 40.7 \mathrm{~kJ} \end{array}$$ calculate the heat of solution of the hydronium ion: $$\mathrm{H}_{3} \mathrm{O}^{+}(g) \stackrel{\mathrm{H}_{3} \mathrm{O}}{\longrightarrow} \mathrm{H}_{3} \mathrm{O}^{+}(a q)$$

In addition to those in Table \(14.3,\) other less stable nitrogen oxides exist. Draw a Lewis structure for each of the following: (a) \(\mathrm{N}_{2} \mathrm{O}_{2}\), a dimer of nitrogen monoxide with an \(\mathrm{N}-\mathrm{N}\) bond (b) \(\mathrm{N}_{2} \mathrm{O}_{2}\), a dimer of nitrogen monoxide with no \(\mathrm{N}-\mathrm{N}\) bond (c) \(\mathrm{N}_{2} \mathrm{O}_{3}\) with no \(\mathrm{N}-\mathrm{N}\) bond (d) \(\mathrm{NO}^{+}\) and \(\mathrm{NO}_{3}^{-}\), products of the ionization of liquid \(\mathrm{N}_{2} \mathrm{O}_{4}\)

An industrial chemist treats solid \(\mathrm{NaCl}\) with concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and obtains gaseous \(\mathrm{HCl}\) and \(\mathrm{NaHSO}_{4}\). When she substitutes solid NaI for \(\mathrm{NaCl}\), she obtains gaseous \(\mathrm{H}_{2} \mathrm{~S},\) solid \(\mathrm{I}_{2},\) and \(\mathrm{S}_{8},\) but no \(\mathrm{HI}\). (a) What type of reaction did the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) undergo with NaI? (b) Why does NaI, but not \(\mathrm{NaCl}\), cause this type of reaction? (c) To produce HI \((g)\) by reacting NaI with an acid, how does the acid have to differ from sulfuric acid?

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