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The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}^{+}(g)+\mathrm{e}^{-} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of Xe. Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.

Short Answer

Expert verified
Yes, we would expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) due to their high reactivity, similar ionization energies, and energetically favorable nature of potential products. The likely product of this reaction is oxygen difluoride (OF₂).

Step by step solution

01

Examine the reactivity of oxygen and fluorine

Oxygen (O₂) and fluorine (F₂) are both highly reactive elements. Oxygen tends to easily form covalent bonds with other elements, while fluorine is the most electronegative element and has a strong tendency to attract and gain electrons. This reactivity suggests that they might have the potential to react with each other.
02

Analyze the ionization energy

The ionization energy given for the oxygen molecule is \(1175 \mathrm{~kJ/mol}\), which is quite similar to the ionization energy of xenon (Xe). This means that removing an electron from an oxygen molecule requires approximately the same amount of energy as removing an electron from a xenon atom. Since reactions generally proceed when there is a decrease in energy, it is important to consider whether the reactions between \(O_2\) and \(F_2\) would lead to a net decrease in energy.
03

Examine the potential products

Based on the reactivity of oxygen and fluorine, one possible reaction between \(O_2\) and \(F_2\) would be the formation of a compound where fluorine gains an electron from oxygen, resulting in a compound like OF₂ (oxygen difluoride). \(O_2 + F_2 \longrightarrow 2\: OF\) The energetics of this reaction are favorable because the highly electronegative fluorine atoms have a strong affinity for electrons, and gaining electrons is an energy-releasing process. Thus, the energy released when fluorine gains an electron compensates the ionization energy required to remove an electron from oxygen, and the reaction proceeds spontaneously.
04

Conclusion

Given that oxygen and fluorine are both highly reactive, the similarities in their ionization energies, and the energetically favorable nature of the potential products, we would expect them to react. The likely product of this reaction is oxygen difluoride (OF₂).

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Key Concepts

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

Reactivity of Elements
Elements vary in their reactivity based on their ability to undergo chemical changes. Reactivity is influenced by the electronic structure of an atom and how easily it can gain or lose electrons.
  • Oxygen is known for its high reactivity, often forming covalent bonds by sharing electrons with other elements.
  • Fluorine, on the other hand, is the most reactive of all the halogens. This high reactivity is due to its electronegativity – a strong tendency to attract and hold onto electrons.
Given this, when two highly reactive elements like oxygen and fluorine encounter each other, a reaction is likely to occur. Their reactivity stems from their tendency to achieve a stable electronic configuration, usually by full outer electron shells.
When elements react, they often do so by rearranging electrons to achieve such stability, forming new compounds in the process. This is the very core of chemical reactivity.
Covalent Bonds
Covalent bonds occur when atoms share pairs of electrons to attain a full outer shell, often resulting in the formation of molecules.
  • Oxygen commonly forms covalent bonds, sharing electrons with various elements, including hydrogen to form water (H₂O) or carbon to form carbon dioxide (CO₂).
  • In a reaction with fluorine, oxygen can form oxygen difluoride (OF₂), a compound where oxygen and fluorine share electrons through covalent bonding.
Covalency allows molecules to have stable configurations by sharing their outer shell electrons. Covalent bonds are essential in organic chemistry and the structure of living organisms.
The sharing of electrons in such bonds helps to lower the potential energy of the compound, making them generally stable unless additional energy inputs prompt a reaction.
Electronegativity
Electronegativity is a chemical property that describes an atom's ability to attract and bind with electrons.
Fluorine is the most electronegative element, which means it has the highest ability to attract electrons towards itself in a chemical bond.
  • The ability of fluorine to attract the shared electron pair more strongly than other elements like oxygen makes it a powerful oxidizing agent.
  • This high electronegativity is also why it forms strong bonds with other elements, including the formation of fluorocarbons and fluorinated compounds.
Fluorine’s electronegativity can influence the formation of polar covalent bonds, where electrons spend more time closer to the fluorine atom than the atom it is bonded with. This property affects the physical and chemical properties of compounds, influencing boiling points, solubility, and reactivity.
Chemical Reactions
Chemical reactions involve the breaking and forming of chemical bonds, resulting in the conversion of reactants into products.
In the interaction between oxygen and fluorine:
  • Oxygen can lose electrons while fluorine readily accepts them due to its high electronegativity.
  • In this specific case, oxygen reacts with fluorine to form oxygen difluoride (OF₂), an example of a chemical reaction where energy is released as the bonds form.
The energy changes involved in chemical reactions stem from the breaking and forming of bonds. An exothermic reaction is one where more energy is released when new bonds are formed than is required to break the original ones.
The reactivity of elements, their electronegativity, and the potential energy changes are all critical factors in determining whether a chemical reaction will occur and what products will be formed.

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Most popular questions from this chapter

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity, (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds." If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}\). Write out the process corresponding to the first ionization energy of hydride. (e) How does the process you wrote in part (d) compare to the process for the electron affinity of elemental hydrogen?

Mercury in the environment can exist in oxidation states 0,+1 , and \(+2 .\) One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.107 ), but instead of using ultraviolet light to eject valence electrons, X-rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's 4 forbitals at \(105 \mathrm{eV},\) from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy. The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV}\), corresponding to the 1 s orbital of oxygen. Overall the researchers concluded that oxidation states were +2 for \(\mathrm{Hg}\) and -2 for \(\mathrm{O} .\) (a) Calculate the wavelength of the X-rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the 1 s electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground- state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case? (d) Use Slater's rules to estimate \(Z_{\text {eff }}\) for the \(4 f\) and valence electrons of \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); assume for this purpose that all the inner electrons with \((n-3)\) or less screen a full + \(1 .\)

Explain the following variations in atomic or ionic radii: (a) \(\Gamma>\mathrm{I}>\mathrm{I}^{+},(\mathrm{b}) \mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}\) (c) \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)

Arrange the following oxides in order of increasing acidity: \(\mathrm{CO}_{2}, \mathrm{CaO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3}, \mathrm{SiO}_{2},\) and \(\mathrm{P}_{2} \mathrm{O}_{5}\)

Compare the elements bromine and chlorine with respect to the following properties: (a) electron configuration, (b) most common ionic charge, (c) first ionization energy, (d) reactivity toward water, (e) electron affinity, (f) atomic radius. Account for the differences between the two elements.

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