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Chapter 7: Problem 100

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
O2 can react with F2 due to the reactivity from double bonds in O2 and the high electronegativity of F2. The product of this reaction is Oxygen Difluoride (OF2), as shown in the chemical equation: \( O_2(g) + 2F_2(g) \longrightarrow 2OF_2(g) \).

Step by step solution

01

Understand the ionization energies and reactivity

Ionization energy is the amount of energy required to remove an electron from an atom or molecule. High ionization energies indicate that the electron is strongly bound and the element is less likely to react. In this case, the ionization energy of O2 and Xe is almost similar. Xe, being a noble gas, has a fully occupied electron orbital which makes it generally unreactive due to high ionization energy. However, oxygen molecule (O2) can be reactive because they form double bonds, seeking other elements to complete their respective electron shells.
02

Consider the reactivity of fluorine

Fluorine (F) is one of the most electronegative and reactive nonmetals in the periodic table. It seeks electrons to complete its electron shell, readily forming bonds with other elements.
03

Evaluate the reaction between O2 and F2

Based on the reactivity of oxygen and fluorine molecules, it would be reasonable to expect that O2 can react with F2. O2, though it has a high ionization energy similar to Xe, is still reactive because of its double bonds. F2, being highly reactive, can form bonds with O2, leading to the formation of new products.
04

Suggest a product or products of this reaction

When O2 reacts with F2, the resulting product is Oxygen Difluoride (OF2), where each oxygen atom takes up two fluorine atoms. The chemical equation for the reaction: \[ O_2(g) + 2F_2(g) \longrightarrow 2OF_2(g) \] This is because oxygen is able to share its electrons with fluorine atoms, forming two single covalent bonds and satisfying their respective electron shell requirements.

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Most popular questions from this chapter

Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\). (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S,\) calculate \(Z_{\text {eff }}\) for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, $S .(\mathbf{d})$ For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Potassium metal is exposed to an atmosphere of chlorine gas. (b) Strontium oxide is added to water. (c) A fresh surface of lithium metal is exposed to oxygen gas. (d) Sodium metal reacts with molten sulfur.

Use electron configurations to explain the following observations: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron affinity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

Discussing this chapter, a classmate says, "An element that commonly forms a cation is a metal." Do you agree or disagree?

Copper and calcium both form +2 ions, but copper is far less reactive. Suggest an explanation, taking into account the ground-state electron configurations of these elements and their atomic radii.
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