Problem 100
Question
The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}{\underline{\phantom{xx}}}^{+}(g)+\mathrm{e}^{-} $$ 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.
Step-by-Step Solution
Verified Answer
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: \( O_2(g) + 2F_2(g) \longrightarrow 2OF_2(g) \).
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
4Step 4: 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.
Key Concepts
ReactivityOxygen DifluorideChemical Reactions
Reactivity
Reactivity is a term used to describe how readily an element or compound engages in a chemical reaction. In general, the reactivity of an element can depend on several factors, such as ionization energy, electronegativity, and the presence of valence electrons. Elements with high ionization energy tend to hold onto their electrons more tightly, making them less likely to participate in reactions. Meanwhile, elements with low ionization energy will more readily surrender or share electrons, participating more eagerly in reactions.
In the case of oxygen ( O_2 ), even though its first ionization energy is comparable to that of Xenon (Xe), a noble gas known for its reluctance to form bonds, oxygen can still be quite reactive. This is because oxygen has two empty slots in its valence shell, allowing it to form double bonds, which can make it more willing to engage in reactions to achieve a stable electron configuration. Likewise, fluorine (F), one of the most electronegative elements, is exceptionally reactive. Because it seeks electrons to complete its valence shell, it consistently forms bonds with other elements.
In the case of oxygen ( O_2 ), even though its first ionization energy is comparable to that of Xenon (Xe), a noble gas known for its reluctance to form bonds, oxygen can still be quite reactive. This is because oxygen has two empty slots in its valence shell, allowing it to form double bonds, which can make it more willing to engage in reactions to achieve a stable electron configuration. Likewise, fluorine (F), one of the most electronegative elements, is exceptionally reactive. Because it seeks electrons to complete its valence shell, it consistently forms bonds with other elements.
- High ionization energy typically means less reactivity.
- Oxygen, despite its higher ionization energy, is reactive because of its capacity for double bonding.
- Fluorine is highly reactive due to its strong tendency to gain electrons.
Oxygen Difluoride
Oxygen Difluoride (OF
_2
) is a fascinating compound that highlights the peculiar reactivity of both oxygen and fluorine. When oxygen reacts with fluorine, they form oxygen difluoride, where each oxygen atom bonds with two fluorine atoms. This molecule is an example of a polar compound, with electrons more attracted to the fluorine due to its higher electronegativity.
The structure of OF _2 involves oxygen in the center, forming two single covalent bonds with the two fluorine atoms. The bending shape of OF _2 is similar to that of water, causing it to be less symmetrical and more polar. This polarity, in turn, influences its properties and behavior in chemical reactions.
The structure of OF _2 involves oxygen in the center, forming two single covalent bonds with the two fluorine atoms. The bending shape of OF _2 is similar to that of water, causing it to be less symmetrical and more polar. This polarity, in turn, influences its properties and behavior in chemical reactions.
- OF _2 has a bent molecular shape, leading to a polar compound.
- It forms through the bonding of one oxygen and two fluorine atoms (O-F-F).
- OF _2 ’s polarity affects its reactivity and interactions with other substances.
Chemical Reactions
Chemical reactions are transformations where substances interact to form new products. During these reactions, chemical bonds are broken and new ones are formed, leading to changes in the composition and properties of the substances involved. Understanding the nature of chemical reactions, such as the one between
O_2
and
F_2
, helps to analyze how new compounds, like OF
_2
, are created.
For any chemical reaction to proceed, certain conditions must be met:
For any chemical reaction to proceed, certain conditions must be met:
- Correct orientation of reacting molecules.
- Sufficient energy to surpass the activation energy barrier.
- A favorable environment, which could involve aspects like temperature and catalysts.
Other exercises in this chapter
Problem 98
The electron affinities, in \(\mathrm{kJ} / \mathrm{mol}\), for the group 11 and group 12 metals are as follows: (a) Why are the electron affinities of the grou
View solution Problem 99
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
View solution Problem 101
It is possible to define metallic character as we do in this book and base it on the reactivity of the element and the ease with which it loses electrons. Alter
View solution Problem 102
Which of the following is the expected product of the reaction of \(\mathrm{Mg}(s)\) and \(\mathrm{N}_{2}(g)\) under heat? (i) \(\mathrm{Mg}_{3} \mathrm{~N}(s)\
View solution