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Why is gaseous OH so much more reactive than \(\mathrm{H}_{2} \mathrm{O}\) vapor?

Short Answer

Expert verified
Answer: Gaseous OH is more reactive than Hâ‚‚O vapor because it is a free radical with an unpaired electron in its electron configuration. This unpaired electron makes OH more likely to react with other substances to achieve a stable electron configuration, while Hâ‚‚O has a stable electron configuration with no unpaired electrons, making it less reactive under normal conditions.

Step by step solution

01

Understand the chemical structure and bonding of OH and Hâ‚‚O

The gaseous hydroxide (OH) is a diatomic molecule with one oxygen atom bonded to one hydrogen atom. The water (Hâ‚‚O) vapor is a triatomic molecule consisting of one oxygen atom bonded to two hydrogen atoms. The OH molecule has a single covalent bond between the oxygen and hydrogen atoms, while the Hâ‚‚O molecule has two covalent bonds between the oxygen atom and the two hydrogen atoms.
02

Analyze the electron configuration of OH and Hâ‚‚O molecules

In an OH molecule, the oxygen atom has six valence electrons, and the hydrogen atom has one valence electron. When they form a bond, they share a pair of electrons, resulting in a total of seven valence electrons. One of these seven valence electrons remains unpaired, making the OH molecule a free radical. In an Hâ‚‚O molecule, the oxygen atom shares two pairs of electrons with the two hydrogen atoms, resulting in a complete octet for the oxygen atom and fulfilling the hydrogen atoms' electron requirements. The electron configuration of Hâ‚‚O is stable, with no unpaired electrons.
03

Understand the role of unpaired electrons in reactivity

Molecules with unpaired electrons, such as free radicals like OH, are highly reactive, as they seek to complete their electron octet by forming chemical bonds with other molecules. The unpaired electron in OH makes it more likely to react with various other substances to achieve a stable electron configuration.
04

Compare the reactivity of OH and Hâ‚‚O

In contrast, the Hâ‚‚O molecule has a stable electron configuration with no unpaired electrons, making it less likely to engage in further chemical reactions under normal conditions. Hâ‚‚O is a relatively stable and unreactive compound, especially when compared to the highly reactive and short-lived OH radical.
05

Conclusion

The gaseous OH molecule is more reactive than Hâ‚‚O vapor because it is a free radical with an unpaired electron in its electron configuration. This unpaired electron makes it more likely to react with other substances, trying to achieve a stable electron configuration. On the other hand, Hâ‚‚O has a stable electron configuration with no unpaired electrons, making it less reactive under normal conditions.

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Key Concepts

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

Chemical Bonding
Chemical bonding is the force that holds atoms together in molecules. The types of chemical bonds include ionic, covalent, and metallic bonds, with covalent being the one that holds the atoms in both OH and Hâ‚‚O together. In a covalent bond, atoms share electrons to achieve greater stability, which is often described by the octet rule. This rule states that atoms tend to form bonds in a way that gives them eight electrons in their valence shell, mimicking the electron configuration of noble gases.

OH (hydroxide) consists of one oxygen atom bonded covalently with one hydrogen atom. This bond forms as oxygen shares one of its six valence electrons with hydrogen's single electron. However, unlike Hâ‚‚O (water), the OH radical does not satisfy the octet rule for the oxygen atom as it has one unpaired electron. This difference in chemical bonding and electron sharing directly contributes to the higher reactivity of OH compared to Hâ‚‚O.

Stability and the Octet Rule

Water, with its two hydrogen atoms, allows the central oxygen atom to share a pair of electrons with each hydrogen atom, satisfying the octet rule and creating a more stable structure. In contrast, the OH radical's unpaired electron pushes it to react with other substances in an attempt to stabilize, making it a more reactive molecule.
Electron Configuration
Electron configuration refers to the arrangement of electrons around the nucleus of an atom. This arrangement is what determines the chemical properties and reactivity of an element or compound. It is particularly significant in understanding why certain substances are more reactive than others.

The valence electrons are the outermost electrons involved in bonding. In the case of Hâ‚‚O, oxygen's six valence electrons form bonds with two electrons from two separate hydrogen atoms, resulting in a stable electron configuration where all electrons are paired. However, with OH, one unpaired electron remains since oxygen can only pair its valence electrons with one electron from a single hydrogen atom.

Unpaired Electrons and Chemical Reactivity

This unpaired electron in the OH molecule's electron configuration is a primary reason for its high reactivity. Unpaired electrons are highly energetic and create a condition of instability as the atom seeks to fill its valence shell to achieve stability. Hence, the OH radical is much more likely to engage in chemical reactions than a molecule like Hâ‚‚O, which has no such unpaired electrons and thus a lower propensity to react.
Free Radicals
Free radicals are molecules or atoms that contain an unpaired valence electron, which makes them extremely reactive. They are often discussed in the context of oxidative stress and damage in biological systems, but they also play a critical role in various chemical processes.

The OH molecule, because of its unpaired electron, qualifies as a free radical. This lone electron drives the radical to seek out other atoms or molecules to bond with, causing it to readily engage in chemical reactions to attain stability. Free radicals generally exist for very short periods due to their high reactivity and often participate in chain reactions, where the radical nature is passed on from one reactant to another.

Reactivity of Free Radicals

The reactivity of free radicals like OH can be attributed to the presence of the unpaired electron seeking to complete the atom's or molecule's electron octet. The unpaired electron provides a strong driving force for chemical change, often making free radicals a key factor in the speed and spontaneity of chemical reactions. In comparison, Hâ‚‚O is not a free radical and thus exhibits far less chemical reactivity under normal circumstances.

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Most popular questions from this chapter

Testing for a Banned Herbicide Sodium chlorate was used in weed-control preparations, but its sale has been banned in EU countries since \(2009 .\) A simple colorimetric test for the presence of the chlorate ion in a solution of herbicide relies on the following reaction: \(2 \mathrm{MnO}_{4}^{-}(a q)+5 \mathrm{ClO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q) \rightarrow\) $$ 2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{ClO}_{4}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The table contains rate data for this reaction. $$\begin{array}{ccccc} & \left[\mathrm{MnO}_{4}^{-}\right]_{0} & \left[\mathrm{ClO}_{3}^{-}\right]_{0} & \left[\mathrm{H}^{+}\right]_{0} & \text { Initial Rate } \\ \text { Experiment } & (M) & (M) & (M) & (M / \mathrm{s}) \\ \hline 1 & 0.10 & 0.10 & 0.10 & 5.2 \times 10^{-3} \\ \hline 2 & 0.25 & 0.10 & 0.10 & 3.3 \times 10^{-2} \\ \hline 3 & 0.10 & 0.30 & 0.10 & 1.6 \times 10^{-2} \\ \hline 4 & 0.10 & 0.10 & 0.20 & 7.4 \times 10^{-3} \\ \hline \end{array}$$ Determine the rate law and the rate constant for this reaction.

A proposed mechanism for the decomposition of hydrogen peroxide consists of three elementary steps: \(\begin{array}{ll}\text { (1) } & \mathrm{H}_{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{OH}(g)\end{array}\) (2) \(\quad \mathrm{H}_{2} \mathrm{O}_{2}(g)+\mathrm{OH}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{HO}_{2}(g)\) (3) \(\quad \mathrm{HO}_{2}(g)+\mathrm{OH}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)\) If the rate law for the reaction is first order in \(\mathrm{H}_{2} \mathrm{O}_{2},\) which step in the mechanism is the rate-determining step?

Write rate laws and determine the units of the rate constant (using the units \(M\) for concentration and s for time) for the following reactions: a. The reaction of oxygen atoms with \(\mathrm{NO}_{2}\) is first order in both reactants. b. The reaction between \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) is second order in NO and first order in \(\mathrm{Cl}_{2}\). c. The reaction between \(\mathrm{Cl}_{2}\) and chloroform \(\left(\mathrm{CHCl}_{3}\right)\) is first order in \(\mathrm{CHCl}_{3}\) and one-half order in \(\mathrm{Cl}_{2}\) "d. The decomposition of ozone \(\left(\mathrm{O}_{3}\right)\) to \(\mathrm{O}_{2}\) is second order in \(\mathrm{O}_{3}\) and an order of -1 in \(\mathrm{O}\) atoms.

For each of the following rate laws, determine the order with respect to each reactant and the overall reaction order. a. Rate \(=k[\mathrm{A}][\mathrm{B}]\) b. Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}]\) c. Rate \(=k[\mathrm{A}][\mathrm{B}]^{3}\)

Write the rate laws for the following elementary steps and identify them as uni-, bi-, or termolecular steps: a. \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) b. \(\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \rightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)\) c. \(2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{NO}_{3}(g)+\mathrm{NO}(g)\)

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