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Arrange the following from lowest to highest ionization energy: \(\mathrm{O}, \mathrm{O}_{2}, \mathrm{O}_{2}^{-}, \mathrm{O}_{2}^{+} .\) Explain your answer.

Short Answer

Expert verified
The species can be arranged in the order of ionization energy as follows: O鈧傗伜 < O鈧傗伝 < O < O鈧. This order reflects the relative ease of removing an electron from each species, considering their stability and electron configurations.

Step by step solution

01

Examine the number of electrons in each species

Determine the number of electrons for each species: - Oxygen atom (O) has 8 electrons. - Oxygen molecule (O鈧) has 16 electrons (8 from each oxygen atom). - Oxygen anion (O鈧傗伝) has 17 electrons (16 from the neutral molecule and one extra electron). - Oxygen cation (O鈧傗伜) has 15 electrons (16 from the neutral molecule minus one electron).
02

Consider the stability of species

The stability of a species depends on its electron configuration. Generally, half-filled or fully-filled orbitals provide greater stability. Here, the neutral oxygen molecule (O鈧) has half-filled orbitals and is more stable compared to the ions.
03

Analyze the ease of electron removal

Examine the ease with which an electron can be removed from each species: - O鈧傗伜 already has 1 electron missing and it has become less stable. Therefore, removing another electron would require less energy. - O鈧 has half-filled orbitals, making it stable. Removing an electron would require more energy. - O鈧傗伝 has gained an additional electron, which adds more electron-electron repulsion. Thus, it would be easier to remove this extra electron (compared to O). - O has a stable electron configuration, but not as stable as O鈧. So, removing an electron from it would require more energy, but not as much as in the case of O鈧.
04

Order the species based on ionization energy

Based on the analysis, the order from lowest to highest ionization energy would be: O鈧傗伜 < O鈧傗伝 < O < O鈧. This order reflects the relative ease of removing an electron from each of these species, taking into account their stability and electron configurations.

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Key Concepts

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

Electron Configuration
Understanding ionization energy requires a fundamental grasp of electron configuration, which is the distribution of electrons in an atom or molecule's atomic or molecular orbitals. Electrons are arranged in orbitals around the nucleus of an atom in a specific pattern, following the principle of increasing energy levels.

Electron configurations are denoted by the principal quantum number, which indicates the energy level or shell, and the letters s, p, d, and f, which represent the different types of orbitals within a shell. For example, the oxygen atom (O) has an electron configuration of 1s虏 2s虏 2p鈦, indicating it has 2 electrons in the innermost shell and 6 electrons in the second shell.

The stability of an electron configuration plays a crucial role in determining ionization energy, with elements generally striving for the most stable electron arrangement possible. A stable electron configuration can mean a full valence shell, like the noble gases, or a half-filled subshell, which is often observed in transition metals. The more stable an electron configuration, the more energy is required to remove an electron, influencing the ionization energy.
Molecular Ions
Molecular ions are molecules that have gained or lost electrons, resulting in a net positive or negative charge. In the context of ionization energy, we're interested in how the addition or removal of electrons affects the stability and energy of the molecule. A molecule can become a cation (positively charged) by losing an electron and an anion (negatively charged) by gaining an electron.

For instance, an oxygen molecule (O鈧) can lose an electron to become an oxygen cation (O鈧傗伜), which disrupts its stable electron configuration and makes it easier to remove another electron. On the other hand, if the oxygen molecule gains an electron, it becomes an oxygen anion (O鈧傗伝), which may create electron-electron repulsion due to the extra electron, potentially making it easier to remove.

Considering molecular ions helps us understand the varying ionization energies because the process to form these ions involves either adding or removing electrons, directly impacting the molecule's stability and the energy required to change its electronic structure.
Electron-Electron Repulsion
The phenomenon of electron-electron repulsion is a key factor when considering ionization energies. As electrons are negatively charged particles, they repel each other. This repulsive force must be overcome to remove an electron from an atom or molecule, and the degree of repulsion can affect how much energy is required to do so.

In a neutral oxygen molecule (O鈧), the electrons are arranged in such a way that they minimize repulsion due to a stable half-filled configuration. However, when an extra electron is added to form the anion (O鈧傗伝), repulsion increases because more electrons are sharing the same space. This makes it slightly easier to remove the most loosely held electron鈥攈ence a lower ionization energy compared to O鈧. Conversely, the cation (O鈧傗伜) with a deficient electron has reduced repulsion among its remaining electrons, and therefore its ionization energy is even lower.

A comprehensive understanding of electron-electron repulsion, along with electron configuration and the nature of molecular ions, gives students a solid foundation to predict and explain trends in ionization energy across different atoms and molecular species.

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Most popular questions from this chapter

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{N}_{2}{ }^{2-}, \mathrm{O}_{2}{ }^{2-}, \mathrm{F}_{2}{ }^{2-}\) b. \(\mathrm{Be}_{2}, \mathrm{~B}_{2}, \mathrm{Ne}_{2}\)

Show how a hydrogen \(1 s\) atomic orbital and a fluorine \(2 p\) atomic orbital overlap to form bonding and antibonding molecular orbitals in the hydrogen fluoride molecule. Are these molecular orbitals \(\sigma\) or \(\pi\) molecular orbitals?

Biacetyl and acetoin are added to margarine to make it taste more like butter. Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

Show how two \(2 p\) atomic orbitals can combine to form a \(\sigma\) or a \(\pi\) molecular orbital.

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in \(\mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{O}}\) : be described in terms of orbitals?

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