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(a) Use orbital diagrams to illustrate what happens when an oxygen atom gains two electrons. (b) Why does O3 not exist?

Short Answer

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(a) The oxygen atom with electronic configuration 1s² 2s² 2p⁴ will have its orbital diagram as: 1s: ↑↓ 2s: ↑↓ 2p: ↑↓ | ↑ | ↑ When it gains two electrons, the 2p orbitals get filled, and the orbital diagram for O²⁻ becomes: 1s: ↑↓ 2s: ↑↓ 2p: ↑↓ | ↑↓ | ↑↓ (b) O³⁻ doesn't exist due to its higher energy and decreased stability when compared to the O²⁻ ion. Gaining a third electron would require occupying the 3s orbital, which is less stable, and increase electron repulsion in the already filled 2p orbitals.

Step by step solution

01

(a) Orbital Diagram of Oxygen Atom and Oxygen Ion with 2 Extra Electrons

First, we need to examine the electronic configuration of an oxygen atom and use orbital diagrams to represent it. Oxygen has 8 electrons with the following electronic configuration: 1s² 2s² 2p⁴. To represent the gained electrons in the oxygen atom, we can show an oxygen ion with two extra electrons in the 2p orbital: Oxygen atom orbital diagram: 1s: ↑↓ 2s: ↑↓ 2p: ↑↓ | ↑ | ↑ When an oxygen atom gains two electrons, these extra electrons will fill into the 2p orbitals. Then, we have: Oxygen ion orbital diagram ( O²⁻ ): 1s: ↑↓ 2s: ↑↓ 2p: ↑↓ | ↑↓ | ↑↓ The oxygen atom has added two electrons to the 2p orbital, resulting in the oxygen ion with a charge of 2- (O²⁻).
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(b) Explanation of Why Oxygen Ion with 3 Extra Electrons (O³⁻) Does Not Exist

In the case of the oxygen atom, it has 6 electrons in its outer shell (2s² 2p⁴). Gaining two electrons will complete its outer shell with 8 electrons, completing its octet, which results in a more stable configuration for the oxygen atom: O²⁻ ion. However, if the oxygen atom were to gain an additional electron to form O³⁻, the incoming electron would not fit into the already completed 2p orbitals. The added electron would have to occupy the 3s orbital, which is more energy-intensive and less stable for the atom. Furthermore, taking in a third electron would also increase the repulsion between the electrons in the already filled 2p orbitals, making the ion even less stable. Thus, an O³⁻ ion would not exist due to its higher energy and decreased stability when compared to the O²⁻ ion.

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Key Concepts

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

Electronic Configuration
Understanding the electronic configuration of an atom is critical for grasping many fundamental concepts in chemistry. Simply put, the electronic configuration is the arrangement of electrons around the nucleus of an atom in its atomic orbitals. For oxygen, with an atomic number of 8, the configuration is noted as 1s² 2s² 2p⁴. This indicates that the first energy level, or shell, contains two electrons in the 1s orbital, and the second shell has two electrons in the 2s orbital and four electrons in the 2p orbitals.

When we represent this using orbital diagrams, we show the orbitals as boxes or lines and the electrons as arrows, indicating the spin of each electron. The principle of minimizing electron repulsion ensures that electrons will fill an empty or half-filled orbital before pairing up. In oxygen's case, the 2p orbital will have unpaired electrons indicated as: ↑ | ↑ | ↑ in the orbital diagram for the atom.

A solid understanding of electronic configuration lays the groundwork for studying chemical reactions, bonding, and more, as the way electrons are arranged determines how atoms interact with one another.
Oxygen Ion
The formation of ions is a common and important process in chemistry. An ion is an atom or molecule that has gained or lost one or more electrons, thereby acquiring a net electric charge. In the case of oxygen, an oxygen ion commonly forms when the atom gains two electrons, resulting in a negative two charge, written as O²⁻. This process can be visualized in an orbital diagram where the extra electrons fill the empty spaces in the oxygen's 2p orbitals, aligning with the goals of achieving lower energy and greater stability.

The orbital diagram after oxygen gains two electrons will show all the 2p orbitals filled: 2p: ↑↓ | ↑↓ | ↑↓. This fuller electronic configuration illustrates how the oxygen atom has achieved a stable, more energetically favorable state. It's essential to appreciate this transition from oxygen atom to oxygen ion as it underlines the atom's drive towards stability through electron gain or loss.
Octet Rule
The octet rule is a chemical principle that reflects the tendency of atoms to prefer to have eight electrons in their valence shell, the outermost electron shell in an atom. This rule is based on the observation that atoms with eight electrons in their valence shell tend to be more stable, resembling the electronic structure of noble gases.

For oxygen, which has six valence electrons, following the octet rule means that gaining two additional electrons to fill its 2p orbitals results in the desirable, stable arrangement of eight valence electrons. This is the driving force behind the oxygen atom's tendency to form the O²⁻ ion. The completion of the octet through the gain of electrons is a recurrent theme in chemical bonding and is at the heart of many ionic and covalent bonding interactions.
Electron Repulsion
Electron repulsion is a principle coming from Coulomb's law, stating that like charges repel each other. In the context of an atom's orbitals, this translates to the behavior that electrons, all negatively charged, will repel one another and seek to minimize this repulsion by occupying different spaces if available.

In oxygen's 2p orbitals, each electron will initially occupy an empty orbital to minimize repulsion: the first three electrons fill as ↑ in each 2p orbital. When oxygen becomes an ion by gaining two additional electrons, to minimize repulsion, these electrons pair up with the existing unpaired electrons, resulting in a fully occupied 2p sublevel with paired spins: ↑↓ | ↑↓ | ↑↓.

Conversely, an oxygen ion with a -3 charge, purportedly O³⁻, would violate the minimized repulsion principle. The extra third electron would have to unnecessarily pair up in a higher energy 3s orbital or contribute to increased repulsion in the filled 2p orbitals, making the O³⁻ ion highly unstable and energetically unfavorable, hence it does not commonly exist.

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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 H. Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?

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Which of the following is the expected product of the reaction of K(s) and H2(g)?(i)KH(s),(ii)K2H(s), (iii) KH2(s), (iv)K2H2(s),or(v)K(s) and H2(g) will not react with one another.

Use electron configurations to explain the following observa tions: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron afnity 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.

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