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(a) Use orbital diagrams to illustrate what happens when an oxygen atom gains two electrons. (b) Why does \(\mathrm{O}^{3-}\) not exist?

Short Answer

Expert verified
When an oxygen atom gains two electrons, its electron configuration becomes \(1s^{2}\, 2s^{2}\, 2p^{6}\) and forms \(\mathrm{O}^{2-}\). The \(\mathrm{O}^{3-}\) ion doesn't exist because all 2p orbitals are already filled, and adding another electron would violate the Pauli Exclusion Principle and create an unstable ion due to strong electron-electron repulsions.

Step by step solution

01

Write the electron configuration for an oxygen atom

An oxygen atom has 8 electrons. Its electron configuration can be written as \(1s^{2}\, 2s^{2}\, 2p^{4}\) which can also be shown by the following orbital diagram: 1s: ↑↓ 2s: ↑↓ 2p: ↑ ↑ ↑ _
02

Adding two electrons to the Oxygen atom

When an oxygen atom gains two electrons, we need to place these additional electrons in its orbitals following the rules: Aufbau principle, Pauli exclusion principle, and Hund's rule. 1s: ↑↓ 2s: ↑↓ 2p: ↑↑ ↑↑ ↑_
03

Write the electron configuration and the ion symbol for the anion

The new electron configuration for the oxygen anion with two additional electrons will be \(1s^{2}\, 2s^{2}\, 2p^{6}\). The ion symbol for the oxygen anion with a -2 charge would be: \(\mathrm{O}^{2-}\). Now, we'll discuss why \(\mathrm{O}^{3-}\) does not exist.
04

Determine the possibility of adding one more electron to \(\mathrm{O}^{2-}\)

If we were to attempt to create an oxygen ion with a -3 charge, we would need to add one more electron to the existing electron configuration. However, all of the orbitals at the 2p subshell are already completely filled with electrons, in accordance with the electron configuration for \(\mathrm{O}^{2-}\): 1s: ↑↓ 2s: ↑↓ 2p: ↑↓ ↑↓ ↑↓
05

Explain why \(\mathrm{O}^{3-}\) does not exist

As seen in the previous step, adding another electron is not possible without breaking the electron configuration rules. Specifically, we would violate the Pauli Exclusion Principle, which states that each orbital can contain a maximum of two electrons with opposite spins. Additionally, placing another electron in an already filled 2p orbital would make the ion highly unstable due to strong electron-electron repulsions. For these reasons, \(\mathrm{O}^{3-}\) does not exist.

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Most popular questions from this chapter

Tungsten has the highest melting point of any metal in the periodic table: \(3422{ }^{\circ} \mathrm{C}\). The distance between \(\mathrm{W}\) atoms in tungsten metal is \(2.74 \AA\). (a) What is the atomic radius of a tungsten atom in this environment? (This radius is called the metallic radius.) (b) If you put tungsten metal under high pressure, predict what would happen to the distance between \(\mathrm{W}\) atoms.

(a) As described in Section \(7.7,\) the alkali metals react with hydrogen to form hydrides and react with halogens- for example, fluorine to form halides, Compare the roles of hydrogen and the halogen in these reactions. How are the forms of hydrogen and halogen in the products alike? (b) Write balanced equations for the reaction of fluorine with calcium and for the reaction of hydrogen with calcium. What are the similarities among the products of these reactions?

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What is the relationship between the ionization energy of an anion with a 1 - charge such as \(\mathrm{F}^{-}\) and the electron affinity of the neutral atom, F?

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