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Chapter 7: Problem 94

(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 \(1s^{2} 2s^{2} 2p^{6}\), with the orbital diagram: 1s: 鈫戔啌 2s: 鈫戔啌 2p: 鈫戔啌鈫戔啌鈫� The ion \(\mathrm{O}^{3-}\) is highly unlikely to exist because adding a third electron would require occupying the 3s orbital, increasing the energy of the system and electron-electron repulsion, making it energetically unfavorable.

Step by step solution

01

Determine the electron configuration of an oxygen atom

The atomic number of an oxygen atom is 8. This means that an oxygen atom has 8 electrons in its various orbitals. The electron configuration of an oxygen atom can be determined by following the Aufbau principle, which states that electrons fill the lowest available energy levels first. As a result, the electron configuration of an oxygen atom is: \[1s^{2} 2s^{2} 2p^{4}\]
02

Use orbital diagrams to illustrate what happens when an oxygen atom gains two electrons

An oxygen atom has 8 electrons, distributed in the 1s, 2s and 2p orbitals. The 2p orbitals have 4 electrons, so when an oxygen atom gains two electrons, these additional electrons will be added to the 2p orbitals. The orbital diagram for an oxygen atom is: 1s: 鈫戔啌 2s: 鈫戔啌 2p: 鈫戔啈鈫� When an oxygen atom gains two electrons, the orbital diagram becomes: 1s: 鈫戔啌 2s: 鈫戔啌 2p: 鈫戔啌鈫戔啌鈫� As a result, the electron configuration for an oxygen atom that has gained two electrons is \(1s^{2} 2s^{2} 2p^{6}\), also written as \[\mathrm{O}^{2-}: [He] 2s^{2} 2p^{6}\]
03

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

In order for the oxygen ion \(\mathrm{O}^{3-}\) to exist, it would need to gain three electrons, which would require electrons to occupy the 3s orbital. However, before an electron occupies a higher energy level, the lower energy level must be completely filled. In the case of the oxygen atom, there are no available spaces in the 2p orbitals after gaining two electrons, and adding one more electron to the 3s orbital would significantly increase the energy of the system. The \(\mathrm{O}^{3-}\) ion would be highly unstable due to the increased electron-electron repulsion and higher energy state. Thus, it is very unlikely for the oxygen ion \(\mathrm{O}^{3-}\) to exist because it would not be energetically favorable.

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Key Concepts

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

Electron Configuration
Every atom has electrons arranged in specific energy levels known as orbitals. Oxygen, with its atomic number of 8, follows the Aufbau principle to fill these orbitals. This principle ensures electrons fill the lowest energy levels available first. For oxygen, this results in an electron configuration of
  • \(1s^2 \)
  • \(2s^2 \)
  • \(2p^4 \)
This means oxygen has a total of 8 electrons distributed across these orbitals.
As electrons fill, they follow additional principles such as the Pauli exclusion principle and Hund's rule. These rules help predict the behavior of electrons, making the configuration both stable and energetically favorable.
Understanding electron configurations is crucial because it provides insights into the element's bonding and reactive properties.
Orbital Diagrams
Orbital diagrams visually represent the arrangement of electrons in an atom鈥檚 orbitals. In these diagrams:
  • Each box represents an orbital.
  • Arrows illustrate electrons, with opposite spins shown using up and down arrows.
Before oxygen gains electrons, its orbital diagram is:
1s: 鈫戔啌
2s: 鈫戔啌
2p: 鈫戔啈鈫�
This configuration shows that the 2p orbitals have 4 electrons. When oxygen gains two electrons, they fill the remaining slots in the 2p orbitals, transforming into:
1s: 鈫戔啌
2s: 鈫戔啌
2p: 鈫戔啌鈫戔啌鈫�
This results in the stable \, \(\text{O}^{2-}\) ion with a full 2p subshell (\(2p^6\)), reflecting increased stability through full electron pairings.
Ion Stability
Ion stability depends greatly on electron configuration. When an oxygen atom gains electrons, it achieves a stable ion configuration similar to the noble gases. For instance, the \, \(\text{O}^{2-}\) ion achieves stability by emulating the electron configuration of neon. However, attempting to form an \, \(\text{O}^{3-}\) ion encounters challenges:
  • Adding a third extra electron would necessitate filling a higher energy orbital (3s).
  • This causes increased electron-electron repulsion and greater energy.
Such a configuration is not energetically favorable. High instability emerges from attempts to occupy higher energy states without fully filling the preceding orbitals.
Understanding why certain ions are stable while others aren't is pivotal in predicting chemical behavior and reactivity.

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Most popular questions from this chapter

Write a balanced equation for the reaction that occurs in each (a) Calcium metal is heated in an atmoof the following cases: sphere of oxygen gas. (b) Copper oxide is heated in an atmosphere of hydrogen gas. (c) Chlorine reacts with nitrogen gas. (d) Boron tribromide reacts with water.

Some ions do not have a corresponding neutral atom that has the same electron configuration. For each of the following ions, identify the neutral atom that has the same number of electrons and determine if this atom has the same electron (d) \(\mathrm{Zn}^{2+}\) configuration. (a) \(\mathrm{Cl}^{-}\), (b) \(\mathrm{Sc}^{3+}\) (c) \(\mathrm{Fe}^{2+}\) (e) \(\mathrm{Sn}^{4+}\)

Write equations that show the processes that describe the first, second, and third ionization energies of a chlorine atom. Which process would require the least amount of energy?

(a) As described in Section 7.7, the alkali metals react with hydrogen to form hydrides and react with halogens to form halides. Compare the roles of hydrogen and halogens in these reactions. Write balanced equations for the reaction of fluorine with calcium and for the reaction of hydrogen with calcium. (b) What is the oxidation number and electron configuration of calcium in each product?

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 274 pm. (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 W atoms.
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