Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(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: ↑↓ 2s: ↑↓ 2p: ↑↓↑↓↑ 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: ↑↓ 2s: ↑↓ 2p: ↑↑↑ When an oxygen atom gains two electrons, the orbital diagram becomes: 1s: ↑↓ 2s: ↑↓ 2p: ↑↓↑↓↑ 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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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’s 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: ↑↓
2s: ↑↓
2p: ↑↑↑
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: ↑↓
2s: ↑↓
2p: ↑↓↑↓↑
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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}{ }^{+}(g)+\mathrm{e}^{-} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of Xe. Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.

Using only the periodic table, arrange each set of atoms in order from largest to smallest: \((\mathbf{a}) \mathrm{Ar},\) As, \(\mathrm{Kr} ;\) (b) \(\mathrm{Cd}, \mathrm{Rb}, \mathrm{Te} ;(\mathbf{c})\) \(\mathrm{C}, \mathrm{Cl}, \mathrm{Cu}\).

Elements in group 17 in the periodic table are called the halogens; elements in group 16 are called the chalcogens. (a) What is the most common oxidation state of the chalcogens compared to the halogens? (b) For each of the following periodic properties, state whether the halogens or the chalcogens have larger values: atomic radii, ionic radii of the most common oxidation state, first ionization energy, second ionization energy.

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, \(\mathrm{F}\) ?

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.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free