Chapter 20: Problem 3

Why is Be more likely to form \(\mathrm{Be}^{2+}\) than \(\mathrm{S}\) is to form \(\mathrm{S}^{2+}\) ?

Short Answer

Expert verified

Beryllium (\(Be\)) is more likely to form \(Be^{2+}\) because it reaches a stable full s-subshell by losing two electrons with less energy requirement. In contrast, for Sulfur (\(S\)) to form \(S^{2+}\), it needs to lose two electrons from its p-shell which requires more energy and doesn't result in a stable electron configuration.

Step by step solution

Electron Configurations

Determine the electron configurations of Beryllium (\(Be\)) and Sulfur (\(S\)). Beryllium has the atomic number of 4, so its electron configuration is \(1s^2 2s^2\). Sulfur has the atomic number of 16, which gives it the electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^4\).

Analyzing the Ionization Energy and Stability

Analyze the ionization energy and the stability of each element. For Beryllium to form \(Be^{2+}\), it loses 2 electrons, which aligns with it having a full s-subshell afterwards, leading to a more stable state. Sulfur, on the other hand, has to lose 2 electrons in the p shell, meaning it won't have a full p-subshell configuration, which is less stable. Thus, forming the \(S^{2+}\) ion would require more energy as compared to the energy required to form \(Be^{2+}\).

Conclusion

Conclude that Beryllium is more likely to form \(Be^{2+}\) because the energy required to remove its electrons is less than that of Sulfur to form \(S^{2+}\), thus making it more likely for Beryllium to form \(Be^{2+}\) than for Sulfur to form \(S^{2+}\).

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.

Start your free trial

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

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Why is Be more likely to form \(\mathrm{Be}^{2+}\) than \(\mathrm{S}\) is to form \(\mathrm{S}^{2+}\) ?

Short Answer

Step by step solution

Electron Configurations

Analyzing the Ionization Energy and Stability

Conclusion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Chemical Analysis

Inorganic Chemistry

Making Measurements

The Earths Atmosphere

Organic Chemistry

Study anywhere. Anytime. Across all devices.