Give the electronic configuration for (a) \(\mathrm{Fe}^{3+}\) (b) \(\mathrm{V}^{2+}\) (c) \(\mathrm{Zn}^{2+}\) (d) \(\mathrm{Cu}^{+}\) (e) \(\mathrm{Mn}^{4+}\)

Short Answer

Expert verified
Question: Determine the electronic configuration for the following ions: Fe\(^{3+}\), V\(^{2+}\), Zn\(^{2+}\), Cu\(^{+}\), and Mn\(^{4+}\). Answer: The electronic configurations are as follows: 1. Fe\(^{3+}\): 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^5\) 2. V\(^{2+}\): 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^3\) 3. Zn\(^{2+}\): 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^{10}\) 4. Cu\(^{+}\): 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^{10}\) 5. Mn\(^{4+}\): 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^3\)

Step by step solution

01

1. Fe\(^{3+}\) Electronic Configuration

The atomic number of iron (Fe) is 26. Thus, its electronic configuration is 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 4s\(^2\) 3d\(^6\). Since Fe\(^{3+}\) lost 3 electrons, we need to remove 3 electrons from its configuration. We'll take 2 electrons from the 4s orbital and 1 from the 3d orbital, leaving us with the electronic configuration 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^5\).
02

2. V\(^{2+}\) Electronic Configuration

The atomic number of vanadium (V) is 23. Thus, its electronic configuration is 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 4s\(^2\) 3d\(^3\). Since V\(^{2+}\) lost 2 electrons, we need to remove 2 electrons from its configuration. We'll take 2 electrons from the 4s orbital, leaving us with the electronic configuration 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^3\).
03

3. Zn\(^{2+}\) Electronic Configuration

The atomic number of zinc (Zn) is 30. Thus, its electronic configuration is 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 4s\(^2\) 3d\(^{10}\). Since Zn\(^{2+}\) lost 2 electrons, we need to remove 2 electrons from its configuration. We'll take 2 electrons from the 4s orbital, leaving us with the electronic configuration 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^{10}\).
04

4. Cu\(^{+}\) Electronic Configuration

The atomic number of copper (Cu) is 29. Thus, its electronic configuration is 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 4s\(^1\) 3d\(^{10}\). Since Cu\(^{+}\) lost 1 electron, we need to remove 1 electron from its configuration. We'll take 1 electron from the 4s orbital, leaving us with the electronic configuration 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^{10}\).
05

5. Mn\(^{4+}\) Electronic Configuration

The atomic number of manganese (Mn) is 25. Thus, its electronic configuration is 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 4s\(^2\) 3d\(^5\). Since Mn\(^{4+}\) lost 4 electrons, we need to remove 4 electrons from its configuration. We'll take 2 electrons from the 4s orbital and 2 from the 3d orbital, leaving us with the electronic configuration 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^6\) 3d\(^3\).

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!

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

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