Write electron configurations for a. the cations \(\mathrm{Sr}^{2+}, \mathrm{Cs}^{+}, \mathrm{In}^{+},\) and \(\mathrm{Pb}^{2+}\) b. the anions \(P^{3-}, S^{2-},\) and \(B r^{-}\)

Short Answer

Expert verified
The electron configurations for the given ions are as follows: a. Cations 1. Sr²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) 2. Cs⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) 3. In⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\) 4. Pb²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10}\) b. Anions 1. P³⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6\) 2. S²⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6\) 3. Br⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\)

Step by step solution

01

1. Sr²⁺ Electron Configuration

Strontium (Sr) has an atomic number of 38. The electron configuration for the neutral Sr atom is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2\] As Sr²⁺ is a cation with a 2+ charge, we will remove two electrons from the 5s orbital. The electron configuration for Sr²⁺ is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\]
02

2. Cs⁺ Electron Configuration

Cesium (Cs) has an atomic number of 55. The electron configuration for the neutral Cs atom is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^1\] As Cs⁺ is a cation with a 1+ charge, we will remove one electron from the 6s orbital. The electron configuration for Cs⁺ is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\]
03

3. In⁺ Electron Configuration

Indium (In) has an atomic number of 49. The electron configuration for the neutral In atom is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^1\] As In⁺ is a cation with a 1+ charge, we will remove one electron from the 5p orbital. The electron configuration for In⁺ is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10}\]
04

4. Pb²⁺ Electron Configuration

Lead (Pb) has an atomic number of 82. The electron configuration for the neutral Pb atom is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^2\] As Pb²⁺ is a cation with a 2+ charge, we will remove two electrons from the 6p orbital. The electron configuration for Pb²⁺ is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10}\] b. Electron configurations for anions
05

1. P³⁻ Electron Configuration

Phosphorus (P) has an atomic number of 15. The electron configuration for the neutral P atom is: \[1s^2 2s^2 2p^6 3s^2 3p^3\] As P³⁻ is an anion with a 3- charge, we will add three electrons in the 3p orbital. The electron configuration for P³⁻ is: \[1s^2 2s^2 2p^6 3s^2 3p^6\]
06

2. S²⁻ Electron Configuration

Sulfur (S) has an atomic number of 16. The electron configuration for the neutral S atom is: \[1s^2 2s^2 2p^6 3s^2 3p^4\] As S²⁻ is an anion with a 2- charge, we will add two electrons in the 3p orbital. The electron configuration for S²⁻ is: \[1s^2 2s^2 2p^6 3s^2 3p^6\]
07

3. Br⁻ Electron Configuration

Bromine (Br) has an atomic number of 35. The electron configuration for the neutral Br atom is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5\] As Br⁻ is an anion with a 1- charge, we will add one electron in the 4p orbital. The electron configuration for Br⁻ is: \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\]

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

Elements in the same family often form oxyanions of the same general formula. The anions are named in a similar fashion. Give the names of the oxyanions of selenium and tellurium: \(\mathrm{ScO}_{4}^{2-}, \mathrm{SeO}_{3}^{2-}, \mathrm{TeO}_{4}^{2-}, \mathrm{TeO}_{3}^{2-}\)

Arrange the following in order of increasing radius and increasing ionization energy. a. \(\mathrm{N}^{+}, \mathrm{N}, \mathrm{N}^{-}\) b. \(\mathrm{Se}, \mathrm{Se}^{-}, \mathrm{Cl}, \mathrm{Cl}^{+}\) c. \(\mathrm{Br}^{-}, \mathrm{Rb}^{+}, \mathrm{Sr}^{2+}\)

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. Al and Cl b. Na and O c. Sr and F d. Ca and Se

For each of the following ions, indicate the total number of protons and electrons in the ion. For the positive ions in the list, predict the formula of the simplest compound formed between each positive ion and the oxide ion. Name the compounds. For the negative ions in the list, predict the formula of the simplest compound formed between each negative ion and the aluminum ion. Name the compounds. a. \(\mathrm{Fe}^{2+}\) b. \(\mathrm{Fe}^{3+}\) c. \(B a^{2+}\) d. \(C s^{+}\) e. \(S^{2-}\) f. \(P^{3-}\) g. \(\mathrm{Br}^{-}\) \(\mathbf{h} . \mathbf{N}^{3-}\)

Arrange the atoms and/or ions in the following groups in order of decreasing size. a. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) b. \(\mathrm{Fe}^{2+}, \mathrm{Ni}^{2+}, \mathrm{Zn}^{2+}\) c. \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}\)

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