Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 7

Give the electron configurations for the following ions: \(\mathrm{Fe}^{2+}, \mathrm{Al}^{3+}, \mathrm{Cu}^{+}, \mathrm{Ba}^{2+}, \mathrm{Br}^{-}\), and \(\mathrm{O}^{2-}\)

Short Answer

Expert verified
Question: Determine the electron configurations for the following ions: \(\mathrm{Fe}^{2+}, \mathrm{Al}^{3+}, \mathrm{Cu}^{+}, \mathrm{Ba}^{2+}, \mathrm{Br}^{-},\) and \(\mathrm{O}^{2-}\). Answer: The electron configurations for the given ions are: \(\mathrm{Fe}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^4\) \(\mathrm{Al}^{3+}\): \(1s^2 2s^2 2p^6 3s^0 3p^0\) \(\mathrm{Cu}^{+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^{10}\) \(\mathrm{Ba}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^0 4d^{10} 5p^6 6s^0\) \(\mathrm{Br}^{-}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) \(\mathrm{O}^{2-}\): \(1s^2 2s^2 2p^6\)

Step by step solution

01

Find the atomic number of each element

Look up the atomic number of each element on the periodic table: \(\mathrm{Fe}\) - atomic number 26, \(\mathrm{Al}\) - atomic number 13, \(\mathrm{Cu}\) - atomic number 29, \(\mathrm{Ba}\) - atomic number 56, \(\mathrm{Br}\) - atomic number 35, \(\mathrm{O}\) - atomic number 8
02

Determine the electron configuration for each neutral atom

Use the Aufbau principle and Pauli exclusion principle to fill the orbitals according to the atomic number. The electron configurations for the neutral atoms are: \(\mathrm{Fe}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6\) \(\mathrm{Al}\): \(1s^2 2s^2 2p^6 3s^2 3p^1\) \(\mathrm{Cu}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}\) \(\mathrm{Ba}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2\) \(\mathrm{Br}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^5\) \(\mathrm{O}\): \(1s^2 2s^2 2p^4\)
03

Modify the electron configurations based on the ion's charge

For positive ions (cations), remove electrons from the highest-energy orbital first. For negative ions (anions), add electrons to the lowest-energy available orbital. The electron configurations for the ions are: \(\mathrm{Fe}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^4\) \(\mathrm{Al}^{3+}\): \(1s^2 2s^2 2p^6 3s^0 3p^0\) \(\mathrm{Cu}^{+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^0 3d^{10}\) \(\mathrm{Ba}^{2+}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^0 4d^{10} 5p^6 6s^0\) \(\mathrm{Br}^{-}\): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) \(\mathrm{O}^{2-}\): \(1s^2 2s^2 2p^6\)

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.

Start your free trial

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

For \(\mathrm{a} \mathrm{K}^{+}-\mathrm{Cl}^{-}\)ion pair, attractive and repulsive energies \(E_{A}\) and \(E_{R}\), respectively, depend on the distance between the ions \(r\), according to $$ \begin{aligned} E_{A} &=-\frac{1.436}{r} \\ E_{R} &=\frac{5.86 \times 10^{-6}}{r^{9}} \end{aligned} $$ For these expressions, energies are expressed in electron volts per \(\mathrm{K}^{+}-\mathrm{Cl}^{-}\)pair, and \(r\) is the distance in nanometers. The net energy \(E_{N}\) is just the sum of the preceding two expressions. (a) Superimpose on a single plot \(E_{N}, E_{R}\), and \(E_{A}\) versus \(r\) up to \(1.0 \mathrm{~nm}\). (b) On the basis of this plot, determine (i) the equilibrium spacing \(r_{0}\) between the \(\mathrm{K}^{+}\)and \(\mathrm{Cl}^{-}\)ions, and (ii) the magnitude of the bonding energy \(E_{0}\) between the two ions. (c) Mathematically determine the \(r_{0}\) and \(E_{0}\) values using the solutions to Problem \(2.14\) and compare these with the graphical results from part (b).

Cite the difference between atomic mass and atomic weight.

(a) Briefly cite the main differences between ionic, covalent, and metallic bonding. (b) State the Pauli exclusion principle.

Chromium has four naturally occurring isotopes: \(4.34 \%\) of \({ }^{50} \mathrm{Cr}\), with an atomic weight of \(49.9460\) amu; \(83.79 \%\) of \({ }^{52} \mathrm{Cr}\), with an atomic weight of \(51.9405 \mathrm{amu} ; 9.50 \%\) of \({ }^{53} \mathrm{Cr}\), with an atomic weight of \(52.9407 \mathrm{amu} ;\) and \(2.37 \%\) of \({ }^{54} \mathrm{Cr}\), with an atomic weight of \(53.9389\) amu. On the basis of these data, confirm that the average atomic weight of \(\mathrm{Cr}\) is \(51.9963 \mathrm{amu}\)

6 Allowed values for the quantum numbers of electrons are as follows: $$ \begin{aligned} n &=1,2,3, \ldots \\ l &=0,1,2,3, \ldots, n-1 \\ m_{l} &=0, \pm 1, \pm 2, \pm 3, \ldots, \pm l \\ m_{s} &=\pm \frac{1}{2} \end{aligned} $$ The relationships between \(n\) and the shell designations are noted in Table 2.1. Relative to the subshells, \(l=0\) corresponds to an \(s\) subshell \(l=1\) corresponds to a \(p\) subshell \(l=2\) corresponds to a \(d\) subshell \(l=3\) corresponds to an \(f\) subshell For the \(K\) shell, the four quantum numbers for each of the two electrons in the \(1 s\) state, in the order of \(n l m_{i} m_{s}\), are \(100 \frac{1}{2}\) and \(100\left(-\frac{1}{2}\right)\) Write the four quantum numbers for all of the electrons in the \(L\) and \(M\) shells, and note which correspond to the \(s, p\), and \(d\) subshells.
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Solid State Physics

Read Explanation

Dynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Thermodynamics

Read Explanation

Conservation of Energy and Momentum

Read Explanation
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