Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 51

How many unpaired electrons are in the following ions? (a) \(\mathrm{Hg}^{2+}\) (b) \(\mathrm{F}^{-}\) (c) \(\mathrm{Sb}^{3+}\) (d) \(\mathrm{Fe}^{3+}\)

Short Answer

Expert verified
Answer: (a) Hg2+ ion has 0 unpaired electrons, (b) F- ion has 0 unpaired electrons, (c) Sb3+ ion has 0 unpaired electrons, and (d) Fe3+ ion has 5 unpaired electrons.

Step by step solution

01

(a) Hg2+ ion

Step 1: Find atomic number of Hg Hg (mercury) is in the periodic table with an atomic number of 80. Step 2: Determine electron configuration of the neutral atom The electron configuration of the neutral Hg atom is \(\mathrm{[Xe] 4f^{14} 5d^{10} 6s^2}\). Step 3: Electron configuration of the ion For the \(\mathrm{Hg}^{2+}\) ion, we need to remove 2 electrons. They are removed from the highest energy level, which is 6s. So, the electron configuration of the ion is \(\mathrm{[Xe] 4f^{14} 5d^{10}}\). Step 4: Count unpaired electrons There are no unpaired electrons in \(\mathrm{Hg}^{2+}\) ion as all the electrons are in fully-filled orbitals.
02

(b) F- ion

Step 1: Find atomic number of F F (fluorine) is in the periodic table with an atomic number of 9. Step 2: Determine electron configuration of the neutral atom The electron configuration of the neutral F atom is \(\mathrm{1s^2 2s^2 2p^5}\). Step 3: Electron configuration of the ion For the \(\mathrm{F}^{-}\) ion, we need to add 1 electron. It will be added to the 2p orbital. So, the electron configuration of the ion is \(\mathrm{1s^2 2s^2 2p^6}\). Step 4: Count unpaired electrons There are no unpaired electrons in \(\mathrm{F}^{-}\) ion as all the electrons are in fully-filled orbitals.
03

(c) Sb3+ ion

Step 1: Find atomic number of Sb Sb (antimony) is in the periodic table with an atomic number of 51. Step 2: Determine electron configuration of the neutral atom The electron configuration of the neutral Sb atom is \(\mathrm{[Kr] 4d^{10} 5s^2 5p^3}\). Step 3: Electron configuration of the ion For the \(\mathrm{Sb}^{3+}\) ion, we need to remove 3 electrons. They are removed first from the 5s orbital, and then from the 5p orbital. So, the electron configuration of the ion is \(\mathrm{[Kr] 4d^{10} 5p^0}\). Step 4: Count unpaired electrons There are no unpaired electrons in \(\mathrm{Sb}^{3+}\) ion as all the electrons are in fully-filled orbitals.
04

(d) Fe3+ ion

Step 1: Find atomic number of Fe Fe (iron) is in the periodic table with an atomic number of 26. Step 2: Determine electron configuration of the neutral atom The electron configuration of the neutral Fe atom is \(\mathrm{[Ar] 3d^6 4s^2}\). Step 3: Electron configuration of the ion For the \(\mathrm{Fe}^{3+}\) ion, we need to remove 3 electrons. They are removed first from the 4s orbital, and then from the 3d orbital since 4s orbital has slightly higher energy level. So, the electron configuration of the ion is \(\mathrm{[Ar] 3d^5}\). Step 4: Count unpaired electrons There are 5 unpaired electrons in \(\mathrm{Fe}^{3+}\) ion as all the electrons are in the 3d orbitals and none of them are fully-filled.

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

Write the ground state electron configuration for (a) \(\mathrm{B}\) (b) \(\mathrm{Ba}\) (c) \(\mathrm{Be}\) (d) \(\mathrm{Bi}\) (e) \(\mathrm{Br}\)

How many electrons in an atom can have each of the following quantum number designations? (a) \(\mathbf{n}=2, \ell=1, \mathbf{m}_{\ell}=0\) (b) \(\mathbf{n}=2, \ell=1, \mathbf{m}_{\ell}=-1\) (c) \(\mathbf{n}=3, \ell=1, \mathbf{m}_{\ell}=0, \mathbf{m}_{s}=+\frac{1}{2}\)

Which of the four atoms \(\mathrm{Rb}, \mathrm{Sr}, \mathrm{Sb}\), or \(\mathrm{Cs}\) (a) has the smallest atomic radius? (b) has the lowest ionization energy? (c) is the least electronegative?

Magnetic resonance imaging (MRI) is a powerful diagnostic tool used in medicine. The imagers used in hospitals operate at a frequency of \(4.00 \times 10^{2} \mathrm{MHz}\left(1 \mathrm{MHz}=10^{6} \mathrm{~Hz}\right) .\) Calculate (a) the wavelength. (b) the energy in joules per photon. (c) the energy in kilojoules per mole.

In the Paschen series, \(\mathbf{n}_{\text {lo }}=3\). Calculate the longest wavelength possible for a transition in this series.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

Nuclear Chemistry

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