Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 102

Using the periodic table as a guide, write the condensed electron configuration and determine the number of unpaired electrons for the ground state of (a) $\mathrm{Cl},(\mathbf{b}) \mathrm{Al},(\mathbf{c}) \mathrm{Zr},(\mathbf{d})\( As, (e) \)\mathrm{Sb},(\mathbf{f}) \mathrm{W}$

Short Answer

Expert verified
The condensed electron configurations and number of unpaired electrons for the given elements are: a) Cl: [Ne] 3s² 3p⁵, 3 unpaired electrons b) Al: [Ne] 3s² 3p¹, 1 unpaired electron c) Zr: [Kr] 5s² 4d², 2 unpaired electrons d) As: [Ar] 4s² 3d¹⁰ 4p³, 3 unpaired electrons e) Sb: [Kr] 5s² 4d¹⁰ 5p³, 3 unpaired electrons f) W: [Xe] 6s² 4f¹⁴ 5d⁴, 4 unpaired electrons

Step by step solution

01

a) Electron Configuration and Unpaired Electrons for Chlorine (Cl)

Chlorine (Cl) has an atomic number (Z) of 17, so it has 17 electrons in its ground state. Following the order in the periodic table: 1s² 2s² 2p⁶ 3s²3p⁵ Condensed Electron Configuration: [Ne] 3s² 3p⁵ Number of Unpaired Electrons: Since the last subshell (3p) has five electrons, there are three unpaired electrons.
02

b) Electron Configuration and Unpaired Electrons for Aluminum (Al)

Aluminum (Al) has an atomic number (Z) of 13, so it has 13 electrons in its ground state. Following the order in the periodic table: 1s² 2s² 2p⁶ 3s² 3p¹ Condensed Electron Configuration: [Ne] 3s² 3p¹ Number of Unpaired Electrons: Since the last subshell (3p) has only one electron, there is one unpaired electron.
03

c) Electron Configuration and Unpaired Electrons for Zirconium (Zr)

Zirconium (Zr) has an atomic number (Z) of 40, so it has 40 electrons in its ground state. Following the order in the periodic table: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d² Condensed Electron Configuration: [Kr] 5s² 4d² Number of Unpaired Electrons: Since the last subshell (4d) has two electrons, there are two unpaired electrons.
04

d) Electron Configuration and Unpaired Electrons for Arsenic (As)

Arsenic (As) has an atomic number (Z) of 33, so it has 33 electrons in its ground state. Following the order in the periodic table: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³ Condensed Electron Configuration: [Ar] 4s² 3d¹⁰ 4p³ Number of Unpaired Electrons: Since the last subshell (4p) has three electrons, there are three unpaired electrons.
05

e) Electron Configuration and Unpaired Electrons for Antimony (Sb)

Antimony (Sb) has an atomic number (Z) of 51, so it has 51 electrons in its ground state. Following the order in the periodic table: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p³ Condensed Electron Configuration: [Kr] 5s² 4d¹⁰ 5p³ Number of Unpaired Electrons: Since the last subshell (5p) has three electrons, there are three unpaired electrons.
06

f) Electron Configuration and Unpaired Electrons for Tungsten (W)

Tungsten (W) has an atomic number (Z) of 74, so it has 74 electrons in its ground state. Following the order in the periodic table: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁴ Condensed Electron Configuration: [Xe] 6s² 4f¹⁴ 5d⁴ Number of Unpaired Electrons: Since the last subshell (5d) has four electrons, there are four unpaired electrons.

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

Identify the specific element that corresponds to each of the following electron configurations and indicate the number of unpaired electrons for each: (a) \(1 s^{2} 2 s^{2}\), (b) \(1 s^{2} 2 s^{2} 2 p^{4}\) (d) \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\) (c) \([\mathrm{Ar}] 4 s^{1} 3 d^{5}\)

Is energy emitted or absorbed when the following electronic transitions occur in hydrogen? (a) from \(n=3\) to \(n=2\), (b) from an orbit of radius \(0.846 \mathrm{nm}\) to one of radius 0.212 \(\mathrm{nm},(\mathbf{c})\) an electron adds to the \(\mathrm{H}^{+}\) ion and ends up in the \(n=2\) shell?

Determine whether each of the following sets of quantum numbers for the hydrogen atom are valid. If a set is not valid, indicate which of the quantum numbers has a value that is not valid: (a) \(n=3, l=3, m_{l}=2, m_{5}=+\frac{1}{2}\) (b) \(n=4, l=3, m_{l}=-3, m_{\mathrm{s}}=+\frac{1}{2}\) (c) \(n=3, l=1, m_{l}=2, m_{s}=+\frac{1}{2}\) (d) \(n=5, l=0, m_{l}=0, m_{\mathrm{s}}=0\) (e) \(n=2, l=1, m_{l}=1, m_{s}=-\frac{1}{2}\)

An experiment called the Stern-Gerlach experiment helped establish the existence of electron spin. In this experiment, a beam of silver atoms is passed through a magnetic field, which deflects half of the silver atoms in one direction and half in the opposite direction. The separation between the two beams increases as the strength of the magnetic field increases. (a) What is the electron configuration for a silver atom? (b) Would this experiment work for a beam of cadmium (Cd) atoms? (c) Would this experiment work for a beam of fluorine (F) atoms?

State where in the periodic table these elements appear: (a) elements with the valence-shell electron configuration \(n s^{2} n p^{5}\) (b) elements that have three unpaired \(p\) electrons (c) an element whose valence electrons are \(4 s^{2} 4 p^{1}\) (d) the \(d\) -block elements [Section 6.9\(]\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

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