Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J.

Slides:



Advertisements
Similar presentations
Trends of the Periodic Table
Advertisements

Sample Exercise 6.1 Concepts of Wavelength and Frequency
Electron Configuration.  In atomic physics and quantum chemistry, electron configuration is the arrangement of electrons of an atom. electrons.
© 2014 Pearson Education, Inc. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro Example 8.1Electron Configurations a. Mg Magnesium has 12 electrons.
1 Chapter 3 Atoms and Elements 3.8 Periodic Trends Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.
Energy levelSublevel# of orbitals/sublevel n = 11s (l = 0)1 (m l has one value) n = 2 2s (l = 0) 1 (m l has one value) 2p (l = 1) 3 (m l has three values)
Chemistry 103 Lecture 8.
Chapter 5 - Electronic Structure and Periodic Trends
Chemistry Jeopardy Electrons, Electron Configuration, Electromagnetic spectrum, free for all.
Chapter 5 Electronic Structure and Periodic Trends
Unit 5: Atomic Structure
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J.
Chapter 6 The Periodic Table 6.3 Periodic Trends
Chapter 8 Review of Quantum Numbers Principal Quantum Number (n) -tells you the energy level -n can be equal to 1, 2, 3, 4, 5, 6, 7… -distance e- is from.
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 1 Elements are  pure substances that cannot be separated into simpler.
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J.
Introduction to bonding. Group 1 Li Na K Rb Cs Fr Group 2 Be Mg Ca Sr Ba Ra Group 7 F Cl Br I At All elements in the same group have the same number of.
Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J.
Introductory Chemistry, 2nd Edition Nivaldo Tro
Chapter 6 The Periodic Table 6.3 Periodic Trends
1 Chapter 4 Atoms and Elements 4.6 Electron Energy Levels · Be · · Mg · · Ca · · Sr · · Ba · Copyright © 2008 by Pearson Education, Inc. Publishing as.
Anything in black letters = write it in your notes (‘knowts’)
Periodic Table and Trends. Organization of the Periodic Table 1. First created by Dmitri Mendeleev A. Organized atoms by atomic masses B. This was ok,
Trends in Periodic Table Properties
Section 11.4 Electron Configurations and Atomic Properties 1.To understand how the principal energy levels fill with electrons in atoms beyond hydrogen.
Chapter 11 Modern Atomic Theory Practice Problems.
Periodic Trends OBJECTIVES:
Tro IC3 1.Frequency 2.Wavelength 3.Amplitude 4.Speed 5.Meter 9.1 The distance between two adjacent crests or troughs in a wave is called the:
Mr. Hollister Holliday Legacy High School Chemistry
1 Chapter 3 Atoms and Elements 3.4 The Atom Copyright © 2009 by Pearson Education, Inc.
Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Introductory Chemistry, 2 nd Edition Nivaldo Tro Chapter 9 Electrons in Atoms and the.
Periodicity questions. Arrange these atoms and ions in order of decreasing size: Mg 2+, Ca 2+, and Ca. Cations are smaller than their parent atoms, and.
Unit 3 – Electrons/PT Exam Review. 1.What is the next atomic orbital in the series: 1s, 2s, 2p, 3s, 3p, ? A. 3d B. 4s C. 4p D. 3f.
Electron Configurations. Electron Configuration __________– the arrangement of electrons in an atom.
Section 11.4 Electron Configurations and Atomic Properties 1.To understand how the principal energy levels fill with electrons in atoms beyond hydrogen.
Chapter 3 Atoms and Elements 3.7 Electron Energy Levels 1.
Steps for Drawing Atoms 1.Find the element on the periodic table. How many protons does it have? 2. Because atoms are neutral, the number of electrons.
Chapter 3 Atoms and Elements 3.7 Electron Energy Levels 1.
Chapter 5 - Electronic Structure and Periodic Trends Electromagnetic Radiation Atomic Spectra and Energy Levels Energy Levels, Sublevels, & Orbitals Orbital.
Chapter 3Atoms and Elements 3.6 Isotopes and Atomic Mass 1  24 Mg 25 Mg 26 Mg Copyright © 2009 by Pearson Education, Inc.
Chapter 4 Atoms and Elements Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.
Objectives To understand how the principal energy levels fill with electrons in atoms beyond hydrogen To learn about valence electrons and core electrons.
The Periodic Table Dimitri Mendeleev. I. Periods The rows ( ) on the periodic table are called PERIODS. The rows ( ) on the periodic table are called.
Copyright ©2008 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry: Concepts & Connections, Fifth.
UNIT 4: THE ATOM CONT. NOTES PACKET. Topic 1: The “Discovery” of the Atom – more detail Complete ChemNotes 4:1 Read article “Taking Apart the Light”;
ELECTRON CONFIGURATION Why are ions more stable than some neutral atoms?
Periodic Table Trends. Remember these groups of the periodic table?
1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 2 Lecture Outline Prepared by Jennifer N. Robertson-Honecker.
Trends on the Periodic Table
Electron Configuration, The Explanatory Power of the Quantum Mechanical Model, and Periodic Trends.
Shells and Subshells The orbitals in an atom are arranged in shells and subshells. Shell: all orbitals with the same value of n Subshell: all orbitals.
Chapter 6 The Periodic Table 6.3 Periodic Trends
Write the orbital notation and electron configuration for the following atoms Be B N F Mg.
Chapter 3 Atoms and Elements
Example Exercise 5.1 Periodic Law
Periodic Trends.
CHEMISTRY Trends and Configurations
Periodic Table.
Chapter 6 The Periodic Table 6.3 Periodic Trends
4.05 Atomic Structure and Electronic Configuration
Ionization Energy Atomic Radius Electron Affinity Electronegativity
Chapter 6 The Periodic Table 6.3 Periodic Trends
Chapter 6 The Periodic Table and Periodic Law.
Many-Electron Atoms We have to examine the balance of attractions and repulsions in the atom to explain why subshells of a given shell have different energies.
The Periodic Table Copyright © 2008 by Pearson Education, Inc.
Chapter 2 Atoms and Elements
Energy Levels & Orbitals
Electron Configuration of Atoms in their Ground State
Electron Configurations
The Periodic Table.
Presentation transcript:

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro (a) wavelength Examine Figure 9.4 to see that X-rays have the shortest wavelength, followed by visible light and then microwaves. EXAMPLE 9.1 Wavelength, Energy, and Frequency Solution : X-rays, visible light, microwaves Arrange the following three types of electromagnetic radiation—visible light, X-rays, and microwaves—in order of increasing: (a) wavelength (b) frequency (c) energy per photon (b) frequency Since frequency and wavelength are inversely proportional— the longer the wavelength, the shorter the frequency—the ordering with respect to frequency is exactly the reverse of the ordering with respect to wavelength. microwaves, visible light, X-rays (c) energy per photon Energy per photon decreases with increasing wavelength but increases with increasing frequency; therefore, the ordering with respect to energy per photon is the same as frequency microwaves, visible light, X-rays

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.1 Wavelength, Energy, and Frequency Continued Arrange the following colors of visible light—green, red, and blue—in order of increasing: (a) wavelength (b) frequency (c) energy per photon SKILLBUILDER 9.1 Wavelength, Energy, and Frequency FOR MORE PRACTICE Example 9.9; Problems 33, 34, 35, 36, 37, 38, 39, 40.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro (a) Magnesium has 12 electrons. Distribute two of these into the 1s orbital, two into the 2s orbital, six into the 2p orbitals, and two into the 3s orbital. You can also write the electron configuration more compactly using the noble gas core notation. For magnesium, we use [Ne] to represent 1s 2 2s 2 2p 6. EXAMPLE 9.2 Electron Configurations Solution : Mg 1s 2 2s 2 2p 6 3s 2 Or Mg [Ne]3s 2 Write electron configurations for each of the following elements. (a) Mg (b) S (c) Ga (b) Sulfur has 16 electrons. Distribute two of these into the 1s orbital, two into the 2s orbital, six into the 2p orbitals, two into the 3s orbital, and four into the 3p orbitals. You can write the electron configuration more compactly by using [Ne] to represent 1s 2 2s 2 2p 6. S 1s 2 2s 2 2p 6 3s 2 3p 4 Or S [Ne]3s 2 3p 4

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.2 Electron Configurations Continued Write electron configurations for each of the following elements. (a) Al (b) Br (c) Sr SKILLBUILDER 9.2 Electron Configurations (c) Gallium has 31 electrons. Distribute two of these into the 1s orbital, two into the 2s orbital, six into the 2p orbitals, two into the 3s orbital, six into the 3p orbitals, two into the 4s orbital, ten into the 3d orbitals, and one into the 4p orbitals. Notice that the d subshell has five orbitals and can therefore hold 10 electrons. You can write the electron configuration more compactly by using [Ar] to represent 1s 2 2s 2 2p 6 3s 2 3p 6. Ga 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 1 Or Ga [Ar]4s 2 3d 10 4p 1

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.2 Electron Configurations Continued FOR MORE PRACTICE Problems 51, 52, 55, 56, 57, 58. FOR MORE PRACTICE Write electron configurations for each of the following ions. (Hint: To determine the number of electrons to include in the electron configuration of an ion, make sure to add or subtract electrons as needed to account for the charge of the ion.) (a) Al 3+ (b) Cl – (c) O 2–

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.3 Writing Orbital Diagrams Write an orbital diagram for silicon. Write an orbital diagram for argon. SKILLBUILDER 9.3 Writing Orbital Diagrams FOR MORE PRACTICE Example 9.10; Problems 53, 54. Solution : Since silicon is atomic number 14, it has 14 electrons. Draw a box for each orbital, putting the lowest-energy orbital (1s) on the far left and proceeding to orbitals of higher energy to the right. Distribute the 14 electrons into the orbitals, allowing a maximum of 2 electrons per orbital and remembering Hund’s rule. The complete orbital diagram is:

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.4 Valence Electrons and Core Electrons Write an electron configuration for selenium and identify the valence electrons and the core electrons. Write an electron configuration for chlorine and identify the valence electrons and core electrons. SKILLBUILDER 9.4 Valence Electrons and Core Electrons FOR MORE PRACTICE Example 9.11; Problems 59, 60, 63, 64. Solution : Write the electron configuration for selenium by determining the total number of electrons from selenium’s atomic number (34) and then distributing them into the appropriate orbitals. Se 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4 The valence electrons are those in the outermost principal shell. For selenium, the outermost principal shell is the n = 4 shell, which contains 6 electrons (2 in the 4s orbital and 4 in the three 4p orbitals). All other electrons, including those in the 3d orbitals, are core electrons.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.5 Writing Electron Configurations from the Periodic Table Use the periodic table to write an electron configuration for arsenic. Use the periodic table to determine the electron configuration for tin. SKILLBUILDER 9.5 Writing Electron Configurations from the Periodic Table Solution : The noble gas that precedes arsenic in the periodic table is argon, so the inner electron configuration is [Ar]. Obtain the outer electron configuration by tracing the elements between Ar and As and assigning electrons to the appropriate orbitals. Remember that the highest n value is given by the row number (4 for arsenic). So, we begin with [Ar], then add in the two 4s electrons as we trace across the s block, followed by ten 3d electrons as we trace across the d block (the n value for d subshells is equal to the row number minus one), and finally the three 4p electrons as we trace across the p block to As, which is in the third column of the p block ( ◄ Figure 9.31). The electron configuration is: As [Ar]4s 2 3d 10 4p 3 FOR MORE PRACTICE Example 9.12; Problems 67, 68, 69, 70.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.6 Atomic Size Choose the larger atom from each of the following pairs. (a) C or O (b) Li or K (c) C or Al (d) Se or I Solution : (a) C or O Carbon atoms are larger than O atoms because, as we trace the path between C and O on the periodic table (see margin), we move to the right within the same period. Atomic size decreases as you go to the right. (b) Li or K Potassium atoms are larger than Li atoms because, as we trace the path between Li and K on the periodic table (see margin), we move down a column. Atomic size increases as you go down a column. (c) C or Al Aluminum atoms are larger than C atoms because, as we trace the path between C and Al on the periodic table (see margin), we move down a column (atomic size increases) and then to the left across a period (atomic size increases). These effects add together for an overall increase. (d) Se or I Based on periodic properties alone, we cannot tell which atom is larger, because as we trace the path between Se and I (see margin) we go down a column (atomic size increases) and then to the right across a period (atomic size decreases). These effects tend to cancel one another.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.6 Atomic Size Continued Choose the larger atom from each of the following pairs. (a) Pb or Po (b) Rb or Na (c) Sn or Bi (d) F or Se SKILLBUILDER 9.6 Valence Electrons and Core Electrons FOR MORE PRACTICE Example 9.13a; Problems 83, 84, 85, 86.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.7 Ionization Energy Choose the element with the higher ionization energy from each of the following pairs. (a) Mg or P (b) As or Sb (c) N or Si (d) O or Cl Solution : (a) Mg or P P has a higher ionization than Mg because, as we trace the path between Mg and P on the periodic table (see margin), we move to the right within the same period. Ionization energy increases as you go to the right. (b) As or Sb As has a higher ionization energy than Sb because, as we trace the path between As and Sb on the periodic table (see margin), we move down a column. Ionization energy decreases as you go down a column. (c) N or Si N has a higher ionization energy than Si because, as we trace the path between N and Si on the periodic table (see margin), we move down a column (ionization energy decreases) and then to the left across a period (ionization energy decreases). These effects sum together for an overall decrease. (d) O or Cl Based on periodic properties alone, we cannot tell which has a higher ionization energy because, as we trace the path between O and Cl (see margin), we go down a column (ionization energy decreases) and then to the right

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.7 Ionization Energy Continued Choose the element with the higher ionization energy from each of the following pairs. (a) Mg or Sr (b) In or Te (c) C or P (d) F or S SKILLBUILDER 9.7 Ionization Energy FOR MORE PRACTICE Example 9.13b; Problems 79, 80, 81, 82.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.8 Metallic Character Choose the more metallic element from each of the following pairs. (a) Sn or Te (b) Si or Sn (c) Br or Te (d) Se or I Solution : (a) Sn or Te Sn is more metallic than Te because, as we trace the path between Sn and Te on the periodic table (see margin), we move to the right within the same period. Metallic character decreases as you go to the right. (b) Si or Sn Sn is more metallic than Si because, as we trace the path between Si and Sn on the periodic table (see margin), we move down a column. Metallic character increases as you go down a column. (c) Br or Te Te is more metallic than Br because, as we trace the path between Br and Te on the periodic table (see margin), we move down a column (metallic character increases) and then to the left across a period (metallic character increases). These effects add together for an overall increase. (d) Se or I Based on periodic properties alone, we cannot tell which is more metallic because as we trace the path between Se and I (see margin), we go down a column (metallic character increases) and then to the right across a period (metallic character decreases). These effects tend to cancel.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.8 Metallic Character Continued Choose the more metallic element from each of the following pairs. (a) Ge or In (b) Ga or Sn (c) P or Bi (d) B or N SKILLBUILDER 9.8 Metallic Character FOR MORE PRACTICE Example 9.13; Problems 87, 88, 89, 90.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.9 Predicting Relative Wavelength, Energy, and Frequency of Light Which of the following types of light—infrared or ultraviolet—has the longer wavelength? Higher frequency? Higher energy per photon? Solution : Infrared light has the longer wavelength (Figure 9.4). Ultraviolet light has the higher frequency and the higher energy per photon.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.10 Writing Electron Configurations and Orbital Diagrams Write an electron configuration and orbital diagram (outer electrons only) for germanium. Solution : Germanium is atomic number 32; therefore, it has 32 electrons. Electron Configuration : Ge 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 2 Or Ge [Ar]4s 2 3d 10 4p 2 Orbital Diagram (Outer Electrons):

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.11 Identifying Valence Electrons and Core Electrons Identify the valence electrons and core electrons in the electron configuration of germanium (given in Example 9.10). Solution :

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.12 Writing Electron Configurations for an Element Based on Its Position in the Periodic Table Write an electron configuration for iodine based on its position in the periodic table. Solution : The inner configuration for I is [Kr]. We begin with the [Kr] inner electron configuration. As we trace from Kr to I, we add 2 5s electrons, 10 4d electrons, and 5 5p electrons. The overall configuration is: I [Kr]5s 2 4d 10 5p 5

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 9.13 Periodic Trends: Atomic Size, Ionization Energy, and Metallic Character Arrange Si, In, and S in order of (a) increasing atomic size, (b) increasing ionization energy, and (c) increasing metallic character. Solution : (a) S, Si, In (b) In, Si, S (c) S, Si, In