Monday March 18, 2024 Day 40 Orbital shapes and Energies |
|
Textbook Readings 7.6: The Shape of Atomic Orbitals |
Course Lectures 7.6 pdf Video Quantum Mechanical Atom 7.9 pdf Video* Atomic Orbitals |
Objectives 1. Describe what is meant by the wavefunction. 2. Distinguish between the probability density graph and the radial probability graph and how they're related. |
The Shapes of Atomic Orbitals |
3. Interpret the probability density graph a.k.a Ψ2 vs. radial distance) and what it suggests about the probablity of finding an electron in an atom. 4.Interpret the radial probability graph (Ψ2r2 vs radial distance) and what it suggests about the probablity of finding an electron in an atom. 5. Be able to identify the shapes and quantum numbers of s, p, d and f orbitals |
|
Homework Problems 42. 1 What is the wavefunction and what is it good for? 42.2 Probability density can be observed at a concert. Where would you expect to find the highest density of people in the audience? Is that the place you'd most likely expect to find someone wearing a red cap? 42.3 When squared, the wavefunction for an electron in an atom yields the probability density (Figure at right). Probability density is volume dependant The rookie chemistry student might look at the probability density graph and suggest incorrectly the electron would be found where in an atom? | |
42.4 When the available space is included in the quantum mechanical atom, the probability of find an electron in the nucleus is practically zero since the nucleus is so small. The Radial Probability is a graph that includes the volume factor (figure at right). Where would the rookie chemistry student correctly predict the highest likelyhood of finding an electron. Draw this graph in your notebook and indicate your answer. |
|
42.5. Orbitals are typically drawn to represent the likelyhood of finding an electron 90% of the time. What percent of the time would we expect to find the electron outside the orbital? 42.6 Nodes are regions within the atom where it is unlikely for an electron to be found. Nodes exist within orbitals and are most easily seen in "s" orbitals. Use your textbook and determine for s orbitals how the number of nodes depends on the n quantum number. |
|
42.7 Identify each of the orbitals at right as "s", "p", "d" or "f". 42.8 For each orbital, write quantum number combinations that could represent the orbital n, l and ml. (Multiple correct answers are possible but they must follow the quantum number rules!) |
By Geek3 - Own work; created with hydrogen-cloud in PythonThis PNG graphic was created with Python., CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=69035277 |
Click and drag the region below for correct answers 42.1 The wavefunction is a solution to the Schodinger equations. The wavefunction includes the particle and wave nature of matter and can be used to, amongst other things, predict the probability or likelyhood of finding an electron in a specific place within an atom. 42.2 My experience is that the highest density of concert-goers is right up next to the stage. People tend to get very tightly packed. However, if you're looking for a red hat, it is much more likely you'll find one if you expand your search to areas away from the stage where most of the people are. 42.3 The rookie chemist would incorrectly pick the place where the curve is highest (i.e. the stage) which is on the nucleus. 42.4 This graph incorporates density and available space. The electron would most likely be found at the graph's peak. However, note that the graph predicts the electron can be found in other locations with lower probability. 42.5 It is 10% likely that an electron will be found outside of the orbital. 42.6 The number of nodes for s orbitals is n - 1 42.7 a. 1s b. 2s c. 2p d. 3s e. 3p f. 3d g. 4s h. 4p i. 4d j. 4f 42.8 Answers may vary but all must follow the rules: a. n = 1 b. l = 0 c. ml = 0 b. n = 2 b. l = 0 c. ml = 0 c. n = 2 b. l = 1 c. ml = -1 d. n = 3 b. l = 0 c. ml = 0 e. n = 3 b. l = 1 c. ml = +1 f. n = 3 b. l = 2 c. ml = -2 g. n = 4 b. l = 0 c. ml = 0 h. n = 4 b. l = 1 c. ml = 0 i. n = 4 b. l = 2 c. ml = -1 j. n = 4 b. l = 3 c. ml = +3 |
Tuesday March 19, 2024 Day 41 Electron Configurations |
|
Textbook Readings 8.1: Nerve Signal Transmission 8.2: The Development of the Periodic Table 8.3: Electron Configurations: How Electrons Occupy Orbitals |
Course Lecture 8.1. pdf Video* Electrons in atoms 8.2. pdf Video* Electron Energies 8.3. pdf Video* Electron Energies and Shielding 8.4. pdf Video* Quantum Mechanical Electron Energies 8.5. pdf Video* Aufbau Principle and Electron Configurations |
Objectives 1. Know how to fill atomic energy levels from the low to high energies utilizing the Aufbau principle, the Pauli Exclusion Principle, and Hund's rule. 2. Utilize the electron spin quantum number, ms, and the Pauli Exclusion Principle to fill suborbitals with electrons. 3. Write electron configurations for neutral atoms from scratch and using the periodic table. |
Quantum Numbers, Atomic Orbitals, and
Electron Configurations (Entire Video) |
4. Given electron an electron configuration, determine the element that belongs with it. 5. Identify the important regions on the periodic table: Metals, Nonmetals, and Semimetals Alkali metals, Alkaline Earth Metals, Halogens and Noble Gases Main Group elements, Transition elements, Inner transition elements |
|
Homework Problems 43.1 What property of electrons gives them their magnetic characteristic? 43.2 What is the maximum number of electrons that can be found in a 3s atomic orbital? 43.3 What is the Pauli Exclusion Principle? 43.4 What are the spin orientations for the electrons in a 3s orbital? ...what are their ms quantum numbers? 43.5 A 3s orbital with two spin up electrons is ... a. Impossible b. discouraged c. possible under extraordinary circumstances d.common 43.6 Draw the orbital diagram and write the electron configuration of a neutral atom that has 16 electrons. 43.7 Draw the orbital diagram and write the electron configuration of a neutral atom that has 19 electrons. 43.8 Draw the orbital diagram and write the electron configuration of a neutral atom that has 22 electrons. 43.9 Draw the orbital diagram and write the electron configuration of a neutral atom that has 50 electrons. (Yes, I know. Ouch.) For the following problems, use your
periodic table to determine the electron configurations
of the elements that are specified. 43.10 Write the *shorthand electron configuration for a neutral nickel atom. 43.11 Write the *shorthand electron configuration for a neutral germanium atom. 43.12 Write the *shorthand electron configuration for a neutral cesium atom. 43.13 Write the *shorthand electron configuration for a neutral bromine atom. * Shorthand refers to
using the nearest noble gas in brackets to abbreviate as much of the
electron configuration as possible. Click and drag the region below for correct answers 43.1 Being charged, electrons spin giving them N and S poles. The poles reverse if the electron spins in the opposite direction. 43.2 The 3s orbital can hold no more than 2 electrons. 43.3 If an orbital is full (2 electrons) the electrons must have opposite spin and ms quantum numbers 43.4 One electron is spin up ↑ the other electron is spin down ↓. ms= +1/2 and ms = -1/2s 43.5 43.6 1s2 2s2 2p6 3s2 3p4 Click here for orbital diagram 43.7 1s2 2s2 2p6 3s2 3p6 4s1 Click here for orbital diagram 43.8 1s2 2s2 2p6 3s2 3p64s2 3d2 Click here for orbital diagram 43.9 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2 Click here for orbital diagram 43.10 [Ar] 4s2 3d8 43.11 [Ar] 4s2 3d104p2 43.12 [Xe] 6s1 43.13 [Ar] 4s2 3d104p5 |
Wednesday March 20, 2024 Day 42 Electron Configurations for Ions, Exceptions and Valence Electrons |
|
Textbook Readings 8.3: Electron Configurations: How Electrons Occupy Orbitals |
Course Lectures: 8.6. pdf Video* Electron Configurations and the Periodic Table: |
Electron configurations for ions |
Electron Configurations, Exceptions, and Valence Electrons |
Objectives 1. To be able to write electron configurations for anions and cations. 2. You will recognize situations where an exception to the Aufbau principle exists. 3. You will know how to derive electron configs for Mo, Cu, Ag, Cr and Au. 4. You will be able to write an electron configuration and from it identify the number of core, outer and valence electrons. |
How to Find the Inner, Outer and Valence
Electrons of an Element |
Homework Problems 44.1 Write the complete electron configuration for a chlorine ion: Cl-. (Source) 44.2 Write the complete electron configuration for an oxygen ion: O2- (Source) 44.3 Write the complete electron configuration for a chromium III ion: Cr3+ (Source) 44.4 Write the complete electron configuration for an iron III ion: Fe3+ (Source) 44.5 For each of the following elements, write out the electron configuation you would expect based upon the simple application of the periodic table, and s/p/d blocks. a. Molybdenum b. Copper c. Silver d. Chromium e. Gold 44.6 Each of the elements from 44.5 is actually an exception to the a straightforward application of the Aufbau principle. For each element write the actual electron configuration and give the reason it is lowest in energy and therefore preferred. 44.7 What are ... a. Core electrons b. Outer electrons c. Valence electrons 44.8 For each of the following neutral atoms, write out the electron configurations using the [Noble Gas] abbreviation. Then identify the number of core electrons, outer electrons and valence electrons. a. Boron b. Chlorine c. Krypton d. Copper e. Iron f. Silver g. Magnesium Click and drag the region below for correct answers 44.1 1s2 2s2 2p6 3s2 3p6 44.2 1s2 2s2 2p6 44.3 1s2 2s2 2p6 3s2 3p6 3d3 44.4 1s2 2s2 2p6 3s2 3p6 3d5 44.5 a. Mo Expected: [Kr] 4d4 5s2 b. Cu Expected: [Ar] 3d9 4s2 c. Ag Expected: [Kr] 4d9 5s2 d. Cr Expected: [Ar] 3d4 4s2 e. Au Expected: [Xe] 4f14 5d9 6s2 Did you remember the 4f electrons in your electron configuration? 44.6 a. Mo Actual: [Kr] 4d5 5s1 moving an electron from 5s to 4d leaves one level "half full" and the other "half full" and this situation is lower in energy. b. Cu Actual: [Ar] 3d10 4s1 moving an electron from 4s to 3d leaves one level "full" and the other "half full" and this situation is lower in energy. c. Ag Actual: [Kr] 4d10 5s1 moving an electron from 5s to 4d leaves one level "full" and the other "half full" and this situation is lower in energy. d. Cr Actual: [Ar] 3d5 4s1 moving an electron from 4s to 3d leaves one level "half full" and the other "half full" and this situation is lower in energy e. Au Actual: [Xe] 4f14 5d10 6s1 moving an electron from 6s to 5d leaves one level "full" and the other "half full" and this situation is lower in energy. 44.7 Core electrons occupy the lost energies and innermost orbitals Outer electrons occupy the largest and outmost orbitals Valence electrons are the outer electrons that can be either transferred or shared with another atom. Typically, these electrons are everything beyond the noble gas core. 44.8 a. Boron [He] 2s2 2p1 2 core electrons, 3 outer electrons, 3 valence electrons b. Chlorine [Ne] 3s2 3p5 10 core electrons, 7 outer electrons, 7 valence electrons c. Krypton [Ar] 3d10 4s2 4p6 18 core electrons, 8 outer electrons, 18 valence electrons d. Copper [Ar] 3d10 4s1 18 core electrons, 1outer electron, 11 valence electrons e. Iron [Ar] 3d6 4s2 18 core electrons, 2outer electrons, 8 valence electrons f. Silver [Kr] 4d10 5s1 36 core electrons, 1 outer electron, 11 valence electrons g. Magnesium [Ne] 3s2 10 core electrons, 2 outer electrons, 2 valence electrons |
Thursday March 21, 2024 Day 43 Trends in Atomic & Ionic Radii and the Effects of Nuclear Shielding |
|
Textbook Reading 8.6: Periodic Trends in the Size of Atoms and Effective Nuclear Charge |
Course Lecture 8.7. pdf Video* Atomic Trends and Ionization Energy 8.9. pdf Video Atomic Trends, Ionic Size & Magentism |
Atomic Radii Trends |
Effective Nuclear Charge, Shielding effect,
& Periodic Properties |
Objectives 1. Describe the methods used to determine the radius of atoms. 2. Use the periodic table and the vertical and horizontal trends to predict relative atomic sizes. 3. Describe how nuclear shielding works and the effects it has on atomic radii. 4. Create isoelectronic series of atoms and ions and order them from smallest to largest. |
Isoelectronic Series and the Relative Sizes
of Ions |
Homework Problems 45.1 What are the three ways atomic radii are "measured?" Draw a picture that demonstrates each. 45.2 In what corner of the periodic table do you find the smallest neutral atoms? In what corner of the periodic table do you find the largest neutral atoms? 45.3 Use your periodic table to predict which of the following would have the largest atomic radius. a. Li b. K c. Rb d. Fr e. Cs 45.4 Use your periodic table to predict which of the following would have the smallest atomic radius. a. Zn b. Sc c. Cr d. Cu e. Ni 45.5 Use your periodic table to predict which of the following would have the largest atomic radius. a. N b. O c. P d. S 45.6 In your own words describe what is meant by "nuclear shielding." 45.7 For each of the following neutral atoms, determine i. Number of protons ii. Number of core electrons iii. Zeff a. Li b. Na c. Be d. Mg 45.8 Generally speaking, each of the following generally produces a larger atom. Why? a. increased shielding of core electrons b. greater numbers of outermost electrons c. Occupancy of larger orbitals (e.g. 1s .... 2s ... etc.) 45.9 For each of the following isoelectronic species, give the electron configuration and calculate Zeff . Based on these results, which should be the smallest? a. F- b. Ne c. Na+ 45.10 For each of the following, determine the electronic configuration and the . Then use these results to arrange them in order of increasing radii. Ca2+ Cl - S2- P3- K+ 45.11 How does orbital "penetration" explain the fact that the 2s orbital is lower in energy than the 2 p orbital? Click and drag below for answers: 45.1 Covalent radius: distance between atom centers connected by covalent bond. Metalic radius: distance between atom centers arranged in a metalic solid. Van der Waals: distance between atom centers in closely packed atomic arrangement an approximation used when solids aren't normally found as in the noble gases. 45.2 The largest atomic species are located in the lower left hand corner of the periodic table. The smallest atomic species are located in the upper right hand corner of the periodic table. 45.3 d 45.4 a 45.5 c 45.6 Nuclear shielding occurs when negatively charged core electrons protect outermost electrons from the full attractive "pull" of the positively charged nucleus. They do this by effectively neutralizing a portion of the nucleus positive charge responsible for the atom's holding onto electrons. 45.7 a. Li: P+ = 3 Core e- = 2 Zeff = +1 b. Na: P+ = 11 Core e- = 10 Zeff = +1 c. Be: P+ = 4 Core e- = 2 Zeff = +2 d. Mg P+ = 12 Core e- = 10 Zeff = +2 45.8 a. Increased shielding means outermost electrons don't experience the full attractive pull of the nucleus. This let's them be further away from the nucleus and the atom is therefore bigger. b. Electrons repel eachother. Adding more electrons to the same orbital introduces more repulsivity within the orbital and it "puffs up" and gets bigger. c. Higher energy orbitals (think 1s ... 2s ... 3s ... etc) are larger. As they're occupied, a larger atom is the result. 45.9 a. F- ... 1s2 2s2 2p6 Zeff = +7 b. Ne ... 1s2 2s2 2p6 Zeff = +8 c. Na+ ... 1s2 2s2 2p6 Zeff = +9 (Larger Zeff produces greater electron attraction and smaller species) 45.10 All are isoelectronic and therefore have the same electron configuration! Ca2+ ... 1s2 2s2 2p6 3s2 3p6 Zeff = +10 (smallest) K+ ... 1s2 2s2 2p6 3s2 3p6 Zeff = +9 Cl - ... 1s2 2s2 2p6 3s2 3p6 Zeff = +7 S2- ... 1s2 2s2 2p6 3s2 3p6 Zeff = +6 P3- ... 1s2 2s2 2p6 3s2 3p6 Zeff = +5 (Largest) 45.11 The 2s orbital has a region of total probabilty that gets in close to the nucleus (labelled penetration on the figure) and this is something the 2p orbital lacks. Since electrons have less potential energy when positioned closer to the nucleus, the 2s orbital is slightly lower in energy than the 2p orbital. |
Friday March 22, 2024 Day 44 Periodic Trends: IONIZATION ENERGY |
|
Textbook Readings 8.7: Ions- Configurations, Magnetic Properties, Radii, and Ionization Energy |
Course Lectures 8.7. pdf Video* Atomic Trends and Ionization Energy |
Ionization Energy* |
The Periodic Table: Atomic Radius,
Ionization Energy, and Electronegativity (Through 5:36) |
Objectives 1. Describe ionization of atoms in words and with "equations." 2. Write first, second, third etc. ionization equations and describe how and why the ionization energies are different. 3. Relate ionization energies to the size of the atom. 4. Describe how ionization energies are affected by 1/2 full and full orbitals. 5. Identify an element based on a list of its ionization energies. |
Practice Problem: Ionization Energy |
Homework Problems 47.1 In your own words describe what ionization energy is and how it depends on the size of an atom or ion. 47.2 In your own words describe how ionization energy depends on the number of electrons in the atomic orbital (i.e. full orbital, half filled orbital, or partially filled orbital). 47.3 Consider the ficticious element "X". Write equations for the first, second and third ionization of this element. Arrange the respective ionization energies (I.E.) in increasing order. 47.4 Using ONLY a periodic table, predict which of the following elements has the highest ionization energy. Explain your answer. a. hydrogen b. helium c. neon d. Calcium 47.5 Using ONLY a periodic table, predict which element in each of the following pairs has the larger first ionization energy. Explain your answer. a. Na or Mg b. Mg or Al c. F or Cl 47.6 What period 3 element would have the following successive ionization energies? IE1 = 786 kJ IE2 = 1580 kJ IE3 = 3230 kJ IE4 = 4360 kJ IE5 = 16000 kJ 47.7 Generally, we expect ionization energies to increase as we consider element left to right on the periodic table. Based on this, we'd expect oxygen's first ionization energy (IE1) to be greater than nitrogen's IE1. Write the electron configurations for neutral oxygen and nitrogen atoms and then explain why oxygen's first ionization energy is actually less than nitrogen's? 47.8 Which of the following atoms would you expect to have the largest IE3 ? Why? a. Na b. Mg c. Al d. Si f. P 47.9 Arrange the folloing in order of decreasing ionization energy: IE1 for Al IE1 for Tl IE2 for Na IE3 for Al Explain your answer. Click and drag the region below for correct answers 47.1 Ionization energy is the energy required to remove an electron. Small atoms generally have high ionization energies 47.2 1/2 or full orbitals are low in energy and preferred. Removing an electron from such an orbital requires a lot of energy in comparison to removing an electron from a partially filled orbital. 47.3 First ionization: X → X+ + e- IE1 Second ionization: X+ → X2+ + e- IE2 Third ioinization: X2+ → X3+ + e- IE3 IE1 < IE2 < IE3 47.4 Helium: Why? 47.5 a. Mg b. Mg c. F Why? 47.6 Silicon Why? 47.7 Oxygen 1s2 2s2 2p4 Nitrogen:1s2 2s2 2p3 Removing an electron from nitrogen means going from a 1/2 full p orbital to a partially filled p orbital. This requires more energy as it isn't preferred. Removing an electron from the oxygen 2p orbital CREATES a 1/2 filled situation and as that's preferred, it's easier and the ionization energy is less. 47.8 IE2 for Na > IE3 for Al > IE1 for Al > IE1 for Tl 47.9 See Dr. Dave's video and practice problem above for the answers to this question. |