Chapter 6 Electronic Structure of Atoms Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 6 Electronic Structure of Atoms John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc.
Light & Waves Light: Crash Course Astronomy #24 Watch this video as an introduction to this unit. Light, Energy, Wavelength & Spectroscopy
Waves To understand the electronic structure of atoms, one must understand the nature of electromagnetic radiation. The distance between corresponding points on adjacent waves is the wavelength () and is measured in meters (m).
Waves The number of waves passing a given point per unit of time is the frequency () and is measured in hertz (Hz) or “per second” (s-1). For waves traveling at the same velocity, the longer the wavelength, the smaller the frequency.
Electromagnetic Radiation All electromagnetic radiation travels at the same velocity: the speed of light (c), 3.00 108 m/s. Therefore, c =
SAMPLE EXERCISE 6.1 Concepts of Wavelength and Frequency Two electromagnetic waves are represented in the image below. (a) Which wave has the higher frequency? (b) If one wave represents visible light and the other represents infrared radiation, which wave is which? Solution (a) The lower wave has a longer wavelength (greater distance between peaks). The longer the wavelength, the lower the frequency ( = c/ ). Thus, the lower wave has the lower frequency, and the upper one has the higher frequency. (b) The electromagnetic spectrum (Figure 6.4) indicates that infrared radiation has a longer wavelength than visible light. Thus, the lower wave would be the infrared radiation.
SAMPLE EXERCISE 6.1 Concepts of Wavelength and Frequency PRACTICE EXERCISE If one of the waves in the image above represents blue light and the other red light, which is which? Answer: The expanded visible-light portion of Figure 6.4 tells you that red light has a longer wavelength than blue light. The lower wave has the longer wavelength (lower frequency) and would be the red light.
SAMPLE EXERCISE 6.2 Calculating Frequency from Wavelength The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation? Solve: Solving Equation 6.1 for frequency gives = c/ . When we insert the values for c and , we note that the units of length in these two quantities are different. We can convert the wavelength from nanometers to meters, so the units cancel: PRACTICE EXERCISE (a) A laser used in eye surgery to fuse detached retinas produces radiation with a wavelength of 640.0 nm. Calculate the frequency of this radiation. (b) An FM radio station broadcasts electromagnetic radiation at a frequency of 103.4 MHz (megahertz; MHz = 106 s–1). Calculate the wavelength of this radiation. Answers: (a) 4.688 1014 s–1, (b) 2.901 m
The Nature of Energy The wave nature of light does not explain how an object can glow when its temperature increases. Max Planck explained it by assuming that energy comes in packets called quanta.
The Nature of Energy Einstein used this assumption to explain the photoelectric effect. He concluded that energy is proportional to frequency: E = h where h is Planck’s constant, 6.63 10−34 J-s.
The Nature of Energy Therefore, if one knows the wavelength of light, one can calculate the energy in one photon, or packet, of that light: c = E = h
SAMPLE EXERCISE 6.3 Energy of a Photon Calculate the energy of one photon of yellow light whose wavelength is 589 nm. Analyze: Our task is to calculate the energy, E, of a photon, given 589 nm. Plan: We can use Equation 6.1 to convert the wavelength to frequency: We can then use Equation 6.3 to calculate energy: Solve: The frequency, , is calculated from the given wavelength, as shown in Sample Exercise 6.2: The value of Planck’s constant, h, is given both in the text and in the table of physical constants on the inside front cover of the text, and so we can easily calculate E: Comment: If one photon of radiant energy supplies 3.37 10–19J, then one mole of these photons will supply This is the magnitude of enthalpies of reactions (Section 5.4), so radiation can break chemical bonds, producing what are called photochemical reactions.
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SAMPLE EXERCISE 6.3 continued PRACTICE EXERCISE (a) A laser emits light with a frequency of 4.69 1014 s–1. What is the energy of one photon of the radiation from this laser? (b) If the laser emits a pulse of energy containing 5.0 1017 photons of this radiation, what is the total energy of that pulse? (c) If the laser emits 1.3 10–2 J of energy during a pulse, how many photons are emitted during the pulse? Answers: (a) 3.11 10–19 J, (b) 0.16 J, (c) 4.2 1016 photons
The Nature of Energy Another mystery involved the emission spectra observed from energy emitted by atoms and molecules.
The Nature of Energy One does not observe a continuous spectrum, as one gets from a white light source. Only a line spectrum of discrete wavelengths is observed. Watch this video: Spectral Lines Video Demonstrate Spectral Tubes in the back store room
The Nature of Energy Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: Electrons in an atom can only occupy certain orbits (corresponding to certain energies). Bohr Atom Video link for review
The Nature of Energy Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: Electrons in an atom can only occupy certain orbits (corresponding to certain energies).
The Nature of Energy Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: Electrons in permitted orbits have specific, “allowed” energies; these energies will not be radiated from the atom.
The Nature of Energy Niels Bohr adopted Planck’s assumption and explained these phenomena in this way: Energy is only absorbed or emitted in such a way as to move an electron from one “allowed” energy state to another; the energy is defined by E = h
The Nature of Energy The energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation: E = −RH ( ) 1 nf2 ni2 - where RH is the Rydberg constant, 2.18 10−18 J, and ni and nf are the initial and final energy levels of the electron.
The Nature of Energy ( ) - E = −RH ( ) 1 nf2 ni2 - Electrons can go from the 2nd level to the 3rd or 4th . Electrons can go from the 1st energy level to the 2nd, 3rd or even 6th.
The Nature of Energy The energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation: E = −RH ( ) 1 nf2 ni2 - Note how “n” correlates to the number in front of the equation; 1 nf2
SAMPLE EXERCISE 6.4 Electronic Transitions in the Hydrogen Atom Predict which of the following electronic transitions produces the spectral line having the longest wavelength: n = 2 to n = 1, n = 3 to n = 2, or n = 4 to n = 3. Solution The wavelength increases as frequency decreases ( = c/. Hence the longest wavelength will be associated with the lowest frequency. According to Planck’s equation, E = h, the lowest frequency is associated with the lowest energy. In Figure 6.13 the shortest vertical line represents the smallest energy change. Thus, the n = 4 to n = 3 transition produces the longest wavelength (lowest frequency) line.
SAMPLE EXERCISE 6.4 Electronic Transitions in the Hydrogen Atom PRACTICE EXERCISE Indicate whether each of the following electronic transitions emits energy or requires the absorption of energy: (a) n = 3 to n = 1; (b) n = 2 to n = 4 . Answers: (a) emits energy, (b) requires absorption of energy
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The Wave Nature of Matter Louis de Broglie posited that if light can have material properties, matter should exhibit wave properties. He demonstrated that the relationship between mass and wavelength was = h mv
SAMPLE EXERCISE 6.5 Matter Waves What is the wavelength of an electron moving with a speed of 5.97 106 m/s? (The mass of the electron is 9.11 10–28 g.) Solve: Using the value of Planck’s constant, and recalling that we have the following: Comment: By comparing this value with the wavelengths of electromagnetic radiation shown in Figure 6.4, we see that the wavelength of this electron is about the same as that of X rays.
SAMPLE EXERCISE 6.5 continued PRACTICE EXERCISE Calculate the velocity of a neutron whose de Broglie wavelength is 500 pm. The mass of a neutron is given in the table on the back inside cover of the text. Answer: 7.92 102 m/s
The Uncertainty Principle Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely is its position known: In many cases, our uncertainty of the whereabouts of an electron is greater than the size of the atom itself! (x) (mv) h 4
The Uncertainty Calculation 4 m (v) (x) v = velocity x uncertainty Use this & page 211 to calculate the uncertainty for problem sets #’s 37 & 38. Here is a video that will help with your calculation: Uncertainty Calculation video Stop watching at the 9:50 minute mark.
Quantum Mechanics Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated. It is known as quantum mechanics.
Quantum Mechanics Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated. It is known as quantum mechanics.
Quantum Mechanics The wave equation is designated with a lower case Greek psi (). The square of the wave equation, 2, gives a probability density map of where an electron has a certain statistical likelihood of being at any given instant in time.
Quantum Numbers Solving the wave equation gives a set of wave functions, or orbitals, and their corresponding energies. Each orbital describes a spatial distribution of electron density. An orbital is described by a set of three quantum numbers.
Principal Quantum Number, n The principal quantum number, n, describes the energy level on which the orbital resides. The values of n are integers ≥ 0.
Azimuthal Quantum Number, l Also called “sublevels”; s, p, d, f This quantum number defines the shape of the orbital. Allowed values of l are integers ranging from 0 to n − 1. We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals.
Azimuthal Quantum Number, l Value of l 1 2 3 Type of orbital s p d f MEMORIZE THESE VALUES!!!
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Review: Principal Quantum Number, n The principal quantum number, n, describes the energy level on which the orbital resides. The values of n are integers ≥ 0.
Azimuthal Quantum Number, l Also called “sublevels”; s, p, d, f This quantum number defines the shape of the orbital. Allowed values of l are integers ranging from 0 to n − 1. We use letter designations to communicate the different values of l and, therefore, the shapes and types of orbitals.
Azimuthal Quantum Number, l Value of l 1 2 3 Type of orbital s p d f MEMORIZE THESE VALUES!!!
Magnetic Quantum Number, ml Psst: That is read as “m sub ell” We called them “orbitals” Describes the three-dimensional orientation of the orbital. These were the “lines” we used to place the electrons; __, __, __ __ __ 1s, 2s, 2p
Magnetic Quantum Number, ml Therefore, on any given energy level, there can be a max of 1s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, etc. s: __ p: __ __ __ d: __ __ __ __ __ f: __ __ __ __ __ __ __
Magnetic Quantum Number, ml Values are integers ranging from -l to l: −l ≤ ml ≤ l. How to read: ANY s sublevel: value = 0 Therefore ml value can be from -0 to +0 Will only have a single orbital Therefore ml value can only be = 0 __
Magnetic Quantum Number, ml Values are integers ranging from -l to l: −l ≤ ml ≤ l. How to read: ANY p sublevel: value = 1 Will have 3 orbitals Therefore ml value can be from -1 to +1 Think number lines: -1, 0, +1 So for each “p orbital” put “0” in the middle and -1 to left and +1 to right __ __ __ -1 0 +1
Magnetic Quantum Number, ml Values are integers ranging from -l to l: −l ≤ ml ≤ l. How to read: ANY d sublevel: value = 2 Will have 5 orbitals Therefore ml value can be from -2 to +2 So for each “d orbital” put “0” in the middle and work left and right from there __ __ __ __ __ -2 -1 0 +1 +2
Magnetic Quantum Number, ml Orbitals with the same value of n form a shell/level. Different orbital types within a shell are subshells/sublevel.
s Orbitals Value of l = 0. Spherical in shape. Radius of sphere increases with increasing value of n.
s Orbitals Observing a graph of probabilities of finding an electron versus distance from the nucleus, we see that s orbitals possess n−1 nodes/standing waves, or regions where there is 0 (ZERO) probability of finding an electron.
p Orbitals Value of l = 1. Have two lobes with a node between them.
d Orbitals Value of l is 2. Four of the five orbitals have 4 lobes; the other resembles a p orbital with a doughnut around the center.
SAMPLE EXERCISE 6.6 Subshells of the Hydrogen Atom (a) Predict the number of subshells in the fourth shell, that is, for n = 4. (b) Give the label for each of these subshells. (c) How many orbitals are in each of these subshells? Analyze and Plan: We are given the value of the principal quantum number, n. We need to determine the allowed values of l and ml for this given value of n and then count the number of orbitals in each subshell. Solution (a) There are four subshells in the fourth shell, corresponding to the four possible values of l (0, 1, 2, and 3). Solution (b) These subshells are labeled 4s, 4p, 4d, and 4f. The number given in the designation of a subshell is the principal quantum number, n; the letter designates the value of the azimuthal quantum number, l: for l = 0, s; for l = 1, p; for l = 2, d; for l = 3, f. Solution (c) There is one 4s orbital (when l = 0, there is only one possible value of ml: 0). There are three 4p orbitals (when l = 1, there are three possible values of ml: 1, 0, and –1). There are five 4d orbitals (when l = 2, there are five allowed values of ml: 2, 1, 0, –1, –2). There are seven 4f orbitals (when l = 3, there are seven permitted values of ml: 3, 2, 1, 0, –1, –2, –3).
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SAMPLE EXERCISE 6.6 Subshells of the Hydrogen Atom PRACTICE EXERCISE (a) What is the designation for the subshell with n = 5 and l = 1? (b) How many orbitals are in this subshell? (c) Indicate the values of ml for each of these orbitals in letter “b”. Answers: (a) 5p; (b) 3; (c) 1, 0, –1
Sample Problem An electron has the quantum numbers: n=2; l=1; ml=-1; ms=1/2. a) What is the subshell designation of this electron? b) What if it were: 3, 0, 1, ½?
Sample Problem An electron has the quantum numbers: n=2; l=1; ml=-1; ms=1/2. a) What is the subshell designation of this electron? n=2 implies 2nd energy level l=1 implies “p” sublevel That is all we need; subshell implies sublevel; Answer = 2p
Sample Problem What if it were: 3, 0, 1, ½? 3 = 3rd energy level (n) 0 = “s” sublevel 1 = orbital with a value of 1 (_-1_ _0_ _+1_) BUT the “s” sublevel can only have ONE orbital and this has 3, so this electron does not exist!!
Energies of Orbitals For a one-electron hydrogen atom, orbitals on the same energy level have the same energy. That is, they are degenerate.
Energies of Orbitals As the number of electrons increases, though, so does the repulsion between them. Therefore, in many-electron atoms, orbitals on the same energy level are no longer degenerate.
Spin Quantum Number, ms In the 1920s, it was discovered that two electrons in the same orbital do not have exactly the same energy. The “spin” of an electron describes its magnetic field, which affects its energy.
Spin Quantum Number, ms This led to a fourth quantum number, the spin quantum number, ms. The spin quantum number has only 2 allowed values: +1/2 and −1/2.
Pauli Exclusion Principle No two electrons in the same atom can have exactly the same energy. For example, no two electrons in the same atom can have identical sets of quantum numbers.
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Electron Configurations Review Electronic Configurations Using The Periodic Table Method (video link)
Electron Configurations Distribution of all electrons in an atom Consist of Number denoting the energy level
Electron Configurations Distribution of all electrons in an atom Consist of Number denoting the energy level Letter denoting the type of orbital
Electron Configurations Distribution of all electrons in an atom. Consist of Number denoting the energy level. Letter denoting the type of orbital. Superscript denoting the number of electrons in those orbitals.
Electron Configurations Distribution of all electrons in an atom. Consist of Number denoting the energy level. Letter denoting the type of orbital. Superscript denoting the number of electrons in those orbitals.
SAMPLE EXERCISE 6.7 Orbital Diagrams and Electron Configurations PRACTICE EXERCISE (a) Write the electron configuration for phosphorus, element 15. (b) How many unpaired electrons does a phosphorus atom possess? Answers: (a) 1s22s22p63s23p3, (b) three
SAMPLE EXERCISE 6.7 Orbital Diagrams and Electron Configurations PRACTICE EXERCISE (a) Write the electron configuration for a phosphide ion. Answers: (a) 1s22s22p63s23p6, remember that nonmetals gain e-’s. So phosphide has 5 valence e-’s and will gain 3 more e-’s giving it a full energy level.
SAMPLE EXERCISE 6.7 Orbital Diagrams and Electron Configurations PRACTICE EXERCISE Write the electron configuration for Iron, element 26. Now write the electron configuration of the ion Iron II. Answers: (a) 1s22s22p63s23p64s23d 5 (b) 1s22s22p63s23p63d 5; Iron II is Fe+2, meaning it lost 2 e-’s from its outer most level. Since 4s was the outer most level it loses 2 e-’s from there first.
Orbital Diagrams Each box represents one orbital. Half-arrows represent the electrons. The direction of the arrow represents the spin of the electron.
Hund’s Rule “For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized.”
Hund’s Rule Not This: “Remember to fill each orbital with a single electron with the SAME SPIN before you go back and pair them up, using OPPOSITE SPIN” This:
SAMPLE EXERCISE 6.7 Orbital Diagrams and Electron Configurations Draw the orbital diagram for the electron configuration of oxygen, atomic number 8. How many unpaired electrons does an oxygen atom possess? Solution Analyze and Plan: Because oxygen has an atomic number of 8, each oxygen atom has 8 electrons. The electrons (represented as arrows) are placed in the orbitals (represented as boxes) beginning with the lowest-energy orbital, the 1s. Each orbital can hold a maximum of two electrons (the Pauli exclusion principle). Because the 2p orbitals are degenerate, we place one electron in each of these orbitals (spin-up) before pairing any electrons (Hund’s rule). Solve: Two electrons each go into the 1s and 2s orbitals with their spins paired. This leaves four electrons for the three degenerate 2p orbitals. Following Hund’s rule, we put one electron into each 2p orbital until all three orbitals have one electron each. The fourth electron is then paired up with one of the three electrons already in a 2p orbital, so that the representation is The corresponding electron configuration is written 1s22s22p4. The atom has two unpaired electrons.
Periodic Table We fill orbitals in increasing order of energy. Different blocks on the periodic table, then correspond to different types of orbitals.
SAMPLE EXERCISE 6.8 Electron Configurations for a Group What is the characteristic valence electron configuration of the group 7A elements, the halogens? Solution Analyze and Plan: We first locate the halogens in the periodic table, write the electron configurations for the first two elements, and then determine the general similarity between them. Solve: The first member of the halogen group is fluorine, atomic number 9. The condensed electron configuration for fluorine is Similarly, that for chlorine, the second halogen, is From these two examples, we see that the characteristic valence electron configuration of a halogen is ns2np5, where n ranges from 2 in the case of fluorine to 6 in the case of astatine.
SAMPLE EXERCISE 6.8 Electron Configurations for a Group PRACTICE EXERCISE Which family of elements is characterized by an ns2np2 electron configuration in the outermost occupied shell? Answer: group 4A
SAMPLE EXERCISE 6.9 Electron Configurations from the Periodic Table (a) Write the full electron configuration for bismuth, element number 83. (b) Write the condensed electron configuration for this element. (c) How many unpaired electrons does each atom of bismuth possess? Solution (a) We write the electron configuration by moving across the periodic table one row at a time and writing the occupancies of the orbital corresponding to each row (refer to Figure 6.29). 1s22s22p63s23p64s23d104p65s24d105p66s24ƒ 145d106p3.
SAMPLE EXERCISE 6.9 Electron Configurations from the Periodic Table (a) Write the electron configuration for bismuth, element number 83. (b) Write the condensed electron configuration for this element. (c) How many unpaired electrons does each atom of bismuth possess? Solution (b) We write the condensed electron configuration by locating bismuth on the periodic table and then moving backward to the nearest noble gas, which is Xe, element 54. Thus, the noble-gas core is [Xe]. The outer electrons are then read from the periodic table as before. Moving from Xe to Cs, element 55, we find ourselves in the sixth row. Moving across this row to Bi gives us the outer electrons. (b Thus, the abbreviated electron configuration is: [Xe]6s24ƒ145d106p3
(c) How many unpaired electrons does each atom of bismuth possess? SAMPLE EXERCISE 6.9 continued (c) How many unpaired electrons does each atom of bismuth possess? Solution c) We can see from the abbreviated electron configuration that the only partially occupied subshell is the 6p. The orbital diagram representation for this subshell is In accordance with Hund’s rule, the three 6p electrons occupy the three 6p. orbitals singly, with their spins parallel. Thus, there are three unpaired electrons in each atom of bismuth.
Answers: (a) [Ar]4s23d7 , (b) [Kr]5s24d105p4 SAMPLE EXERCISE 6.9 continued PRACTICE EXERCISE Use the periodic table to write the condensed electron configurations for (a) Co (atomic number 27), (b) Te (atomic number 52). Answers: (a) [Ar]4s23d7 , (b) [Kr]5s24d105p4
Some Anomalies Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row.
Some Anomalies For instance, the electron configuration for chromium is [Ar] 4s1 3d 5 rather than the expected [Ar] 4s2 3d 4. This is because a ½ -full “s & d” sublevel are more stable than a full “s” sublevel
Some Anomalies Predict: What other element in the same period will do the same thing? Did you say Copper? Remember ½ of full sublevels are more stable The electron configuration for copper is [Ar] 4s1 3d 10 rather than [Ar] 4s2 3d 9.
Some Anomalies This occurs because the 4s and 3d orbitals are very close in energy. Others elements to know: Mo & Ag Find them and compare their locations to the others These anomalies occur in f-block atoms, as well. Don’t worry about these.
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