Chapter 6. Electronic. Electronic Structure of Atoms Pearson Education

Transcription

1 Chapter 6

2 Laser: step-like energy transition 6.1 The Wave Nature of Light 6.2 Quantized Energy and Photons 6.3 Line Spectra and the Bohr Model 6.4 The Wave Behavior of Matter 6.5 Quantum Mechanics and Atomic Orbitals 6.6 Representations of Orbitals 6.7 Many-Electron Atoms 6.8 Electron Configurations 6.9 Electron Configurations and the Periodic Table 2014 Pearson Education, Inc.

3 This chapter is all about electronic structure the arrangement and energy of electrons. It may seem odd to start by talking about waves. However, extremely small particles have properties that can only be explained in this manner!

4 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 ( ).

5 Waves The number of waves passing a given point per unit of time is the frequency ( ). For waves traveling at the same velocity, the longer the wavelength, the smaller the frequency. If the time associated with the lines to the left is one second, then the frequencies would be 2 s 1 and 4 s 1, respectively. s 1 : hertz (Hz)

6 Electromagnetic Radiation All electromagnetic radiation travels at the same velocity: The speed of light (c) is m/s. c =

7 Sample Exercise 6.1 Concepts of Wavelength and Frequency Two electromagnetic waves are represented in the margin. (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) Wave 1 has a longer wavelength (greater distance between peaks). The longer the wavelength, the lower the frequency (ν = c λ). Thus, Wave 1 has the lower frequency, and Wave 2 has the higher frequency. (b) The electromagnetic spectrum (Figure 6.4) indicates that infrared radiation has a longer wavelength than visible light. Thus, Wave 1 would be the infrared radiation. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

8 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? Solution Analyze We are given the wavelength, λ, of the radiation and asked to calculate its frequency, ν. Plan The relationship between the wavelength and the frequency is given by Equation 6.1. We can solve for ν and use the values of λ and c to obtain a numerical answer. (The speed of light, c, is m s to three significant figures.) Solve Solving Equation 6.1 for frequency gives (ν = c λ). When we insert the values for c and λ, 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: Check The high frequency is reasonable because of the short wavelength. The units are proper because frequency has units of per second, or s 1. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

9 The Nature of Energy The wave nature of light does not explain how an object can glow when its temperature increases. E = h, 2h, 3h

10 The Nature of Energy Quanta Max Planck explained it by assuming that energy comes in packets called quanta (singular: quantum). E = h, 2h, 3h 1918 Nobel Prize in Physics

11 The Photoelectric Effect Einstein used quanta to explain the photoelectric effect. Each metal has a different energy at which it ejects electrons (work function). At lower energy, electrons are not emitted. He concluded that energy is proportional to frequency: E = h where h is Planck s constant, J s Nobel Prize in Physics

12 Sample Exercise 6.3 Energy of a Photon Calculate the energy of one photon of yellow light that has a wavelength of 589 nm. Solution Analyze Our task is to calculate the energy, E, of a photon, given λ = 589 nm. Plan We can use Equation 6.1 to convert the wavelength to frequency: ν = c λ We can then use Equation 6.3 to calculate energy: E = hν Solve The frequency, ν, is calculated from the given wavelength, as shown in Sample Exercise 6.2: ν = c λ = s 1 The value of Planck constant, h, is given both in the text and in the table of physical constants on the inside back cover of the text, and so we can easily calculate E: E = ( J-s)( s 1 ) = J Comment If one photon of radiant energy supplies J, we calculate that one mole of these photons will supply: ( photons mol)( J photon) = J mol Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

13 Atomic Emissions Another mystery in the early twentieth century involved the emission spectra observed from energy emitted by atoms and molecules.

14 Continuous vs. Line Spectra For atoms and molecules, one does not observe a continuous spectrum (the rainbow ), as one gets from a white light source. Only a line spectrum of discrete wavelengths is observed. Each element has a unique line spectrum.

15 The Hydrogen Spectrum Johann Balmer (1885) discovered a simple formula relating the four lines to integers. Johannes Rydberg advanced this formula. Neils Bohr explained why this mathematical relationship works Nobel Prize in Physics

16 The Bohr Model Niels Bohr adopted Planck s assumption and explained these phenomena in this way: 1. Only orbits of certain radii, corresponding to certain specific energies, are permitted for the electron in a hydrogen atom. 2. An electron in a permitted orbit is in an allowed energy state. An electron in an allowed energy state does not radiate energy and, therefore, does not spiral into the nucleus. 3. Energy is emitted or absorbed by the electron only as the electron changes from one allowed energy state to another. This energy is emitted or absorbed as a photon that has energy E = h

17 The Bohr Model Excited state The energy absorbed or emitted from the process of electron promotion or demotion can be calculated by the equation E = hcr H 1 n 2 E = hcr ( ) H 1 n f 2 1 n i 2 where R H is the Rydberg constant, m 1, and n i and n f are the initial and final energy levels of the electron. Ground state n: principal quantum number

18 Sample Exercise 6.4 Transitions in the Hydrogen Atom Using Figure 6.12, predict which of these 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. E = hcr ( ) H 1 n f 2 1 n i 2 Solution The wavelength increases as frequency decreases (λ = c ν). Hence, the longest wavelength will be associated with the lowest frequency. According to Planck s equation, E = hv, the lowest frequency is associated with the lowest energy. In Figure 6.12 the shortest vertical line represents the smallest energy change. Thus, the n = 4 to n = 3 transition produces the longest wavelength (lowest frequency) line. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

19 Important Ideas from the Bohr Model Points that are incorporated into the current atomic model include the following: 1. Electrons exist only in certain discrete energy levels. 2. Energy is involved in the transition of an electron from one level to another. It only works for hydrogen. Classical physics would result in an electron falling into the positively charged nucleus. Bohr simply assumed it would not! Circular motion is not wave-like in nature. Limitations

20 The Wave Nature of Matter Louis de Broglie theorized that if light can have material properties, matter should exhibit wave properties (matter waves). He demonstrated that the relationship between mass and wavelength was The wave nature of light is used to produce this electron micrograph. = h mv p

21 Sample Exercise 6.5 Matter Waves What is the wavelength of an electron moving with a speed of m s? The mass of the electron is kg. Solution Analyze We are given the mass, m, and velocity, v, of the electron, and we must calculate its de Broglie wavelength, λ. Plan The wavelength of a moving particle is given by Equation 6.8, so λ is calculated by inserting the known quantities h, m, and v. In doing so, however, we must pay attention to units. Solve Using the value of Planck constant: we have the following: h = J-s Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

22 The Uncertainty Principle Heisenberg showed that the more precisely the momentum of a particle is known, the less precisely is its position is known: ( x) ( mv) h 4 Electron (5 x 10 6 m/s, 1% uncertainty)

23 Quantum Mechanics Erwin Schrödinger developed a mathematical treatment into which both the wave and particle nature of matter could be incorporated. This is known as quantum mechanics. The solution of Schrödinger s wave equation is designated with a lowercase Greek psi ( ). The square of the wave equation, 2, gives the electron density, or probability of where an electron is likely to be at any given time.

24 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 (n, l, m l ).

25 Principal Quantum Number (n) The principal quantum number, n, describes the energy level on which the orbital resides. The values of n are integers 1. These correspond to the values in the Bohr model.

26 Angular Momentum Quantum Number (l) 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.

27 Magnetic Quantum Number (m l ) The magnetic quantum number describes the threedimensional orientation of the orbital. Allowed values of m l are integers ranging from l to l: l m l l Therefore, on any given energy level, there can be up to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, and so forth.

28 Shell and Subshell 1. Orbitals with the same value of n form an electron shell: different orbital types within a shell are subshells. 2. The shell with principal quantum number n consists of exactly n subshells. 3. Each subshell consists of a specific number of orbitals ( l m l l). 4. The total number of orbitals in a shell is n 2.

29 Sample Exercise 6.6 Subshells of the Hydrogen Atom (a) Without referring to Table 6.2, 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? Solution Analyze and Plan We are given the value of the principal quantum number, n. We need to determine the allowed values of l and m l for this given value of n and then count the number of orbitals in each subshell. Solve There are four subshells in the fourth shell, corresponding to the four possible values of l (0, 1, 2, and 3). 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 angular momentum quantum number, l: for l = 0, s; for l = 1, p; for l = 2, d; for l = 3, f. There is one 4s orbital (when l = 0, there is only one possible value of m l : 0). There are three 4p orbitals (when l = 1, there are three possible values of m l : 1, 0, 1). There are five 4d orbitals (when l = 2, there are five allowed values of m l : 2, 1, 0, 1, 2). There are seven 4f orbitals (when l = 3, there are seven permitted values of m l : 3, 2, 1, 0, 1, 2, 3). Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

30 s Orbitals The value of l for s orbitals is 0. They are spherical in shape. The radius of the sphere increases with the value of n.

31 s Orbitals For an ns orbital, the number of peaks is n. For an ns orbital, the number of nodes (where there is zero probability of finding an electron) is n 1. As n increases, the electron density is more spread out and there is a greater probability of finding an electron further from the nucleus.

32 p Orbitals The value of l for p orbitals is 1. m l : -1, 0, 1 They have two lobes with a node between them.

33 d Orbitals The value of l for a d orbital is 2. m l : -2, -1, 0, 1, 2 Four of the five d orbitals have four lobes; the other resembles a p orbital with a doughnut around the center.

34 f Orbitals Very complicated shapes Seven equivalent orbitals in a sublevel l = 3 m l : -3, -2, -1, 0, 1, 2, 3

35 Energies of Orbitals Hydrogen For a one-electron hydrogen atom, orbitals on the same energy level have the same energy. Chemists call them degenerate orbitals.

36 Energies of Orbitals Many-electron Atoms As the number of electrons increases, so does the repulsion between them. Therefore, in atoms with more than one electron, not all orbitals on the same energy level are degenerate. Orbital sets in the same sublevel are still degenerate. Energy levels start to overlap in energy (e.g., 4s is lower in energy than 3d.)

37 Spin Magnetic Quantum Number, m s 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. This led to the spin quantum number, m s. The spin quantum number has only two allowed values, +½ and ½.

38 Pauli Exclusion Principle No two electrons in the same atom can have exactly the same energy. Therefore, no two electrons in the same atom can have identical sets of quantum numbers. This means that every electron in an atom must differ by at least one of the four quantum number values: n, l, m l, and m s. An orbital can hold a maximum of two electrons and they must have opposite spins.

39 Electron Configurations 4p 5 The way electrons are distributed in an atom is called its electron configuration. The most stable organization is the lowest possible energy, called the ground state. Each component consists of a number denoting the energy level; a letter denoting the type of orbital; a superscript denoting the number of electrons in those orbitals.

40 Orbital Diagrams Each box in the diagram represents one orbital. Half-arrows represent the electrons. The direction of the arrow represents the relative spin of the electron.

41 Hund s Rule For degenerate orbitals, the lowest energy is attained when the number of electrons with the same spin is maximized. This means that, for a set of orbitals in the same sublevel, there must be one electron in each orbital before pairing and the electrons have the same spin, as much as possible.

42 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 eight electrons. Figure 6.25 shows the ordering of orbitals. The electrons (represented as half 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 orbital diagram is The corresponding electron configuration is written 1s 2 2s 2 2p 4. The atom has two unpaired electrons. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

43 Condensed Electron Configurations Elements in the same group of the periodic table have the same number of electrons in the outer most shell. These are the valence electrons. The filled inner shell electrons are called core electrons. These include completely filled d or f sublevels. We write a shortened version of an electron configuration using brackets around a noble gas symbol and listing only valence electrons.

44 Outer-Shell and Valence Electrons

45 Periodic Table We fill orbitals in increasing order of energy. Different blocks on the periodic table correspond to different types of orbitals: s = blue, p = pink (main-group elements), d = orange (transition elements), f = tan (lanthanides and actinides, or inner transition elements)

46 , Inc. Se: [Ar]3d 10 4s 2 4p 4

47 , Inc.

48 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 the configurations. Solve The first member of the halogen group is fluorine (F, element 9). Moving backward from F, we find that the noble-gas core is [He]. Moving from He to the element of next higher atomic number brings us to Li, element 3. Because Li is in the second period of the s block, we add electrons to the 2s subshell. Moving across this block gives 2s 2. Continuing to move to the right, we enter the p block. Counting the squares to F gives 2p 5. Thus, the condensed electron configuration for fluorine is F: [He]2s 2 2p 5 The electron configuration for chlorine, the second halogen, is Cl: [Ne]3s 2 3p 5 From these two examples, we see that the characteristic valence electron configuration of a halogen is ns 2 np 5, where n ranges from 2 in the case of fluorine to 6 in the case of astatine. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

49 Sample Exercise 6.9 Electron Configurations from the Periodic Table (a) Based on its position in the periodic table, write the condensed electron configuration for bismuth, element 83. (b) How many unpaired electrons does a bismuth atom have? Solution (a) Our first step is to write the noble-gas core. We do this by locating bismuth, element 83, in the periodic table. We then move backward to the nearest noble gas, which is Xe, element 54. Thus, the noble-gas core is [Xe]. Next, we trace the path in order of increasing atomic numbers from Xe to Bi. Moving from Xe to Cs, element 55, we find ourselves in period 6 of the s block. Knowing the block and the period identifies the subshell in which we begin placing outer electrons, 6s. As we move through the s block, we add two electrons: 6s 2. As we move beyond the s block, from element 56 to element 57, the curved arrow below the periodic table reminds us that we are entering the f block. The first row of the f block corresponds to the 4f subshell. As we move across this block, we add 14 electrons: 4f 14. With element 71, we move into the third row of the d block. Because the first row of the d block is 3d, the second row is 4d and the third row is 5d. Thus, as we move through the ten elements of the d block, from element 71 to element 80, we fill the 5d subshell with ten electrons: 5d 10. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

50 Sample Exercise 6.9 Electron Configurations from the Periodic Table Continued Moving from element 80 to element 81 puts us into the p block in the 6p subshell. (Remember that the principal quantum number in the p block is the same as that in the s block.) Moving across to Bi requires three electrons: 6p 3. The path we have taken is Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

51 Sample Exercise 6.9 Electron Configurations from the Periodic Table Continued Putting the parts together, we obtain the condensed electron configuration: [Xe]6s 2 4f 14 5d 10 6p 3. This configuration can also be written with the subshells arranged in order of increasing principal quantum number: [Xe]4f 14 5d 10 6s 2 6p 3. Finally, we check our result to see if the number of electrons equals the atomic number of Bi, 83: Because Xe has 54 electrons (its atomic number), we have = 83. (If we had 14 electrons too few, we would realize that we have missed the f block.) (b) We see from the condensed electron configuration that the only partially occupied subshell is 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 the bismuth atom. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

52 Chromium as an Anomaly For instance, the electron configuration for chromium is [Ar] 4s 1 3d 5 rather than the expected [Ar] 4s 2 3d 4. This occurs because the 4s and 3d orbitals are very close in energy. These anomalies occur in f-block atoms with f and d orbitals, as well.

53 Some Anomalies Some irregularities occur when there are enough electrons to half-fill s and d orbitals on a given row. Expected Cr = [Ar]4s 2 3d 4 Cu = [Ar]4s 2 3d 9 Mo = [Kr]5s 2 4d 4 Ru = [Kr]5s 2 4d 6 Pd = [Kr]5s 2 4d 8 Found Experimentally Cr = [Ar]4s 1 3d 5 Cu = [Ar]4s 1 3d 10 Mo = [Kr]5s 1 4d 5 Ru = [Kr]5s 1 4d 7 Pd = [Kr]5s 0 4d 10

54 Sample Integrative Exercise Putting Concepts Together Boron, atomic number 5, occurs naturally as two isotopes, 10 B and 11 B, with natural abundances of 19.9% and 80.1%, respectively. (a) In what ways do the two isotopes differ from each other? Does the electronic configuration of 10 B differ from that of 11 B? (b) Draw the orbital diagram for an atom of 11 B. Which electrons are the valence electrons? (c) Indicate three major ways in which the 1s electrons in boron differ from its 2s electrons. (d) Elemental boron reacts with fluorine to form BF 3, a gas. Write a balanced chemical equation for the reaction of solid boron with fluorine gas. (e) ΔH f for BF 3 (g) is kj mol. Calculate the standard enthalpy change in the reaction of boron with fluorine. (f) Will the mass percentage of F be the same in 10 BF 3 and 11 BF 3? If not, why is that the case? Solution (a) The two isotopes of boron differ in the number of neutrons in the nucleus. (Sections 2.3 and 2.4) Each of the isotopes contains five protons, but 10 B contains five neutrons, whereas 11 B contains six neutrons. The two isotopes of boron have identical electron configurations, 1s 2 2s 2 2p 1, because each has five electrons. (b) The complete orbital diagram is The valence electrons are the ones in the outermost occupied shell, the 2s 2 and 2p 1 electrons. The 1s 2 electrons constitute the core electrons, which we represent as [He] when we write the condensed electron configuration, [He]2s 2 2p 1. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

55 Sample Integrative Exercise Putting Concepts Together Continued (c) The 1s and 2s orbitals are both spherical, but they differ in three important respects: First, the 1s orbital is lower in energy than the 2s orbital. Second, the average distance of the 2s electrons from the nucleus is greater than that of the 1s electrons, so the 1s orbital is smaller than the 2s. Third, the 2s orbital has one node, whereas the 1s orbital has no nodes (Figure 6.19). Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.

56 Sample Integrative Exercise Putting Concepts Together Continued (d) The balanced chemical equation is 2 B(s) + 3 F 2 (g) 2 BF 3 (g) (e) ΔH = 2( ) [0 + 0] = kj. The reaction is strongly exothermic. (f) As we saw in Equation 3.10 (Section 3.3), the mass percentage of an element in a substance depends on the formula weight of the substance. The formula weights of 10 BF 3 and 11 BF 3 are different because of the difference in the masses of the two isotopes (the isotope masses of 10 B and 11 B are and amu, respectively). The denominators in Equation 3.10 would therefore be different for the two isotopes, whereas the numerators would remain the same. Chemistry: The Central Science, 13th Edition Brown/LeMay/Bursten/Murphy/Woodward/Stoltzfus, Inc.