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1/7/2013 Chemistry: Atoms First Julia Burdge & Jason Overby Chapter 3 Quantum Theory and the Electronic Structure of Atoms Kent L. McCorkle Cosumnes River College Sacramento, CA Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 3 Quantum Theory and the Electronic Structure of Atoms 3.7 Quantum Numbers Principal Quantum Number (n) Angular Momentum Quantum Number (l) Magnetic Quantum Number (ml) Electron Spin Quantum Number (ms) 3.8 Atomic Orbitals s Orbitals p Orbitals d Orbitals and other High-Energy Orbitals Energies of Orbitals 3.9 Electron Configuration Energies of Atomic Orbitals in Many-Electron Systems The Pauli Exclusion Principle Aufbau Principle Hund’s Rule General Rules for Writing Electron Configurations 3.10 Electron Configurations and the Periodic Table Forms of Energy Potential energy is the energy possessed by an object by virtue of its position. There are two forms of potential energy of great interest to chemists: Chemical energy is energy stored within the structural units of chemical substances. Electrostatic energy is potential energy that results from the interaction of charged particles. 3 Quantum Theory and the Electronic Structure of Atoms 3.1 Energy and Energy Changes Forms of Energy Units of Energy 3.2 The Nature of Light Properties of Waves The Electromagnetic Spectrum The Double-Slit Experiment 3.3 Quantum Theory Quantization of Energy Photons and the Photoelectric Effect 3.4 Bohr’s Theory of the Hydrogen Atom Atomic Line Spectra The Line Spectrum of Hydrogen 3.5 Wave Properties of Matter The de Broglie Hypothesis Diffraction of Electrons 3.6 Quantum Mechanics The Uncertainty Principle The Schrödinger Equation The Quantum Mechanical Description of the Hydrogen Atom 3.1 Energy and Energy Changes Energy is the capacity to do work or transfer heat. All forms of energy are either kinetic or potential. Kinetic energy (Ek) is the energy of motion. m is the mass of the object u is its velocity One form of kinetic energy of particular interest to chemists is thermal energy, which is the energy associated with the random motion of atoms and molecules. Energy and Energy Changes Kinetic and potential energy are interconvertible – one can be converted to the other. Although energy can assume many forms, the total energy of the universe is constant. Energy can neither be created nor destroyed. When energy of one form disappears, the same amount of energy reappears in another form or forms. This is known as the law of conservation of energy. Q1 and Q2 represent two charges separated by the distance, d. 1 1/7/2013 Units of Energy Worked Example 3.1 The SI unit of energy is the joule (J), named for the English physicist James Joule. Calculate the kinetic energy of a helium atom moving at a speed of 125 m/s. It is the amount of energy possessed by a 2-kg mass moving at a speed of 1 m/s. Strategy Use Ek = ½mu2 to calculate the kinetic energy of an atom. Note that for units to cancel properly, giving Ek in joules, the mass of the helium atom (4.003 amu) must be in kilograms. The factor for conversion of amu to g is 1.661×10-24 g/1 amu. Therefore, the mass of a helium atom in kilograms is: Ek = ½ mu2 = ½(2 kg)(1 m/s)2 = 1 kg∙m2/s2 = 1 J The joule can also be defined as the amount of energy exerted when a force of 1 newton (N) is applied over a distance of 1 meter. 1J=1N·m Because the magnitude of a joule is so small, we often express large amounts of energy using the unit kilojoule (kJ). 4.003 amu × 1 kg 1.661×10-24 g × = 6.649×10-27 kg 1 amu 1×103 g Solution Ek = ½mu2 = ½(6.49×10-27 kg)(125 m/s) = 5.19×10-23 kg∙m2/s2 = 5.19×10-23 J Think About It We expect the energy of a single atom, even a fast-moving one, to be extremely small. 1 kJ = 1000 J 3.2 The Nature of Light Worked Example 3.2 How much greater is the attraction between charges of +2 and -2 than attraction between charges of +1 and -1 if the opposite charges are separated by the same distance. QQ Strategy Use Eel α 1d 2 to compare the magnitudes of the two Eel values. Because the distance between charges in the same in both cases, we can solve for the ratio of Eel values without actually knowing the distance. Both the distance and the proportionality constant cancel in the solution. Think About It Doubling both charges causes a four-fold in the magnitude of the electrostatic energy Setup increase Eel(+2,-2) = c Q1Q2 , Q1 = +2 and Q2 = -2 d particles. between charged Eel(+1,-1) = c Q1Q2 , Q1 = +1 and Q2 = -1 Visible light is only a small component of the continuum of radiant energy known as the electromagnetic spectrum. d Solution 2×(-2) d 1×(-1) c× d c× =4 The attraction between charges of +2 and -2 is four times as large as the attraction between charges +1 and -1. Properties of Waves All forms of electromagnetic radiation travel in waves. Waves are characterized by: Wavelength (λ; lambda) – the distance between identical points on successive waves The Nature of Light The speed of light (c) through a vacuum is a constant: c = 2.99792458×108 m/s Normally rounded to, c = 3.00×108 m/s. Speed of light, frequency and wavelength are related: Frequency (ν; nu) – the number of waves that pass through a particular point in 1 second. Amplitude – the vertical distance from the midline of a wave to the top of the peak or the bottom of the trough. λ is expressed in meters v is expressed in reciprocal seconds (s−1) s-1 is also known as hertz (Hz) 2 1/7/2013 The Electromagnetic Spectrum An electromagnetic wave has both an electric field component and a magnetic component. The electric and magnetic components have the same frequency and wavelength. The Double-Slit Experiment When light passes through two closely spaced slits, an interference pattern is produced. Constructive interference is a result of adding waves that are in phase. Destructive interference is a result of adding waves that are out of phase. This type of interference is typical of waves and demonstrates the wave nature of light. 3.3 Worked Example 3.3 A laser commonly used in the treatment of vascular skin lesions has a wavelength of 532 nm. What is the frequency of this radiation? Strategy We must convert the wavelength to meters and solve for frequency using c = λν. c Setup Rearranging the equation to solve for frequency gives ν = λ . The speed of light, c, is 3.00×108 m/s. Solution 1×10-9 m λ = 532 nm× 1 nm ν= = 5.32×10-7 m 3.00×108 m/s = 5.64×1014 s-1 5.32×10-7 m Think About It Make sure your units cancel properly. A common error in this type of problem is neglecting to convert wavelength to nanometers. Quantum Theory The energy E of a single quantum of energy is h is called Planck’s constant: 6.63×10−34 J∙s The idea that energy is quantized rather than continuous is like walking up a staircase or playing the piano You cannot step or play anywhere (continuous), you can only step on a stair or play on a key (quantized). Quantum Theory When a solid is heated, it emits electromagnetic radiation, known as blackbody radiation, over a wide range of wavelengths. The amount of energy given off at a certain temperature depends on the wavelength. Classical physics failed to completely explain the phenomenon. Assumed that radiant energy was continuous; that is, could be emitted or absorbed in any amount. Max Planck suggested that radiant energy is only emitted or absorbed in discrete quantities, like small packages or bundles. A quantum of energy is the smallest quantity of energy that can be emitted (or absorbed). Photons and the Photoelectric Effect Albert Einstein used Planck’s theory to explain the photoelectric effect. Electrons are ejected from the surface of a metal exposed to light of a certain minimum frequency, called the threshold frequency. The number of electrons ejected is proportional to the intensity. Below the threshold frequency no electrons were ejected, no matter how bright (or intense) the light. 3 1/7/2013 Photons and the Photoelectric Effect Photons and the Photoelectric Effect Einstein proposed that the beam of light is really a stream of particles. Shining light onto a metal surface can be thought of as shooting a beam of particles – photons – at the metal atoms. These particles of light are now called photons. Each photon (of the incident light) must posses the energy given by the equation: If the ν of the photons equals the energy the binds the electrons in the metal, then the light will have enough energy to knock the electrons loose. If we use light of a higher ν, then not only will the electrons be knocked loose, but they will also acquire some kinetic energy. Photons and the Photoelectric Effect This is summarized by the equation KE is the kinetic energy of the ejected electron W is the binding energy of the electron Worked Example 3.4 Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00×104 nm (infrared region) and (b) a photon with a wavelength of 52 nm (ultraviolet region). (c) Calculate the maximum kinetic energy of an electron ejected by the photon in part (b) from a metal with a binding energy of 3.7 eV. Strategy We must use c = λν and E = hν to determine the energy of each photon. In part (c), we will use hν = KE + W to find the kinetic energy of an ejected electron. The binding energy, given in eV, must be converted to J. Solution 1×10-9 m (a) 5.00×104 nm × = 5.00×10-5 m 1 nm c 3.00×108 m/s ν= = = 6.00×1012 s-1 λ 5.00×10-5 m E = hν = (6.63×10-34 J∙s)(6.00×10-12 s-1) = 3.98×10-21 J Worked Example 3.4 (cont.) Calculate the energy (in joules) of (a) a photon with a wavelength of 5.00×104 nm (infrared region) and (b) a photon with a wavelength of 52 nm (ultraviolet region). (c) Calculate the maximum kinetic energy of an electron ejected by the photon in part (b) from a metal with a binding energy of 3.7 eV. Think About It Remember that frequency and wavelength Solution -9 m are inversely Thus, as wavelength decreases, 1×10proportional. (b) 52 nm × = 5.2×10-8 m frequency 1and nmenergy increase. Note that in part (c), subtracting the 8binding energy made a relatively small c 3.00×10 m/s 15 s-1 ν= = =of5.8×10 -8 m change to the energy the incident photon. If the incident λ 5.2×10 photon had been in the X-ray region of the spectrum, the E = hν = (6.63×10-34 J∙s)(5.8×1015 s-1) = 3.8×10-18 J difference between its energy and the kinetic energy of the ejected electron would have been negligible. 1.602×10-19 J = 5.9×10-19 J (c) W = 3.7 eV × 1 eV KE = hv - W = 3.8×10-18 J - 5.9×10-19 J = 3.2×10-18 J 3.4 Bohr’s Theory of the Hydrogen Atom Sunlight is composed of various color components that can be recombined to produce white light. The emission spectrum of a substance can be seen by energizing a sample of material with some form of energy. The “red hot” or “white hot” glow of an iron bar removed from a fire is the visible portion of its emission spectrum. The emission spectrum of both sunlight and a heated solid are continuous; all wavelengths of visible light are present. 4 1/7/2013 Atomic Line Spectra Bohr’s Theory of the Hydrogen Atom Line spectra are the emission of light only at specific wavelengths. Every element has its own unique emission spectrum. The Line Spectrum of Hydrogen Bohr’s Theory of the Hydrogen Atom The Rydberg equation can be used to calculate the wavelengths of the four visible lines in the emission spectrum of hydrogen. Neils Bohr attributed the emission of radiation by an energized hydrogen atom to the electron dropping from a higher-energy orbit to a lower one. As the electron dropped, it gave up a quantum of energy in the form of light. R∞ is the Rydberg constant (1.09737317 x 107 m−1) Bohr showed that the energies of the electron in a hydrogen atom are given by the equation: λ the wavelength of a line in the spectrum n1 and n2 are positive integers where n2 > n1. The Line Spectrum of Hydrogen As an electron gets closer to the nucleus, n decreases. En is the energy n is a positive integer The Line Spectrum of Hydrogen Bohr’s theory explains the line spectrum of the hydrogen atom. Radiant energy absorbed by the atom causes the electron to move from the ground state (n = 1) to an excited state (n > 1). En becomes larger in absolute value (but more negative) as n gets smaller. En is most negative when n = 1. Called the ground state, the lowest energy state of the atom For hydrogen, this is the most stable state Conversely, radiant energy is emitted when the electron moves from a higher-energy state to a lower-energy excited state or the ground state. The quantized movement of the electron from one energy state to another is analogous to a ball moving and down steps. The stability of the electron decreases as n increases. Each energy state in which n > 1 is called an excited state. nf is the final state ni is the initial state 5 1/7/2013 Bohr’s Theory of the Hydrogen Atom Suppose an electron is initially in an excited state, ni. During emission, the electron drops to a lower energy state, nf. The energy difference between the initial and final states is Bohr’s Theory of the Hydrogen Atom Bohr’s Theory of the Hydrogen Atom To calculate wavelength, substitute c/λ for ν and rearrange: nf is the final state ni is the initial state 3.5 Worked Example 3.5 Wave Properties of Matter Calculate the wavelength (in nm) of the photon emitted when an electron transitions from the n = 4 state to the n = 2 state in a hydrogen atom. Louis de Broglie reasoned that if light can behave like a stream of particles (photons), then electrons could exhibit wavelike properties. Setup According to de Broglie, electrons behave like standing waves. Think About It Look again at the line spectrum of h = 6.63×10-34 J∙s and c = 3.00×108 m/s hydrogen and make sure your result matches one of them. Note that for an emission, ni, is always greater than nf, and -18 Solution 1 the 1equation gives2.18×10 a positive Jresult. - 12 = λ (6.63×10-34 J∙s )(3.00×108 m/s) 22 4 = 2.055×106 m-1 λ = 4.87×10-7 m × 1 nm 1×10-9 m = 487 nm Only certain wavelengths are allowed. At a node, the amplitude of the wave is zero. 6 1/7/2013 Wave Properties of Matter Worked Example 3.6 De Broglie deduced that the particle and wave properties are related by the following expression: λ is the wavelength associated with the particle m is the mass (in kg) u is the velocity (in m/s) The wavelength calculated from this equation is known as the de Broglie wavelength. Calculate the de Broglie wavelength of the “particle” in the following two cases: (a) a 25-g bullet traveling at 612 m/s and (b) an electron (m = 9.109×10-31 kg) moving at 63.0 m/s. Setup Think About It While you are new at solving these problems, always write out the units of Planck’s constant (J∙s) as kg∙m2/s. This will enable you to check your unit -34 J∙s, or 6.63×10-34 kg∙m2/s; Remember m must be expressed in h = 6.63×10 cancellations and detect common errors such as expressing kg. mass in grams rather than kilograms. Note that the calculated wavelength of a macroscopic object, even one Solution as small as a bullet, is extremely small. An object must be 1 kg = 0.025 kg (a) 25 g × at least asgsmall as a subatomic particle in order for its 1000 wavelength to be large enough for us to observe. 6.63×10-34 kg∙m2/s h = λ = = 4.3×10-35 m (0.025 kg)(612 m/s) mu (b) λ = 3.6 Diffraction of Electrons Experiments have shown that electrons do indeed possess wavelike properties: h mu = 6.63×10-34 kg∙m2/s = 1.16×10-5 m (9.109×10-31 kg)(63.0 m/s) Quantum Mechanics The Heisenberg uncertainty principle states that it is impossible to know simultaneously both the momentum p and the position x of a particle with certainty. Δx is the uncertainty in position in meters Δp is the uncertainty in momentum X-ray diffraction pattern of aluminum foil Electron diffraction pattern of aluminum foil. Worked Example 3.7 An electron in a hydrogen atom is known to have a velocity of 5×106 m/s + 1 percent. Using the uncertainty principle, calculate the minimum uncertainty in the position of the electron and, given that the diameter of the hydrogen atom is less than 1 angstrom (Å), comment on the magnitude of this uncertainty compared to the size of the atom. Strategy The uncertainty in the velocity, 1 percent of 5×106 m/s, is Δu. Calculate Δx and compare it with the diameter of they hydrogen atom. Setup The mass of an electron is 9.11×10-31 kg. Planck’s constant, h, is 6.63×10-34 kg∙m2/s. Δu is the uncertainty in velocity in m/s m is the mass in kg Worked Example 3.7 An electron in a hydrogen atom is known to have a velocity of 5×106 m/s + 1 percent. Using the uncertainty principle, calculate the minimum uncertainty in the position of the electron and, given that the diameter of the hydrogen atom is less than 1 Think angstrom (Å), comment on theerror magnitude of this uncertainty to About It A common is expressing the masscompared of the sizethe of particle the atom.in grams instead of kilograms, but you should discover this inconsistency if you check your unit cancellation Solution carefully. Remember if = one uncertainty is small, the other 6 m/s 4 m/s Δu = 0.01 × 5×10that 5×10 must be large. The uncertainty principle applies in a practical h way only Δx = to submicroscopic particles. In the case of a 4πobject, ∙ mΔu where the mass is much larger than that macroscopic of an electron, small uncertainties, relative to the size of the -34 kg∙m2/s 6.63×10 object, both position and velocity. Δxare = possible for-31 > 1×10-9 m 4π(9.11×10 kg)(5×104 m/s) The minimum uncertainty in the position x is 1×10-9 m = 10Å. The uncertainty is 10 times larger than the atom! 7 1/7/2013 Quantum Mechanics Quantum Mechanics Erwin Schrödinger derived a complex mathematical formula to incorporate the wave and particle characteristics of electrons. The Schrödinger equation specifies possible energy states an electron can occupy in a hydrogen atom. Wave behavior is described with the wave function ψ. The probability of finding an electron in a certain area of space is proportional to ψ2 and is called electron density. The energy states and wave functions are characterized by a set of quantum numbers. Instead of referring to orbits as in the Bohr model, quantum numbers and wave functions describe atomic orbitals. 3.7 Quantum Numbers Quantum Numbers Quantum numbers are required to describe the distribution of electron density in an atom. The principal quantum number (n) designates the size of the orbital. There are three quantum numbers necessary to describe an atomic orbital. Larger values of n correspond to larger orbitals. The allowed values of n are integral numbers: 1, 2, 3 and so forth. The principal quantum number (n) – designates size The angular moment quantum number (l) – describes shape The magnetic quantum number (m l) – specifies orientation Quantum Numbers The value of n corresponds to the value of n in Bohr’s model of the hydrogen atom. A collection of orbitals with the same value of n is frequently called a shell. Quantum Numbers The angular moment quantum number (l) describes the shape of the orbital. The values of l are integers that depend on the value of the principal quantum number The magnetic quantum number (m l) describes the orientation of the orbital in space. The values of ml are integers that depend on the value of the angular moment quantum number: – l,…0,…+l The allowed values of l range from 0 to n – 1. Example: If n = 2, l can be 0 or 1. l 0 1 2 3 Orbital designation s p d f A collection of orbitals with the same value of n and l is referred to as a subshell. 8 1/7/2013 Quantum Numbers Worked Example 3.8 Quantum numbers designate shells, subshells, and orbitals. What are the possible values for the magnetic quantum number (m l) when the principal quantum number (n) is 3 and the angular quantum number (l) is 1? Strategy Recall that the possible values of ml depend on the value of l, not on the value of n. Setup The possible values of ml are – l,…0,…+l. Solution The possible values of ml are -1, 0, and +1. Think About It Consult Table 3.2 to make sure your answer is correct. Table 3.2 confirms that it is the value of l, not the value of n, that determines the possible values of ml. Quantum Numbers Quantum Numbers The electron spin quantum number (m s ) is used to specify an electron’s spin. A beam of atoms is split by a magnetic field. Statistically, half of the electrons spin clockwise, the other half spin counterclockwise. There are two possible directions of spin. Allowed values of ms are +½ and −½. 3.8 Quantum Numbers To summarize quantum numbers: All s orbitals are spherical in shape but differ in size: principal (n) – size 1s < 2s < 3s angular (l) – shape Required to describe an atomic orbital magnetic (ml) – orientation principal quantum number (n = 2) principal (n = 2) 2px Atomic Orbitals related to the magnetic quantum number (ml ) angular momentum (l = 1) electron spin (ms) direction of spin 2s angular momentum quantum number (l = 0) Required to describe an electron in an atomic orbital ml = 0; only 1 orientation possible 9 1/7/2013 Atomic Orbitals Atomic Orbitals The p orbitals: The d orbitals: Five orientations: Three orientations: l = 1 (as required for a p orbital) ml = –1, 0, +1 l = 2 (as required for a d orbital) ml = –2, –1, 0, +1, +2 Energies of Orbitals Worked Example 3.9 The energies of orbitals in the hydrogen atom depend only on the principal quantum number. List the Think values of n, l, and ml for each the orbitals in a to 4dverify subshell. About It Consult theoffollowing figure your answers. Strategy Consider the significance of the number and the letter in the 4d designation and determine the values of n and l. There are multiple values for ml, which will have to be deduced from the value of l. 3s subshell 3; l = 0) 3rd shell (n (n ==3p 3)subshell (n3d = subshell 3; l = 1) (n = 3; l = 2) Setup The integer at the beginning of the orbital designation is the principal quantum number (n). The letter in an orbital designation gives the value of the angular momentum quantum number (l). The magnetic quantum number (ml) can have integral values of – l,…0,…+l. Solution 2s 2ndsubshell shell (n = 2)2p subshell (n = 2; l = 1) (n = 2; l = 0) 4d principal quantum number, n = 4 angular momentum quantum number, l = 2 Possible ml are -2, -1, 0, +1, +2. 3.9 Electron Configurations Electron Configurations The electron configuration describes how the electrons are distributed in the various atomic orbitals. If hydrogen’s electron is found in a higher energy orbital, the atom is in an excited state. In a ground state hydrogen atom, the electron is found in the 1s orbital. A possible excited state electron configuration of hydrogen Ground state electron configuration of hydrogen 2s 1s 2p 2p 2p 1s1 2s1 number of electrons in the orbital or subshell angular momentum (l = 0) The use of an up arrow indicates an electron with ms = + ½ Energy Energy principal (n = 1) 2s 2p 2p 2p 1s 10 1/7/2013 Electron Configurations Electron Configurations The helium emission spectrum is more complex than the hydrogen spectrum. In a multi-electron atoms, the energies of the atomic orbitals are split. There are more possible energy transitions in a helium atom because helium has two electrons. Splitting of energy levels refers to the splitting of a shell (n=3) into subshells of different energies (3s, 3p, 3d) Electron Configurations Electron Configurations According to the Pauli exclusion principle, no two electrons in an atom can have the same four quantum numbers. The Aufbau principle states that electrons are added to the lowest energy orbitals first before moving to higher energy orbitals. The ground state electron configuration of helium 2p 2p The ground state electron configuration of Li 1s2 2s 2p Quantum number Principal (n) 1s describes the 1s orbital Angular moment (l) Magnetic (ml) describes the electrons in the 1s orbital Electron spin (ms) 1 0 0 +½ 1 0 0 ‒½ Energy Energy 2p Li has a total of 3 electrons 2s 1s 2p 2p 1s22s1 The third electron must go in the next available orbital with the lowest possible energy. The 1s orbital can only accommodate 2 electrons (Pauli exclusion principle) Electron Configurations Electron Configurations The Aufbau principle states that electrons are added to the lowest energy orbitals first before moving to higher energy orbitals. The Aufbau principle states that electrons are added to the lowest energy orbitals first before moving to higher energy orbitals. Be has a total of 4 electrons B has a total of 5 electrons The ground state electron configuration of Be 1s 2p 2p 1s22s2 2p Energy Energy 2p 2s The ground state electron configuration of B 2p 2p 1s22s22p1 2s 1s 11 1/7/2013 Electron Configurations Electron Configurations According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized. According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized. C has a total of 6 electrons The ground state electron configuration of C N has a total of 7 electrons 1s22s22p2 2p 1s22s22p3 2p 2p 2s The 2p orbitals are of equal energy, or degenerate. 1s Put 1 electron in each before pairing (Hund’s rule). Energy Energy 2p The ground state electron configuration of N 2p 2p 2s The 2p orbitals are of equal energy, or degenerate. 1s Put 1 electron in each before pairing (Hund’s rule). Electron Configurations Electron Configurations According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized. According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized. O has a total of 8 electrons The ground state electron configuration of O F has a total of 9 electrons 1s22s22p4 1s 2p 1s22s22p5 2p 2p Once all the 2p orbitals are singly occupied, additional electrons will have to pair with those already in the orbitals. Energy Energy 2p 2s 2s 1s 2p 2p When there are one or more unpaired electrons, as in the case of oxygen and fluorine, the atom is called paramagnetic. Electron Configurations Electron Configurations According to Hund’s rule, the most stable arrangement of electrons is the one in which the number of electrons with the same spin is maximized. General rules for writing electron configurations: Ne has a total of 10 electrons Energy 2p 2s 1s 2p The ground state electron configuration of F The ground state electron configuration of Ne 1) Electrons will reside in the available orbitals of the lowest possible energy. 1s22s22p6 2) Each orbital can accommodate a maximum of two electrons. 2p When all of the electrons in an atom are paired, as in neon, it is called diamagnetic. 3) Electrons will not pair in degenerate orbitals if an empty orbital is available. 4) Orbitals will fill in the order indicated in the figure. 12 1/7/2013 3.10 Worked Example 3.10 Electron Configurations and the Periodic Table Write the electron configuration and give the orbital diagram of a calcium (Ca) atom (Z = 20). The electron configurations of all elements except hydrogen and helium can be represented using a noble gas core. Setup Because Z = 20, Ca has 20 electrons. They will fill in according to the diagram at right. Each s subshell can contain a maximum of two electrons, whereas each p subshell can contain a maximum of six electrons. The electron configuration of potassium (Z = 19) is 1s22s22p63s23p64s1. Solution Because 1s22s22p63s23p6 is the electron configuration of argon, we can simplify potassium’s to [Ar]4s1. Ca 1s22s22p63s23p64s2 The ground state electron configuration of K: 1s2 2s2 2p6 3s2 3p6 4s2 Think About It Look at the figure again to make sure you have filled the orbitals in the right order and that the sum of electrons is 20. Remember that the 4s orbital fills before the 3d orbitals. 1s22s22p63s23p64s1 [Ar] [Ar]4s1 Electron Configurations and the Periodic Table Electron Configurations and the Periodic Table Elements in Group 3B through Group 1B are the transition metals. Following lanthanum (La), there is a gap where the lanthanide (rare earth) series belongs. Electron Configurations and the Periodic Table Electron Configurations and the Periodic Table After actinum (Ac) comes the actinide series. 13 1/7/2013 Electron Configurations and the Periodic Table Electron Configurations and the Periodic Table There are several notable exceptions to the order of electron filling for some of the transition metals. There are several notable exceptions to the order of electron filling for some of the transition metals. Chromium (Z = 24) is [Ar]4s13d5 and not [Ar]4s23d4 as expected. Copper (Z = 29) is [Ar]4s13d10 and not [Ar]4s23d9 as expected. The reason for these anomalies is the slightly greater stability of d subshells that are either half-filled (d5) or completely filled (d10). Cr Chromium (Z = 24) is [Ar]4s13d5 and not [Ar]4s23d4 as expected. Copper (Z = 29) is [Ar]4s13d10 and not [Ar]4s23d9 as expected. The reason for these anomalies is the slightly greater stability of d subshells that are either half-filled (d5) or completely filled (d10). [Ar] Cu 4s 3d 3d 3d 3d 3d [Ar] 4s Greater stability with half-filled 3d subshell 3 Write the electron configuration for an arsenic atom (Z = 33) in the ground state. The order of filling beyond the noble gas core is 4s, 3d, and 4p. Fifteen electrons go into these subshells because there are 33 – 18 = 15 electrons in As beyond its noble gas core. 2 2 6 2 6 2 3 Solution As [Ar]4s23d104p3 Think About It Arsenic is a p-block element; therefore, we should expect its outermost electrons to reside in a p subshell. 3 3d 3d 3d 3d Greater stability with filled 3d subshell Worked Example 3.11 Setup The noble gas core for As is [Ar], where Z = 18 for Ar. 3d 10 Chapter Summary: Key Points Forms of Energy The Nature of Light Properties of Waves The Electromagnetic Spectrum The Double-Slit Experiment Quantum Theory Quantization of Energy Photons and the Photoelectric Effect Bohr’s Theory of the Hydrogen Atom Atomic Line Spectra The Line Spectrum of Hydrogen Wave Properties of Matter The de Broglie Hypothesis Diffraction of Electrons Quantum Mechanics The Uncertainty Principle The Schrödinger Equation The Quantum Mechanical Description of the Hydrogen Atom Quantum Numbers (n,l,ml,ms) Atomic Orbitals s orbitals, p orbitals, d orbitals and other High-Energy Orbitals Chapter Summary: Key Points Energies of Orbitals Electron Configuration Energies of Atomic Orbitals in Many-Electron Systems The Pauli Exclusion Principle The Aufbau Principle Hund’s Rule General Rules for Writing Electron Configurations Electron Configurations and the Periodic Table Group Quiz #3 • Calculate the energy of a photon that has a wavelength of 35.6 nm (in the xray region). (Hint: Watch units!!!) 14