Chemistry part 10, Julia Burdge,2e (2009) doc

214 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms Think About It Consult Table 6.2 to make sure your answer is correct. Table 6.2 confirms that it is the value of C, not the value of n, that determines the possible values of me. When n • e can be When e is be IS me can 1 only 0 0 only 0 2 o or 1 0 only 0 1 -1 ,0,or+1 3 0,1,or2 0 only 0 1 -1,0,or+1 2 -2, -1,0,+1,or+2 , 4 0, 1, 2, or 3 0 only 0 1 -1,0,or+1 2 -2, -1,0, +1, or +2 3 -3, -2, -1,0, +1, +2, or +3 • • • • • • • • • • • • e = 3 BBB[D1 +1 ~ +2~ +3 ~e = 31fsubshenl e = 2 1-2 ~ - llQ]EJEJ BBCDI +1 ~ +2 ~ e = 21 dsubShelll e = 1 BI 0 ~ +1 ~ BQ]EJ B[DB e = 11 p subshelll e = 0 0 [D Q] [D e = 01 S subshenl n = 1 n=2 n = 3 n=4 Figure 6.15 Illustration of how quantum numbers designate shells, subshells, and orbitals. . · S~mplePr~blem6.7 . • _ .' - -0 • What are the possible values for the magnetic quantum number (m e) when the principal quantum number (n) is 3 and the angular momentum quantum number (C) is 17 Strategy Use the rules governing the allowed values of me. Recall that the possible values of me depend on the value of C, not on the value of n. Setup The possible values of me are - C, ,0, , +C. Solution The possible values of me are -1, 0, and + 1. Practice Problem A What are the possible values for me when the principal quantum number (n) is 2 and the angular momentum quantum number (C) is 07 Practice Problem B What are the possible values for me when the principal quantum number (n) is 3 and the angular momentum quantum number (C) is 27 Electron Spin Quantum Number (ms) Whereas three quantum numbers are sufficient to describe an atomic orbital, an additional quan- tum number becomes necessary to describe an electron that occupies the orbital. Experiments on the emission spectra of hydrogen and sodium atoms indicated that each line in the emission spectra could be split into two lines by the application of an external mag- netic field. The only way physicists could explain these results was to assume that electrons act like tiny magnets. If electrons are thought of as spinning on their own axes, as Earth does, their magnetic properties can be accounted for. According to electromagnetic theory, a spinning charge SECTION 6.6 Quantum Numbers 215 generates a magnetic field, and it is this motion that causes an electron to behave like a magnet. Figure 6.16 shows the two possible spinning motions of an electron, one clockwise and the other counterclockwise. To specify the electron's spin, we use the electron spin quantum number (m s ). Because there are two possible directions of spin, opposite each other: m; has ' two possible vaiiies:" +~ and -~. Conclusive proof of electron spin was established by Otto Stern 13 and Walther Gerlach 14 in 1924. Figure 6.17 shows the basic experimental arrangement. A beam of gaseous atoms generated in a hot furnace passes through a nonuniform magnetic field. The interaction between an electron and the magnetic field causes the atom to be deflected from its straight-line path. Because the direction of spin is random, the electrons in half of the atoms will be spinning in one direction. Those atoms will be deflected in one way. The electrons in the other half of the atoms will be spin- . ning in the opposite direction. Those atoms will be deflected in the other direction. Thus, two spots of equal intensity are observed on the detecting screen. To summarize, we can designate an orbital in an atom with a set of three quantum number s. These three quantum numbers indicate the size (n), shape (€), and orientation (me) of the orbital. A fourth quantum number (ms) is necessary to designate the spin of an electron in the orbital. Checkpoint 6.6 Quantum Numbers 6.6.1 Which of the following is a legitimate 6. 6.3 set of three quantum numbers: n, e, and me 7 (Select all that apply.) a) 1, 0, ° b) 2,0,0 c) 1, 0, + 1 d) 2, 1, + 1 e) 2,2, - 1 6.6.4 6.6.2 How many orbitals are there in a subshell designated by the quantum number s n = 3, e = 27 a) 2 b) 3 c) 5 d) 7 e) 10 Atom beam Detecting screen Magnet How many subshells are there in the shell designated by n = 37 a) 1 , b) 2 c) 3 d) 6 e) 9 What is the total number of orbitals in the shell designated by n = 37 a) 1 b) 2 c) 3 d) 6 e) 9 Slit screen Oven 13 . Otto Stern (1888-1969). German physicist. He made important contributions to the study of the magnetic properties of atoms and the kinetic theory of gases. Stern was awarded the Nobel Prize in Physics in 1943. l ·t Walther Gerlach (1889-1979). German physicist. Gerlach's main area of research was in quantum theor y. Two electron s in the same orbital with opposite spins are referred to as "paired." (a) (b) Figure 6.16 (a) Clockwise and (b) counterclockwise spins of an electron. The magnetic fields generated by these two spinning motions are analogous to those from the two magnets. The upward and downward arrows are used to denote the direction of spin. Figure 6.17 Experimental arrangement for demonstrating the spinning motion of electrons. A beam of atoms is directed through a magnetic field. When a hydrogen atom, with a sing le electron, passes through the field, it is deflected in one direction or the ot her, depending on the direction of the electron's spin. In a stream consisting of many atoms, there will be equal distributions of the two kinds of spins, so two spots of equal intensity are detected on the screen. • 21 6 CHAPTER 6 Qu ant um Theory and th e Electronic Structure of Atoms The radial probability distribution can be tho ug ht of as a map of "where an electron spends most of i ts time." ~ '" ~ ~ >- ~ '" " " '0 >- ~ - .0 '" .0 0 A: ~ '" ~ - - '" 4-; 0 a ::l '" ~ " 0 ~ ::l .0 l:l '" '- '0 0 . ~ - .0 '" .0 e 0 - '" '0 '" ~ n= 1 £=0 o 2 4 r (10-lO m) n = 1 £=0 0 2 4 r (lO - lO m) Is orbital (a) Atomic Orbitals Strictly speaking, an atomic orbital does not have a well-defined shape because the wave function characterizing the orbital extends from the nucleus to infinity. In that sense, it is difficult to say what an orbital looks like. On the other hand, it is certainly useful to think of orbitals as having specific shapes. Being able to visualize atomic orbitals is essential to understanding the formation of chemical bonds and molecular geometry, which are discussed in Chapters 8 and 9. In this sec- tion, we will look at each type of orbital separately. s Orbitals For any value of the principal quantum number (n), the value 0 is possible for the angular momen- tum quantum number (f), corresponding to an s subshell. Furthermore, when f = 0, the magnetic quantum number (me) has only one possible value, 0, corresponding to an s orbital. Therefore, there is an s subshell in every shell, and each s subshell contains just one orbital, an s orbital. Figure 6.18 illustrates three ways to represent the distribution of electrons: the probability den SIty,' ' di e 'spheriC' <il 'di' stn 6ut!o'n' of erectioii density;' arid : the' radial probability distribution (the ~ '" ~ ~ >- ~ '" ~ - " '0 0 - .0 '" .0 0 p o o , n=2 £= 0 2 4 6 r (10- lO m) n = 2 £=0 2 4 6 r (10-lO m) 2s orbital (b) 8 8 /"'< '" ~ ~ 0 .~ '" " " '0 0 .~ - .0 '" n=3 .0 e £ = p o 2 o 2 - - 4 6 8 10 12 14 r (10- lO m) n = 3 £=0 4 6 8 10 12 14 r (10- 10 m) 3s orbital (c) Figure 6.18 From top to bottom, the probability density and corresponding relief map, the distribution of electron density represented spherically with shading corresponding to the,relief map above, and the radial probability di stribution for (a) the Is , (b) the 2s, and (c) the 3s orbitals of hydrogen. SECTION 6.7 Atomic Orbitals 217 z z z z y y y x x x x Pz P x Py (a) (b) probability of finding the electron as a function of distance from the nucleus) for the Is , 2s, and 3s orbitals of hydrogen. The boundary surface (the outermost surface of the spherical representation) is a common way to represent atomic or bitals, incorporating the volume in which there is about a 90 percent probability of finding the electron at any given time. All s orbitals are spherical in shape but differ in size, which increases as the principal quan- tum number increases. The radial probability distribution for the Is orbital exhibits a maximum at o 52.9 pm (0.529 A) from the nucleus. Interestingly, this distance is equal to the radius of the n = 1 orbit in the Bohr model of the hydrogen atom. The radial probability distribution plots for the 2s and 3s orbitals exhibit two and three maxima , respectively, with the greatest probability occ urring at a greater distance from the nucleus as n in c rea ses. Bet ween the two maxima for the 2s orbital there is a point on the plot where the probability drops to zero. Thi s corresponds to a node in the electron density, where the standing wave has zero amplitude. There a re two such nodes in the radial probability distribution plot of the 3s orbital. Although the details of electron density variation within each boundary surface are lost, the most important features of atomic orbitals are their overall shapes and relative sizes, which are adequately represented by boundary surface diagrams. p Orbitals When the principal quantum number (n) is 2 or greater, the value 1 is po ss ible for the angular mom en- tum quantum number (C), corresponding to a p subshell. And, when C = 1, the magnetic quantum num- ber (me) has three possible values: -1,0, and + 1, each corresponding to a differentp orbital. Therefore, there is a p subshell in every shell for which n :> 2, and each p subshell contains three p orbitals. These three p orbitals are labeled 2px, 2py, and 2pz (Figure 6.19), with the subscripted letters indicating the axis along which each orbital is oriented. These three p orbitals are identical in size, shape, and energy; they differ from one another only in orientation. Note, however, that there is no simple relation between the values of me and the x, y, and z directions. For our purpose, you need only remember th at because there are three possible values of me, there are three p orbitals with different orientations. The boundary surface diagrams of p orbitals in Figure 6.19 show that each p orbital can be thought of as two lobes on opposite sides of the nucleu s. Like s orbitals, p orbitals increase in size f rom 2p to 3p to 4p orbital and so o n. d Orbitals and Other Higher-Energy Orbitals When the principal quantum number (n) is 3 or greater, the value 2 is possible fo r the angular momentum quantum number (C), corresponding to a d subshell. When C = 2, the magnetic quan- tum number (m e) ha s five possible values, - 2, - 1, 0, + I , and +2 , each corresponding to a different d orbital. Again there is no direct correspondence between a given ori en tation and a particular me value. All the 3d orbitals in an atom are identical in energy and are labeled with subscripts den ot - ing their orientation with respect to the x, y, and .G axes and to the planes defined by them. Th e d orbitals that have higher principal quantum numb ers (4d, 5d, etc.) have s hape s similar to those shown for the 3d orbitals in Figure 6.20. z z z z z J Y Y Y y y x x x x '- x Figure 6.19 (a) Electron distribution in a p orbital. (b) Boundary surfaces for the Px' P l" and pz orbitals. Figure 6.20 Boundary surfaces for the d orbit al s. 218 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms Think About It Consult Figure 6.15 to verify your answers. Figure 6. 21 Orbital energy levels in the hydrogen atom. Each box represents one orbital. Orbitals with the same principal quantum number (n) all have the same energy. The! orbitals are important when accounting for the behavior of elements with atomic num- bers greater than 57, but their shapes are difficult to represent. In general chemistry we will not concern ourselves with the shapes of orbitals having e values greater than 2. Sample Problem 6.8 shows how to label orbitals with quantum numbers. Sample Problem 6.8 List the va lues of n, e, and me for each of the orbitals in a 4d subshell. Strategy Consider the significance of the number and the letter in the 4d designation and determine the va lues of nand e. There are multiple possible values for me, which will have to be deduced from the value of e. Setup The integer at the beginning of an orbital designation is the principal quantum number (n). The letter in an orbital designation gives the value of the angular momentum quantum number (e). The magnetic quantum number (m s) can have integral values of -e, , 0, . , + e. Solution The values of nand e are 4 and 2, respectively, so the possible values of me are -2, -1,0, +1, and +2. Practice Problem A Give the values of n, e, and me for the orbitals in a 3d subshell. Practice Problem B Using quantum numbers, explain why there is no 2d subshell. Energies of Orbitals The energies of orbitals in the hydrogen atom depend only on the value of the principal quantum number (n), and energy increases as n increases. For this reason, orbitals in the same shell have the same energy regardless of their subshell (Figure 6.21). Is < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f Thus, all four orbitals (one 2s and three 2p) in the second shell have the same energy; all nine orbitals (one 3s, three 3p, and five 3d) in the third shell have the same energy; and all sixteen orbitals (one 4s, three 4p, five 4d, and seven 4j) in the fourth shell have the same energy. The energy picture is more complex for many-electron atoms than' it is for hydrogen, as is discussed in Section 6.8. -~-14P ~ 4p ~@ -14 d ~ 4d ~ 4d ~ 4d ~B -1 4f ~ 4f ~ 4f ~ 4f ~0~~- -G-1 3P ~ 3p ~~-13d ~ 3d ~ 3d ~ 3d ~~ -~-12P ~ 2p ~~ -~ SECTION 6.8 Electron Configuration 219 Checkpoint 6.7 Atomic Orbitals 6.7.1 6.7.2 How many orbitals are there in the Sf subshell? a) 5 b) 7 c) 14 d) 16 e) 28 The energy of an orbital in the hydrogen atom depends on • a) n, e, and me b) nand e c) n only d) e only e) me only Electron Configuration 6. 7.3 In a hydrogen atom, which orbitals are high er in ene rgy than a 3s orbital? (Select all that apply.) a) 3p b) 4s c) 2p d) 3d e) 4p 6. 7.4 Which of the following sets of quantum number s, n, f, and me, corresponds to a 3p orbital? a) 3,0,0 b) 3,1 , 0 c) 3, 2, - 1 d) 1, 1, -2 e) 1, 3, 1 The hydrogen atom is a particularly simple system because it contains only one electron. The electron may reside in the Is orbital (the ground state), or it may be found in some higher-energy orbital (an excited state). With many-electron systems, we need to know the ground-state electron configuration that is, how the electrons are distributed in the various atomic orbitals. To do this, we need to know the relative energies of atomic orbitals in a many-electron system, which differ from those in a one-electron system such as hydrogen. Energies of Atomic Orbitals in Many-Electron Systems Consider the two emission spectra shown in Figure 6.22. The spectrum of helium contains more lines than that of hydrogen. This indicates that there are more possible transitions, correspond- ing to emission in the visible range, in a helium atom than in a hydrogen atom. This is due to the splitting' o(eiiergy 'ieveis ' c' aus ecl ' by ' eiec ' tro s' tatlc ' inter actIons ' betwe ' en heilum ;s' two · eiectron · s.' · . Figure 6.23 shows the general order of orbital energies in a many-electron atom. In contrast to the hydrogen atom, in which the energy of an orbital depends only on the value of n (Figure 6.21), the energy of an orbital in a many-electron system depends on both the value of n and the value of e. For example, 3p orbitals all have the sa me energ y, but they are higher in energy than the 3s orbital and lower in energy than the 3d orbitals. In a many-electron atom, for a given value of n, the energy of an orbital increases with increasing value of e. One important consequence of me splitting of energy levels is the relative energies of d orbitals in one shell and the s orbital in the next higher shell. As Figure 6.23 shows, the 4s orbital is lower in energy than the 3d orbitals. Like- wise, the 5s orbital is lower in energy than the 4d orbital, and so on. This fact becomes important when we determine how the electrons in an atom populate the atomic orbitals. I 400nm I 400nm 500 500 H ydrogen Helium I 600 I 600 I 700 I 700 "Splitting" of energy levels refers to the splitting of a shell into su bshells of different energies, as shown in Figure 6.23. Figure 6.22 Comparison of the emission spectra of H and He. 220 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms Figure 6.23 Orbital energy levels in many-electron atoms. For a given value of n, orbital energy increases with the value of e. 15 ' is read as " one 5 one." T he groun d state for a ma ny-e lectron atom is the one in whi ch all the e lect ron s o cc upy o rb i ta ls of t he lowe st pos s ibl e ene r gy. ~ 4d ~ 4d r~~B - - ~- ~~~- - n ______________ ~ 3d ~ 3d r~~~ - -~ ~~~ - ~- ~~~ -~ -~ The Pauli Exclusion Principle According to the Pauli 15 exclusion principle, no two electrons in an atom can have the same four quantum numbers. If two electrons in an atom have the same n, e, and me values (meaning that they occupy the same orbital), then they must have different values of ms; that is, one must have ms = +~ and the other must have ms = - ~. Because there are only two possible values for m s , and no two electrons in the same orbital may have the same value for m s , a maximum of two electrons may occupy an atomic orbital, and these two electrons must have opposite spins. We can indicate the arrangement of electrons in atomic orbitals with labels that identify each orbital (or subshell) and the number of electrons in it. Thus, we could describe a hydrogen atom in . . . . . . . . . . I the ground state using Is . ~ denotes the number of electrons lsI in the orbital or subshell denotes the principal / ~ denotes the angular momentum quantum number n quantum number e We can also represent the arrangement of electrons in an atom using orbital diagrams, in which each orbital is represented by a labeled box. The orbital diagram for a hydrogen atom in the ground state is The upward arrow denotes one of the two possible spins (one of the two possible ms values) of the electron in the hydrogen atom (the other possible spin is indicated with a downward arrow). Under certain circumstances, as we will see shortly, it is useful to indicate the explicit locations of electrons. . . . . . . . . . . . . . . . . . . . The orbital diagram for a helium atom in the ground state is 15. Wolfgang Pauli (1900-195 8) . Austrian physicis t. One of the founders of quantum mechanics, Pauli was awarded the Nobel Prize in Ph ys ics in 1945. SECTION 6.8 Electron Configuration 221 The label Ii · i~di~~t~ ~ · th~;~ . ~~ . t;;~ . ~ l~~tr~~ ~ · i~ · the i; · ~~bit~l · . · N~te · ~ so that· th~ ~~~~ · i · i~ · the · bo~ · . . . point in opposite directions, representing opposite electron spins. Generally when an orbital dia- gram includes an orbital with a single electron, we represent it with an upward arrow although we could represent it equally well with a downward arrow. The choice is arbitrary and has no effect on the energy of the electron. The Aufbau Principle We can continue the process of writing electron configurations for elements based on the order of orbital energies and the Pauli exclusion principle. This proce ss is based on the Aujbau principle, which makes it possible to "build" the periodic table of the elements and determine their electron configurations by steps. Each step involves adding one proton to the nucleus and one electron to the appropriate atomic orbital. Through this proce ss we gain a detailed knowledge of the electron configurations of the elements. As we will see in later chapters, knowledge of electron configura- tions helps us understand and predict the properties of the elements. It also explains why the ele- ments fit into the periodic table the way they do. After helium, the next element in the periodic table is lithium, which ha s three electrons. Because of the restrictions imposed by the Pauli exclusion principle, an orbital can accommodate no more than two electrons. Thu s, the third electron cannot reside in the Is orbital. Instead, it must reside in the next available orbital with the lowest po ss ible energy. According to Figure 6.23, this is me 2s orbital. Therefore, the electron configuration of lithium is l i2s1, and the orbital diagram is irnilarly, we can write the electron configuration of beryllium as li2i and represent it with the orbital diagram ith both the Is and the 2s orbitals filled to capacity, the next electron, which is needed for the electron configuration of boron, must reside in the 2p subshell. Because all three 2p orbitals are of ~ ual energy, or degenerate, the electron can occupy anyone of them. By convention, we usually ,h ow the first electron to occupy the p subshell in the first empty box in the orbital diagram. 11 1 Hund's Rule • Till the sixth electron, which is needed to represent the electron configuration of carbon, reside in iIe 2p orbital that is already half occupied, or will it reside in one of the other, empty 2p orbitals? \ccording to Hund'sl6 rule, the most stable arrangement of electrons in orbitals of equal energy l.S me one in which the number of electrons with the sa me spin is ma ximized. As we have seen, no 0':0 electrons in any orbital may have the same spin, so maximizing the number of electrons with iIe same spin requires putting the electrons in separate orbitals. Accordingly, in any subshell, an :: lectron will occupy an empty orbital rather than one that already contains an electron. The electron configuration of carbon is, therefore, l i2i2p 2, and it s orbital diagram is ' . Frederick Hund (1896-1997). German physicis t. Hund's work was ma inly in quantum mec hanics. He also helped to .:e elop the mol ec ular orbital theory of chemical bonding. 1 S2 is read as "o ne s two " n ot as "one s , squared. " 222 CHAPTER 6 Quantum Theory and the Electronic Structure of Atoms Is 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p Figure 6. 24 A simple way to remember the order in which orbitals fill with electrons. Remember that in this context, de g enerate means "of equal energy. " Orbitals in the same s ubshell are degenerate. • • • . ".,., . ~==~. Multimedia Atomic Structure electron configurations. Think About It Look at Figure 6.23 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. Similarly, the electron configuration of nitrogen is l i2i2 p 3, and its orbital diagram is 111111 1 2 p 3 Once all the 2p orbitals are singly occupied, additional electrons will have to pair with those already in the orbitals. Thus, the electron configurations and orbital diagrams for 0, F, and Ne are 0 ls22s22p4 0] 0] IHI1 11 I ls2 2s2 2p4 F ls 2 2s 2 2 p 5 0] 0] IH11~11 I ls 2 2s2 2 p 5 Ne IHIHIHI 2 p 6 General Rules for Writing Electron Configurations Ba sed on the preceding examples we can formulate the following general rules for determining the electron configuration of an element in the ground state: 1. Electrons will reside in the available orbitals of the lowest possible energy. 2. Each orbital can accommodate a maximum of two electrons. " "' :' ( . Eiectrons ' wiii ' not ' paIr ' i"n ' degenerate orbitals if an empty orbital is available. 4. Orbitals will fill in the order indicated in Figure 6.23. Figure 6.24 provides a simple way for you to remember the proper order. Sample Problem 6.9 illustrates the procedure for determining the ground· state electron con· figuration of an atom . Sample Problem 6.9 Write the electron configuration and give the orbital diagram of a calcium (Ca) atom (Z = 20). Strategy Use the general rules given and the Aufbau principle to "build" the electron configuration of a calcium atom and represent it with an orbital diagram. Setup Because Z = 20, we know that a Ca atom has 20 electrons. They will fill orbitals in the order designated in Figure 6.23, obeying the Pauli exclusion principle and Hund's rule. Orbitals will fill in the following order: I s, 2s, 2p, 3s, 3p, 4s. Each s subshell can contain a maximum of two electrons, whereas each p subshell can contain a maximum of six electrons. Solution 0] 0] IHIHIHI 0] IHIHIHI 0] ls2 2s2 2 p 6 3s 2 3 p 6 4s 2 Practice Problem A Write the electron configuration and give the orbital diagram of a rubidium (Rb) atom (Z = 37). Practice Problem B Write the electron configuration and give the orbital diagram of a bromine (Br) atom (Z = 35). SECTION 6.9 Electron Configurations and the Periodic Table 223 Checkpoint 6.8 Electron Configuration 6.8.1 Which of the following electron configurations correctly 6.8.3 Which orbital diagram is correct for the ground state S atom? represents the Ti atom? a) Is 2 2S2 2 p 6 3i 3l3ct [IT] [IT] b) I s2 2i 2l3s 2 3l4i 3d 2 a) ls2 2s2 c) ls 2 2i 2 p 6 3i 3l4s 2 3dIO Ii 2i 2 p 6 3s 2 3p6 3dIO d) b) [IT] [IT] Ii 2s2 2 p 6 3i 3p6 4s 4 e) ls2 2s2 6.8.2 What element is repre se nted by the following electron [IT] [IT] configuration? Ii 2S2 2 p 6 3s 2 3 p 6 4s 2 3d 10 4p4 c) a) Br l s2 2s2 b) As [IT] [IT] c) S d) ls2 2s2 d) Se e) Te e) [IT] [IT] ls2 2s2 Electron Configurations and the Periodic Table The electron configurations of all elements except hydrogen and helium can be represented usi ng a noble gas core, which shows in brackets the electron configuration of the noble gas element that most recently precedes the element in question, followed by the electron configuration in the out- ermost occupied subshells. Figure 6.25 gives the ground-state electron configurations of elements from H (Z = 1) through Rg (Z = 111). Notice the similar pattern of electron configurations in the elements lithium (Z = 3) through neon (Z = 10) and those of sodium (Z = 11) through argon (Z = 18 ). Both Li and Na, for example, have the configuration ns 1 in their outermost occupied subshells. For Li, n = 2; for Na, n = 3. Both F and CI have electron configuration nin/, where n = 2 for F and n = 3 for CI, and so on. As mentioned in Section 6.8, the 4s subshell is filled before the 3d subshell in a many- electron atom (see Figure 6.23). Thu s, the electron configuration of potassium (Z = 19) is li 2i2p 6 3s 2 3 p 64s1. Because 1 i2i 2p 6 3s 2 3 p 6 is the electron configuration of argon, we can sim- plify the electron configuration of potassium by writing [Ar]4 s1, where [Ar] denotes the "a rgon core." ~ [Ar] • [Ar] 4s 1 The placement of the outermost electron in the 4s orbital (rather than in the 3d orbital) of potassium is strongly supported by experimental evidence. The physical and chemical proper- ties of potassium are very similar to those of lithium and sod ium , the first two alkali metals. In both lithium and sodium, the outermost electron is in an s orbital (there is no doubt that the ir outermost electrons occupy s orbitals because there is no 1d or 2d subshell). Ba sed on its similarities to the other alkali metals, we expect potassium to have an analogous electron onfiguration; that is, we expect the last electron in potass ium to occupy the 4s rather than the 3d orbital. The elements from Group 3B through Group 1B are transition metals [ ~~ Section 2.4] . Transition metals either have incompletely filled d subshells or readily give rise to cations that have incompletely filled d subshells. In the first transition metal series, from scandium (Z = 21) IHI HI HI [IT] IHIHI 2 p 6 3s 2 3p4 I HIH I HI [IT] I H 11 11 2 p 6 3s 2 3p4 IHIHI 2p4 I H 11 11 I 2p4 1 1~IHIH I 2 p 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Although zinc and the other elements in Group 2B sometimes are included under the heading "t rans ition metals," they neither have nor readily acquire partially filled d subshells. Strictly speaking, they are not transition metals. through copper (Z = 29), additional electrons are placed in the 3d orbitals according to Hund's rule. However, there are two anomalies. The electron configuration of chromium (Z = 24) is [Ar]4S13d 5 and not [Ar]4i3d 4 , as we might expect. A similar break in the pattern is observed for I [...]... uncertainty in measuring the momentum is 1.0 X 10- 7 of the momentum, calculate the uncertainty in the baseball's position where u and c are the speeds of the particle and light, respectively (a) In particle accelerators, protons, electrons, and other charged particles are often accelerated to speeds close to the speed of light Calcu late the wavelength (in nm) of a proton moving at 50.0 percent the speed of... electrons Section 6.4 • A node is a point at which a standing wave has zero amplitude • Having observed that light could exhibit particle-like behavior, de Broglie proposed that matter might also exhibit wavelike behavior The de Broglie wavelength is the wavelength associated with a particle of very small mass Soon after de Broglie's proposal, experiments showed that electrons could exhibit diffraction-a... astronomers able to determine the temperature at the surface of stars in genera l? Calculate the wavelength and frequency of an emitted gamma particle having the energy of3.14 X lOl l li mo! 6.128 According to Einstein's special theory of relativity, the mass of a moving particle, mmoving, is related to its mass at mrest , by the following equation In,.est A microwave oven operating at 1.22 X 10 8 nm is used... photon _l 2· Section 6.7 • Section 6.3 • An emission spectrum is the light given off by an object when it is excited thermally Emission spectra may be continuous, including all the wavelengths within a particular range, or they may be line spectra, consisting only of certain discrete wavelengths Two electrons that occupy the same atomic orbital in the ground state mu st have different electron spin quantum... characterized by the filling of the 6d subshell With few exceptions, you should be able to write the electron configuration of any ele=ent, using Figure 6.23 (or Figure 6.24) as a guide Elements that require particular care are the =eview Questions 1 - What is a wave? Using a diagram, define the following terms associated... radiation in the ultraviolet region after absorbing visible light? Explain your answer 6.31 What are photons? What role did Einstein's explanation of the photoelectric effect play in the development of the particle-wave interpretation of the nature of electromagnetic radiation? 6.29 Explain how astronomers are able to tell which elements are present in distant stars by analyzing the electromagnetic radiation... frequency 6.18 When copper is bombarded with high-energy electrons, X rays are emitted Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm 6.20 6.21 A particular form of electromagnetic radiation has a frequency of 9.87 X 10 15 Hz (a) What is its wavelength in nanometers? In meters? (b) To what region of the electromagnetic spectrum would you assign... how many photons does this energy correspond to? 6.23 The radioactive 60Co isotope is used in nuclear medicine to treat certain types of cancer Calculate the wavelength and frequency of an emitted gamma particle having the energy of 1.29 X lOll J/mol 10- 19 J E 2 : -10 X 10- 19 J E I : -15 X 10- 19 J (a) What is the wavelength of the photon needed to excite an electron from EI to E4? (b) What is the energy . could exhibit particle-like behavior, de Broglie propo se d that matter might also exhibit wavelike behavior. The de Broglie wavelength is the wavelength associated with a particle of. with atomic num- bers greater than 57, but their shapes are difficult to represent. In general chemistry we will not concern ourselves with the shapes of orbitals having e values greater than. orbital? a) 3,0,0 b) 3,1 , 0 c) 3, 2, - 1 d) 1, 1, -2 e) 1, 3, 1 The hydrogen atom is a particularly simple system because it contains only one electron. The electron may reside in the