Chapter 06 Electron wang

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1 Chapter 6 Electronic Structure and The Periodic Table 1

2 v Ch.6 Electronic Structure and The Periodic Table v v v v v v v v v v v v v v v v v v v v v v v v v v v 6-1: Light, photon energies, and atomic spectra 1. The Wave Nature of Light:Wavelength and Frequency 2. The particle Nature of Light ;Photon Energies 3. Atomic Spectra 6-2: The Hydrogen atom 氫原子模型 1. Bohr Model 2. Quantum Mechanical Model 6-3: Quantum number 量子數 1. First Quantum Number, n;principal Energy Levels 2. Second Quantum Number,l;Sublevels(s, P,d,f) 3. Third Quantum Number,ml;Orbitals 4. Fourth Quantum Number,Ms ;Electron Spin 5. Pauli Exclusion Principle 6-4 :Atomic orbitals;shapes and size 6-5 :Electron configurations in atoms 1. Electron Configuration from Sublevel Energies 2. Filling of Sublevels and the Periodic Table 3. Electron Configuration from the periodic Table 6-6: Orbital diagrams of atoms 6-7 Electron arrangements in monatomic ions 1. Ions with Noble-Gas Structures 2. Transition Metal Cations 6-8 Periodic trends in the properties of atoms 1. Atomic Radius 2. Ionic Radius 3. Ionization Energy 4. Electronegativity 2

3 6-1 Light, Photon Energies, and atomic spectra 1.The wave Nature of light :Wavelength and Frequency Wavelength (λ) The distance between two consecutive crests or troughs, most often measured in meters or nanometers(1nm=10-9 m) Amplitude (Ψ) 水平點到波峰或波谷的高度 Frequency (ν): The number of wave cycles that pass a given point in unite time. 3 ν= 10 8 / s = 10 8 Hz ( 單位時間內所行經的波週數以秒的倒數 (1/s) 或赫茲 (Hterz, Hz) 表示 )

4 1.Wave nature of Light The speed at which a wave moves through space can be found by multiplying the length of a wave cycle by the number of cycles passing a point in unit time The speed (c) of the wave = λ x ν c = λ x ν = 長度波波時間 = 長度時間 Speed of light (c) in vacuum = x 10 8 m/s 4

5 Ex6.1:The red light associated with the aurora borealis is emitted by excited oxygen atom at 630.0nm.What is the frequency of the light? Sol: λ υ = 1m = 630 nm 9 10 nm m = s m HZ = = / m s 5

6 2.The Particle Nature of Light ; Photon Energies Max Planck(1900)-Blackbody radiation 波動性 Albert Einstein(1905)-Photoelectric effect 粒子性 E=hν= hc λ h: Planck s constant 蒲朗克常數 = J s 6

7 Ex6-2:The wavelength is m (a) The energy, in Joules, of a photon emitted by an excited oxygen atom (b) The energy, in kilojoules, of a mole of such photons ( 以仟焦耳 (KJ) 表示 1mol 光子的能量 ) (a) = hc E = hν = λ 34 ( J s)( m 8 m/ s) = J 1KJ E = J / 個 10 3 J = KJ / mol mol 23 個 7

8 The energy, in Joules, of a photon emitted (a) λ= nm ( 紅外線 ) (b) λ= nm (x-ray) E = hν hc = λ = J s m 8 m / s = J E = hν hc = λ = J s m 8 m / s = J 8

9 某光子的能量為 J,What is λ? E = hν = hc λ = = E hc λ J s J 8 m / s = m 9

10 Fig 6.2 The electromagnetic spectrum

11 11

12 3.Atomic Spectra( 原子光譜 ) v Sir Isaac Newton showed that visible (white) light from the sum can be broken sown into its various color components by a prism. The Spectrum obtained is continuous ; it contains essentially all wavelengths between 400 and 700 nm. v 原子發射光譜則是不連續的, 僅發生在某些特定的光譜 ( 線光譜 Line Spectra) v The situation with high-energy atoms of gaseous elements is quite different(fig6.3) 12

13 Line Emission Spectrum of Hydrogen Atoms nm nm

14 圖 6.3 Continuous and Line emission spectra 14

15 Chemistry in Action: Discovery of the Noble Gases Sir William Ramsay 15

16 6.2 The Hydrogen atom Bohr s Model of the Atom (1913) 1. 電子的能量是特定值, 被量子化的 2. 當電子從高能階移向低能階時, 光被放射出來相對量, 決定為吸收或發射光譜 1 E n = -R H ( ) n 2 n (principal quantum number) = 1,2,3, R H (Rydberg constant) = x J

17 E = hν E = hν

18 1. Bohr designated zero energy as the point at which the proton and electron are completely separated E = R n H 2 2. The hydrogen electron is in its lowest energy (ground state),n=1,electron absorbs enough energy, it moves to a higher, (excited state) 3. An excited electron gives off energy as a photon of light, it drops back to a lower energy state. E = hν = E hi E lo hν = R H 1 ( n ) hi 2 1 ( n ) lo 2 ν = R H h ( n 1 lo ) 2 ( n 1 hi ) 2 18

19 n i = 3 n i = 2 n i = 3 n f = 2 E photon = E = E f -E i 1 E f = -R H ( ) n 2 f 1 E i = -R H ( ) n 2 i 1 1 n 2 E = R H ( ) i n 2 f n f f =

20 Ex6.3:Calculate the wavelength in nanometers of the line in the Balmer series that results from the transition n=4 至 n=2 ν ( n 18 RH J = 2 2 = = h lo) ( nhi) J s 2 4 / s λ c = ν = m / s / s 9 10 nm m = nm 20

21 Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. E photon = E photon = 2.18 x J x (1/25-1/9) E photon = E = x J E photon = h x c / λ E = R H ( ) λ = h x c / E photon λ = 6.63 x (J s) x 3.00 x 10 8 (m/s)/1.55 x J λ =1280 nm 1 1 n 2 i n 2 f

22 Chemistry in Action: Electron Microscopy λ e = nm STM image of iron atoms on copper surface 22

23 2.Quantum Mechanical Model 量子力學模型 Bohr s theory for the structure of the hydrogen atom was highly successful 0.1 % error. When Bohr s theory is applied to the helium atom have 5%error v The Kinetic energy of an electron is inversely related to the volume of the region to which it is confined.( 電子的動能和其所侷限的活動範圍的體積成反比 ) v It is impossible to specify the precise position of an electron in an atom at a given instant. v ( 在特定的範圍內要正確指出電子在原子中真正的位置是不可能的 ) 23

24 Fig6-4 Two different ways of showing the electron distribution in the ground state of the hydrogen atom.( 兩種不同的方式表示氫原子電子在基態時的分佈 ) 24

25 6.3 Quantum numbers Schrodinger Wave Equation Ψ 電子波在空間中每一點的振幅 Ψ = fn(n,, m l, m s ) Principal energy levels( 電子與原子核的距離 n=1 n=2 n=

26 1.First Quantum Number,n ; (Principle quantum number) v 標明主能階 principal quantum number) n v n 愈大電子能量愈大離子原核愈遠 v 為 1 以上之整數 v n = 1, 2, 3, 4,. 26

27 Second Quantum Number 角動量量子數 (angular momentum quantum number) Schrodinger Wave Equation n = 1, = 0 n = 2, = 0 or 1 n = 3, = 0, 1, or 2 Ψ = fn(n,, m l, m s ) for a given value of n, l = 0, 1, 2, 3, n-1 共有 n 個 = 0 = 1 = 2 = 3 s orbital p orbital d orbital f orbital sharp pricipal diffuse fundamental Shape of the volume of space that the e - occupies ns < np < nd < nf

28 = 2 (d orbitals) = 0 (s orbitals) = 1 (p orbitals)

29 Third Quantum Number 磁量子數 (magnetic quantum number) m (Orbitals) Schrodinger Wave Equation Ψ = fn(n,, m, m s ) for a given value of l m = -,., 0,. + if = 1 (p orbital), m = -1, 0, or 1 if = 2 (d orbital), m = -2, -1, 0, 1, or 2 orientation of the orbital in space

30 m = -1 m = 0 m = 1 m = -2 m = -1 m = 0 m = 1 m =

31 Fourth Quantum Number 旋量子數 (spin quantum number) m s Schrodinger Wave Equation Ψ = fn(n,, m l, m s ) Electron Spin m s = +½ or -½ m s = +½ m s = -½

32 2.Pauli exclusion principle ( 庖立不相容原理 ) v No two electrons in an atom can have the same set of four quantum numbers. v ( 一個原子中的任二個電子不能同時存在兩組完全相同的四種量子數 ) v If two electrons occupy the same orbital, they must gave opposed spins. Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time 32

33 Ex6.4: consider the following sets of quantum numbers (n,l,ml,ms) Which ones could not occur? For the valid sets, identify the orbital involved. (a) (3,1,0,+1/2) (b) (1,1,0,-1/2) (c) (2,2,0,+1/2) (d) (4,3,2,+1/2) (e) (2,1,0,0) 33

34 Ex6.5: (A) What is the capacity for electrons of an S sublevel? A d sublevel? f sublevel? (B) What is the total capacity for electrons of the fourth principal level? SOL: (a) s sublevel: 1orbital 2e - / orbital=2e - p sublevel: 3orbital 2e - / orbital=6e - d sublevel: 5orbital 2e - / orbital=10e - f sublevel: 7orbital 2e - / orbital=14e - (b) 2e - (4s)+ 6e - (4p)+ 10e - (4d)+ 14e - (4f)=32e - 34

35 電子在原子中可能的量子數組 v table.pdf 35

36 主層次層與軌域的容納量 v 量子數 n 的每一主層包含了 n 個次層共 2n 2 個電子 v 量子數的每一次層包含了總數 2 +1 個軌域 ; 共有 2(2 +1) 個電子一個 s 次層 ( =0) 包含 1 個軌域一個 p 次層 ( =1) 包含 3 個軌域一個 d 次層 ( =2) 包含 5 個軌域一個 f 次層 ( =3) 包含 7 個軌域 v 任何一個軌域最多僅能容納 2 個電子, 且旋轉的方向必須是相反的 36

37 Schrodinger Wave Equation Ψ = fn(n, l, m l, m s ) Shell electrons with the same value of n Subshell electrons with the same values of n and l Orbital electrons with the same values of n, l, and m l How many electrons can an orbital hold? If n, l, and m l are fixed, then m s = ½ or - ½ Ψ = (n, l, m l, ½) or Ψ = (n, l, m l, -½) An orbital can hold 2 electrons

38 How many 2p orbitals are there in an atom? n=2 2p l = 1 If l = 1, then m l = -1, 0, or +1 3 orbitals How many electrons can be placed in the 3d subshell? n=3 If l = 2, then m l = -2, -1, 0, +1, or +2 3d 5 orbitals which can hold a total of 10 e - l =

39 6.4 Atom orbitals ;Shapes and sizes They differ from one another only in size.as n increases, the radius of the orbital becomes larger. A P orbital consists of two lobes along an axis (x,y,z), there is zero probability of finding an electron at the origin, at the nucleus or the atom. 39

40 6.5 Electron configurations in Atoms Shows the number of electrons, indicated by a superscript, in each sublevel. 電子組態用以描述原子中電子的排列情形, 將每一層的電子數呈現在該次層的右上角在軌域或次層的電子數主量子數 n 1s 1 角動量量子數 l Orbital diagram H 1s

41 Single electron orbital energy level 能階由主量子數 n 決定 n=3 n=2 1 E n = -R H ( ) n 2 n=

42 多電子原子的軌域能階能階由 n 及 l 決定 n=3 l = 2 n=3 l = 0 n=2 l = 0 n=3 l = 1 n=2 l = 1 n=1 l =

43 Fig 6.8Electron energy sublevels in the order of increasing energy. 43

44 Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

45 原子基態的電子組態遵循原則 v 構築原理電子進入軌域的順序由低能階開始 v 庖立不相容原理不個軌域內最多只能容納兩個電子, 且在同一軌域內的兩個電子自旋方向必相反 v 韓德法則 (Hund s rule) 整個電子要填入同能階的同型軌域 ( 如 2Px, 2Py, 2Pz) 時, 必須先以相同自旋方式填入不同方位的軌域而不成對, 等各軌域均有一個電子時, 才能填入自旋方式相反的電子進入而成對 45

46 Fill up electrons in lowest energy orbitals (Aufbau principle)?? Li Be B C electrons B Be Li 1s 1s 2 2s 2 2s 2 2p 12 1 H He 12 electrons He H 1s

47 The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund s rule). C F Ne O electrons Ne C N O F 1s 2 2 2s 2 2 2p

48 Ex6.6:Find the electron configurations of the sulfur (S) and (Ni) atoms? S 16 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p = 16 electrons abbreviate [Ne]3s 2 3p 4 [Ne] 1s 2 2s 2 2p 6 Ni 28 electrons 1s < 2s < 2p < 3s < 3p < 4s < 3d 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d = 28 electrons abbreviate [Ar]4s 2 3d 8 [Ar] 1s 2 2s 2 2p 6 3s 2 3p 6 48

49 What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s = 12 electrons Abbreviated as [Ne]3s 2 [Ne] 1s 2 2s 2 2p 6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons 1s 2 2s 2 2p 6 3s 2 3p 5 Last electron added to 3p orbital 1s < 2s < 2p < 3s < 3p < 4s = 17 electrons n = 3 l = 1 m l = -1, 0, or +1 m s = ½ or -½

50 Ex6.7:Find the iodine atom. write (a) The electron configurations (b) the abbreviated electron configuration. (a) I 53 electrons 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p = 53 electrons [Kr] 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 (b) Abbreviate electron configuration [Kr]5s 2 4d 10 5p 5 50

51 Filling of Sublevels and the Periodic Table v The atoms of elements in a group of the periodic table have the same distribution of electrons in the outermost principal energy level. (1) The elements in groups 1 and 2 are filling an s sublevel. (2) The elements in groups 13 through 18 fill p sublevels. (3) The transition metals, in the center of the periodic table, fill d sublevels. (4) The two sets of 14 elements listed separately at the bottom of the table are filling f sublevels with a principal quantum number two less than the period number. 51

52 Outermost sub shell being filled with electrons

53 53

54 特例 v 3d and4s orbitals have very similar energies. it has been suggested that there is a slight increase in stability with a half- filled (Cr) or completely filled (cu )3d sublevel. v 因此半滿或全滿的 3d 次層可稍微的提高穩定度 Cu 29 electrons 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 9 error = 29 electrons 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 Abbreviated electron configuration [Ar]4s 1 3d 10 54

55 Cr 24 electrons 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 4 error = 24 electrons 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 Abbreviated electron configuration [Ar]4s 1 3d 5 55

56 6.6 Orbital diagrams of atoms v It is useful to go a step further and show how electrons are distribute among orbitals. v Orbital diagrams each orbital is represented by ( ),and electrons are shown by arrows written( ),or ( ),depending on spin. 56

57 1. 在所有已填滿的軌域, 同一軌域內的 2 個電子必須是相反的旋轉 2. 根據韓德法則, 在所給予的次層內儘可能維持最多的半滿軌域 57

58 Hund s rule 韓德法則 v When several orbitals of equal energy are available, as in a given sublevel, electrons enter singly with parallel spin. v Hund s rule is based on experiment. Notice v 1.In all filled orbitals, the two electrons have opposed spins. v 2. In accordance with Hund s rule, within a given sublevel there are as many half filled orbitals as possible. 58

59 順磁 Paramagnetic unpaired electrons 逆磁 Diamagnetic all electrons paired 2p 2p

60 v Paramagnetic 順磁當電子自旋相為或時, 則兩電子的淨磁場會加強而呈順磁即可被磁鐵吸引的物質 v Diamagnetic 逆磁若電子自旋配對為或時, 則電子的磁效應會相互抵消而呈逆磁指會受到磁場些微排斥的物質 60

61 Ex6.8:Construct orbital diagrams fro atoms of sulfur and iron atom S 16 electrons 1s 2 2s 2 2p 6 3s 2 3p 4 1s 2s 2p 3s 3p 16S ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Fe 26 electrons 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6 1s 2s 2p 3s 3p 26Fe ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4s 3d ( ) ( ) ( ) ( ) ( ) ( ) 61

62 ns 1 Ground State Electron Configurations of the Elements d 1 d 5 ns 2 ns 2 np 1 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 d 10 4f 5f 62

63 6.7 Electron arrangements in Monatomic ions v When a monatomic ion is formed from an atom, electrons are added to or removed from sublevels in the highest principal energy level. 1.Ions with Noble-Gas Structures v Elements close to a noble gas in the periodic table form ions that have the same number of electrons as the noble- gas atom. v 7 N(1s 2 2s 2 2p 3 ) + 3e - 7 N 3- (1s 2 2s 2 2p 6 ) v 8 O(1s 2 2s 2 2p 4 ) + 2e - 8 O 2- (1s 2 2s 2 2p 6 ) v 9 F(1s 2 2s 2 2p 5 ) + e - 9 F - (1s 2 2s 2 2p 6 ) v 11 Na(1s 2 2s 2 2p 6 3s 1 ) 11 Na + (1s 2 2s 2 2p 6 ) + e - v 12 Mg(1s 2 2s 2 2p 6 3s 2 ) 12 Mg +2 (1s 2 2s 2 2p 6 ) + 2e - v 13 Al(1s 2 2s 2 2p 6 3s 2 3p 1 ) 13 Al +3 (1s 2 2s 2 2p 6 )+3e - 63

64 Na + : [Ne] Al 3+ : [Ne] F - : 1s 2 2s 2 2p 6 or [Ne] O 2- : 1s 2 2s 2 2p 6 or [Ne] N 3- : 1s 2 2s 2 2p 6 or [Ne] Na +, Al 3+, F -, O 2-, and N 3- 與 Ne 為等電子 ( isoelectronic ) What neutral atom is isoelectronic with H -? H - : 1s 2 same electron configuration as He 64

65 Electron Configurations of Cations of Transition Metals When a cat ion is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n 1)d orbital's. Fe: [Ar]4s 2 3d 6 Fe 2+ : [Ar]4s 0 3d 6 or [Ar]3d 6 Mn: [Ar]4s 2 3d 5 Mn 2+ : [Ar]4s 0 3d 5 or [Ar]3d 5 Fe 3+ : [Ar]4s 0 3d 5 or [Ar]3d 5 65

66 Electron Configurations of Cations and Anions Of Representative Elements Na [Ne]3s 1 Ca [Ar]4s 2 Al [Ne]3s 2 3p 1 Na + [Ne] Ca 2+ [Ar] Al 3+ [Ne] Atoms lose electrons so that cation has a noble-gas outer electron configuration. Atoms gain electrons so that anion has a noble-gas outer electron configuration. H 1s 1 H - 1s 2 or [He] F 1s 2 2s 2 2p 5 F - 1s 2 2s 2 2p 6 or [Ne] O 1s 2 2s 2 2p 4 O 2-1s 2 2s 2 2p 6 or [Ne] N 1s 2 2s 2 2p 3 N 3-1s 2 2s 2 2p 6 or [Ne]

67 Cations and Anions Of Representative Elements

68 68

69 Example 6.9 Give the electron configuration of (a) 30 Zn Zn:1s2 2s 2 2p 6 3s 2 3p 6 4s 2 3d Zn+2 :1s 2 2s 2 2p 6 3s 2 3p 6 4s 0 3d 10 (b) 34 Se Se:1s2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4 34 Se-2 :1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 69

70 6.8 Periodic tends in the properties of atoms 原子之週期性趨勢 v The chemical and physical properties of elements are a periodic function of atomic number. v 元素的物理化學性質與原子序週期性有關 v Elements within a given vertical group resemble one another chemically because chemical properties repeat themselves at regular intervals of 2,8,18,32. 70

71 atomic radius The atomic radius is taken to be one half the distance of closest approach between atoms in an elemental substance. 1.Decrease across a period from left to right in the periodic table. 2.Increase down a group in the periodic table. 71

72 同一週期原子半徑由左至右逐漸變小同一族中隨原子序增加而原子半徑變大 72

73 Effective Nuclear Charge (Z eff ) 在多電子原子中, 當外層電子接近原子核時 increasing Z, 會受到其餘電的遮蔽效應, 使電子與原子核的正電荷之靜電引力變小此外電 eff 子間的排斥力也會抵消部份, 因荷電荷增加的引力 increasing Z eff 當有效荷電荷增加, 則原子半徑變小 73

74 74

75 Atomic Radii

76 Ionic Radius Positive ions are smaller than the metal atoms from which they are formed. Negative ions are larger than the nonmetal atoms from which they are formed. 76

77 Cation is always smaller than atom from which it is formed. Anion is always larger than atom from which it is formed

78 78 6.7

79 79

80 80 8.3

81 v 同一原子形成陽離子時, 隨著其失去電子愈多, 則其半徑愈小 Fe > Fe +2 > Fe +3 v 同一原子形成陰離子時, 隨著其獲得電子愈多, 則其半徑愈大 S -2 > S 81

82 Example 6.10 Using only the periodic table, arrange each of the following sets of atoms and ions in order of increasing size. (a) Mg, Al, Ca Ca > Mg > Al Mg, Al, 為同週期 12 Mg, 13 Al, 故原子半徑 Mg > Al Mg, Ca, 為同族 12 Mg, 20 Ca, 故原子半徑 Ca > Mg (b) S, Cl, S -2 S, Cl, 為同週期 16 S, 17 Cl, 故原子半徑 S > Cl S, S -2, 陰離子半徑較原子大, 故原子半徑 S -2 > S (c) Fe, Fe +2, Fe +3 Fe 3+ 2 S -2 > S > Cl Fe + Fe 82

83 ionization energy v It is a measure of how difficult it is to remove an electron from a gaseous atom v Energy must always be absorbed to bring about ionization, so ionization energies are always positive quantites. 須吸收能量才足以產生離子態, 因此游離能均為正值 v M(g) M + (g) + e - E 1 =first ionization energy v M + (g) M +2 (g) + e - E 2 =Second ionization energy E 1 < E 2 v Increases across the periodic table from left to right. v Decreses moving down the periodic table 83

84 General Trend in First Ionization Energies Increasing First Ionization Energy Increasing First Ionization Energy 84

85 85

86 Filled n=1 shell Filled n=2 shell Filled n=3 shell Filled n=4 shell Filled n=5 shell 86

87 例外同週期中游離能隨原子序增加而增加同族中游離能隨主量子數增加而減少 87

88 Example 6.11: Consider the three elements C,N, and Si. Using only the periodic table, predict which of the three elements has (a) the largest atomic radius; the smallest atomic radius. (b) the largest ionization energy ; the smallest ionization energy. Sol: (a) C is larger than N but smaller than Si. Silicon must be the largest atom and nitrogen the smallest. (b) C has a smaller ionization energy than N but a larger ionization energy than Si. 88

89 週期表預測下列三個元素 B, C 及 Al 之 (a) 何者具最大及最小原子半徑 Al > B > C B, C, 為同週期 5 B, 6 C, 故原子半徑 B > C B, Al, 為同族 5 B, 13 Al, 故原子半徑 Al > B (b) 何者具有最大及最小游離能 C > B > Al B, C, 為同週期 5 B, 6 C, 故游離能 C > B B, Al, 為同族 5 B, 13 Al, 故游離能 B > Al 89

90 electronegativity v 由 Linus Pouling 提出, 為相對觀念 v Atom is a measure of its tendency to lose electrons, the larger the ionization energy,the more difficult is to remove an electron. v 陰電性值 v 同一週期由左而右增加, 同一族由上而下變小 90

91 General Trend in Electronegativity Increasing First Ionization Energy Increasing First Ionization Energy 91

92 92

93 93

94 94

95 Classification of bonds by difference in electronegativity Difference Bond Type 0 Covalent 2 Ionic 0 < and <2 Polar Covalent Increasing difference in electronegativity Covalent share e - Polar Covalent partial transfer of e - Ionic transfer e - 95

96 Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H 2 S; and the NN bond in H 2 NNH 2. Cs 0.7 Cl = 2.3 Ionic H 2.1 S = 0.4 Polar Covalent N 3.0 N = 0 Covalent 96

97 Electron affinity is the negative of the energy change that occurs when an electron is accepted by an atom in the gaseous state to form an anion. X (g) + e - X - (g) F (g) + e - X - (g) H = -328 kj/mol EA = +328 kj/mol O (g) + e - O - (g) H = -141 kj/mol EA = +141 kj/mol 97

98 98

99 99 8.5

100 100

Group 1A Elements (ns 1, n 2) M M +1 + 1e - 2M (s) + 2H 2 O (l) 4M

101 Group 1A Elements (ns 1, n 2) M M e - 2M (s) + 2H 2 O (l) 4M (s) + O 2(g) 2MOH (aq) + H 2(g) 2M 2 O (s) Increasing reactivity 101

102 Group 2A Elements (ns 2, n 2) M M e - Be (s) + 2H 2 O (l) Mg (s) + 2H 2 O (g) M (s) + 2H 2 O (l) No Reaction Mg(OH) 2(aq) + H 2(g) M(OH) 2(aq) + H 2(g) M = Ca, Sr, or Ba Increasing reactivity 102

103 Group 3A Elements (ns 2 np 1, n 2) 4Al (s) + 3O 2(g) 2Al (s) + 6H + (aq) 2Al 2 O 3(s) 2Al 3+ (aq) + 3H 2(g) 103

104 Group 4A Elements (ns 2 np 2, n 2) Sn (s) + 2H + (aq) Pb (s) + 2H + (aq) Sn 2+ (aq) + H 2(g) Pb 2+ (aq) + H 2(g) 104

105 Group 5A Elements (ns 2 np 3, n 2) N 2 O 5(s) + H 2 O (l) 2HNO 3(aq) P 4 O 10(s) + 6H 2 O (l) 4H 3 PO 4(aq) 105

106 Group 6A Elements (ns 2 np 4, n 2) SO 3(g) + H 2 O (l) H 2 SO 4(aq) 106

107 Group 7A Elements (ns 2 np 5, n 2) X + 1e - X -1 X 2(g) + H 2(g) 2HX (g) Increasing reactivity 107

108 Group 8A Elements (ns 2 np 6, n 2) Completely filled ns and np subshells. Highest ionization energy of all elements. No tendency to accept extra electrons. 108

109 Properties of Oxides Across a Period basic acidic

110 110

近代物理

近代物理 Ch6. Quantum Theory of the Hydrogen Atom Introduction 1945 Schrodinger Schrodinger equation : If only I knew more mathematics wave function analytical solution Schrodinger Equation for the Hydrogen Atom