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91536105 Fundamentals of Organic and Inorganic Chemistry Tosapol Maluangnont, Ph.D. Lecture 1 – November 05, 2013 Lecture 1 Atomic Structure Required reading: Shriver & Atkins pp8-12, 14-20 In this Chapter, we will be covering : • Hydrogenic atoms & Rydberg formula • Basic principles of quantum mechanics • Four quantum numbers and atomic orbitals • Many-electron atoms: shielding and Zeff • Pauli’s exclusion principle, Hund’s rule, and Aufbau principle Page 1 of 33 What kind of information can we gather from these symbols? 19 F 9 • Fluorine 12 13 14 C C C 6 6 6 • Three isotopes of Carbon. • Atomic number Z = number of proton = 9 • Z=6 • The number of e- = 9 • The number of e- = 6. • Atomic mass = 19. (This is what you use in calculation of the molar mass etc.) • Atomic mass is 12, 13 and 14 respectively. • The number of neutron = 19 – 9 = 10 • The number of neutron is 6, 7, and 8 respectively. Page 2 of 33 What kind of information can we gather from these symbols? 14 C 6 14 vs N 7 • These two elements have the same atomic mass (p+n) = 14. 1 2 H H 1 1 proton deuterium 3 H 1 tritium • Hydrogen is the only element where each isotope has its own symbol: H, D, and T. • For carbon, Z = 6, whereas for nitrogen, Z = 7. • Different elements can have the same number of (p+n). Elements with different atomic number are different elements. Page 3 of 33 Try this Fill in this Table. Symbol 1 H 1 7 Li 3 No. of proton 4 No. of electron No. of neutron 5 Page 4 of 33 Hydrogen-like atom • Hydrogen atom has one electron (1s1). • The presence of only one atom makes it free from complicating effects of electron-electron repulsions. • Hydrogen-like atom is also called hydrogenic atoms. • Examples are H, He+ and C5+ (why?) H He C Z=1 Z=2 Z=6 1s1 1s2 1s22s22p2 So, He+ 1s1 So, C5+ 1s1 Reminder : Z is the atomic number, which is the number of proton. The number of electron (in a neutral atom) is equal to that of proton. http://www.youtube.com/watch?v=OJzW2RoZq1Y Page 5 of 33 Rydberg Formula R = Rydberg constant = 1.097×107 m-1 l = wavelength (m) n1 = 1, 2, … n2 = n1+1, n1+2, … Series with n1 = 1 Lyman series (in UV) Series with n1 = 2 Balmer series (in visible) Series with n1 = 3 Paschen series (IR) Series with n1 = 4 Brackett series (IR) The energy is not continuous, but dependent on n which is an integer. Don’t forget E = hn = hc/ l (h = Planck’s constant = 6.626×10-34 m2 kg s-1 n = frequency, c = speed of light = 2.997×108 m s-1, l = wavelength) All info in this slide will be given in the exam. Page 6 of 33 Example Calculate the wavelength of the first transition in the visible region in the atomic spectrum of H. จงคำนวณควำมยำวคลืน ่ ทีส ่ อดคลองกั บกำรเปลีย ่ นระดับพลังงำนแรก ้ ในช่วง visible light ในสเปคตรัมของอะตอมไฮโดรเจน In the visible region, this is the Balmer’s series with n1 = 2. The first transition therefore will begin at n2 = 3. Substituting these values together with R (= 1.097×107 m-1) gives 1/l = 1.524 x 10-3 nm-1 l = 656.3 nm (recall 1 nm = 10-9 m) (Don’t just read this – make sure you actually get this number.) Page 7 of 33 Try this Calculate the wavelength of the other transitions (in addition to that at 656 nm) in the visible region in the atomic spectrum of H. จงคำนวณควำมยำวคลืน ่ ทีส ่ อดคลองกั บกำรเปลีย ่ นระดับพลังงำนอืน ่ ๆ ้ (นอกเหนือจำกที่ 656 nm) ในช่วง visible light ในสเปคตรัมของอะตอม ไฮโดรเจน Page 8 of 33 Basic principles of quantum mechanics • Electrons can behave as particles or as waves “wave-particle duality”. Here, we will concentrate on the fact that they are wave. • A wave can be described by a mathematical equation (recall a sin-wave, cos-wave, etc.). • Wavefunction y (psi) – a function of the position x, y, z which describes the behavior of an electron • Schrödinger equation where me = the mass of an electron V = the potential energy of the electron E = the total energy of the electron Solution can be obtained only for certain values of E. That is, the energy is quantized. Page 9 of 33 • Like other waves, wavefunctions can interact. • There are regions of positive and negative amplitude (sign). • Constructive interference - Enhanced probability of finding the electrons in that region. • Destructive interference - Reduced probability of finding the electrons in that region. • This will be used to explain chemical bonding later. Page 10 of 33 • The probability of finding an electron at a given location is proportional to y2 at that point. • There is a high probability of finding the electron where y2 is large. • The electron will not be found where y2 is zero • y2 is called probability density. Page 11 of 33 Quantum Number n – principal quantum number • This is the shell of electrons. n = 1 is called K-shell, n = 2 is called Lshell, and so on. • The allowed energies are specified by n. For a hydrogen-like atom with the atomic number Z, with n R hcR = = = Compare this n vs the n in slide #6. 1, 2, 3, … 1.097×107 m-1 13.6 eV • The zero of energy (E = 0) means that electron and nucleus is widely separated. • The negative energy means that, electron in a bound state is lower than widely separated electron and nucleus. Page 12 of 33 Example What is the ratio of the energy of a ground-state H atom to that of a He+ ion? จงคำนวณอัตรำส่ วนของพลังงำนที่สภำวะพืน้ ของ อะตอมไฮโดรเจนเทียบกับอะตอมของ He+ Note that H atom and He+ ion are both hydrogen-like atoms. Therefore, we can use the equation En = -hcRZ2/n2. For the ground state, n = 1. Therefore, E(H)/E(He+) = (-hcRZH2)/ (-hcRZHe2) = ZH2/ ZHe+2 = 12 / 22 = 0.25 Page 13 of 33 Example The ionization energy of H atom is 13.6 eV. What is the difference in energy between n = 1 and n = 6 levels? กำหนดให้ พลังงำนไอออไนเซชันของอะตอมไฮโดรเจนเท่ ำกับ 13.6 eV จง คำนวณควำมแตกต่ ำงของพลังงำนที่ระดับ n = 1 เทียบกับระดับที่ n = 6. The question gives E(n=1) = -13.6 eV, and asks for what is E(n=1) – E(n=6)? Similar to the previous example, E(n=1)/E(n=6) = (-hcRZ2/12)/(-hcRZ2/62)/ = (1/12)/(1/62)/ = 36 This gives E(n=6) = E(n=1)/36 = -13.6 eV/ 36 = -0.378 eV Therefore E(n=1) – E(n=6) = (-13.6) – (-0.378) eV = -13.2 eV Page 14 of 33 Atomic orbital – wavefunction of an electron in an atom. Electrons in an atom are classified by a set of four quantum number: n, l, ml, ms We have seen n already when we talked about the electronic structure of hydrogenic atom. But, what’s about the rest? No electrons in the same atom will have all four of the quantum numbers the same! Page 15 of 33 Radial and angular part of wavefunctions • Orbitals are easily described with spherical polar coordinates. • The wavefunction of hydrogenic atom is the product of the part with radius, and another part with the angle. • R(r) of the radius depends on n and l. • Y(q,F) of the angular depends on l and ml. Page 16 of 33 n – principal quantum number: the shell of electrons. n = 1 is called K-shell, n = 2 is called L-shell, and so on l=0 l=1 l=2 l=3 l – orbital angular momentum quantum number: angular dependence of orbitals the subshell of electrons (s, p, d, etc.) ml – magnetic quantum number: orientation in space (component of l around the axis) ms – magnetic quantum number: Electrons can be either spin up ms = +1/2 or ↑ or a spin down ms = -1/2 or↓ or b (The fifth quantum number is the spin of an electron s, and is always +1/2.) n l 0 1 2 3 Subshell s p d f l = 0, ml = 0 (s) l = 1, ml = -1, 0, +1 (there are 3 p orbitals) l = 2, ml = -2, -1, 0, +1, +2 (there are 5 d orbitals) Page 17 of 33 Q: What are the orbitals having n = 4 and l = 1. How many orbitals are there in this set? ออร์ ปิตัลที่ n = 4 และ l = 1 คือออร์ ปิตัลใด และออร์ ปิตัลนีป้ ระกอบด้ วยกี่ออร์ ปิ ตัลย่ อย A: n = 4 is the number of the shell. It can have l = 0, 1, 2 and 3. l = 1 means that these are the p orbitals. For l = 1, there are three values of ml (-1, 0, +1) So, these orbitals are 4p orbitals (there are 3 of them, namely 4px, 4py, and 4pz.) ออรปิ์ ตล ั ที่ n = 4 และ l = 1 คือ ออรปิ์ ตล ั 4p ซึง่ ประกอบดวยสำมออร ปิ์ ตล ั คือ 4px, 4py และ 4pz ้ Page 18 of 33 n, l, ml, ms Q: What is this orbital: 3, 2, 0, +1/2 A: n = 3 is the number of the shell. It can have l = 0, 1, and 2 l = 2 means that these are the d orbitals. For l = 2, there are five values of ml (-2, -1, 0, +1, +2) So, the all five orbitals below are the correct answer to this question. (In the exam, giving just one answer is ok.) 3dxy, 3dxz, 3dyz, 3dz2, or 3dx2-y2 (Note that sometimes it is not defined with ml belongs to which orbitals.) Page 19 of 33 • The quantum number l describes the angular variation of orbitals (= shape) • The quantum number ml describes the orientation in space (direction). • Boundary surface – region with high probability of finding an electron • Spherically symmetrical • Equal probability of finding electrons within the sphere • For an s-orbital, l = 0 and ml = 0. s orbital (spherical) I will expect that you can sketch the shape of s, p and d orbitals with correct labeling of the axis. Page 20 of 33 p orbitals (dumb-bell shape) • px directs itself along the x-axis (and so on for py and pz). • Nodal plane – the plane cutting through a nucleus, separating the region of + and – sign of the wavefunction. (Note the bright and dark lope which is opposite to each other.) • No electrons at nodal plane • There are three ml values (-1, 0, and +1) for p-orbitals, representing px, py and pz. I will expect that you can sketch the shape of s, p and d orbitals with correct labeling of the axis. Page 21 of 33 dzx dxy dyz dx2-y2 dz 2 d orbitals • dzx is on the zx plane (and so is dyzand dxy). • dx2-y2 is along the x and y axis • dz2 looks like a cone along the z-axis. • There are two nodal planes for each d orbital. • There are five values of ml for d orbitals: -2, -1, 0, +1, and +2. I will expect that you can sketch the shape of s, p and d orbitals with correct labeling of the axis. Page 22 of 33 • The d orbitals (l = 2) can be found for elements of period n > 2. The d orbitals mostly describe the chemistry of transition metals. Reminder: n = 1 l = 0 (s) n = 2 l = 0, 1 (s and p) n = 3 l = 0, 1 and 2 (s, p and d) • The absence of d orbitals in the first member of group 15, 16, and 17, makes them different from the rest. This is called anomalies. (Anomaly แปลวำ่ สิ่ งทีแ ่ ปลก ประหลำด) • Example: Coordination number (= the number of ions (atoms) surround that ion (atom).) neutral: NCl3 vs PCl3, PCl5 ion: NH4+ vs PCl6- Page 23 of 33 Page 24 of 33 Many-electron atoms Consider the ground state e- configuration of the following atoms H: 1s1 He: 1s2 1s orbital of He is more compact than that of H. (There are 2 protons in He’s nucleus which draw ecloser.) Li: 1s22s1 (NOT 1s3!) “Pauli exclusion principle” • no more than two electrons may occupy a single orbital. • If they do, their spins must be paired. Page 25 of 33 • An electron at different shell feels reduced (and unequal) amount of charge at the nucleus due to shielding by electrons closer to nucleus. (Think about the pressure you experience in the classroom, which is probably a function of where you sit!) • The magnitude of shielding is depending on which electrons you are looking at. • Because of shielding, the “effective nuclear charge Zeff” is lower than the true nuclear charge. • Zeff (usually) increases across the period. • Zeff = Z – s, where Z is the atomic number, and s is the shielding constant Page 26 of 33 I will expect that you can explain the trend in Zeff. Page 27 of 33 • 1s e- of H atom: it is attracted by the proton at the nucleus, and it does not interact with any other electrons. In this case, Zeff = Z = 1. • 1s e- of Li atom: Because of interaction between 2 e- in the 1s orbital (Li 1s2), one electron shields another electron. Each electron does not fully feel the presence of Z = 3. Instead, Zeff is 2.69. H 1s2 Li 1s22s1 • 2s e- of Li atom: This is electron is further away from the nucleus. Even more, it is shielded by 2ein the 1s orbital. So, instead of Z= 3, Zeff is only 1.28. Page 28 of 33 Example Use data from Table 1.2 to calculate the shielding constants for the outermost electron in the elements Li and Be. จงใช้ ข้อมูลจำกตำรำง 1.2 คำนวณ shielding constant ของอิเลคตรอนวงนอก สุดของอะตอมตัง้ แต่ Li และ Be Li: Be: s = Z – Zeff s = 3 - 1.28 = 1.72 s = 4 – 1.91 = 2.09 Try this Use data from Table 1.2 to calculate the shielding constants for the outermost electron in the elements from Be to F. Page 29 of 33 I will expect that you know this in the exam. Aufbau principle 1. General order of occupation: 1s 2s 2p 3s 3p 4s 3d 4p 2. Hund’s rule – when more than one orbital has the same energy, electrons occupy separate orbitals, with parallel spins Z=1 Z=2 Z=3 Z=4 Z=5 Z=6 Z=7 Z=8 2p1 H He Li Be B C N O 1s1 1s2 1s22s1, or [He]2s1 [He]2s2 [He]2s22p1 [He]2s22p2 [He]2s22p3 [He]2s22p4 2p2 2p3 closed shell 2p4 Page 30 of 33 Aufbau principle (continued) 3.1 For incompletely filled d subshell, e-configuration is mostly in the form 4s23dn 3.2 Spin correlation a case where half-filled or filled shell is more stable for 4s23dn. Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Z = 18 1s22s22p63s23p6 or [Ne]3s23p6 closed shell [Ar] Z = 19 [Ar]4s1 Z = 20 [Ar]4s2 Z = 21 [Ar]3d14s2 Z = 22 [Ar]3d24s2 Z = 23 [Ar]3d34s2 Z = 24 [Ar]3d54s1 (not 3d44s2!) spin correlation Z = 25 [Ar]3d54s2 (not 3d64s1) spin correlation Z = 26 [Ar]3d64s2 Z = 27 [Ar]3d74s2 Z = 28 [Ar]3d84s2 Z = 29 [Ar]3d104s1 (not 3d94s2!) spin correlation Page 31 of 33 C: [He]2s22p2 N: [He]2s22p3 O: [He]2s22p4 Q: We previously mentioned that Zeff increases across the period. However, going from C to N, the increase in Zeff is 0.69, while going from N to O, the increase in Zeff is only 0.62. Why? A: From C to N, electrons add into an empty p orbital. From N to O, the eadd into occupied p orbital. There is more repulsion in the latter case, making the increase in Zeff less. Page 32 of 33 Exercises Do the exercises in Shriver & Atkins on p32. 1.8 1.11 1.12 1.13 1.14 1.15 Page 33 of 33