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Atomic Structure and the Periodic Table Development of the Periodic Law Development of atomic weights Dalton's Atomic Theory Elements consist of atoms Each atom of the same element is identical Atoms of different elements are different Molecules of compounds are formed by the combination of two or more different atoms, in fixed whole-number atom ratios In a chemical reaction, atoms are not created or destroyed, but only rearranged into different combinations The Consequences Atomic (combining) weights Atomic (combining) weights Avogadro's combining volumes Cannizzaro's atomic weights if a reaction is balanced as: A + B --> C then x grams of A and y grams of B must form z grams of C, where x, y, and z are the atomic or molecular weights of A, B, and C. therefore if 1 g of hydrogen reacts with 8 g of oxygen to form 9 g of water, and the atomic weight of hydrogen is 1, then the atomic weight of oxygen is 8. 1 Avogadro's Combining Volumes Ordering schemes Hydrogen gas + Oxygen gas → Water 2 volumes 1 volume 2 volumes Simple gaseous elements must be diatomic molecules Combining weights are not always atomic weights Dobereiner's Triads Telluric Helix Similar elements occur in three's and the atomic weight of the second is approximately equal to the average of the atomic weight of the first and third. A vertical cylinder with 16 equidistant lines drawn parallel to its axis. Draw a helix at 45 to the axis and arrange the elements on the spiral in order of increasing atomic weight. In some cases, elements that lie directly under each other show similarity. The properties of the elements are the properties of numbers Cl 35; Br 80; I 127; average 81 S 32; Se 79; Te 128; average 80 Li 7; Na 23; K 39; average 23 Telluric Helix The Law of Octaves Correlate the chemical properties of the elements with their atomic weights 2 The Law of Octaves H Li G B C Mendeleev's periodic law N O F Na Mg Al Si P S Cl K Ca Cr Ti Mn Fe Co & Ni Cu Zn Y In As Se Br Rb Sr Ce & La Zr Di & Mo Ro & Ru Pd Ag Cd U Sn Sb Te I Cs Ba & V Ta W Nb Au Pt & Ir Tl Pb Hg Bi Os Th If the elements are ordered by atomic weight, there is a similarity between every eighth element. The properties of the elements can be represented as periodic functions of their atomic weights. emphasize group similarities, reversing the order of atomic weights where necessary. Leave vacancies for undiscovered elements, predicting properties. Development of a model of the atom Mosley X-ray spectra of the elements ν = K(Z - k) where K and k are constants, and Z is the atomic number Line Spectra Thomson Model n wavelength frequency 3 656.3 nm 4.57 x 1014 Hz 4 486.1 6.17 x 1014 5 434.0 6.91 x 1014 6 410.2 7.31 x 1014 sphere with uniform positive charge negatively charged electrons imbedded in the sphere, with a uniform distribution. ν = R(1/22 - 1/n2) where R is a constant (Rydberg constant), and n is an integer greater than 2 which corresponds to a line. 3 Thomson Model Thomson Model Line spectra are due to the oscillation of the electrons Electron emission by active metals due to loss of an imbedded electron. Problems Rutherford Model scattering of alpha particles Using this model, there should be only very small angle scattering of alpha particles. experimentally, small angle and large angle scattering is observed positive charge is concentrated in a small nucleus at the center of the atom electrons moving around the nucleus Problems Bohr Model Classical physics electrons in orbits around the nucleus electrons are moving at a high velocity, such that the Centrifugal force balances the electrostatic attraction. m v2 e2 = r r2 4 Bohr Model Bohr Model Only certain orbitals are possible, meaning that only discrete energy states are allowed Energy is absorbed or emitted only when there is a change of energy state If a change of state occurs, then the energy associated with it must be Bohr Model Bohr Model the energy of an electron in the nth level is found to be the light emission is En = ∆E = E2 - E1 = hν21 ν21 = (E2 - E1)/h 2π me Z n2 h2 2 4 2 2 π 2 m e4 Z 2 1 1 - 2 ν= 2 2 ch n2 n1 this means that 2π2me4/ch3 must be the Rydberg constant Problems Bohr-Sommerfeld Model Fine structure in line spectra Allow orbitals to be elliptical The two degrees of freedom requires two quantum numbers. The second describing the elongation of the ellipse. 5 Problems New Approach Zeeman Effect—the splitting of spectral lines in the presence of a magnetic field. deBroglie Relationship Schrodinger Equation deBroglie considered equating the energy of light with that of a moving particle. Using deBroglies idea, consider an electron as a wave disturbance rather than as a particle. δ 2ψ + δ 2ψ + δ 2ψ + 8 π 2 m (E - V )ψ = 0 2 δ x 2 δ y2 δ z 2 h Polar Coordinates Separation of Variables assumption Ψ(r,θ,φ) = R(r)Y(θ,φ) 6 Radial Distribution Function Radial Density Function 2Z (n - l - 1)! - ρ/2 ρ 2 l+1 ρ Ln +1 ( ρ ) R n ,l (r) = 3 e n a o 2n [(n + 1)!] 3 ao = h2 4 π 2 µ e2 determines relationship between n and l quantum numbers n = 1, 2, 3,... l = 0, 1, 2,... (n-1) (n - l - 1)! >= 0 so l cannot be .GE. n Spherical Harmonics Spherical Harmonics (1) θ ml = Yml(θ,φ) = θml(θ) φ m (φ) (2l + 1)(l − m )! 2(l + m )! P (cosθ ) m l determines relationship between l and m quantum numbers l = 0, 1, 2, 3,... m = 0, ±1, ±2, ±3,..., ±l (l - |m|)! >= 0 so |m| cannot be .GE. l Spherical Harmonics (2) Φm = Orbital Shapes-Total Density Function 1 ± imφ e 2π m = 0, 1, 2, 3,... 7 Quantum Number Relationships Orbital Filling Within a given atom, the lower the value of n, the more stable (or lower in energy) the orbital There are n types of orbitals in the nth energy level There are 2l + 1 orbitals of each type There are n - l - 1 nodes in the radial distribution functions of all orbitals (spherical nodes) There are l nodal surfaces in the angular distribution function of all the orbitals There are n - 1 nodes in the total distribution functions of all the orbitals Aufbau principle--Starting with the hydrogen atom, and successively adding a proton to the nucleus and an electron to the electron cloud. Empirical Rules Electron Spin Orbital energies increase as (n+l) increases For two orbitals of the same (n+l), the one with the smaller n lies lower in energy There still were unexplained doublets in the line spectra where single lines were expected. explaination by allowing the electron to spin with a spin angular momentum of 1/2 in units of h/2π Pauli Exclusion Principle Correlation of Electron Motion No two electrons may have the same four quantum numbers spin correlation(exchange energy)--energy is lower when two electrons have the same spin Charge correlation energy--lower energy if electrons have different angular functions 8 Electron Configuration Aufbau Diagram Ambiquities in the Aufbau Diagram Writing Electron Configurations Cr 3d54s1 Mo 4d55s1 W 5d36s2 Using quantum numbers Using spdf notation Cu Ag Au 3d104s1 4d105s1 5d106s1 Term Symbols The energy of the individual orbitals are dependent on n and l Which is also influenced by the coupling between the orbital momentum and spin momentum And replusion between the electron Total Orbital Angular Momentum Quantum Number L Assume that all of the angular momenta of the different electrons, li, in an atom couple together. L = l1 + l2, l1 + l2 - 1, l1 + l2 - 2 ,..., |l1 - l2| L is derived from Ml, which is the vectorial sum of the Ml values for all electrons. for n electrons Ml = ml(1) + ml(2) + ... ml(n) and Ml = L, L-1, L-2, ..., -L the number of possible values for Ml will be 2L+1 9 Total Spin Angular Momentum Quantum Number S Assume that the individual electron spins couple together. S is derived from Ms, which is the algebraic sum of the s values for all electrons for n electrons Ms = ms(1) + ms(2) + ... ms(n) and Ms = S, S-1, S-2, ..., -S the number of possible values for Ms will be 2S+1 Determining Term Symbols Closed shells and subshells are ignored. Term Symbols Derive the term symbol for ground state Li (1s22s1) The term symbol is constructed as follows 2S+1X where 2S+1 is the spin multiplicity and comes from S, and X is a letter that comes from the value of L for L = 0, X is S L = 1, X is P L = 2, X is D L = 3, X is F Discount all closed shells, ie the 1s Find the possible values for Ml Ml = ml(1) = 0 Find the possible values for L Ml = 0, therefore L = 0 Find the possible values for Ms Ms = ms, ms = +1/2 or -1/2 Ms = +1/2 or -1/2 Derive the term symbol for ground state Li (1s22s1) Derive the term symbol for ground state C (1s22s22p2) Find the possible values for S Ms = +1/2 or -1/2 S = 1/2 The term symbol since L = 0 then X is S S is 1/2 therefore the ground state of Li is 2S Draw the individual microstates 10 Derive the term symbol for ground state C (1s22s22p2) Derive the term symbol for ground state C (1s22s22p2) The possible values of Ml is 2, 1, 0, -1, -2, and values for Ms are 1, 0, -1 Use + and - to indicate spin and 1, 0, or -1 to indicate the orbital The microstate with Ml = 2 and Ms = 1 must be eliminated due to the Pauli exclusion principle. To have Ml of 2 means that the electrons share the same orbital, and a Ms of 1 indicates that they have the same spin. Derive the term symbol for ground state C (1s22s22p2) Derive the term symbol for ground state C (1s22s22p2) If we start at the top, there is a microstate with Ml = 1 and Ms = 1, which is indicative of a state that has L = 1 and S = 1, the 3P state. A 3P state has a total of 9 microstates Ml = 0, Ms = 1, Ml = -1, Ms = 1, Ml = 1, Ms = 1 Ml = 0, Ms = 0, Ml = -1, Ms = 0, Ml = 1, Ms = 0 Ml = 0, Ms = -1, Ml = -1, Ms = -1, Ml = 1, Ms = -1 remove them from the array If we look at the very top, there is a microstate with Ml = 2 and Ms = 0, which is indicative of a state that has L = 2 and S = 0, the 1D state. A 1D state has a total of 5 microstates Ml = 1, Ms = 0; Ml = 0, Ms = 0; Ml = -1, Ms = 0 Ml = -2, Ms = 0; Ml = 2, Ms = 0 Derive the term symbol for ground state C (1s22s22p2) Derive the term symbol for ground state C (1s22s22p2) What remains is a single microstate with Ml = 0 and Ms = 0. This microstate indicates that there is a term having L = 0 and S = 0, or 1S. Thus we have three possible term symbols for the ground state configuration of carbon, 3P, 1D, and 1S. Which one is the ground state? 11 Hund's First Rule Order of Excited States the term with the maximum spin multiplicity will be lowest in energy, therefore the term symbol for the ground state of carbon is 3P. The 1D and 1S terms will be excited states Hund's second rule: If more than one term has the same multiplicity, the term with the highest L value will be lower in energy. Thus, the 1D state will be the first excited state. The Total Angular Quantum Number J Hund's Third Rule J is determined by taking the sum and difference between L and S, that is L + S and L - S, and then take the sequence of numbers from L + S to |L - S| in integral steps. Therefore, for the 3P state there are three more specific states 3P0, 3P1, and 3P 2 the term with the lowest J value lies lowest in energy if the partially filled subshell is less than half-filled. Splitting of Orbital Energy An Easier Method 1S 1 microstate 1D 5 microstates 3P 9 microstates p2 15 microstates no electronelectron repulsions electronelectron repulsions 1S _ 0 1 microstate 1D _ 2 5 microstates 3P 0 will be the ground state Partial Terms D. H. McDaniel, J. Chem. Ed. 1977, 54 147. 3P _ 2 5 microstates 3P _ 1 3 microstates 3P _ 0 1 microstate spin-orbit coupling 12 Derive the term symbol for ground state C (1s22s22p2) If we just consider the electron spin, in a p2 configuration, three spin configurations are possible p0a p2b p1a p1b p2a p0b Where pa has a spin of +½ and pb has a spin of -½ Orbital Occupancy Orbital Set s p d f g 0 1 2 3 S S S S S 4 5 P D F G P PF PFH PFHK S PF D S SDFGI SDFGI PFH PF(2)G SD(2)F HIKM G(2)HI(2) KLN 6 7 F S Derive the term symbol for ground state C (1s22s22p2) Derive the term symbol for ground state C (1s22s22p2) Each spin configuration may be identified with one or more partial terms p2a p0b P × S => P P × P => S P D p1a p1b To determine the multiplicity, we look at the maximum Ms value for each partial term. P maximum Ms = 1, so the term symbol is 3P D maximum Ms = 0, so the term symbol is 1D P maximum Ms = 0, so the term symbol is 1P S maximum Ms = 0, so the term symbol is 1S product of partial terms gives terms having L values in integer runs from |La-Lb| through |La+Lb| Derive the term symbol for ground state C (1s22s22p2) Derive the term symbol for ground state C (1s22s22p2) Flaw in partial terms Some microstates will be assigned to more than one term symbol Thus we have three possible term symbols for the ground state configuration of carbon, 3P, 1D, and 1S. For each term symbol of a higher multiplicity, one may eliminate one of the same value of L with a lower multiplicity. 13 Determining the number of microstates number of microstate s = n! e! n! where n is the number of possible positions for an electron, e is the number of electrons and h is the number of leftover positions (n - e). f3 configuration three spin configurations are possible f2a f1b f3a f0b and a total of 364 microstates are possible (14!/(3!*11!) f3 configuration f3 configuration f3a f0b (SDFGI) x S => 4S, 4D, 4F, 4G, 4I Term Symbols: 4I, 4G, 4F, 4D, 4S, 2L, 2K, 2H(2), 2G(2), 2F(2), 2D(2), 2P Ground State: 4I f2a f1b (PFH) x F => 2G, 2F, 2D 2I, 2H, 2G, 2F, 2D, 2P, 2S 2L, 2K, 2I, 2H, 2G, 2F, 2D Polyelectron factors Slater's Rules Penetration (shielding): outer electrons do not "see" the same nuclear charge as electrons below them. ie a different effective nuclear charge. The electrons are grouped in the order 1s, 2s & 2p, 3s & 3p, 3d, 4s and 4p, 4d, 4f, etc., with the s and p orbitals grouped together. Electrons in groups lying above that of a particular electron do not shield it at all. A shielding of 0.35 is contributed by each other electron in the same group, except for a 1s electron which contributes 0.30 to the shielding of the other 1s electron For d and f electron the shielding from underlying groups is 1.00 for each electron in the underlying group. For s and p electrons the shielding from the immediately underlying shell (n - 1) is 0.85 for each electron, and that from groups further in is 1.00 for each electron. 14 Slater's Rules Ionization Energies Energy level calculation Outer-most electron leaves first. Electron with largest n lost first Within the nth shell the electron with the largest l. − ( Z − S ) 213 .6eV E= n2 Trends in Ionization Energies Ionization Energy (MJ) 2.50 Energy released when an electron is added to a neutral atom. X + e- ---> X- + energy He Ne 2.00 F N 1.50 H 1.00 Be 0.50 Li C Ar Cl P O S Mg B Na Si Al 0.00 0 5 10 Atomic Number Atomic Radii Electron Affinity 15 20 E. A. decreases down a group Cl (3.61) > Br (3.36) E. A. increases across a group Covalent Radii One half the bond length in a non-polar homonuclear bond Factors effecting: bond polarity, # of ein outermost orbital, nuclear charge 15 Ionic Radii 16