Atomic Structure and the Periodic Table. Development of the Periodic Law. The Consequences. Atomic (combining) weights

Transcription

1 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 Avogadro's combining volumes Cannizzaro's atomic weights Atomic (combining) 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.

2 Avogadro's Combining Volumes Ordering schemes Hydrogen gas + Oxygen gas Water volumes 1 volume volumes Simple gaseous elements must be diatomic molecules Combining weights are not always atomic weights Dobereiner's Triads 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. Cl 35; Br 80; I 17; average 81 S 3; Se 79; Te 18; average 80 Li 7; Na 3; K 39; average 3 Telluric Helix 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 Telluric Helix The Law of Octaves Correlate the chemical properties of the elements with their atomic weights

3 The Law of Octaves H Li G B C 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 Th Hg Bi Os If the elements are ordered by atomic weight, there is a similarity between every eighth element. Mendeleev's periodic law 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. Mosley Development of a model of the atom 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 3 wavelength nm frequency 4.57 x Hz sphere with uniform positive charge negatively charged electrons imbedded in the sphere, with a uniform distribution x x x ν = R(1/ -1/n ) where R is a constant (Rydberg constant), and n is an integer greater than which corresponds to a line.

4 Thomson Model Line spectra are due to the oscillation of the electrons Thomson Model 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 Classical physics Bohr Model electrons in orbits around the nucleus electrons are moving at a high velocity, such that the Centrifugal force balances the electrostatic attraction. e r = mv r

5 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 Bohr Model If a change of state occurs, then the energy associated with it must be E = E -E 1 = hν 1 ν 1 = (E -E 1 )/h Bohr Model the energy of an electron in the nth level is found to be E n = m 4 π e Z nh Bohr Model the light emission is 4 ν = π me Z c h 1 n 1 - n 1 this means that π me 4 /ch 3 must be the Rydberg constant Problems Fine structure in line spectra Bohr-Sommerfeld Model Allow orbitals to be elliptical The two degrees of freedom requires two quantum numbers. The second describing the elongation of the ellipse.

6 Problems New Approach Zeeman Effect the splitting of spectral lines in the presence of a magnetic field. debroglie Relationship debroglie considered equating the energy of light with that of a moving particle. Schrodinger Equation Using debroglies idea, consider an electron as a wave disturbance rather than as a particle. δ ψ 8 m + δ ψ + δ ψ + π δ x δ y δ z h ( E - V) ψ = 0 Polar Coordinates Separation of Variables assumption Ψ(r,θ,φ) = R(r)Y(θ,φ)

7 Radial Distribution Function 3 ( n - l -1)! [( n +1 )! ] Z -ρ/ +1 R, (r) = e L n+ 1 ( ) 3 ρ ρ l n l ρ n ao n a o = h 4 π µ e determines relationship between n and l quantum numbers n = 1,, 3,... l = 0, 1,,... (n-1) (n - l - 1)! >= 0 so l cannot be.ge. n Radial Density Function Spherical Harmonics Y m l (θ,φ) = θm l (θ) φ m (φ) Spherical Harmonics (1) ( l +1)( l m )! ( l + m )! ( ) m θ = P m l cosθ l determines relationship between l and m quantum numbers l = 0, 1,, 3,... m = 0, ±1, ±, ±3,..., ±l (l - m )! >= 0 so m cannot be.ge. l Spherical Harmonics () Φ m = 1 e π ± imφ Orbital Shapes-Total Density Function m = 0, 1,, 3,...

8 Quantum Number Relationships 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 l + 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 Orbital Filling Aufbau principle--starting with the hydrogen atom, and successively adding a proton to the nucleus and an electron to the electron cloud. Empirical Rules 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 Electron Spin 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/ in units of h/π Pauli Exclusion Principle No two electrons may have the same four quantum numbers Correlation of Electron Motion 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

9 Electron Configuration Aufbau Diagram Ambiquities in the Aufbau Diagram Cr 3d 5 4s 1 Cu 3d 10 4s 1 Mo 4d 5 5s 1 Ag 4d 10 5s 1 W 5d 3 6s Au 5d 10 6s 1 Writing Electron Configurations Using quantum numbers Using spdf notation 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, l i, in an atom couple together. L = l 1 + l, l 1 + l -1, l 1 + l -,..., l 1 -l L is derived from M l, which is the vectorial sum of the M l values for all electrons. for n electrons M l = m l (1) + m l () +... m l (n) and M l = L, L-1, L-,..., -L the number of possible values for M l will be L+1

10 Total Spin Angular Momentum Quantum Number S Assume that the individual electron spins couple together. S is derived from M s, which is the algebraic sum of the s values for all electrons for n electrons M s = m s (1) + m s () +... m s (n) and M s = S, S-1, S-,..., -S the number of possible values for M s will be S+1 Determining Term Symbols Closed shells and subshells are ignored. Term Symbols The term symbol is constructed as follows S+1 X where S+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 =, X is D L = 3, X is F state Li (1s s 1 ) Discount all closed shells, ie the 1s Find the possible values for M l M l = m l (1) = 0 Find the possible values for L M l = 0, therefore L = 0 Find the possible values for M s M s = m s, m s = +1/ or -1/ M s = +1/ or -1/ state Li (1s s 1 ) Find the possible values for S M s = +1/ or -1/ S = 1/ The term symbol since L = 0 then X is S S is 1/ therefore the ground state of Li is S state C (1s s p ) Draw the individual microstates

11 state C (1s s p ) The possible values of M l is, 1, 0, -1, -, and values for M s are 1, 0, -1 Use + and - to indicate spin and 1, 0, or -1 to indicate the orbital state C (1s s p ) The microstate with M l = and M s = 1 must be eliminated due to the Pauli exclusion principle. To have M l of means that the electrons share the same orbital, and a M s of 1 indicates that they have the same spin. state C (1s s p ) 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 3 P state. A 3 P 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 state C (1s s p ) If we look at the very top, there is a microstate with M l = and M s = 0, which is indicative of a state that has L = and S = 0, the 1 D state. A 1 D state has a total of 5 microstates M l = 1, M s = 0; M l = 0, M s = 0; M l = -1, M s = 0 M l = -, M s = 0; M l =, M s = 0 state C (1s s p ) What remains is a single microstate with M l = 0 and M s = 0. This microstate indicates that there is a term having L = 0 and S = 0, or 1 S. state C (1s s p ) Thus we have three possible term symbols for the ground state configuration of carbon, 3 P, 1 D, and 1 S. Which one is the ground state?

12 Hund's First Rule the term with the maximum spin multiplicity will be lowest in energy, therefore the term symbol for the ground state of carbon is 3 P. The 1 D and 1 S terms will be excited states Order of 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 1 D state will be the first excited state. The Total Angular Quantum Number J 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 3 P state there are three more specific states 3 P 0, 3 P 1, and 3 P Hund's Third Rule the term with the lowest J value lies lowest in energy if the partially filled subshell is less than half-filled. 3 P 0 will be the ground state Splitting of Orbital Energy 1 S 1 S 0 _ 1 micro- 1 microstate state 1 D 1 D _ 5 micro- 5 microstates states An Easier Method Partial Terms D. H. McDaniel, J. Chem. Ed. 1977, p 15 microstates 3 P _ 5 microstates 3 P 3 P 1 _ 9 micro- 3 microstates states 3 P 0 _ 1 microstate no electron- electron- spin-orbit electron electron coupling repulsions repulsions

13 state C (1s s p ) If we just consider the electron spin, in a p configuration, three spin configurations are possible p a p0 b p 0 a p b p 1 a p1 b Where p a has a spin of +½ and p b has a spin of -½ Orbital Occupancy Orbital Set s S p S P P S d S D PF PF D S f S F PFH SDFGI SDFGI PFH F S g S G PFHK PF()G SD()F HIKM G()HI() KLN state C (1s s p ) Each spin configuration may be identified with one or more partial terms p a p0 b P S => P P P => S P D p 1 a p1 b product of partial terms gives terms having L values in integer runs from L a -L b through L a +L b state C (1s s p ) To determine the multiplicity, we look at the maximum M s value for each partial term. P maximum M s = 1, so the term symbol is 3 P D maximum M s = 0, so the term symbol is 1 D P maximum M s = 0, so the term symbol is 1 P S maximum M s = 0, so the term symbol is 1 S state C (1s s p ) Flaw in partial terms Some microstates will be assigned to more than one term symbol state C (1s s p ) Thus we have three possible term symbols for the ground state configuration of carbon, 3 P, 1 D, and 1 S. For each term symbol of a higher multiplicity, one may eliminate one of the same value of L with a lower multiplicity.

14 Determining the number of microstates n! number of microstate s = 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). f 3 configuration three spin configurations are possible f 3 a f0 b f a f1 b and a total of 364 microstates are possible (14!/(3!*11!) f 3 configuration f 3 configuration f 3 a f0 b (SDFGI) x S => 4 S, 4 D, 4 F, 4 G, 4 I Term Symbols: 4 I, 4 G, 4 F, 4 D, 4 S, L, K, H(), G(), F(), D(), P Ground State: 4 I f a f1 b (PFH) x F => G, F, D I, H, G, F, D, P, S L, K, I, H, G, F, D 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, s & p, 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.

15 Slater's Rules Ionization Energies Energy level calculation E = ( Z S) 13.6eV n Outer-most electron leaves first. Electron with largest n lost first Within the nth shell the electron with the largest l. Trends in Ionization Energies Ionization Energy (MJ).50 He Ne.00 F 1.50 N Ar H Cl O P C 1.00 Be S Mg B Si Li Al 0.50 Na Atomic Number Electron Affinity Energy released when an electron is added to a neutral atom. X + e - ---> X - + energy E. A. decreases down a group Cl (3.61) > Br (3.36) E. A. increases across a group Atomic Radii Covalent Radii One half the bond length in a non-polar homonuclear bond Factors effecting: bond polarity, # of e- in outermost orbital, nuclear charge

16 Ionic Radii