Where Are We Going…? Many Electron Atoms For any 2 e - atom or ion, the Schrödinger equation cannot be solved for • • Week 10: Orbitals and Terms every electron: � Russell-Saunders coupling of orbital and spin angular momenta Σ Σ Σ Σ Σ Σ Σ Σ Σ Σ Σ Σ Σ Σ Σ Σ Ze Ze � Free-ion terms for p 2 � Free-ion terms for p 2 H H-like = 2 + − − − − ½ mv i • Week 11: Terms and ionization energies r i i � Free-ion terms for d 2 i � Ionization energies for 2p and 3d elements Treatment leads to configurations • • Week 12: Terms and levels � for example: He 1s 2 , C 1s 2 2s 2 2p 2 � Spin-orbit coupling � Total angular momentum � Total angular momentum • • Inclusion of interelectron repulsion leads to terms Inclusion of interelectron repulsion leads to terms Week 13: Levels and ionization energies • � for example: p 2 1 D, 3 F and 1 S e 2 Σ Σ � j-j coupling Σ Σ � characterized by S and L quantum numbers − − − − � Ionization energies for 6p elements � energy given by Hund’s 1 st and 2 nd rules r ij i � j � (2S+1)(2L+1) degenerate Slide 1/16 Magnetism Due To Spin Magnetism Due To Orbit • Electron(s) with orbital angular momentum generate a magnetic field • Electron(s) with spin angular momentum generate a magnetic field perpendicular to plane of loop perpendicular to plane of loop � magnitude related to L � magnitude related to L � magnitude related to S � magnitude related to S � direction related to M L � direction related to M S Slide 3/16 Slide 4/16
Orbital Magnetism Orbital Magnetism • Electrons generate magnetism through their orbital motion rotation of a p x orbital by 90 � gives a p y orbital and vice versa: • This is associated with an ability to rotate an orbital about an axis into generating magnetism about the z-direction an identical and degenerate orbital. • To be able to do this: � the orbitals involved must have the same energy � there must not be an electron in the second orbital with the same z z spin as that in the first orbital. If there is, the electron cannot orbit without breaking the Pauli principle. y y x x x x free orbitals free orbital available no free orbital available rotation of a p x orbital by 90 � gives a p y orbital and vice versa: available for for electron to hop for electron to hop into: generating magnetism about the z-direction electron to hop into: into: no orbital magnetism orbital magnetism orbital magnetism L = 1 L = 1 L = 0 Slide 6/16 Spin Orbit Coupling Russell – Saunders Coupling • There is a magnetic interaction between the magnetism generated • The magnetic interaction increases with the atomic number by the spin and orbital motions � for most of the periodic table, electrostatic >> magnetic � results in different values for the total angular momentum , J Treat electrostatic to give terms characterized by L and S • � l 1 + l 2 + � = L , s 1 + s 2 + � = S • Then treat spin-orbit second to give levels : � L + S = J spin magnetism orbital magnetism orbital magnetism � J is the total angular momentum � J is the total angular momentum e 2 Σ Σ Σ Σ − − − − H = H H-like + + � L.S r ij i � j lowest energy configurations terms levels highest energy Slide 7/16
Hund’s 3 rd Rule Russell – Saunders Coupling For each L and S value: For less than half-filled shells, smallest J lies lowest • • � J = L + S , L + S � 1, L + S � 2 � . L � S � p 2 : ground term is 3 P with S = 1 and L = 1 � Each level, M J = J, J -1, J - 2, � - J (2 J +1 values) � Each level, M J = J, J -1, J - 2, � - J (2 J +1 values) � J = 2, 1 and 0 � J = 2, 1 and 0 � less than half-filled: 2 S +1 L J 3 P 2 3 P 3 P 1 3 P 0 Slide 9/16 Slide 10/16 Hund’s 3 rd Rule Magnetism For more than half-filled shells, highest J lies lowest • • The magnetic moment is given by: 12 � p 4 : ground term is 3 P with S = 1 and L = 1 = g[J(J + 1)] � eff � J = 2, 1 and 0 � J = 2, 1 and 0 � more than half-filled: � where g is the Landé splitting factor , [S(S 1) L(L 1)] + + 3 3 P 0 g = + 2 2J(J 1) + 3 P 1 3 P p 2 : ground level is 3 P 0 with J = 0, S = 1, L = 1 • 3 P � � eff = 0 (p 2 is diamagnetic, at least at low temperature) p 4 : ground level is 3 P 2 with J = 2, S = 1, L = 1 • � g = 3/2 and � eff = 3.68 B.M. (B.M. = � Bohr Magnetons � ) 3 P 2 Slide 11/16 Slide 12/16
Ionization Energies: (iii) Hund’s 3 rd Rule j-j Coupling • For very heavy elements, magnetic coupling becomes large p-block ionization energies: M � � � � M + 12 25 11 • Treat spin-orbit first to give spin-orbitals for each electron: 2p 6p y (eV) 20 y (eV) 10 ionization energy (e � j = l + s each value is (2 j +1) degenerate � j = l + s each value is (2 j +1) degenerate ionization energy ( 3p 15 9 4p 5p Then add individual j values together to give J 8 • 6p 10 7 � j 1 + j 2 + � = J 5 6 5 0 For p-electrons, l = 1 and s = 1/2 • 1 2 3 4 5 6 p n 1 2 3 4 5 6 p n � j = 1/2 and 3/2 with former lowest in energy � j = 1/2 and 3/2 with former lowest in energy For 6p, there is a decrease between p 2 and p 3 • � j = 3/2 No half-filled shell effect! j = 1/2 Slide 13/16 Slide 14/16 j-j Coupling Summary For p-electrons, l = 1 and s = 1/2 • Spin and orbital magnetism � j = 1/2 and 3/2 with former lowest in energy • Electrons have intrinsic magnetism due to spin Electrons may also have orbital magnetism Electrons may also have orbital magnetism • • j = 3/2 Spin-orbit coupling • Usually weak magnetic coupling between spin and orbit Characterized by levels with total angular momentum, J j = 1/2 • Hund’s 3 rd Rule Lowest J lies lowest for < 1/2 filled shells • • • If electrostatic >> magnetic If electrostatic >> magnetic Highest J lies lowest for > 1/2 filled shells Highest J lies lowest for > 1/2 filled shells • • � overall increase due to increasing nuclear charge Consequences � decrease in ionization energy for p 4 due to pairing (1 st rule) Magnitude of magnetism due to J , L and S • Stabilization of p 1 and p 2 , destabilization of p 4 – p 6 • If magnetic > electrostatic • � overall increase due to increasing nuclear charge Task! � decrease in ionization energy for p 3 due to repulsive magnetic Work out ground levels and magnetism for f n elements interaction (3 rd rule) • Slide 15/16
Recommend
More recommend
