Presentation on theme: "1 He and hydrogenoid ions The one nucleus-electron system."— Presentation transcript:
1 He and hydrogenoid ions The one nucleus-electron system
2 topic Mathematic required. Schrödinger for a hydrogenoid Orbital s Orbital p
3 Two prerequisites Our world is 3D! We need to calculate integrals and derivatives in full space (3D). A system of one atom has spherical symmetry. Spherical units are appropriate. r rather than x,y,z except that was defined in cartesian units.
4 Spherical units x = r sin . cos y = r sin . sin z = r cos
10 Integration in space Integration over , and r gives V = 4/3 r 4 The integration over and gives the volume between two spheres of radii r and r+dr: dV = 4 r 2 dr Volume of a sphere Volume between two concentric spheres
13 Radial density dP/dR = 4 r 2 * It is the density of probability of finding a particule (an electron) at a given distance from a center (nucleus) It is not the density of probability per volume dP/dV= It is defined relative to a volume that increases with r. The unit of is L -3/2
14 Schrödinger for a hydrogenoid (1 nucleus – 1 electron) The definition of an orbital atomic orbital: any function e (x,y,z) representing a stationary state of an atomic electron. Born-Oppenheimer approximation: decoupling the motion of N and e (x N,y N,z N,x e,y e,z e )= (x N,y N,z N ) (x e,y e,z e ) m H =1846 m e : When e - covers 1m H covers 2.4 cm, C 6.7 mm and Au 1.7 mm
15 Schrödinger for a hydrogenoid (1 nucleus – 1 electron)
16 Schrödinger for a hydrogenoid (1 nucleus – 1 electron)
17 Schrödinger for a hydrogenoid (1 nucleus – 1 electron) We first look for solutions valid for large r
18 Solution for large r Which of the 2 would you chose ?
19 Solution e - r still valid close to the nucleus Already set to zero by taking Ne - r New To be set to zero leading to a condition on : a quantification due to the potential
20 The quantification of is a quantification on E This energy is negative. The electron is stable referred to the free electron
21 Energy units 1 Rydberg = 21.8 10 -19 Joules =14.14 10 5 J mole -1 1 Rydberg = 13.606 eV = 0.5 Hartree (atomic units) 1eV (charge for an electron under potential of 1 Volt) 1eV = 1.602 10 -19 Joules = 96.5 KJoules mole -1 (→ = 8065.5 cm -1 ) 1eV = = 24.06 Kcal mole -1.
22 Atomic units The energy unit is that of a dipole +/- e of length a 0 It is the potential energy for H which is not the total energy for H (-1/2 a.u.) (E=T+V) Atomic units : h/2 =1 and 1/4 0 =1 –The Schrödinger equation becomes simpler lengthchargemassenergy a 0, Bohr radius e, electron charge 1.602 10 -19 C m, electron mass 9.11 10 -41 Kg Hartree
23 Normalization of Ne - r From math textbooks The density of probability is maxima at the nucleus and decreases with the distance to the nucleus.
24 Radial density of probability a 0 /Z is the most probable distance to the nucleus; it was found by Niels Bohr using a planetary model. The radial density close to zero refers to a dense volume but very small; far to zero, it corresponds to a large volume but an empty one
28 Excited states We have obtained a solution using e - r ; it corresponds to the ground state. There are other quantified levels still lower than E=0 (classical domain where the e is no more attached to the nucleus)o We can search for other spherical function N n P n (r)e - r where P n (r) is a polynom of r of degree n-1
29 Orbital ns E ns = Z 2 /n 2 E 1s (H) E 2s = Z 2 /4 E 1s (H) Nodes: spheres for solution of equation P n =O n-1 solutions Average distance Principal quantum number
30 Orbital 2s = E 2s = Z 2 /4 E 1s (H) Average distance 5 a 0 /Z A more diffuse orbital: One nodal surface separating two regions with opposite phases: the sphere for r=2a 0 /Z. Within this sphere the probability of finding the electron is only 5.4%. The radial density of probability is maximum for r=0.764a 0 /Z and r=5.246a 0 /Z. Between 4.426 et 7.246 the probability of finding the electron is 64%.
33 Resolution of Schrödinger equation in , Solving the equation in and leads to define two other quantum numbers. They also can be defined using momentum instead of energy. For : Angular Momentum (Secondary, Azimuthal) Quantum Number (l): l = 0,..., n-1. For : Magnetic Quantum Number (ml): m l = -l,..., 0,..., +l.
34 Resolution of Schrödinger equation in ,the magnetic quantum number
35 Quantum numbers Principal Quantum Number (n): n = 1, 2,3,4, …, ∞ Specifies the energy of an electron and the size of the orbital (the distance from the nucleus of the peak in a radial probability distribution plot). All orbitals that have the same value of n are said to be in the same shell (level). Angular Momentum (Secondary, Azimunthal) Quantum Number (l): l = 0,..., n-1. Magnetic Quantum Number (m): m = -l,..., 0,..., +l. Spin Quantum Number (m s ): m s = +½ or -½. Specifies the orientation of the spin axis of an electron. An electron can spin in only one of two directions (sometimes called up and down).
36 Name of orbitals 1s 2s 2p (2p +1, 2p 0, 2p -1 ) 3s 3p (3p +1, 3p 0, 3p -1 ) 3d (3p +2, 3p +1, 3p 0, 3p -1 3p -2 ) 4s 4p (4p +1, 4p 0, 4p -1 ) 4d (4d +2, 4d +1, 4d 0, 4d -1 4d -2 ) 4f (4f +3, 4f +2, 4f +1, 4f 0, 4f -1 4f -2, 4f -3 The letter indicates the secondary Quantum number, l The index indicates the magnetic Quantum number The number indicates the Principal Quantum number
37 Degenerate orbitals E= Z 2 /n 2 (E 1sH ) 1s 2s 2p (2p +1, 2p 0, 2p -1 ) 3s 3p (3p +1, 3p 0, 3p -1 ) 3d (3p +2, 3p +1, 3p 0, 3p -1 3p -2 ) 4s 4p (4p +1, 4p 0, 4p -1 ) 4d (4d +2, 4d +1, 4d 0, 4d -1 4d -2 ) 4f (4f +3, 4f +2, 4f +1, 4f 0, 4f -1 4f -2, 4f -3 Depends only on the principal Quantum number 1 function 4 functions 9 functions 16 functions Combination of degenerate functions: still OK for hydrogenoids. New expressions (same number); real expressions; hybridization.
39 symmetry of 2p Z Nodes: No node for the radial part (except 0 and ∞ ) cos = 0 corresponds to = /2 : the xy plane or z/r=0 : the xy plane The 2p z orbital is antisymmetric relative to this plane cos(- )=-cos The z axis is a C ∞ axis
40 Directionality of 2p Z This is the product of a radial function (with no node) by an angular function cos . It does not depend on and has the z axis for symmetry axis. The angular contribution to the density of probability varies like cos 2 Within cones: Full space 2 cones: a diabolo
41 angle Density inside The half cone (1-cos( l ) 3 )/2 Part of the volume (1-cos l )/2 15° 0.049 0.017 30°0.1750.067 45°0.3230.146 60°0.4370.25 75°0.4910.37 90°0.5
42 Directionality of 2p Z This is the product of a radial function (with no node) by an angular function cos . It does not depend on and has the z axis for symmetry axis. The angular contribution to the density of probability varies like cos 2 Within cones: angle Density inside The half cone (cos l ) 3 -1)/2 Part of the volume (cos l -1)/2 15° 0.049 0.017 30°0.1750.067 45°0.3230.146 60°0.4370.25 75°0.4910.37 90°0.5 Probability is 87.5% in half of the space 22.5% in the other half
43 Spatial representation of the angular part. Let us draw all the points M with the same contribution of the angular part to the density The angular part of the probability is OM = cos 2 All the M points belong to two spheres that touch at O
48 One electron equally distributed on the three 2p levels 2 is proportional to x 2 /r 2 + y 2 /r 2 + z 2 /r 2 =1 and thus does not depend on r: spherical symmetry An orbital p has a direction, like a vector. A linear combination of 3 p orbitals, is another p orbital with a different axis: The choice of the x,y, z orbital is arbitrary
49 orbitals = N radial function angular function r l (polynom of degree n-l-r) n-l-r nodes l nodes
50 d orbitals Clover, the forth lobe is the lucky one; clubs have three
53 Compare the average radius of 1s for the hydrogenoides whose nuclei are H and Pb. Pb 207 82 Make comments on the 1s orbital the atom Pb?
54 Name these orbitals Spherical coordinates x = r sin cos ; y = r sin sin ; z = r cos
55 Paramagnetism Is an atom with an odd number of electron necessarily diamagnetic? Is an atom with an even number of electron necessarily paramagnetic? What is the (l, m l ) values for Lithium? Is Li dia or para? Why?