chem3340-2.pdf

CHEM-3340 Physical Chemistry II Atomic structure and atomic spectra Electronic Structure Description of arrangement of electrons around the nucleus in atoms Simplest case: hydrogenic atoms - one electron atom of general atomic number Z. For example: H, He+, Li2+, etc. Many-electron atom: more than o...

CHEM-3340 Physical Chemistry II Atomic structure and atomic spectra Electronic Structure Description of arrangement of electrons around the nucleus in atoms Simplest case: hydrogenic atoms - one electron atom of general atomic number Z. For example: H, He+, Li2+, etc. Many-electron atom: more than one electron Importance of hydrogenic atoms: Schrödinger equation can be solved analytically Set of concepts to be applied to many-electron atoms and molecules Emission Spectrum An electric charge is passed through a gas (or vapor) or an element is exposed to a hot flame Atoms are produced (e.g., H2 -> 2 H), and by absorbing energy they get into ‘excited’ states The atoms discard the energy by emitting electromagnetic radiation at discrete frequencies as they return to the ground state (lowest energy) The frequency (wavelength, wavenumber) of the emitted radiation is recorded: emission spectrum Emission spectrum of hydrogen Johannes Rydberg There are series of lines Rydberg: all these lines can be described by the following mathematical formula: ⇤ = RH 1 n21 1 n22 ⇥ n1 =1, Lyman series n1 =2, Balmer series n1 =3, Paschen series n1 =4, Brackett series :wavenumber RH=109677 1/cm : Rydberg constant n1, n2: integers n1 =1,2,... n2= n1+1, n1+2,... Discrete Emission Spectrum: Quantized Energy When atom changes energy level from one to another (transition), the difference is carried away by a photon of frequency ν ∆E=hν The emission spectrum provides experimental information about the energy levels of the atoms Can quantum mechanics interpret these energies? Schrödinger Equation:Potential Energy and Reduced Mass Hydrogenic Atom Potential Energy: Coulombic (electrostatic) interaction between the positively charged nucleus and the negatively charged electron at distance r q1 q2 Charge of nucleus: q = Ze 1 V = 4⇥ 0 r Charge of electron: q2 = -e e: charge of electron ε0: vacuum permittivity V = Ze2 4⇥ 0 r The mass of proton is much larger (1830x) than the mass of electron: Nuclear model: the nucleus is in the center (0,0,0), and the electron moves around the center Mathematical consequence: the mass of the electron needs to be replaced by the reduced mass, μ 1 1 1 = + µ me mN me mN µ= me + mN mN: mass of nucleus Note: we usually take μ = me Schrödinger Equation for Hydrogenic Atoms d d d + + dx2 dy 2 dz 2 2 2 2 2µ 2 ⇥ Ze2 4⇤ 0 r =E Multiply each side by 4πε0 /e2 1 4⇤ 2 µe2 2 0 d d d + + 2 2 dx dy dz 2 2 2 ⇥ Z r 4⇤ 0 = 2 E e 4⇥ 0 a0 = = 52.9pm 2 me e 2 Bohr radius: 2 Defines a length-scale for the wavefunction! General Solution Expressed in polar co-ordinates (r, ϕ, θ) (r, ⇥, ) = R(r)Y (⇥, ) Separation of variables! Boundary conditions for R(r) R → 0 as r→∞ This introduces a principal quantum number, n R is proportional to exp[-r/(na0)] → n is related to distance from nucleus Y (ϕ, θ): same as rotation on a sphere Two quantum numbers: Orbital angular momentum quantum number: l Magnetic quantum number: ml Quantum numbers Principal quantum number n=1,2,... determines the energy Orbital angular momentum quantum number l=0,1,2,..., n-1 determines the magnitude of angular momentum: [l (l+1)]0.5 ħ Magnetic quantum number ml=0, ±1, ±2,..., ± l determines the z-component of angular momentum: mlħ Spin quantum number: ms = ± 1/2 (s =1/2) (inherent property of electron) To specify the state of the electron, we need to give these four numbers! Solution of Schrödinger Equation: Energy En = Z 2 µe4 32⇤ 2 20 2 n2 n=1,2,... They depend only on the principal quantum number, n (do not depend on l, ml, ms ) They depend on 1/n2 Spacing is large for small n They are all negative Separation will depend on Z2: larger atomic number nuclei have lower energies Rydberg constant Change of energy when an electron drops from principal quantum number n1 to n2 En1 = h⇥ = hc/ = hc⇥ E = E n2 ⇤⇤ = E/hc = 1 Z µe hc 32⌅ 2 20 2 1 n22 4 2 1 n21 ⇥ wavenumber of accompanying photon Rydberg constant For hydrogen: Z =1, μ = me 4 RH me e = 2 3 8 0h c Exact numerical confirmation of experimental value! Energy with Rydberg Constant En = 2 4 Z µe 32⇤ 2 20 2 n2 ⇤⇤ = E/hc = 4 2 Rydberg constant, RH By comparing the two equations: En = 1 Z µe hc 32⌅ 2 20 2 hcRH Z 2 n2 First energy level: E1 = -hcRH Second energy level: E2 = -hcRH/4 1 n22 1 n21 ⇥ Ionization Energy The minimum energy needed to remove an electron completely from an atom: I For H: Raise the electron from the lowest energy to zero level I = hcRH=2.180x10-18 J This corresponds to 1312 kJ/mol, 13.59 eV Shells and subshells We will look at the ‘shapes’ of wavefunctions Easier to give them names: Principal quantum number: largely determines the average distance from the nucleus: shell n= 1, 2, 3, 4,.... K, L, M, N shells In a shell there are many subshells l= 0, 1, s, p, 2, d, 3,.... f subshells In a shell, there are n2 orbitals (wavefunctions): each energy level is n2 fold degenerate Hydrogenic Atoms: Wavefunctions n,l,ml (r, ⇥, ) = Rn,l (r)Yl,ml (⇥, ) R: radial wavefunction-depends on n and l Y: angular wavefunction-depends on l and ml Radial Wavefunction General form: Rn,l(r) = N x (polynomial in r) x exp(-ρ/2) ρ =2Zr/(na0) Decay in r N: normalization constant Exponential decay with some humps Radial Solutions Radial Solutions Properties: as r →∞, R→0, boundary condition For l=0: maximum at r=0 (no angular momentum) for |l|>0: r=0, R=0: because of rotation Nodes: when a wavefunction is zero. Radial nodes: where R is zero ns orbitals have (n-1) nodes 1s 3s 2s 2p 3p 3d Mean radius of an orbital Because R depends on n and l, the mean radius is expected to depend on these only ⇥ l(l + 1) a0 2 < r >= n 1 + 0.5(1 ) 2 n Z The average radius: increases with n Decreases with the atomic number For the electron of H: =3/2 a0 Orbitals: 1s n=1, l=0, ml =0 1,0,0 1 e = 3 0.5 ( a0 ) r/a0 Independent of angle: spherical symmetrical Decays exponentially from from a maximum value at the nucleus (r=0) Probability density of finding an electron at a point: P(r) = Ψ Ψ* Probability of finding the electron in a volume of ∆V: P(r) ∆V Example: the probability of finding the electron at the nucleus in a 1 pm3 volume: p = |Ψ| 2=1/(πa03) x 1 pm3 = 2.2x10-6. Once in 455,000 observations What to plot? Ψ2 tells us the probability density Probability density of s orbitals 1s Density of points corresponds to Ψ2 Boundary Surface The surface that captures a high proportion (90%) of the electron probability Radial Distribution Function What is the probability of finding the electron at a distance r? What is the probability density finding the electron in a thin shell at distance r and r+dr? Volume: 4r2π dr P(r) = 4r2π Ψ Ψ* P(r): radial distribution function P(r) = r2 R(r)2 Most probable radius: a0 n=2, l=1, ml =0 p orbitals⇥ (2, 1, 0) = 1 32⇥a50 0.5 rcos e Short notation: Ψ=r cosθ f(r)= z f(r) Also called pz orbital At the center Ψ = 0: because of nonzero angular momentum (l 0) At the plane: cos θ = 90o,(x,y) plane Ψ = 0 +/- sign denotes the sign of Ψ r/(2a0 ) px and py orbitals n=2, l=1, ml =+-1 The solutions are complex Usually linear combinations are shown (Schrödinger cats): also solutions Ψpx=x f(r) =Ψ2,1,1 - Ψ2,1,-1 Ψpy=y f(r) =iΨ2,1,1 - iΨ2,1,-1 d orbitals n=3, l=2, ml =-2,-1,0,1,2 The five orbitals are combined to give another five orbitals of the form dxy = xy f(r) dyz = yz f(r) dzx = zx f(r) dx2-y2 = 0.5(x2-y2 )f(r) dz2 = 30.50.5 (3z2-r2 )f(r) Spectroscopic transitions and selection rules When a transition occurs between two energy levels, a photon is emitted At each level the electron can be in any state l, ml The photon has ‘inherent’ angular momentum (s=1) -> not all transition between states (n,l,ml) are allowed (certain transitions are forbidden) Selection rules for allowed transitions: ∆n = any, ∆l = ±1, ∆ml = 0, ±1 During the transition the angular momentum shall be conserved Grotrian Diagram Selection rules for allowed transitions: ∆n = any, ∆l = ±1, ∆ml = 0, ±1 The Structure of Many Electron Atoms The Schrödinger equation is difficult to solve Individual orbitals: similar to those of hydrogenic orbitals, but with nuclear charges that are modified by the presence of the other electrons We are looking for electron configuration: list of occupied orbitals The electron occupies the lowest energy levels We have many electrons with wave functions Ψk The overall wave function is the product of atomic orbitals: Ψ= Ψ1Ψ2.... H: 1s1. One electron on the s orbital Pauli Exclusion Principle How many electrons can we put on an orbital? Pauli Exclusion Principle: No more than two electrons may occupy any given orbital, and if two electrons do occupy one orbital, then their spins must be paired He: 1s2 , with paired spins of electrons (one electron has ms =1/2, ↑, the other has ms =-1/2, ↓ ) How about Li (3 electrons)? 2s, 2px, 2py, 2pz Effective Nuclear Charge An electron at distance r from the nucleus experiences repulsion from all the electrons within sphere of radius r This repulsion is equivalent to a point negative charge located in the nucleus: shielding constant,σ Effective charge: Zeff = Z -σ Penetration and shielding The p and s electrons have different shielding constants s electrons are most likely to be found closer to the nucleus than the p electrons: greater penetration Greater penetration: the s electrons experience less shielding, larger effective nuclear charge, and lower energy Similar effects for s, p, d, f electrons: order of shielding in a given subshell s