Chemistry 431Chemistry 431Lecture 7

Chemistry 431Chemistry 431

The Schrödinger equation for hydrogeng q y gThe electronic structure of atoms

Absorption spectra of atomsp pMany-electron atoms

NC State University

The Solar SpectrumThe Solar Spectrum

There are gaps in the solar emissionThere are gaps in the solar emissioncalled Frauenhofer lines.called Frauenhofer lines.The gaps arise from specific atoms in The gaps arise from specific atoms in the sun that absorb radiation.the sun that absorb radiation.

Experimental observation of Experimental observation of hydrogen atomhydrogen atomhydrogen atom hydrogen atom

Hydrogen atom emission is “quantized”. It Hydrogen atom emission is “quantized”. It y g qy g qoccurs at discrete wavelengths (and therefore occurs at discrete wavelengths (and therefore at discrete energies).at discrete energies).The Balmer series results from four visible The Balmer series results from four visible lines at lines at 410 nm410 nm, , 434 nm434 nm, , 496 nm496 nm and and 656 nm656 nm..The relationship between these lines was The relationship between these lines was shown to follow the Rydberg relation.shown to follow the Rydberg relation.

Atomic spectraAtomic spectraAtomic spectraAtomic spectraAtomic spectra consist of series of narrow lines.Atomic spectra consist of series of narrow lines.Empirically it has been shown that the Empirically it has been shown that the wavenumber of the spectral lines can be fit bywavenumber of the spectral lines can be fit by

)(111~1222 nnR >−==

⎟⎟⎟⎟⎟⎞

⎜⎜⎜⎜⎜⎛

λν

21 nn ⎟⎟

⎠⎜⎜

⎝λ

where R is the Rydberg constant and n and nwhere R is the Rydberg constant, and n1 and n2are integers.

El t i St t f H dEl t i St t f H dElectronic Structure of HydrogenElectronic Structure of Hydrogen

The Schrödinger equation for hydrogenThe Schrödinger equation for hydrogenSeparation of variables Separation of variables

Radial and angular partsRadial and angular partsHydrogen atom wavefunctionsHydrogen atom wavefunctionsExpectation valuesExpectation valuesExpectation valuesExpectation valuesSpectroscopy of atomic hydrogenSpectroscopy of atomic hydrogen

Schrödinger equation for hydrogen:Schrödinger equation for hydrogen:The kinetic energy operatorThe kinetic energy operator

The Schrödinger equation in three dimensions is:The Schrödinger equation in three dimensions is:

h22Ψ VΨ EΨ– h

2μ ∇ 2Ψ + VΨ = EΨTh t d lTh t d l d id iThe operator delThe operator del--squared is:squared is:

∇ 2 = ∂2+ ∂2

+ ∂2

The procedure uses a spherical polar coordinate system.The procedure uses a spherical polar coordinate system.

∇∂x2 +

∂y 2 +∂z2

p p p yp p p yInstead of x, y and z the coordiantes are Instead of x, y and z the coordiantes are θθ, , φφ and r.and r.

Schrödinger equation for hydrogen:Schrödinger equation for hydrogen:g q y gg q y gThe form of the potentialThe form of the potential

The Coulomb potential between the electron and The Coulomb potential between the electron and the proton isthe proton is V = V = --ZeZe22/4/4πεπε00rrThe hamiltonian for both the proton and electron The hamiltonian for both the proton and electron is:is:

H h22 h2

2 V

Separation of nuclear and electronic variables Separation of nuclear and electronic variables

H = – h2mN

∇ N2– h

2me∇ e

2 + V

Separation of nuclear and electronic variables Separation of nuclear and electronic variables results in an electronic equation in the centerresults in an electronic equation in the center--ofof--mass coordinates:mass coordinates: H =H = hh22/2/2μ∇μ∇22 ZeZe22/4/4πεπε rrmass coordinates:mass coordinates: H = H = --hh22/2/2μ∇μ∇22 -- ZeZe22/4/4πεπε00rr

(1/(1/μμ=1/m=1/mee+1/m+1/mNN))..

Separation of ariablesSeparation of ariablesSeparation of variablesSeparation of variables

It would be impossible to solve the equationIt would be impossible to solve the equationwith all three variables simultaneously.with all three variables simultaneously.yyInstead a procedure known as separation of variablesInstead a procedure known as separation of variablesis used. is used.

The steps are:The steps are:1 Substitute in 1 Substitute in ΨΨ(r(r θθ φφ) = R(r)Y() = R(r)Y(θθ φφ))1. Substitute in 1. Substitute in ΨΨ(r,(r,θθ,,φφ) = R(r)Y() = R(r)Y(θθ,,φφ))2. Divide both sides by R(r)Y(2. Divide both sides by R(r)Y(θθ,,φφ))3. Multiply both sides by 23. Multiply both sides by 2μμrr22p y yp y y μμ

Separation of variablesSeparation of variablesIt is not possible to solve the equationIt is not possible to solve the equationfor all three variables simultaneously.for all three variables simultaneously.yyThe steps for the separation of variables are:The steps for the separation of variables are:1. Substitute in 1. Substitute in ΨΨ(r,(r,θθ,,φφ) = R(r)Y() = R(r)Y(θθ,,φφ))2 Di id b th id b R( )Y(2 Di id b th id b R( )Y(θθ φφ))2. Divide both sides by R(r)Y(2. Divide both sides by R(r)Y(θθ,,φφ))3. Multiply both sides by 23. Multiply both sides by 2μμrr22

– h2

R(r)ddr r2dR

dr + 2μr2

h2e2

4πε0r+ E R(r)R(r) dr dr h 4πε0r

h2 1 d sinθ dY + 1 d 2Y = 0– Y(θ,φ) sinθ dθ sinθ dθ + sin2θ dφ2 = 0

Radial and angular equationsRadial and angular equationsg qg qSeparate into two equations by setting each Separate into two equations by setting each equation equal to a separation constant, equation equal to a separation constant, ββ..The radial equation is:The radial equation is:

– h2

R(r)ddr r2dR

dr + 2μr2

h2e2

4πε r + E R(r) = –βR(r) dr dr h2 4πε0r( ) β

The angular equation is:The angular equation is:

– h2

Y(θ,φ)1

sinθddθ sinθ dY

dθ + 1sin2θ

d 2

dφ2 = β( ,φ) dφ

Separation of angular variablesSeparation of angular variablesThe sperical harmonics arise from the productof Θ(θ)Φ(φ) after substituting Y(θ,φ) = Θ(θ)Φ(φ)

1Φ(φ)

∂2Φ∂ϕ2 + sin2θβ+ sinθ

Θ(θ)∂

∂θ sinθ∂Θ∂θ = 0

( ) (φ) g ( ,φ) ( ) (φ)

The θ and φ equations are obtained with separationconstant m2

Φ(φ) ∂ϕ Θ(θ) ∂θ ∂θ

constant m .

sin2θβ+ sinθΘ(θ)

∂∂θ sinθ∂Θ

∂θ = m2Θ(θ) ∂θ ∂θ

1 ∂2Φ 21Φ(φ)

∂ Φ∂ϕ2 = – m2

Solution of theSolution of the φφ equationequationSolution of theSolution of the φφ equationequationThe φ equation has a simple form:

∂2Φ (φ)∂ 2 = – m2Φ(φ)∂ϕ2 (φ)

The solution of this equation is:

This is an oscillatory function since:Φ = e±imφ

This is an oscillatory function since:

e±ix = cos (x) ± i sin(x)

Solutions of the angular equationSolutions of the angular equationSolutions of the angular equationSolutions of the angular equation

The The wavefunctionwavefunction solutions of the angular solutions of the angular equation are spherical harmonics equation are spherical harmonics YYllmm ((θθ,,φφ). ). These functions describe the angular distribution These functions describe the angular distribution of atomic of atomic orbitalsorbitals and are the and are the wavefunctionswavefunctions for for the rigid rotor of polyatomic molecules.the rigid rotor of polyatomic molecules.The degeneracy of a given orbital is 2The degeneracy of a given orbital is 2ll+1 and +1 and The degeneracy of a given orbital is 2The degeneracy of a given orbital is 2ll+1 and +1 and the angular momentum of the electron is the angular momentum of the electron is √√ll((ll + 1)+ 1)hh√√ll((ll + 1)+ 1)hh..

MAPLE MAPLE k h h i l h ik h h i l h iworksheet on spherical harmonicsworksheet on spherical harmonics

The form of the spherical harmonics YThe form of the spherical harmonics Ynnll ((θθ,,φφ) is quite ) is quite familiar. The shape of the sfamiliar. The shape of the s--orbital resembles the first orbital resembles the first ppspherical harmonic Yspherical harmonic Y0000. . Attached to this lecture are three MAPLE worksheets that Attached to this lecture are three MAPLE worksheets that Attached to this lecture are three MAPLE worksheets that Attached to this lecture are three MAPLE worksheets that illustrate the s, p and d orbitals respectively. The idea is to illustrate the s, p and d orbitals respectively. The idea is to obtain an interactive picture of the mathematical form and obtain an interactive picture of the mathematical form and obtain an interactive picture of the mathematical form and obtain an interactive picture of the mathematical form and the plots of the functions.the plots of the functions.

YY thth bit lbit lYY00 or the sor the s--orbitalorbital

YY0000

YY or the por the p orbitalsorbitalsYY11 or the por the p--orbitalsorbitals

YY11 11 YY1010

YY th dth d bit lbit lYY22 or the dor the d--orbitalsorbitals

YY2222 YY21 21 YY2020YY2222 YY21 21 YY2020

The effective potential: result of The effective potential: result of th l ti f th l tth l ti f th l tthe solution of the angular partthe solution of the angular part

The solutions for the angular part result in a term in The solutions for the angular part result in a term in potential energy equal to:potential energy equal to:p gy qp gy q

V’ = V’ = hh22ll((ll+1)/2+1)/2μμrr22

This term contains the contributions to the energy from This term contains the contributions to the energy from This term contains the contributions to the energy from This term contains the contributions to the energy from angular terms. angular terms.

Together with the Coulomb potential the effective Together with the Coulomb potential the effective Together with the Coulomb potential the effective Together with the Coulomb potential the effective potential energy is:potential energy is:

VV ffff == -- ZeZe22/4/4πεπε00rr ++ hh22ll((ll+1)/2+1)/2μμrr22VVeff eff ZeZe /4/4πεπε00rr ++ hh ll((ll+1)/2+1)/2μμrr

The radial equation for hydrogenThe radial equation for hydrogenThe radial equation for hydrogenThe radial equation for hydrogen

Making the above approximations we have anMaking the above approximations we have anradial hamiltonian (energy operator)radial hamiltonian (energy operator)

– h2

2μ∇ 2R – Ze2

4πε0rR –

h2 l (l +1)2μr 2 R = ER2μ 4πε0r 2μr

The solutions have the form: The solutions have the form: RRn,n,ll = = NNn,n,ll ρρll ee--ρρ/2/2 LLn,n,ll(r)(r)wherewhere ρρ = 2Zr/na= 2Zr/na00 andand aa00 = 4= 4πεπε00hh22/me/me22

NNn,n,l l is the normalization constant.is the normalization constant.LL (r)(r) is an associated Laguerre polynomialis an associated Laguerre polynomialLLn,n,ll(r) (r) is an associated Laguerre polynomial.is an associated Laguerre polynomial.

The Bohr radiusThe Bohr radiusThe Bohr radiusThe Bohr radius

The quantityThe quantity 44 hh22// 22 is known as the Bohr is known as the Bohr The quantityThe quantity aa00 == 44πεπε00hh22/me/me2 2 is known as the Bohr is known as the Bohr radius. The Bohr radius is radius. The Bohr radius is aa00 = 0.529 Å. Since it = 0.529 Å. Since it emerges from the calculation of the wave functions andemerges from the calculation of the wave functions andemerges from the calculation of the wave functions andemerges from the calculation of the wave functions andenergies of the hydrogen atom it is a fundamental unit.energies of the hydrogen atom it is a fundamental unit.In soIn so--called atomic units the unit of length is the Bohrcalled atomic units the unit of length is the Bohrggradius. So 1 Å is approximately 2 Bohr radii.radius. So 1 Å is approximately 2 Bohr radii.

Y h ld d di i l ( it) l i d ifY h ld d di i l ( it) l i d ifYou should do a dimensional (unit) analysis and verifyYou should do a dimensional (unit) analysis and verifythat that aa00 has units of length!has units of length!

Solutions of the radial equationSolutions of the radial equationqq

1/2

The normalization constant depends on n andThe normalization constant depends on n and ll..

Nn,l = –n –l – 1 !

2n n + 1 !3

1/2

2na0

l + 3/2

2n n + 1 !

The associated Laguerre polynomials are:The associated Laguerre polynomials are:1 l 0 L1( ) 1n = 1 l = 0 L 1

1(x) = – 1

n = 2 l = 0 L 21(x) = – 2!(2 – x)

n = 2 l = 1 L3(x) = 3!

These are givenThese are givenfor completeness.for completeness.

n = 2 l = 1 L 3(x) = – 3!

n = 3 l = 0 L 31(x) = – 3!(3 – 3x + 1/2x2)

n = 3 l = 1 L3(x) = – 4!(4 – x)n = 3 l = 1 L 4(x) = – 4!(4 – x)

n = 3 l = 2 L 55(x) = – 5!

Normalization of the radial functionsNormalization of the radial functionsNormalization of the radial functionsNormalization of the radial functions

Each of the radial equation solutions is a polynomialEach of the radial equation solutions is a polynomialmultiplying an exponential. The normalization ismultiplying an exponential. The normalization isobtained from the integral:obtained from the integral:obtained from the integral:obtained from the integral:

R l* R l r

2dr = 1∞

Rnl Rnl r dr 10

The volume element here isThe volume element here is rr22drdr which is the “r”which is the “r”The volume element here isThe volume element here is rr22dr dr which is the rwhich is the rpart of the spherical coordinate volume elementpart of the spherical coordinate volume elementrr22sinsinθθdrddrdθθddφφ..φφ

MAPLE worksheetMAPLE worksheetNormalization of the radial functionsNormalization of the radial functions

A MAPLE worksheet attached to this lecture illustratesA MAPLE worksheet attached to this lecture illustratesthe normalization of the first three radial functions. Thethe normalization of the first three radial functions. Theworksheet includes plots of the functions.worksheet includes plots of the functions.

Wh i i l t k i i d th t l tWh i i l t k i i d th t l tWhen examining a plot keep in mind that you can plotWhen examining a plot keep in mind that you can plotthe wave function or the square of the wave function.the wave function or the square of the wave function.We often plot the square of the wave function becauseWe often plot the square of the wave function becauseWe often plot the square of the wave function, becauseWe often plot the square of the wave function, becausethe integral of the square of the wave function gives thethe integral of the square of the wave function gives theprobability.probability.p yp y

Energy levels of hydrogen atomEnergy levels of hydrogen atomEnergy levels of hydrogen atomEnergy levels of hydrogen atomThe energy levels of the hydrogen atom are The energy levels of the hydrogen atom are The energy levels of the hydrogen atom are The energy levels of the hydrogen atom are specified by the principal quantum number n:specified by the principal quantum number n:

4

E = – μe4

32π2ε02h2

1n2

All states with the same quantum number n All states with the same quantum number n have the same energy.have the same energy.

0

All states of negative energy are bound states, All states of negative energy are bound states, states of positive energy are unbound and are states of positive energy are unbound and are p gyp gypart of the continuum. part of the continuum.

The Rydberg ConstantThe Rydberg ConstantThe Rydberg ConstantThe Rydberg ConstantThe energy levels calculated using the The energy levels calculated using the The energy levels calculated using the The energy levels calculated using the Schrödinger equation permit calculation of the Schrödinger equation permit calculation of the Rydberg constantRydberg constantRydberg constant.Rydberg constant.One major issue is units. Spectroscopists often One major issue is units. Spectroscopists often use units of wavenumber or cmuse units of wavenumber or cm--11 At first this At first this use units of wavenumber or cmuse units of wavenumber or cm 11. At first this . At first this seems odd, but hseems odd, but hνν = hc/= hc/λλ = hc= hcνν where where νν is the is the value of the transition in wavenumbersvalue of the transition in wavenumbers

~~

value of the transition in wavenumbers.value of the transition in wavenumbers.

R = 1 μe4

in cm-1R = hc 32π2ε02h2 in cm 1

Shells and subshellsShells and subshells

All of the orbitals of a given value of n for a shellAll of the orbitals of a given value of n for a shellAll of the orbitals of a given value of n for a shell.All of the orbitals of a given value of n for a shell.n = 1, 2, 3, 4 .. correspond to shells K, L, M, N…n = 1, 2, 3, 4 .. correspond to shells K, L, M, N…Orbitals with the same value of n and different Orbitals with the same value of n and different values of values of ll form subshells.form subshells.ll = 0, 1, 2, ... correspond to subshells s, p, d …= 0, 1, 2, ... correspond to subshells s, p, d …Using the quantum numbers that emerge from Using the quantum numbers that emerge from Using the quantum numbers that emerge from Using the quantum numbers that emerge from solution of the Schrödinger equation the solution of the Schrödinger equation the subshells can be described as orbitalssubshells can be described as orbitalssubshells can be described as orbitals.subshells can be described as orbitals.

Hydrogenic orbitalsHydrogenic orbitalsy gy g

s orbitals are spherically symmetrical. The 1s s orbitals are spherically symmetrical. The 1s s orbitals are spherically symmetrical. The 1s s orbitals are spherically symmetrical. The 1s wavefunction decays exponentially from a wavefunction decays exponentially from a maximum value of (1/maximum value of (1/ππaa00

33))1/21/2 at the nucleus. at the nucleus. maximum value of (1/maximum value of (1/ππaa00 )) at the nucleus. at the nucleus. p orbitals have zero amplitude at r=0, and the p orbitals have zero amplitude at r=0, and the electron possesses an angular momentum of electron possesses an angular momentum of electron possesses an angular momentum of electron possesses an angular momentum of hh√√ll((ll +1+1).). The orbital with m = 0 has zero angular The orbital with m = 0 has zero angular momentum about the z axis The angular momentum about the z axis The angular momentum about the z axis. The angular momentum about the z axis. The angular variation is cosvariation is cosθθ which can be written as z/r which can be written as z/r leading to the name pleading to the name p orbitalorbitalleading to the name pleading to the name pzz orbital.orbital.

Hydrogen 1s radial wavefunctionHydrogen 1s radial wavefunctionHydrogen 1s radial wavefunctionHydrogen 1s radial wavefunction2.0

The 1s orbital has no The 1s orbital has no nodes and decays nodes and decays exponentiallyexponentially

1.5exponentially.exponentially.RR1s1s = 2(1/a= 2(1/a00))3/23/2ee--ρρ/2/2

ll1.0

R1,

0

Ψ1s

n = 1 andn = 1 and ll = 0 = 0 are are the quantum numbers the quantum numbers for this orbital for this orbital

0.5

for this orbital. for this orbital. 1086420

ρρ

The Radial Distribution in The Radial Distribution in Hydrogen 2s and 2p orbitalsHydrogen 2s and 2p orbitals

0.6

ΨΨ2p

0.4Rn,

l

Ψ2sp

0.2

0.0

1086420ρ

Expectation valuesExpectation valuesExpectation valuesExpectation valuesThe expectation (or average) value of an observable <r> is given The expectation (or average) value of an observable <r> is given

< r > = Ψ *rΨdτ∞

p ( g ) gp ( g ) gby the general formula.by the general formula.

< r > Ψ rΨdτ0

As written the above integral describes the expectation valueAs written the above integral describes the expectation valuef th l f th di I t ti th l f th l f th di I t ti th l of the mean value of the radius, r. Integration over the angular of the mean value of the radius, r. Integration over the angular

part gives 1 because the spherical harmonics are normalized.part gives 1 because the spherical harmonics are normalized.The volume element can be written dThe volume element can be written dτ τ = r= r22dr. The mean value is:dr. The mean value is:

< r > = Ψ *rΨ r 2dr = Ψ *Ψ r 3dr∞∞

r Ψ rΨ r dr Ψ Ψ r dr00

MAPLE worksheet onMAPLE worksheet onExpectation (Average) valuesExpectation (Average) values

There is a MAPLE worksheet attached to this lecture There is a MAPLE worksheet attached to this lecture that illustrates the use of expectation values. The that illustrates the use of expectation values. The example of the position <r> is given for the 1s, 2s and example of the position <r> is given for the 1s, 2s and 3s radial wave functions. The expectation value or 3s radial wave functions. The expectation value or average value of r gives the average distance of anaverage value of r gives the average distance of anaverage value of r gives the average distance of anaverage value of r gives the average distance of anelectron from the nucleus in a particular orbital. Sinceelectron from the nucleus in a particular orbital. Sincethe 2s orbital has one radial node and the 3s orbital hasthe 2s orbital has one radial node and the 3s orbital hasthe 2s orbital has one radial node and the 3s orbital hasthe 2s orbital has one radial node and the 3s orbital hastwo radial nodes the average distance of the electrontwo radial nodes the average distance of the electronfrom the nucleus is shown to increase as:from the nucleus is shown to increase as:

<r> for 3s > <r> for 2s > <r> for 1s<r> for 3s > <r> for 2s > <r> for 1s

Spectroscopy of atomic hydrogenSpectroscopy of atomic hydrogenSpectroscopy of atomic hydrogenSpectroscopy of atomic hydrogen

Spectra reported in wavenumbers, Spectra reported in wavenumbers, Rydberg fit all of the series of hydrogen spectra Rydberg fit all of the series of hydrogen spectra

ν = 1λy g y g py g y g p

with a single equation, with a single equation, Absorption or emission of a photon of frequencyAbsorption or emission of a photon of frequency

ν = R 1n1

2 – 1n2

2

Absorption or emission of a photon of frequencyAbsorption or emission of a photon of frequencyνν occurs in resonance with an energy change, occurs in resonance with an energy change, ΔΔE = hE = hνν (Bohr frequency condition).(Bohr frequency condition).ΔΔE hE hνν (Bohr frequency condition).(Bohr frequency condition).Solutions of Schrödinger equation result in Solutions of Schrödinger equation result in further selection rulesfurther selection rulesfurther selection rules.further selection rules.

Spectroscopic transitionsSpectroscopic transitionsSpectroscopic transitionsSpectroscopic transitions

A transition requires a transfer from one state A transition requires a transfer from one state with its quantum numbers (nwith its quantum numbers (n11,, ll11, m, m11) to another ) to another q (q ( 11,, 11,, 11))state (nstate (n22, , ll22, m, m22).).Not all transitions are possible: there are Not all transitions are possible: there are Not all transitions are possible: there are Not all transitions are possible: there are selection rules, selection rules, ΔΔll = = ±± 1, m = 0, 1, m = 0, ±± 11These rules demand conservation of angular These rules demand conservation of angular momentum. Since a photon carries an intrinsic momentum. Since a photon carries an intrinsic angular momentum of 1.angular momentum of 1.

ManyMany--electron atoms:electron atoms:The orbital approximationThe orbital approximation

The orbital approximation to the total manyThe orbital approximation to the total manyThe orbital approximation to the total manyThe orbital approximation to the total many--electron wavefunction electron wavefunction ΨΨ(r(r11, r, r22, ...) is to rewrite , ...) is to rewrite it as a product of “oneit as a product of “one--electron” waveelectron” wave--ppfunctions functions ψψ(r(r11))ψψ(r(r22)… so that )… so that

ΨΨ(r(r11, r, r22, ...) = , ...) = ψψ(r(r11))ψψ(r(r22)… )… (( 11 22 )) ψψ(( 11))ψψ(( 22))The configuration of an atom is the list of The configuration of an atom is the list of occupied orbitals.occupied orbitals.The Pauli exclusion principle states that the The Pauli exclusion principle states that the spins must be paired if two electrons are to spins must be paired if two electrons are to occupy one orbital and no more than two occupy one orbital and no more than two electrons may occupy an orbital. electrons may occupy an orbital.

Penetration and shieldingPenetration and shieldingggIn manyIn many--electron atoms the s, p, d etc. orbitals electron atoms the s, p, d etc. orbitals of each shell are not degenerate.of each shell are not degenerate.An electron distance r from the nucleus An electron distance r from the nucleus experiences the nuclear charge Z shielded by experiences the nuclear charge Z shielded by all of the other electrons. The effective charge all of the other electrons. The effective charge is Zis Zis Zis Zeffeff..ZZeffeff = Z = Z -- σσ where where σσ is the shielding parameter.is the shielding parameter.An s electron has a greater penetration thro ghAn s electron has a greater penetration thro ghAn s electron has a greater penetration through An s electron has a greater penetration through inner shells than a p electron of the same shell inner shells than a p electron of the same shell because the p electron has a node at thebecause the p electron has a node at thebecause the p electron has a node at the because the p electron has a node at the nucleus.nucleus.

The Aufbau principleThe Aufbau principleThe Aufbau principleThe Aufbau principleAufbau means “buildingAufbau means “building--up” in German.up” in German.The configurations of atoms are built up by The configurations of atoms are built up by population of the hydrogenic orbitals.population of the hydrogenic orbitals.p p y gp p y gImagine a bare nucleus with charge Z and Imagine a bare nucleus with charge Z and then add Z electrons to the orbitals in the then add Z electrons to the orbitals in the following order: following order: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s etc.1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s etc.p p p pp p p pHund’s rule an atom in its ground state Hund’s rule an atom in its ground state adopts a configuration with the greatestadopts a configuration with the greatestadopts a configuration with the greatest adopts a configuration with the greatest number of unpaired spins. number of unpaired spins.

Why is 4s lower than 3d?Why is 4s lower than 3d?Why is 4s lower than 3d?Why is 4s lower than 3d?Shielding of 3d electrons is less effective Shielding of 3d electrons is less effective than 4s. Thus, an occupied 3d wave than 4s. Thus, an occupied 3d wave function is pulled closer to the nucleus than function is pulled closer to the nucleus than a 4s wave function.a 4s wave function.The more confined volume for 3d electrons The more confined volume for 3d electrons means that they have greater electronmeans that they have greater electron--electron repulsion than 4s.electron repulsion than 4s.To relieve electronTo relieve electron--electron repulsion there electron repulsion there is a general trend that (n+1)s is is a general trend that (n+1)s is occupliedoccupliedbefore before ndnd in the periodic table (n > 2).in the periodic table (n > 2).

The problem of multiple electronsThe problem of multiple electronsThe central difficulty with application of the The central difficulty with application of the Schrödinger equation is the presence ofSchrödinger equation is the presence ofSchrödinger equation is the presence of Schrödinger equation is the presence of electronelectron--electron interaction terms in the electron interaction terms in the potential energy:potential energy:potential energy:potential energy:

V = – Ze2

4πε r + e2

4πε rΣi j 1

I,J

Σi 1

I

No analytical solutionsNo analytical solutions

4πε0ri 4πε0riji, j = 1i = 1

electron – nuclear electron – electronNo analytical solutions. No analytical solutions. HartreeHartree--Fock procedure. Find solutions that Fock procedure. Find solutions that optimize the electron in each orbital in theoptimize the electron in each orbital in theoptimize the electron in each orbital in the optimize the electron in each orbital in the presence of the field of all of the other orbitals.presence of the field of all of the other orbitals.

HartreeHartree Fock procedureFock procedureHartreeHartree--Fock procedureFock procedure

As an e ample for He e can rite the t o electronAs an example, for He we can write the two electronwave function as a product of orbitals.

Ψ(r r ) = φ(r )φ(r )Ψ(r1,r2) = φ(r1)φ(r2)The probability distribution, φ*(r2)φ(r2)dr2 for electron2 corresponds to a charge density in classical2 corresponds to a charge density in classical physics. Therefore, we can say that the effective potential felt by electron 1 is: p y

V 1eff (r1) = φ*(r2) 1

r12φ*(r2)dr2r12

The antiThe anti--symmetry symmetry y yy yrequirementrequirement

Th f ti t b tiTh f ti t b ti t it iThe wave function must be antiThe wave function must be anti--symmetric symmetric with respect of electron exchange. This with respect of electron exchange. This means that is must change sign if any twomeans that is must change sign if any twomeans that is must change sign if any two means that is must change sign if any two electrons change places. electrons change places. Slater came up with a general way to do thisSlater came up with a general way to do thisSlater came up with a general way to do this Slater came up with a general way to do this using a determinant. Here, we show it for using a determinant. Here, we show it for two electrons:two electrons:two electrons:two electrons:

Ψ r1,r2 = 12

φa(r1) φb(r1)φ (r2) φb(r2)

= 12

φa(r1)φb(r2) – φa(r2)φb(r1)2 φa(r2) φb(r2) 2

The Slater determinantThe Slater determinantThe twoThe two--electron case is pretty easy. The electron case is pretty easy. The general case for N general case for N orbitalsorbitals is:is:

φ ( ) φ ( ) φ ( ) φ ( )

Ψ r r r r = 1

φa(r1) φb(r1) φc(r1) φd(r1)...φa(r2) φb(r2) φc(r2) φd(r2)...φa(r3) φb(r3) φc(r3) φd(r3)...Ψ r1,r2,r3,r4.... =

N! φa(r4) φb(r4) φc(r4) φd(r4).........

A N x N determinant is painful to do by A N x N determinant is painful to do by hand but is very easy to automate for usehand but is very easy to automate for usehand, but is very easy to automate for use hand, but is very easy to automate for use in computer programs. in computer programs.

The selfThe self--consistent field methodconsistent field methodThe selfThe self--consistent field methodconsistent field methodThe effective or average potential can be used in aone electron hamiltonian operator for electron 1

H1eff (r1) = – 1

2∇ 12 – 2

r + V1eff (r1)H1 (r1) 2∇ 1 r V1 (r1)

The Schrödinger equation is solved for electron 1

H1eff φ(r1) = ε1φ(r1)

Start with a trial function φ(r ) and solve for φ(r )Start with a trial function φ(r2) and solve for φ(r1).Using φ(r1) calculate an effective potential for 2 and solve for φ(r2). Continue until convergence is φ( 2) greached.

Atomic term symbolsAtomic term symbolsAtomic term symbolsAtomic term symbols

The letter indicates the total orbital angular The letter indicates the total orbital angular momentum quantum number of all electrons momentum quantum number of all electrons in an atom L =in an atom L = ll ++ ll ll ++ ll 1 |1 | ll ll ||in an atom L =in an atom L = ll11 + + ll22 , , ll11 + + ll22 -- 1, ….| 1, ….| ll11 -- ll22 ||The left superscript gives the multiplicity The left superscript gives the multiplicity 2S+1 where S = s2S+1 where S = s + s+ s + s+ s ++2S+1, where S = s2S+1, where S = s11 + s+ s22 + s+ s33 + ….+ ….The right subscript gives the total angular The right subscript gives the total angular momentum J = L + S L + Smomentum J = L + S L + S -- 1 |L1 |L -- S|S|momentum J = L + S, L + S momentum J = L + S, L + S -- 1, … |L 1, … |L -- S| S| The term symbol isThe term symbol is

X 2S+1LX LJ

Hund’sHund’s rules determine the rules determine the ground stateground state termterm symbolsymbol

Each state is designated by a term symbol corresponds Each state is designated by a term symbol corresponds to a wave function that is an to a wave function that is an eigenfunctioneigenfunction of Lof L22 and Sand S22

ith i ith i with unique energy.with unique energy.The state with the larges value of S is the most stable.The state with the larges value of S is the most stable.For states with the same value of S, the state with the For states with the same value of S, the state with the largest value of L is the most stable.largest value of L is the most stable.If th t t h th l f L d S If th t t h th l f L d S If the states have the same value of L and S If the states have the same value of L and S -- subshellsubshell less than half filled, smallest J is most stable less than half filled, smallest J is most stable -- subshellsubshell more than half filled the state with the largest more than half filled the state with the largest -- subshellsubshell more than half filled, the state with the largest more than half filled, the state with the largest value of J is the most stable.value of J is the most stable.

Explicit Rules for DeterminingExplicit Rules for DeterminingTerm SymbolsTerm SymbolsTerm SymbolsTerm Symbols

To determine the states(terms) of a given atom or ion: To determine the states(terms) of a given atom or ion: Write down the electronic configuration (ignore closedWrite down the electronic configuration (ignore closedWrite down the electronic configuration (ignore closed Write down the electronic configuration (ignore closed subshellsubshell electrons)electrons)Determine the number of distinct microstates that can Determine the number of distinct microstates that can represent that configuration. If you have represent that configuration. If you have ee electrons in a electrons in a single open single open subshellsubshell of 2l+1 of 2l+1 orbitalsorbitals, this value is , this value is #microstates = (2(2l+1))!/#microstates = (2(2l+1))!/ ee!(2(2l+1)!(2(2l+1)-- ee)!)!#microstates = (2(2l+1))!/#microstates = (2(2l+1))!/ ee!(2(2l+1)!(2(2l+1)-- ee)! )! Tabulate the number of microstates that have a given MTabulate the number of microstates that have a given MLLand Mand MSS

Decompose your table into terms by elimination Decompose your table into terms by elimination Test the total degeneracy of the resultant terms to account Test the total degeneracy of the resultant terms to account f ll th i t t t d i t 2 d 3f ll th i t t t d i t 2 d 3for all the microstates counted in parts 2 and 3 for all the microstates counted in parts 2 and 3 Determine the lowest term using Determine the lowest term using HundsHunds Rules. Rules.

ExamplesExamplesppH(1sH(1s11))ground state term symbol is ground state term symbol is 22SS1/21/2

He(1sHe(1s22))Ground state term symbol is Ground state term symbol is 11SS00

He(1sHe(1s112s2s11) An Excited State Configuration) An Excited State ConfigurationTerms: Terms: 11SS00 , , 33SS11 {There is no {There is no 33SS00 oror 33SS--11 term }term }

B(1sB(1s222s2s222p2p11))T b lT b l 22PP 22PPTerm symbol: Term symbol: 22PP1/21/2, , 22PP3/23/2The spin orbit splitting is The spin orbit splitting is regularregular so so

22P1/2. is the ground state. P1/2. is the ground state.

ExamplesExamples

C(1sC(1s222s2s222p2p22))Calculate the number of possible electronCalculate the number of possible electronCalculate the number of possible electron Calculate the number of possible electron

arrangements in the given configuration arrangements in the given configuration Th 6!/4!2! 15 i t t t dTh 6!/4!2! 15 i t t t dThere are 6!/4!2! = 15 microstates expected There are 6!/4!2! = 15 microstates expected Write down all these possibilities Write down all these possibilities T b l t th t t l b b MT b l t th t t l b b M d Md MTabulate the total numbers by MTabulate the total numbers by MLL and Mand MSS

Decompose this table into terms Decompose this table into terms

ExamplesExamples

This listing of configurations isThis listing of configurations isd i d b t ti lld i d b t ti llderived by systematically derived by systematically considering every possibleconsidering every possiblearrangement of 2 electrons arrangement of 2 electrons in six possible positions.in six possible positions.3 3 orbitalsorbitals + 2 spins+ 2 spinsDegeneracy = Degeneracy = (2l + 1)(2s + 1) = 2(2l + 1)(2l + 1)(2s + 1) = 2(2l + 1)

ExamplesExamples

This is an example of a This is an example of a t bl f i t tt bl f i t ttable of microstates. table of microstates.

These are the terms These are the terms