quantum mechanics and atomic structure

General Chemistry I 1

QUANTUM MECHANICSAND ATOMIC STRUCTURE

51 The Hydrogen Atom

52 Shell Model for Many-Electron Atoms

53 Aufbau Principle and Electron Configurations

54 Shells and the Periodic Table Photoelectron Spectroscopy

55 Periodic Properties and Electronic Structure

5CHAPTER

General Chemistry I


Colors ofFireworks

from atomic emission

red from Sr

orange from Ca

yellow from Nayellow from Na

green from Ba

blue from Cu


51 THE HYDROGEN ATOM

- The hydrogen atom is the simplest example of a one-electron atom or ion (ie H He+ Li2+ hellip)

- For solution of the Schroumldinger equation

spherical coordinates r

Cartesian coordinates x y z

to express angular orientation


Energy Levels

For a hydrogen atom V(r) = Coulomb potential energy

Solutions of the Schroumldinger equation

1 rydberg = 218times10-18 J

Principal quantum number n indexes the individual energy levels

E = En =Z2e4me

8 02n2h2

_ n = 1 2 3


Coordinate Systems

2 2 2 2

2 2 2 2( ) ( ) ( )

8

hV x y z x y z E x y z

m x y z

2 2 2 2 2

2 2 2 2 2 2 2( ) ( )

8 ( )

h ex y z E x y z

m x y z x y z

- Cartesian Coordinates

- Spherical Polar Coordinates (SPC)2 2 2

22 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r

Spherical polar coordinates (r θ Ф) vs Cartesian coordinates (x y z)

V(x y z) ndash can be expressed in various coordinate systems

For Coulomb potential V convenient to express

in SPC (V = -e2r) Cartesian V = -e2[x2 + y2 +z2]12


Quantization of the Angular Momentum

any integral value from 0 to n-1

value of l 0 1 2 3

orbital type s p d f orbitals


For every value of n n2 sets of quantum numbers


These sets of state are said to be degenerate

Energy level diagram for the H atom


Boundary Conditions

Boundary Conditions yield quantum numbers


Wave Functions

radial part angular part spherical harmonics


( ) ( ) ( )r R r Y

- Spherical Polar Coordinates and the Wave Function

2 2 22

2 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r

2 2 22

2 2 2 2 2 2

1 1 1( ) sin ( ) 0

8 sin sin

h er E x y z x y z

m r r r r r r

2 2 2 22 2

2 2 2

1 1( ) sin ( ) 0

8 sin sin

h e rr Er x y z x y z

m r r r

Depends only on r Depends only on

- Schroumldinger wave function

rarr factored into radial (R) and angular (Y) functions


Orbitals

EXAMPLE 51


Specification of the Wave Functions Orbitals Schroumldingerrsquos Wave Function

ie) n = 3 (only one orbit in the Bohr- de Broglie model)

32 (nine) different ways the electron can vibrate to form a standing wave rarr 9 degenerate quantum states

Orbitals

replacement of Bohrrsquos orbit for 3D Schroumldingerrsquos wave function

- Radial part Rnℓ(r)

Laguerre polynomial (rn-1) x e-r(na0

)

- Angular part Yℓm(θФ)

Legendre function (sin θ amp cos θ) x eimФ

a0 larr Bohrrsquos radius


SphericalSymmetryno angularnodes

1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3


Sizes and Shapes of Orbitals

Graphical representations of the orbitals

1) Slicing up 3D space into various 2D and 1D regions and examining the value of wave function at each point


2) Looking only at the radial behavior (ldquovertical slicerdquo)

R100(r)



s orbitals

s orbital

constant Y

All s orbitals arespherically symmetric


a function of r only

spherically symmetrical

exponentially decaying

no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)

zero at r = 2a0 = 106Aring

nodal sphere or radial node

[rlt2ao Ψgt0 positive] [rgt2ao Ψlt0

negative]


isosurface1s2s

3s

wave functionplot

radialprobability

density(or function)

plot

A surface of pointswith the same valueof wave functions


Radial Distribution Functions RDFs

Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces

- No clear boundary of an atom

- Defining the size of atom as the extent of a ldquoballoon skinrdquo inside which 90 of the probability density of the electron is contained

1s 141 Aring 2s 483 Aring 3s 1029 Aring

An ns orbital has n-1 radial nodes


p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)

middot Ф = 0 rarr cylindrical symmetry about the z-axis

middot R21(r) rarr ra0 no radial nodes except at the origin

middot cos θ rarr angular node at θ = 90o x-y nodal plane

middot r cos θ rarr z-axis 2p0 rarr labeled as 2pz

2


2p+1 and 2p-1 orbitals R21Y11 R21Y1-1 (n = 2 ℓ = 1 m = plusmn1)

middot Y11(θФ) rarr eplusmniФ = cosФ plusmn i sinФ larr Eulerrsquos formula was used

middot taking linear combinations of these gave two real orbitals

called 2px and 2py

- px and py differ from pz only in the angular

factors (orientations)


2p 3p

Wave function and RDF plots for 2p and 3p orbitals

r2Rnl2



d Orbitals

d-orbitals



Orbital Shapes and Sizes General Notes



Stern-Gerlach Experiment experimental evidence for the existence of electron spin

Stern and Gerlach (1926) Ground state Na atoms were passed through a

strong inhomogeneous magnetic field

All spin magnetisms in inner orbitals

rarr cancelled except an electron in 3s

orbital

Y (θФ)(3s) rarr no rotation

The intrinsic magnetism of the 3s electron

rarr only possible to respond to the external

field

3s electrons rarr line up with their north

poles either along or against the main

north pole of the magnet

The lack of a ldquostraight-throughrdquo beam

with the magnet rarr clear evidence of two-

valued electronrsquos magnetism


Electron Spin

- An electron has two spin states as uarr(up) and darr(down) or a and b

ms spin magnetic quantum number

- the values of ms only +12 and -12


52 SHELL MODEL FOR MANY-ELECTRON ATOMS

- In many-electron atoms Coulomb potential energy equals the sum of nucleus-electron attractions and electron-electron repulsions

- No exact solutions of Schroumldinger equation

- In a helium atom

r1 = the distance of electron 1 from the nucleus


r12 = the distance between the two electrons


Interparticle distance in He

Screening of one electron by another

r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)

r12 equiv r 1 ndash r2

interelectron distance

Repulsion between electronsThis new cross-term makes the Schrodinger equation impossible to solve exactly and causes the Bohr model to fail for He and other atoms

11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals

Self-consistent field (SCF) orbital approximation method by Hartree Fock Slater and others

- Building up an effective nuclear charge Zeffe

1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r

- all electrons were treated as independent meaning that we

neglect the repulsion term

ie) for He

Zeff(He) = 16875 lt 2


Three simplifying assumptions by Hartree

The wave function becomes a product of these one-electron orbitals

3 The effective field is spherically symmetric that is it has no angular dependence

The equations for the unknown effective field and the unknownone-electron orbitals must be solved by iteration until a self-consistent solution is obtained


1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0

Eg for the ground state of He

1s orbital1s orbital

1s2

occupancy


The ground state of Ar Z = 18


sum of the radial probability density functionsof all the occupied orbitals


Shielding Effects

- Energy-level diagrams for many-electron atoms

1) The degeneracy of the p d and f orbitals is removed due to the difference of Zeff from the Coulomb field

2) Energy values are distinctly shifted from the values of corresponding H orbitals due to the strong attraction by nuclei with Z gt 1

- Each electron attracted by the nucleus and repelled by the other electrons

rarr shielded from the full nuclear attraction by the other electrons


- effective nuclear charge Zeff e lt Ze

- Hartree orbital energy (rydbergs)

Eg For Ar (Z =18) Zeff(1) ~ 16 Zeff(2) ~ 8 Zeff(3) ~ 25


Energy level diagram for many-electron atoms


Penetration

s-electron ndash very close to the nucleushighly penetrates through the inner shell

p-electron ndash penetrates less than an s-orbitaleffectively shielded from the nucleus

In a many-electron atom because of the effects ofpenetration and shielding the order of energiesof orbitals in a given shell is s lt p lt d lt f

Zeff(s) gt Zeff(p) gt Zeff(d) gt Zeff(f)

Order of orbital shielding (for fixed value of n)


The Building-up (Aufbau) Principle is a set of rules that allows us to construct ground state electron configurationsof the elements

1 Assume electrons lsquooccupyrsquo orbitals in such a way as to minimize the total energy (lowest energy first)

2 Assume a maximum of two electrons can lsquooccupyrsquo an orbital and these must have opposite spins (Paulirsquos Exclusion Principle no two electrons in an atom can have the same set of quantum numbers)

3 Assume electrons occupy unoccupied degenerate subshell orbitals first and with parallel spins (Hundrsquos rule)

In building-up electron configurations in order of energy subshell energy overlap must be taken into account (see later)

53 AUFBAU PRINCIPLE AND ELECTRON CONFIGURATIONS


H 1s1

He 1s2

Li 1s22s1 or [He]2s1

Electron configuration of an atom is a list of all its occupied orbitals with the numbers of electrons that each one contains

Electron configurations of first ten elements (H to Ne)

Standard notation Box and arrow notation(showing electron spins)

Core electrons

Valence electron


Be 1s22s2 or [He]2s2

B 1s22s22p1 or [He]2s22p1

C 1s22s22p2 or [He]2s22p2

parallel spinsby Hundrsquos rulerarr 1s22s22px

12py1


- Elements in Period 2 (from Li to Ne) the valence shell with n = 2

- p-block elements N to Ne filling of p orbitals

s-block elements H to Be filling of s orbitals


Magnetic properties as a test of electronic configurations

- paramagnetic a substance attracted into a magnetic fieldwith one or more unpaired electrons

- diamagnetic a substance pushed out of a magnetic fieldwith all electrons paired

Examples paramagnetic H Li B Cdiamagnetic He Be

C



n = 3 Na [He]2s22p63s1 or [Ne]3s1 to Ar [Ne]3s23p6

n = 4 from Sc (scandium Z = 21) to Zn (zinc Z = 30) the next 10 electrons enter the 3d-orbitals (d-block elements)

The (n+l) rule Order of filling subshells in neutral atoms is

determined by filling those with the lowest

values of (n+l) first Subshells in a group with

the same value of (n+l) are filled in order

of increasing n due to orbital screening

order 1s lt 2s lt 2p lt 3s lt 3p lt 4s lt 3d lt 4p lt 5s lt 4d lt hellip

n = 5 5s-electrons followed by the 4d- electrons n = 6 Ce (cerium [Xe]4f15d16s2)


Alternative Pictorial Representation of Orbital Energy Order


Anomalous Configurations

Exceptions to the Aufbau principle


Filling of electron subshells in relationto the periodic table


All atoms in a given period have the same type of core with the same n

All atoms in a given group have analogous valence electron configurationsthat differ only in the value of n

Period 2

Period 3

Period 4

Group 1IA Group 18VIII

Period 5

Period 6

Period 7


54 SHELLS AND THE PERIODIC TABLE PHOTOELECTRON SPECTROSCOPY

A shell is defined precisely as a set of orbitals that have the same principal quantum number

The photoelectron spectroscopy (PES) principledetermining the energy level of each orbital by measuring theionization energy required to remove each electron from the atom


Basic design of a PEspectrometer



- Koopmansrsquo approximation

- with the frozen orbital approximation

Eg for Ne with 1s2 2s2 2p6

the orbital energies are the samein the ion despite the loss of anelectron


528 Photoelectron spectroscopic studies of silicon atoms excited by X-rays of wavelength9890 x 10-10 m show four peaks in which the electrons have speeds (all x 107 ms)of (1) 2097 (2) 2093 (3) 2014 and (4) 1971(a) Calculate the ionization energy of the electrons corresponding to each peak(b) Assign each peak to an orbital of the silicon atom1eV = 16022 x 10-19 J c = 299792 x 108 ms h = 662607 x 10-34 J sme = 910938 x 10-31 kg

Solutions

(a) E(X-rays) = h = hc =(662607 x 10-34 J s)(299792 x 108 ms)

9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV

Using the PES equation IE = h - KE = h - 12mev2

IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV

(b) The ground-state electron configuration of silicon is 1s22s22p63s23p2 Peak 1corresponds to removal of siliconrsquos 3p electrons which are its least tightly boundelectronsPeaks 2 3 and 4 correspond to removal of Sirsquos 3s 2p and 2s electronsrespectively The 1s electron does not give a peak It is probably at an energylower than ndash12536 eV and inaccessible with the X-radiation used in thisexperiment


(Fig 525)Energies of atomicsubshellsdeterminedby PES


55 PERIODIC PROPERTIES AND ELECTRONIC STRUCTURE

Atomic radius defined as half the distance between the centers ofneighboring atoms

- r decreases from left to right across a period (effective nuclear charge increases)

- r increases from top to bottom down a group (change in n and size of valence shell)

General trends

Ionic radius its share of the distance between neighboring ions in an ionic solid


- The rate of increase changes considerably

ie Li+ Na+ to K+

substantial change

to Rb+ Cs+

small change due to filling d-orbitals

Lanthanide contraction

filling of the 4f orbitals(poor shielders) radii of 3rd row D-block elementssimilar to those of 2nd row


Molar volumes (cm3 mol-1) of atoms in the solid phase

= size of the atoms + geometry of the bonding


Periodic Trends in Ionization Energies

From He to Li a large reduction in IE1

2s e- further than 1s e- and 2s e- seesa net +1 charge

From Be to B slight reduction in IE1

fifth e- in a higher energy 2p orbital

From N to O slight reduction in IE1

2 e- in the same 2p orbital leading togreater repulsion

From left to right generally increase in IE1 due to the increase of Zeff

From top to bottom generally decrease in IE1 due to the increase of n



Electron Affinity

(kJ mol-1)

- The periodic trends in EA parallel those in IE1 with one unit lower shifteg F to F- large EA because of its closed-shell configuration


10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


Colors ofFireworks

from atomic emission

red from Sr

orange from Ca

yellow from Nayellow from Na

green from Ba

blue from Cu









Energy Levels





E = En =Z2e4me

8 02n2h2

_ n = 1 2 3


Coordinate Systems

2 2 2 2

2 2 2 2( ) ( ) ( )

8


m x y z

2 2 2 2 2

2 2 2 2 2 2 2( ) ( )

8 ( )

h ex y z E x y z

m x y z x y z



22 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r








value of l 0 1 2 3








Boundary Conditions



Wave Functions



( ) ( ) ( )r R r Y


2 2 22

2 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r

2 2 22

2 2 2 2 2 2

1 1 1( ) sin ( ) 0

8 sin sin

h er E x y z x y z

m r r r r r r

2 2 2 22 2

2 2 2

1 1( ) sin ( ) 0

8 sin sin


m r r r





Orbitals

EXAMPLE 51





Orbitals




)






1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56









Energy Levels





E = En =Z2e4me

8 02n2h2

_ n = 1 2 3


Coordinate Systems

2 2 2 2

2 2 2 2( ) ( ) ( )

8


m x y z

2 2 2 2 2

2 2 2 2 2 2 2( ) ( )

8 ( )

h ex y z E x y z

m x y z x y z



22 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r








value of l 0 1 2 3








Boundary Conditions



Wave Functions



( ) ( ) ( )r R r Y


2 2 22

2 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r

2 2 22

2 2 2 2 2 2

1 1 1( ) sin ( ) 0

8 sin sin

h er E x y z x y z

m r r r r r r

2 2 2 22 2

2 2 2

1 1( ) sin ( ) 0

8 sin sin


m r r r





Orbitals

EXAMPLE 51





Orbitals




)






1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


Coordinate Systems

2 2 2 2

2 2 2 2( ) ( ) ( )

8


m x y z

2 2 2 2 2

2 2 2 2 2 2 2( ) ( )

8 ( )

h ex y z E x y z

m x y z x y z



22 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r








value of l 0 1 2 3








Boundary Conditions



Wave Functions



( ) ( ) ( )r R r Y


2 2 22

2 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r

2 2 22

2 2 2 2 2 2

1 1 1( ) sin ( ) 0

8 sin sin

h er E x y z x y z

m r r r r r r

2 2 2 22 2

2 2 2

1 1( ) sin ( ) 0

8 sin sin


m r r r





Orbitals

EXAMPLE 51





Orbitals




)






1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56




value of l 0 1 2 3








Boundary Conditions



Wave Functions



( ) ( ) ( )r R r Y


2 2 22

2 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r

2 2 22

2 2 2 2 2 2

1 1 1( ) sin ( ) 0

8 sin sin

h er E x y z x y z

m r r r r r r

2 2 2 22 2

2 2 2

1 1( ) sin ( ) 0

8 sin sin


m r r r





Orbitals

EXAMPLE 51





Orbitals




)






1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56







Boundary Conditions



Wave Functions



( ) ( ) ( )r R r Y


2 2 22

2 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r

2 2 22

2 2 2 2 2 2

1 1 1( ) sin ( ) 0

8 sin sin

h er E x y z x y z

m r r r r r r

2 2 2 22 2

2 2 2

1 1( ) sin ( ) 0

8 sin sin


m r r r





Orbitals

EXAMPLE 51





Orbitals




)






1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


Wave Functions



( ) ( ) ( )r R r Y


2 2 22

2 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r

2 2 22

2 2 2 2 2 2

1 1 1( ) sin ( ) 0

8 sin sin

h er E x y z x y z

m r r r r r r

2 2 2 22 2

2 2 2

1 1( ) sin ( ) 0

8 sin sin


m r r r





Orbitals

EXAMPLE 51





Orbitals




)






1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


( ) ( ) ( )r R r Y


2 2 22

2 2 2 2 2 2

1 1 1sin ( ) ( )

8 sin sin

h er x y z E x y z

m r r r r r r

2 2 22

2 2 2 2 2 2

1 1 1( ) sin ( ) 0

8 sin sin

h er E x y z x y z

m r r r r r r

2 2 2 22 2

2 2 2

1 1( ) sin ( ) 0

8 sin sin


m r r r





Orbitals

EXAMPLE 51





Orbitals




)






1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


Orbitals

EXAMPLE 51





Orbitals




)






1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56





Orbitals




)






1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56



1 angularnode

2 angular nodes

n = 1

n = 2

n = 3

No radial node

1 radial node

2 radial nodes

No radial node

1 radial node

n = 3

n = 2

No radial noden = 3







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56







R100(r)



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56



s orbitals

s orbital

constant Y






no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56





no nodes

- 1s (n = 1 ℓ = 0 m = 0) R10(r) and Y00(θФ)

- 2s (n = 2 ℓ = 0 m = 0) R20(r) and Y00(θФ)




negative]


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


isosurface1s2s

3s

wave functionplot

radialprobability


plot




Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56



Prob(r) = r2[R(r)]2

Ψ2 x 4πr2 dr

2 2 2

0 0

2 22 2

2 22

222

22

sin

4 4 ( ) ( )

4 ( ) ( )

4 ( ) 1 2

( )

r d d

r r R r Y

r R r Y

r R r

r R r


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


Boundary Surfaces






p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


p Orbitals

2p0 orbital R21 Y10 (n = 2 ℓ =1 m = 0)





2





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56





called 2px and 2py




2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


2p 3p


r2Rnl2



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56



d Orbitals

d-orbitals











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56











orbital




field








Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


Electron Spin








- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56





- In a helium atom







r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56




r1equiv (r1θ1Ф1)

r2equiv (r2θ2Ф2)




11 1 1 2 2 2 2( ) ( )r r r r


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


Hartree Orbitals



1 11 1 1 2 2 2 2 2( ) ( ) ( ) ( )a br r r r r r



ie) for He

Zeff(He) = 16875 lt 2







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56







1 0 2 0

1 0 2 0

1 0 2

3 2 3 2 0 0

1 12 2

3 2 3 2

0 0

3

30

1 1( ) ( ) ( )

1 1eff eff

eff eff

Zr a Zr aa b

Z r a Z r a

eff eff

Z r a Z r aeff

a ar r r r e e

Z Z

a ae e

Z Z

Ze e

a

0



1s2

occupancy






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56






Shielding Effects













Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56








Penetration














H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56









H 1s1

He 1s2





Core electrons

Valence electron






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56






12py1










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56










C






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56






















Period 2

Period 3

Period 4


Period 5

Period 6

Period 7















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56















Solutions


9890 x 10-10 m

= 20085 x 10-16 J =1 eV

16022 x 10-19 J(20085 x 10-16 J)

= 12536 eV


IE(1) = 12536 eV _ (910938 x 10-31 kg)(2097 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 35 eV


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


IE(2) =12536 eV _ (910938 x 10-31 kg)(2093 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 83 eV

IE(3) =12536 eV _ (910938 x 10-31 kg)(2014 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1005 eV

IE(4) =12536 eV _ (910938 x 10-31 kg)(1971 x 107 ms)2

2

1 eV

16022 x 10-19 J

= 1492 eV









General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56








General trends




ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56



ie Li+ Na+ to K+

substantial change

to Rb+ Cs+



















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56
















Electron Affinity

(kJ mol-1)



10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56


10 Problem Sets

For Chapter 5

6 18 22 28 32 38 44 46 48 56

quantum mechanics and atomic structure

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