Chapter 7 Atomic Structure. Chapter goals Describe the properties of electromagnetic radiation.Describe the properties of electromagnetic radiation. Understand.

Chapter 7Chapter 7

Atomic StructureAtomic Structure

Chapter goalsChapter goals

• Describe the properties of electromagnetic Describe the properties of electromagnetic radiation.radiation.

• Understand the origin of light from excited Understand the origin of light from excited atoms and its relationship to atomic atoms and its relationship to atomic structure.structure.

• Describe experimental evidence for wave-Describe experimental evidence for wave-particle duality.particle duality.

• Describe the basic ideas of quantum Describe the basic ideas of quantum mechanics.mechanics.

• Define the three quantum numbers (Define the three quantum numbers (nn, , ll, and , and mmll) and their relationship to atomic structure.) and their relationship to atomic structure.

Electromagnetic RadiationElectromagnetic Radiation

• lightlight• dual nature: wave and particledual nature: wave and particle• transverse wave: perpendicular oscillating transverse wave: perpendicular oscillating

electric and magnetic fieldselectric and magnetic fields

• longitudinal wave: alternating areas of longitudinal wave: alternating areas of compression and decompression. compression and decompression. The The direction of the wave is along the direction direction of the wave is along the direction of propagationof propagation

• soundsound

Transverse WavesTransverse Waves• lightlight• do not require medium for propagationdo not require medium for propagation

AmplitudeAmplitudeheight of wave at maximumheight of wave at maximum

YY

ZZ

XX

AmplitudeAmplitude

Wavelength, Wavelength, lambda)lambda)distance traveled by wave in 1 complete oscillation;distance traveled by wave in 1 complete oscillation;distance from the top (crest) of one wave to the topdistance from the top (crest) of one wave to the topof the next wave.of the next wave.

YY

ZZ

XX

Amplitude & WavelengthAmplitude & Wavelength

7

• measured in m, cm, nm, measured in m, cm, nm, Å (angstrom)Å (angstrom)• 1 1 Å = 1 Å = 1 −10−10 m = m = 1 1 −8−8 cm cm • frequency, frequency, (nu)(nu), measured in s, measured in s−1−1 (hertz) (Hz): (hertz) (Hz): number of number of complete oscillations or cyclescomplete oscillations or cycles

passing a point per unit time (s)passing a point per unit time (s)• speed of propagation, speed of propagation, distance traveled by ray per unit timedistance traveled by ray per unit time in vacuum, all electromagnetic radiation in vacuum, all electromagnetic radiation

travels at same ratetravels at same rate c = 2.998 x 10c = 2.998 x 101010 cm/s (speed of light) cm/s (speed of light) = 2.998 x 10= 2.998 x 1088 m/s m/s ((slower in airslower in air)) mm c (c () = ) = ss11))m)m) ss

Speed of Energy TransmissionSpeed of Energy Transmission

9Tro: Chemistry: A Molecular Approach, 2/e

What is the wavelength in nm of orange light, What is the wavelength in nm of orange light, which has a frequency of 4.80 x 10which has a frequency of 4.80 x 101414 s s−1−1??

c = c =

c 2.998c 2.998101088 m m ss−1−1 = = = = = 6.25 = 6.25 10 10−7−7 mm 4.80 4.80 10 101414 ss−1−1

1 nm1 nm6.25 6.25 10 10−7−7 m m = 625 nm = 625 nm 1 1 10 10−9−9 m m

Names to rememberNames to remember• Max Planck: quantized energy E = hMax Planck: quantized energy E = h ~1900 ~1900

• Albert Einstein: photoelectric effect ~1905Albert Einstein: photoelectric effect ~1905

• Niels Bohr: 2-D version of atomNiels Bohr: 2-D version of atom

• EEnn=(-R=(-RHH)(1/n)(1/n22) Balmer, 1885, then Bohr, 1913) Balmer, 1885, then Bohr, 1913

• Louis de Broglie: Wavelike properties of Louis de Broglie: Wavelike properties of matter ~1915matter ~1915

• Werner Heisenberg: Uncertainty Principle Werner Heisenberg: Uncertainty Principle ~1923 ~1923

• Erwin Schrödinger: Schrödinger Equation Erwin Schrödinger: Schrödinger Equation ~1926~1926

Planck’s equationPlanck’s equation• Planck studied black body radiation, such Planck studied black body radiation, such

as that of a heated body, and realized that to as that of a heated body, and realized that to explain the energy spectrum he had to explain the energy spectrum he had to assume that:assume that:1.1. An object can gain or lose energy by An object can gain or lose energy by

absorbing or emitting radiant energy in absorbing or emitting radiant energy in QUANTAQUANTA of specific frequency ( of specific frequency ())

2.2. light has particle characterlight has particle character ( (photonsphotons)) cc

• Planck’s equation is Planck’s equation is E = h E = h = h = h ──── E = energy of one photon E = energy of one photon h = Planck’s constant = 6.626h = Planck’s constant = 6.6261010−34−34 J Js/photons/photon

Electromagnetic SpectrumElectromagnetic Spectrum

0.01nm0.01nm

-rays-rays

Electromagnetic SpectrumElectromagnetic Spectrum

0.01nm0.01nm

-rays-rays

x-raysx-rays

1nm1nm

Electromagnetic SpectrumElectromagnetic Spectrum

0.01nm0.01nm

-rays-rays

x-raysx-rays

1nm1nm200nm200nm

vacuum UVvacuum UV

Electromagnetic SpectrumElectromagnetic Spectrum

0.01nm0.01nm

-rays-rays

x-raysx-rays

1nm1nm200nm200nm

vacuum UVvacuum UV

400nm400nm

UVUV

Electromagnetic SpectrumElectromagnetic Spectrum

0.01nm0.01nm

-rays-rays

x-raysx-rays

1nm1nm200nm200nm

vacuum UVvacuum UV

400nm400nm

UVUV

800nm800nm

Vis.Vis.

Electromagnetic SpectrumElectromagnetic Spectrum

0.01nm0.01nm

-rays-rays

x-raysx-rays

1nm1nm200nm200nm

vacuum UVvacuum UV

400nm400nm

UVUV

800nm800nm

Vis.Vis.

2525mm

nearnearinfraredinfrared

Electromagnetic SpectrumElectromagnetic Spectrum

0.01nm0.01nm

-rays-rays

x-raysx-rays

1nm1nm200nm200nm

vacuum UVvacuum UV

400nm400nm

UVUV

800nm800nm

Vis.Vis.

2525mm

nearnearinfraredinfrared

1mm1mm

far IRfar IR

Electromagnetic SpectrumElectromagnetic Spectrum

0.01nm0.01nm

-rays-rays

x-raysx-rays

1nm1nm200nm200nm

vacuum UVvacuum UV

400nm400nm

UVUV

800nm800nm

Vis.Vis.

2525mm

nearnearinfraredinfrared

1mm1mm

far IRfar IR

100mm100mm

-waves-waves

Electromagnetic SpectrumElectromagnetic Spectrum

0.01nm0.01nm

-rays-rays

x-raysx-rays

1nm1nm200nm200nm

vacuum UVvacuum UV

400nm400nm

UVUV

800nm800nm

Vis.Vis.

2525mm

nearnearinfraredinfrared

1mm1mm

far IRfar IR

100100mm

-waves-waves

radio wavesradio waves

ElectroElectromagneticmagnetic SpectrumSpectrum

c = c = = c= cE = E = hh

E = h c/E = h c/

Compact disk players use lasers that emit Compact disk players use lasers that emit red light with a wavelength of 685 nm. What red light with a wavelength of 685 nm. What is the energy of one photon of this light? is the energy of one photon of this light? What is the energy of one mole of photons What is the energy of one mole of photons of that red light?of that red light?

, nm , nm , m , m , s, s−1−1 E, J/photon E, J/photon E, J/mole E, J/mole

1010−9−9 m c Avogadro’s m c Avogadro’s = = E = h E = h nm nm number number

1010−9−9 m m 685 nm 685 nm = 6.85 = 6.85 10 10−7−7 m m 1 nm 1 nm

c 2.998c 2.998101088 m m ss−1−1 = = = = = 4.38 = 4.38 10 101414 ss−1−1

6.85 6.85 10 10−7−7 m m

E = hE = h = (6.626 = (6.6261010−34−34 J Js/photon)s/photon)4.38 4.38 10101414 ss−1−1

= 2.90= 2.901010−19−19 J/photon J/photon

= (2.90= (2.901010−19−19 J/photon) J/photon)6.0226.02210102323 photons/mol photons/mol

= 1.75 = 1.75 10 1055 J/mol J/mol

Example: Calculate the number of photons in a Example: Calculate the number of photons in a laser pulse with wavelength 337 nm and total laser pulse with wavelength 337 nm and total energy 3.83 mJ energy 3.83 mJ

Solve:

E=hc/, 1 nm = 10−9 m, 1 mJ = 10−3 J, Etotal=Ephoton # photons

Conceptual Plan:

Relationships:

= 337 nm, Epulse = 3.83 mJ

number of photons

Given:

Find:

(nm) (m) Ephotonnumberphotons

25

The Photoelectric EffectThe Photoelectric Effect• Light can strike the surface of some metals Light can strike the surface of some metals

causing electrons to be ejected.causing electrons to be ejected.• It demonstrates the particle nature of light.It demonstrates the particle nature of light.

The Photoelectric EffectThe Photoelectric Effect• What are some practical uses of the What are some practical uses of the

photoelectric effect?photoelectric effect?• Electronic door openersElectronic door openers• Light switches for street lightsLight switches for street lights• Exposure meters for camerasExposure meters for cameras• Albert Einstein explained the effectAlbert Einstein explained the effect

– Explanation involved light having particle-Explanation involved light having particle-like behavior.like behavior.

The minimum energy needed to eject the eThe minimum energy needed to eject the e−− is isE = h E = h (Planck’s equation) (Planck’s equation)It is also called ‘threshold’ or binding energy.It is also called ‘threshold’ or binding energy.– Einstein won the 1921 Nobel Prize in Physics Einstein won the 1921 Nobel Prize in Physics

for this work.for this work.

Prob.: An energy of 2.0Prob.: An energy of 2.0101022 kJ/mol is required to cause a Cs kJ/mol is required to cause a Cs atom on a metal surface to loose an electron. Calculate the atom on a metal surface to loose an electron. Calculate the longest possible longest possible of light that can ionize a Cs atom. of light that can ionize a Cs atom.

From the value of energy we calculate the frequency (From the value of energy we calculate the frequency () and, with this ) and, with this we calculate lambda (we calculate lambda ().).Firstly, we need to calculate the energy in J per atom; it is given in kJ Firstly, we need to calculate the energy in J per atom; it is given in kJ per mol of atoms...per mol of atoms... kJ 1000 J 1 molkJ 1000 J 1 mol2.02.0101022 ──── x x ────────── x x ────────────────────────── = 3.3 = 3.31010−19−19 Joule per atom Joule per atom mol kJ 6.022 mol kJ 6.022 10102323 atoms atoms

E 3.3 E 3.3 1010−19−19 Joule Joule E = h E = h = = ──── = = ────────────────────────── = 5.0 = 5.010101414 s s-1-1

h 6.626 h 6.626 1010−34−34 J s J s

c 2.998c 2.998101088 m s m s-1-1

Now, speed of light, c = Now, speed of light, c = = = ──── = = ──────────────────────= 6.0= 6.01010−7−7 m m 5.05.010101414 s s-1-1

1 nm1 nm6.06.01010−7−7 m m ──────────────── = 600 nm (Visible light) = 600 nm (Visible light) 111010−9−9 m m

Prob. : A switch works by the photoelectric effect. The metal Prob. : A switch works by the photoelectric effect. The metal you wish to use for your device requires 6.7you wish to use for your device requires 6.71010−19−19 J/atom J/atom to to remove an electron. Will the switch work if the light falling remove an electron. Will the switch work if the light falling on the metal has a on the metal has a = 540 nm or greater? Why?= 540 nm or greater? Why?

The energy of photon is The energy of photon is calculated with Planck’s Equationcalculated with Planck’s Equation c c E = h E = h = h = h ── ── If calculated E If calculated E 6.7 6.71010−19−19

J, J, the switch willthe switch will work. work.

111010−9−9 m m 540 nm 540 nm ──────────── = 5.40 = 5.401010−7−7 m m nmnm 2.9982.998101088 m s m s−1−1 E = 6.626E = 6.6261010−34−34 J Js s ──────────────────── = 3.68 = 3.681010−19−19 J J 5.405.401010−7−7 m m The switch won’t open, because E < The switch won’t open, because E < 6.76.71010−19−19 J. J. has to has to be less than 540 nm. The shorter be less than 540 nm. The shorter the higher the higher and E and E

Atomic Line Spectra and the Bohr AtomAtomic Line Spectra and the Bohr Atom(Niels Bohr, (Niels Bohr, 1885-19621885-1962))

• AnAn emission spectrumemission spectrum is formed by an is formed by an electric electric current passing through a gas in a vacuum tube (at current passing through a gas in a vacuum tube (at very low pressure) which causes the gas to emit light.very low pressure) which causes the gas to emit light.

– Sometimes called aSometimes called a bright bright line spectrum.line spectrum.

Atomic Line Spectra and the Bohr Atomic Line Spectra and the Bohr AtomAtom

• The Rydberg The Rydberg equation is an equation is an empirical equation empirical equation that relates the that relates the wavelengths of the wavelengths of the lines in the hydrogen lines in the hydrogen spectrum. Lines are spectrum. Lines are due to transitions due to transitions

nn22 ──── upper level ──── upper level

nn11 ──── lower level ──── lower level

nn11, n, n22 = 1, 2, 3, 4, 5,… = 1, 2, 3, 4, 5,…

hydrogen of spectrumemission

in the levelsenergy theof

numbers therefer to sn’

n n

m 10 1.097 R

constant Rydberg theis R

n

1

n

1R

1

21

1-7

22

21

Example 5-8. What is the wavelength in angstroms Example 5-8. What is the wavelength in angstroms of light emitted when the hydrogen atom’s energy of light emitted when the hydrogen atom’s energy changes from n = 4 to n = 2?changes from n = 4 to n = 2?

22 1-7

22

21

12

4

1

2

1m 10 1.097

1

n

1

n

1R

1

2n and 4n

1 1 1 1 ÅÅ = = ────────────────── = 4.862 = 4.862 10 10−7−7 m m ───── = 4862 ───── = 4862 ÅÅ 2.0572.057101066 m m−1−1 10 10−10−10 m m

That corresponds to the green line in H spectrumThat corresponds to the green line in H spectrum

1-6

1-7

1-7

m 10 2.057 1

1875.0m 10 1.097 1

0625.0250.0m 10 1.097 1

The electron in a hydrogen atom relaxes from n=7 with the emission of photons of light with a wave length of 397 nm. What is the final level of the electron after the transition?The transition of the electron goes from ni = 7 to nf = ?

 _____________ ni = 7 │ 1x 10−9 m │ photon (E = h ) = 397nm ———— = │ 1 nm ___________ nf = ? = 3.97x 10−7 m   1 1 1— = 1.097 x 107 m−1(— − —) nf

2 72

CONTD: The electron in a hydrogen atom relaxes from n=7 by emitting photons of light with a wave length of 397 nm. What is the final level of the electron after the transition?   1 1 ————————————— = — − 2.041x10−2 3.97 x 10−7 m x 1.097 x 107 m−1 nf

2   1 —— = 0.2296 + 0.02041 nf

2   1 1 —— = 0.25 nf

2 = —— = 4 nf = 2 nf

2 0.25

Atomic Line Spectra and the Bohr Atomic Line Spectra and the Bohr AtomAtom

• An An absorption spectrumabsorption spectrum is formed by shining a is formed by shining a beam of white light through a sample of gasbeam of white light through a sample of gas..

– Absorption spectra indicate the wavelengths of Absorption spectra indicate the wavelengths of light that have beenlight that have been absorbedabsorbed..

Every element has a unique spectrum. Every element has a unique spectrum. Thus we can use spectra to identify Thus we can use spectra to identify elements.elements.This can be done in the lab, stars, This can be done in the lab, stars, fireworks, etc.fireworks, etc.

Bohr Model of the AtomBohr Model of the Atom

• planetary modelplanetary model

• considers only the particle nature of the considers only the particle nature of the electronelectron

• pp++ & n packed tightly in ‘ & n packed tightly in ‘tinytiny’ nucleus’ nucleus

• electrons traveling in circular paths, electrons traveling in circular paths, orbits, in space surrounding nucleusorbits, in space surrounding nucleus

• size, energy, and esize, energy, and e–– capacity of orbits capacity of orbits increase as does distance from nucleus increase as does distance from nucleus (orbital radius)(orbital radius)

• orbits orbits quantizedquantized (only certain levels exist) (only certain levels exist)

PP++

nn

ee––

ee––

ee––

ee––

ee––

ee––

ee––

n=1n=1

n=2n=2

n=3n=3n=4n=4n=5n=5n=6n=6

Energy Levels

E

-1 -1-1 -1EE ── ── EE ── ── r nr n22

n=1n=1

n=2n=2

n=3n=3n=4n=4n=5n=5n=6n=6

Exciting the electron from ground level (n = 1) to upper levels (n > 1)Energy is absorbed

E

n=1n=1

n=2n=2

n=3n=3n=4n=4n=5n=5n=6n=6

hh

One One photonphotonper per transitiontransition

Decay of the electron from upper levels to lower levels:Energy is emittedEmission of Photons

E

n=1n=1

n=2n=2

n=3n=3n=4n=4n=5n=5n=6n=6

hh

Balmer Series, nf = 2, for hydrogen.There are other series.

E

Calculating E Difference Between two LevelsCalculating E Difference Between two LevelsA school teacher was the first to find this! A school teacher was the first to find this!

Johann BalmerJohann Balmer

1 11 1E= E= ||EEfinalfinal − E− Einitialinitial| = R| = RHH(── − ──)(── − ──) nnff

22 n nii22

• RRHH= 2.18 x 10= 2.18 x 10-18 -18 J/atom = 1312 kJ/molJ/atom = 1312 kJ/mol

• nni i andand nnf f = principal quantum numbers of = principal quantum numbers of

the initial and final states: nthe initial and final states: nff < n < nii

• 1,2,3,4….1,2,3,4….

Problem: Calculate Problem: Calculate E and E and for the violet line for the violet line of Balmer series of H. nof Balmer series of H. ninitialinitial = 6 n = 6 nfinalfinal = 2 = 2

1 11 1 E= E= RRHH(── − ──) (── − ──) RRHH= 2.18 x 10= 2.18 x 10-18 -18 J/atomJ/atom nnff

22 n nii22

1 11 1E= 2.18 x 10E= 2.18 x 10-18 -18 JJ(── − ──) = 4.84 x 10(── − ──) = 4.84 x 10−19−19 J J 2222 6 622

E = hE = h = c/ = c/ Then, Then, E = hc/E = hc/

hc 6.626hc 6.6261010−34−34 J Js s 2.998x10 2.998x1088 ms ms−1−1 = = ── = ───────────────────── ── = ───────────────────── E E 4.84 x 104.84 x 10−19−19 J J

= 4.104 x 10= 4.104 x 10−7−7 m m ( (1 1 Å/10Å/10−10−10m) = 4104 m) = 4104 Å = 410.4 Å = 410.4 nmnm

Bohr Model of the AtomBohr Model of the Atom

• Bohr’s theory correctly explains the H Bohr’s theory correctly explains the H emission spectrum and those of hydrogen-emission spectrum and those of hydrogen-like ions (Helike ions (He++, Li, Li2+2+ … 1e … 1e−− species) species)

• The theory fails for atoms of all other The theory fails for atoms of all other elements because it is not an adequate elements because it is not an adequate theory: it doesn’t take into account the fact theory: it doesn’t take into account the fact that the (that the (very smallvery small) electron can be ) electron can be thought as having wave behavior.thought as having wave behavior.

The Wave Nature of the ElectronThe Wave Nature of the Electron

• In 1925 Louis de In 1925 Louis de Broglie published Broglie published his Ph.D. his Ph.D. dissertation.dissertation.– A crucial element A crucial element

of his dissertation of his dissertation is that electrons is that electrons have wave-like have wave-like properties.properties.

– The electron The electron wavelengths are wavelengths are described by the described by the de Broglie de Broglie relationship.relationship.

particle of velocity v

particle of mass m

constant sPlanck’ h mv

h

The Wave-Particle Duality of the The Wave-Particle Duality of the ElectronElectron

• Consequently, we now know that Consequently, we now know that electrons (in fact - all particles) have both electrons (in fact - all particles) have both a particle and a wave like character.a particle and a wave like character.

• This wave-particle duality is a This wave-particle duality is a fundamental property of submicroscopic fundamental property of submicroscopic particles (particles (not for macroscopic onesnot for macroscopic ones.).)

The Wave-Particle Duality of the ElectronThe Wave-Particle Duality of the Electron• Example: Determine the wavelength, in m and Example: Determine the wavelength, in m and ÅÅ, ,

of an electron, with mass 9.11 x 10of an electron, with mass 9.11 x 10-31-31 kg, having a kg, having a velocity of 5.65 x 10velocity of 5.65 x 1077 m/s. m/s.

h = 6.626 x 10h = 6.626 x 10−−3434 Js = 6.626 x 10 Js = 6.626 x 10−−3434 kg m kg m22/s/s

h 6.626h 6.6261010−34−34 kg mkg m22ss−1−1 = = ── = ──────────────────── ── = ──────────────────── mv 9.11mv 9.111010−31−31kgkg 5.65x10 5.65x1077 ms ms−1−1 = 1.29 = 1.29 10 10−11 −11 mm Good:Good: 1 1 Å Å

withinwithin = = 1.29 1.29 10 10−11 −11 mm ────── = 0.129 ────── = 0.129 Å Å atomicatomic 1010−10−10 m m dimensionsdimensions

The Wave-Particle Duality of the ElectronThe Wave-Particle Duality of the Electron

• Example: Determine the wavelength, in m, of a Example: Determine the wavelength, in m, of a 0.22 caliber bullet, with mass 3.89 x 100.22 caliber bullet, with mass 3.89 x 10 -3-3 kg, kg, having a velocity of 395 m/s, ~ 1300 ft/s.having a velocity of 395 m/s, ~ 1300 ft/s.

h = 6.626 x 10h = 6.626 x 10−−3434 Js = 6.626 x 10 Js = 6.626 x 10−−3434 kg m kg m22/s/s

h 6.626h 6.6261010−34−34 kg mkg m22ss−1−1 = = ── = ──────────────── ── = ──────────────── mv 3.89mv 3.891010−3−3kgkg 395 ms 395 ms−1−1

= 4.31 = 4.31 10 10−34 −34 mm = 4.31 = 4.31 10 10−24 −24 Å Å too too small! It doesn’t apply to macro-objects! small! It doesn’t apply to macro-objects!

Quantum Mechanical Model of the AtomQuantum Mechanical Model of the Atom

considers both particle and wave nature of electronsconsiders both particle and wave nature of electrons

• Heisenberg and Born in 1927 developed the concept Heisenberg and Born in 1927 developed the concept of the of the Uncertainty Principle:Uncertainty Principle:

It is impossible to determine simultaneously both the It is impossible to determine simultaneously both the position (x,y,z) and momentum (mv) of an electron (or position (x,y,z) and momentum (mv) of an electron (or any other small particle).any other small particle).

• Consequently, we must speak of the electrons’ Consequently, we must speak of the electrons’ positions about the atom in terms of positions about the atom in terms of probability probability functionsfunctions, i.e., wave equation written for each electron., i.e., wave equation written for each electron.

• These probability functions are represented as These probability functions are represented as orbitalsorbitals in quantum mechanics. They are the wave equations in quantum mechanics. They are the wave equations squared and plotted in 3 dimensions.squared and plotted in 3 dimensions.

The Uncertainty PrincipleThe Uncertainty Principle

Heisenberg showed Heisenberg showed that the more that the more precisely the precisely the momentum (or the momentum (or the velocity) of a particle velocity) of a particle is known, the less is known, the less precisely its position precisely its position is known and vice-is known and vice-versa:versa:

(x) (mv) h4

x: uncertainty in uncertainty in

position position

mv): uncertainty uncertainty in momentumin momentum

h: Planck’s constant Planck’s constant

The Uncertainty PrincipleThe Uncertainty PrincipleExample: Determine the uncertainty in the position Example: Determine the uncertainty in the position of an electron, with mass 9.11 x 10of an electron, with mass 9.11 x 10-31-31 kg, having a kg, having a velocity of 5×10velocity of 5×1066 m/s. Assume that uncertainty in m/s. Assume that uncertainty in velocity is 1%. h = 6.626 x 10velocity is 1%. h = 6.626 x 10−−3434 kg m kg m22/s/s

(mv) = m (mv) = m v and v and v = (1/100) 5×10v = (1/100) 5×1066 m/s =5x10 m/s =5x1044

h 6.626h 6.6261010−34−34 kg mkg m22ss−1−1 x = x = ───── = ──────────────────── ───── = ──────────────────── 44m m v 4v 4 9.11 9.111010−31−31kgkg 5x10 5x1044 ms ms−1−1 x = 1×10x = 1×109 9 m = 10 Å !!! m = 10 Å !!! That means the That means the uncertainty in the position is much bigger than the uncertainty in the position is much bigger than the real size of an atom whose diameter is 1-2 Å.real size of an atom whose diameter is 1-2 Å.We cannot say where the electron is !!!We cannot say where the electron is !!!

Schrödinger’s Model of the AtomSchrödinger’s Model of the AtomBasic Postulates of Quantum TheoryBasic Postulates of Quantum Theory

1.1. Atoms and molecules can exist only in certain Atoms and molecules can exist only in certain energy states. In each energy state, the atom or energy states. In each energy state, the atom or molecule has a definite energy. When an atom or molecule has a definite energy. When an atom or molecule changes its energy state, it must emit or molecule changes its energy state, it must emit or absorb just enough energy to bring it to the new absorb just enough energy to bring it to the new energy state (the quantum condition).energy state (the quantum condition).

2.2. Atoms or molecules emit or absorb radiation (light) Atoms or molecules emit or absorb radiation (light) as they change their energies. The frequency of as they change their energies. The frequency of the light emitted or absorbed is related to the the light emitted or absorbed is related to the energy change by a simple equation.energy change by a simple equation.

hc

h E

Schrödinger’s Model of the AtomSchrödinger’s Model of the Atom

3.3. The allowed energy states of atoms and The allowed energy states of atoms and molecules can be described by sets of molecules can be described by sets of numbers called numbers called quantum numbersquantum numbers..

• Quantum numbers are the solutions of the Quantum numbers are the solutions of the Schrödinger, Heisenberg & Dirac equations.Schrödinger, Heisenberg & Dirac equations.

• FourFour quantum numbers are necessary to quantum numbers are necessary to describe energy states of electrons in describe energy states of electrons in atoms.atoms.

EV8

b

equationdinger oSchr

2

2

2

2

2

2

2

2

..

zyxm

OrbitalOrbital• region of space within which one can expect to region of space within which one can expect to

find an electronfind an electron• no solid boundariesno solid boundaries• electron capacity electron capacity of of 2 per orbital2 per orbital• space surrounding nucleus divided up into space surrounding nucleus divided up into

large volumes called large volumes called shellsshells• shells subdivided into smaller volumes called shells subdivided into smaller volumes called

subshellssubshells• orbitals located in subshellsorbitals located in subshells• as shells get further from nucleus, energy, size, as shells get further from nucleus, energy, size,

and electron capacity increaseand electron capacity increase• shells, subshells, and orbitals described by shells, subshells, and orbitals described by

quantum numbersquantum numbers

Quantum NumbersQuantum Numbers

• The principal quantum number has the The principal quantum number has the symbol nsymbol n

• n = 1, 2, 3, 4, ... indicates n = 1, 2, 3, 4, ... indicates shellshell

• K, L, M, N, … K, L, M, N, … shellsshells

• as n increases, so does size, energy, and as n increases, so does size, energy, and electron capacityelectron capacity

The electron’s The electron’s energy dependsenergy depends principally principally on non n . .

Quantum NumbersQuantum Numbers• The angular momentum (The angular momentum (azimuthalazimuthal) quantum ) quantum

number has the symbolnumber has the symbol . . It indicates It indicates subshell.subshell.

= 0, 1, 2, 3, 4, 5, .......(n-1)= 0, 1, 2, 3, 4, 5, .......(n-1)

from 0 to maximum (n-1) for each nfrom 0 to maximum (n-1) for each n

= = s, p, d, f, g, h, ....... s, p, d, f, g, h, ....... SubshellsSubshells tells us the shape of the orbitals.tells us the shape of the orbitals.

• These orbitals are the volume around the These orbitals are the volume around the atom that the electrons occupy 90-95% of the atom that the electrons occupy 90-95% of the time.time.

This is one of the places where Heisenberg’s This is one of the places where Heisenberg’s Uncertainty principle comes into play.Uncertainty principle comes into play.

Magnetic Quantum Number, mMagnetic Quantum Number, mll

• The symbol for the magnetic quantum number is mThe symbol for the magnetic quantum number is m. . It specifies the orientation of the orbital.It specifies the orientation of the orbital.

For a given For a given l,l,

mm = - = - , (- , (- + 1), (- + 1), (- +2), .....0, ......., ( +2), .....0, ......., ( -2), ( -2), ( - -1), 1),

• IfIf = 0 (or an s orbital), then m = 0 (or an s orbital), then m = = 0. 0. – Notice that there is only 1 value of Notice that there is only 1 value of mm..

This implies that there is one s orbital per n This implies that there is one s orbital per n value. n value. n 1 1

• IfIf = 1 (or a p orbital), then m = 1 (or a p orbital), then m = = -1,0,+1.-1,0,+1.– There are 3 values of There are 3 values of mm..

Thus there are three p orbitals per n value. Thus there are three p orbitals per n value. n n 2 2

Magnetic Quantum Number, mMagnetic Quantum Number, m

• IfIf = 2 = 2 (or a d orbital), then m (or a d orbital), then m = = -2, -1, 0, +1, +2.-2, -1, 0, +1, +2.– There are 5 values of There are 5 values of mm..Thus there are five d orbitals per n value. n Thus there are five d orbitals per n value. n 3 3

• IfIf = 3 = 3 (or an f orbital), then (or an f orbital), then

mm = = -3, -2, -1, 0, +1, +2, +3. -3, -2, -1, 0, +1, +2, +3. – There are 7 values of There are 7 values of mm. . 2x2x +1 orbitals +1 orbitalsThus there are seven f orbitals per n value, n Thus there are seven f orbitals per n value, n 4 4

• Theoretically, this series continues on to g, h, i, Theoretically, this series continues on to g, h, i, etc. orbitals.etc. orbitals.– Practically speaking atoms that have been Practically speaking atoms that have been

discovered or made up to this point in time only discovered or made up to this point in time only have electrons in s, p, d, or f orbitals in their have electrons in s, p, d, or f orbitals in their ground state (ground state (unexcitedunexcited) ) configurations.configurations.

nn shellshell ll subshellsubshell mmll

#orbitals#orbitals 22ll + 1 + 1 #e#e––

11 KK 00 ss 00 11 2222 LL 00 ss 00 11 22

11 pp ––1,0,11,0,1 33 6633 MM 00 ss 00 11 22

11 pp ––1,0,11,0,1 33 66

88

22 dd -2,-1,0,1,2-2,-1,0,1,2 55 10101818

44 NN 00 ss 00 11 2211 pp ––1,0,11,0,1 33 6622 dd -2,-1,0,1,2-2,-1,0,1,2 55 101033 ff -3,-2,-1,0,1,2,3-3,-2,-1,0,1,2,3 77 1414

3232

n2

1

4

9

16

Max

Maximum two electrons per orbital

Electrons Indicated by Shell Electrons Indicated by Shell and Subshelland Subshell

SymbolismSymbolism

nnll##

principal numberprincipal number letter: letter: s, p, d,.. orbitals, p, d,.. orbital

#electrons#electrons

3s3s22 5p5p44 4f4f55

there are 4 electrons in the 5p orbitals

4s4s11 4f4f14144d4d1212 3p3p77

The Shape of The Shape of Atomic OrbitalsAtomic Orbitals

• s orbitals are spherically symmetric.s orbitals are spherically symmetric.

A plot of the surface density as a function A plot of the surface density as a function of the distance from the nucleus for an of the distance from the nucleus for an

s orbital of a hydrogen atom s orbital of a hydrogen atom

• It gives the probability of finding the electron It gives the probability of finding the electron at a given distance from the nucleusat a given distance from the nucleus

A node (nodal A node (nodal

point or surface)point or surface)

is a point whereis a point where and theand the

probability probability

density (density (22) are) are

zerozero

© 2012 Pearson Education, Inc.

s s OrbitalsOrbitals

Observing a graph of Observing a graph of probabilities of finding probabilities of finding an electron versus an electron versus distance from the distance from the nucleus, we see that nucleus, we see that ss orbitals possess orbitals possess n n − − 1 1 nodes, or regions nodes, or regions where there is 0 where there is 0 probability of finding an probability of finding an electron.electron.

p orbitalsp orbitals• p orbital properties:p orbital properties:

– The first p orbitals appear in the n = 2 The first p orbitals appear in the n = 2 shell.shell.

• p orbitals are peanut or dumbbell shaped p orbitals are peanut or dumbbell shaped volumes.volumes.– They are directed along the axes of a They are directed along the axes of a

Cartesian coordinate system.Cartesian coordinate system.• There are 3 p orbitals per n level. There are 3 p orbitals per n level.

– The three orbitals are named pThe three orbitals are named pxx, p, pyy, p, pzz..– They have anThey have an = 1. = 1.– mm = = -1,0,+1 3 values of m-1,0,+1 3 values of m

p Orbitalsp Orbitalsyy

zz

xx

yyzz

xxyy

zz

xxppxx ppyy

ppzz

d orbitalsd orbitals• d orbital properties:d orbital properties:

– The first d orbitals appear in the n = 3 shell.The first d orbitals appear in the n = 3 shell.• The five d orbitals have two different shapes:The five d orbitals have two different shapes:

– 4 are clover leaf shaped.4 are clover leaf shaped.– 1 is peanut shaped with a doughnut around it.1 is peanut shaped with a doughnut around it.– The orbitals lie directly on the Cartesian axes The orbitals lie directly on the Cartesian axes

or are rotated 45or are rotated 45oo from the axes. from the axes.• There are 5 d orbitals per n

level.– The five orbitals are named – They have an = 2.

– m = -2,-1,0,+1,+2

5 values of m

222 zy-xxzyzxy d ,d ,d ,d ,d

d Orbitalsd Orbitals

yy

zz

xx

yy

zz

xx

yy

zz

xx

xx

yy

zz

xx

zz

yy

ddxx22

–y–y22

ddzz22

ddxyxy

ddxzxz ddyzyz

f orbitalsf orbitals• f orbital properties:f orbital properties:

– The first f orbitals appear in the n = 4 shell.The first f orbitals appear in the n = 4 shell.• The f orbitals have the most complex The f orbitals have the most complex

shapes.shapes.• There are seven f orbitals per n level.There are seven f orbitals per n level.

– The f orbitals have complicated names.The f orbitals have complicated names.– They have anThey have an = 3= 3– mm = = -3, -2, -1,0, +1,+2, +3 7 values of m-3, -2, -1,0, +1,+2, +3 7 values of m

– The f orbitals have important effects in the The f orbitals have important effects in the lanthanide and actinide elements.lanthanide and actinide elements.

f orbitalsf orbitals

• f orbital shapesf orbital shapes

Why are Atoms Spherical?Why are Atoms Spherical?

72

Prob: A possible excited state for the H atom has Prob: A possible excited state for the H atom has an electron in a 5d orbital. List all possible sets of an electron in a 5d orbital. List all possible sets of quantum numbers n, quantum numbers n, ll, and m, and mll for this electron for this electron

n = 5, n = 5, ll = 2 five possible m = 2 five possible mll: -2, -1, 0, +1, +2: -2, -1, 0, +1, +2

Then, there are five sets of (n, Then, there are five sets of (n, ll , m, mll))

(5, 2, (5, 2, -2)-2)

(5, 2, (5, 2, -1)-1)

(5, 2, 0)(5, 2, 0)

(5, 2, (5, 2, +1)+1)

(5, 2, (5, 2, +2)+2)

Prob: Which of the following represent valid sets Prob: Which of the following represent valid sets of quantum numbers? For a set that is invalid, of quantum numbers? For a set that is invalid, explain briefly why it is not correct.explain briefly why it is not correct.

a)a) n = 3, n = 3, ll = 3, m = 3, mll: 0 No: maximum : 0 No: maximum ll = n-1 = n-1

ll = 0, 1, and 2, no 3 = 0, 1, and 2, no 3

b) n = 2, b) n = 2, ll = 1, m = 1, mll: 0 : 0

c) n = 6, c) n = 6, ll = 5, m = 5, mll: : −−11

d) n = 4, d) n = 4, ll = 3, m = 3, mll: : −−4 No: minimum value of m4 No: minimum value of mll: : −−l,l,

that is mthat is mll: : −3−3

Prob: What is the maximum number of orbitals Prob: What is the maximum number of orbitals that can be identified by each of the following that can be identified by each of the following sets of quantum numbers? When “none” is the sets of quantum numbers? When “none” is the correct answer, explain the reason.correct answer, explain the reason.

Answer Why?Answer Why?

a) n = 4, a) n = 4, ll = 3 = 3 Seven mSeven mll: -3, -2, -1, 0, 1, 2, 3 : -3, -2, -1, 0, 1, 2, 3

b) n = 5, b) n = 5, 25 25 n n22

c) n = 2, c) n = 2, ll = 2 = 2 None maximum None maximum ll = n - 1 = n - 1

d) n = 3, d) n = 3, ll = 1, m = 1, mll: -1: -1 One, just that described #s.One, just that described #s.

Prob: State which of the following are incorrect Prob: State which of the following are incorrect designations for orbitals according to the designations for orbitals according to the quantum theory: 3p, 4s, 2f, and 1pquantum theory: 3p, 4s, 2f, and 1p

Answer Why?Answer Why?

a) 3p correct n = 3, a) 3p correct n = 3, ll = 1 = 1 maximum maximum ll = 3-1=2 = 3-1=2

b) 4s b) 4s correctcorrect

c) 2f c) 2f incorrect maximum incorrect maximum ll = 1 ( = 1 (ll = 3 for f) = 3 for f)

d) 1pd) 1p incorrectincorrect maximum maximum ll = 0 ( = 0 (ll = 1 for p) = 1 for p)