Chapter 8: The Quantum Mechanical Atom

Chapter 8: The Quantum

Mechanical Atom

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Electromagnetic Energy Electromagnetic Radiation

Light energy or wave

Travels through space at speed of light in vacuum

c = speed of light = 2.9979 x 108 m/s

Successive series of these waves or oscillations

Waves or Oscillations Systematic fluctuations in intensities of

electrical and magnetic forces

Varies rhythmically with time

Exhibit wide range of energy2


Properties of Waves Wavelength ()

Distance between two successive peaks or troughs Unit = meter

Frequency () number of waves per second that pass a given point

in space Unit = Hertz (Hz) = cycles/sec = 1/sec = s1

Related by = c

3


Properties of Waves Amplitude

Intensity of wave Maximum and minimum height Varies with time as travels through space

Nodes Points of zero amplitude Place where wave goes though axis Distance between nodes is always same

4

nodes


Learning Check: Converting from Wavelength to Frequency

The bright red color in fireworks is due to emission of light when Sr(NO3)2 is heated. If the wavelength is ~650 nm, what is the frequency of this light?

5

m10650

m/s109979.29

8

c

= 4.61 × 1014 s–1 = 4.61 × 1014 Hz


Your Turn!WCBS broadcasts at a frequency of 880 kHz. What is the wavelength of their signal?

A.341 m

B.293 m

C.293 mm

D.341 km

E.293 mm

6

s/10880

m/s 1000.33

8

c


Electromagnetic Spectrum

7

high energy, short waves

low energy, long waves

Comprised of all frequencies of light Divided into regions according to

wavelengths of radiation


Electromagnetic SpectrumVisible Light

Band of ’s that human eyes can see 400 to 700 nm Make up spectrum of colors 700 nm ROYGBIV 400 nm

8

White light Equal amount of all these colors Can separate by passing through prism


Important Experiments in Atomic TheoryLate 1800’s:

Classical physics incapable of describing atoms and molecules

Matter and energy believed to be distinct Matter: made up of particles Energy: light waves

Beginning of 1900’s: Several experiments proved this idea incorrect Experiments showed that electrons acted like:

Tiny charged particles in some experiments Waves in other experiments

9


Particle Theory of Light Max Planck and Albert Einstein (1905)

Electromagnetic radiation is stream of small packets of energy

Quanta of energy or photons Each photon travels with velocity = c Pulses with frequency =

Energy of photon of electromagnetic radiation is proportional to its frequency Energy of photon E = h h = Planck’s constant

= 6.626 x 1034 J·s

10


Learning CheckWhat is the frequency, in sec–1, of radiation which has an energy of 3.371 x 10–19 joules per photon?

11

hE

s J10626.6

J10371.334

19

= 5.087×1014 s–1


Your Turn!A microwave oven uses radiation with a frequency of 2450 MHz (megahertz, 106 s–1) to warm up food. What is the energy of such photons in joules?

A.1.62 x 10–30 J

B.3.70 x 1042 J

C.3.70 x 1036 J

D.1.62 x 1044 J

E.1.62 x 10–24 J

12

hE

MHzs101

MHz 2450sJ10626.6E16

34


Photoelectric Effect Shine light on metal surface Below certain frequency ()

Nothing happens Even with very intense light

Above certain frequency () number of electrons ejected increases as

intensity increases Kinetic energy (KE) of ejected electrons

increases as increases

KE = h – BE h = energy of light shone on surface BE = binding energy of electron

13


Means that Energy is Quantized Can occur only in discrete units of size h

1 photon = 1 quantum of energy Energy gained or lost in whole number multiples

of h

E = nh If n = NA, then one mole of photons gained or lost

E = NAhIf light required to start reaction

Must have light above certain frequency to start reaction

Below minimum threshold E, brightness is NOT important

14


Learning CheckHow much energy is contained in one mole of photons, each with frequency 2.00 × 1013?

E = NAh

15

E = (6.02×1023 mol–1)(6.626×10–34 J∙s)(2.00×1013 s–1)

E = 7.98 × 103 J/mol


Your Turn!If a mole of photons has an energy of 1.60 × 10–3 J/mol, what is the frequency of each photon? Assume all photons have the same frequency.

A.8.03 × 1028 Hz

B.2.12 × 10–14 Hz

C.3.20 × 1019 Hz

D.5.85 × 10–62 Hz

E.1.33 × 105 Hz

16

hNE

A

)sJ10626.6)(mol 1002.6(

mol/J1060.134123

3


Ex. Photosynthesis If you irradiate plants with IR and MW

radiation No photosynthesis Regardless of light intensity

If you irradiate plants with Visible Light Photosynthesis occurs Brighter light now means more

photosynthesis

17


Electronic Structure of Atom Most information comes from:

1. Study of light absorption Electron absorbs energy

Moves to higher energy “excited state”

2. Study of light emission e loses photon of light

Drops back down to lower energy “ground state”

18

ground state

excited state

+h

h

excited state

ground state


Continuous Spectrum Continuous unbroken spectrum of all colors

i.e., visible light through a prism

Consider light given off when spark passes through gas under vacuum

Spark (electrical discharge) excites gas molecules (atoms)

19

+ gas


Line Spectrum Spectrum that has only a few discrete lines Also called atomic spectrum or emission

spectrum Each element has unique emission spectrum

20


Atomic Spectra Atomic line spectra are rather complicated Line spectrum of hydrogen is simplest

Single electron 1st success in explaining quantized line spectra 1st studied extensively

J.J. Balmer Found empirical equation to fit lines in visible

region of spectrum

J. Rydberg More general equation explains all emission

lines in H atom spectrum (IR, Vis, and UV)21


Rydberg Equation

Can be used to calculate all spectral lines of hydrogen

RH = 109,678 cm1 = Rydberg

constant = wavelength of light emittedn1 & n2 = whole numbers

(integers) from 1 to where n2 > n1

If n1 = 1, then n2 = 2, 3, 4, …

22

22

21

111

nnRH

Corresponds to allowed energy levels for atom


Learning Check: Using Rydberg Equation

Consider the Balmer series where n1 = 2. Calculate (in nm) for the transition from n2 = 6 down to n1 = 2.

23

361

41

cm678,1096

1

2

11 122HR

m101

1nm100cm

1mcm101029.4

cm9.372,24

19

51

= 410.3 nm Violet line in spectrum

=24,373cm–1


Learning CheckA photon undergoes a transition from nhigher down to n = 2 and the emitted light has a wavelength of 650.5 nm?

)n1

41(

13455.71

22

24

)n1

21(cm678,091

cm105.6501

22

21

7

n2 = 3

110.0.134557

141

n122

10911001

n22 .

.

cm105.650nm1

cm101nm5.650 7

7


Your Turn!What is the wavelength of light (in nm) that is emitted when an excited electron in the hydrogen atom falls from n = 5 to n = 3?

A.1.28 × 103 nm

B.1.462 × 104 nm

C.7.80 × 102 nm

D.7.80 × 10–4 nm

E.3.65 × 10–7 nm

25

221

5

1

3

1cm 678,109

1

1cmnm101

cm 7997

1 7

1

1cm 77991


Significance of Atomic Spectra Atomic line spectra tells us

When excited atom loses energy Only fixed amounts of energy can be lost Only certain energy photons are emitted Electron restricted to certain fixed energy levels

in atoms

Energy of electron is quantized Simple extension of Planck's Theory

Any theory of atomic structure must account for Atomic spectra Quantization of energy levels in atom

26


What Does Quantized Mean?

Energy is quantized if only certain discrete values are allowed

Presence of discontinuities makes atomic emission quantized

27

Potential Energy of Rabbit


Bohr Model of Atom 1st theoretical model of atom to

successfully account for Rydberg equation Quantization of Energy in Hydrogen atom

Correctly explained Atomic Line Spectra

Proposed that electrons moved around nucleus like planets move around sun Move in fixed paths or orbits Each orbit has fixed energy

28


Energy for Bohr Model of H Equation for energy of e in H atom

Ultimately b relates to RH by b = RHhc

OR

where b = RHhc = 2.1788 x 1018 J/atom

Allowed values of n = 1, 2, 3, 4, … n = Quantum number Used to identify orbit

29

2

1

nE

2

422

h

meb

22 n

hcR

n

bE H


Energy Level Diagram for H Atom Absorption of

photon Electron raised to

higher energy level

Emission of photon Electron falls to

lower energy level

30

E s are quantized Every time e drops from n = 3 to n

= 1 Same frequency photon is emitted Yields line spectra


Bohr Model of H E is negative number Reference point E = 0 when n =

e not attached to nucleus

Sign arises from Coulombic attraction between + and – charges (oppositely charged bodies)

Coulomb's Law Attractive force

Stronger attractive force = more negative E

31

them between distance

B) on A)(chargeon (chargeE


Bohr Model of H n = 1 1st Bohr orbit

Most stable E state = ground state = Lowest E state

Electron remains in lowest E state unless disturbed

How to disturb the atom? Add E = h e raised higher n orbit n = 2, 3, 4, … Higher n orbits = excited states = less stable So e quickly drops to lower E orbit and emits

photon of E equal to E between levels

E = Eh – El h = higher l = lower

32


Bohr’s Model Fails Theory could not explain spectra of multi- electron atoms Theory doesn’t explain collapsing atom paradox If e– doesn’t move,

atom collapses

Positive nucleus should easily capture e–

Vibrating charge should radiate and lose energy

33


Your Turn!In Bohr's atomic theory, when an electron moves from one energy level to another energy level more distant from the nucleus,A.energy is emitted.

B.energy is absorbed.

C.no change in energy occurs.

D.light is emitted.

E.none of these

34


Light Exhibits Interference

Constructive interference Waves “in-phase” lead to greater amplitude Add

Destructive interference Waves “out-of-phase” lead to lower amplitude Cancel out

35


Diffraction and Electrons Light

Exhibits interference Has particle nature

Electrons Known to be particles Also demonstrate interference

36


Standing vs. Traveling WavesTraveling wave

Produced by wind on surfaces of lakes and oceans

Standing wave Produced when guitar string

is plucked Center of string vibrates Ends remain fixed

37


Bead on a Wire

Any energy is possible, even zero Same chance of finding bead anywhere

on wire Can know exact position and velocity of

bead simultaneously

38


Wave on a Wire Integer number (n) of peaks and troughs

is required Wavelength is quantized:

39

n2L


How Do We Describe an Electron? Has both wave and particle properties

Confining electron makes its behavior more wavelike

Free electrons behave more like particles Energy of moving electron is E=½ mv2

40


Electron on Wire—Theories Standing wave Half-wavelength must occur

integer number of times along wire’s length

de Broglie’s equation links these m = mass of particle v = velocity of particle

Combining gives:

2

22

8

mLhnE

41

mvh

n2L


de Broglie Explains Quantized Energy

Electron energy quantized Depends on integer n

Energy level spacing changes when positive charge in nucleus changes Line spectra different for

each element

Lowest energy allowed is for n =1

Energy cannot be zero, hence atom cannot collapse

42


Learning Check: Calculate for e-What is the deBroglie wavelength

associated with an electron of mass 9.11 x 10–31 kg traveling at a velocity of 1.0 x 107 m/s?

43

1J/sm1kg

kg109.11m/s101.0

sJ106.626 22

317

34

= 7.27 x 10–11 m


Your Turn!Calculate the deBroglie wavelength of a baseball with a mass of 0.10 kg and traveling at a velocity of 35 m/s.

A.1.9 × 10–35 m

B.6.6 × 10–33 m

C.1.9 × 10–34 m

D.2.3 × 10–33 m

E.2.3 × 10–31 m

44

J1s/mkg1

kg10.0s/m35sJ10626.6 2234


Wave FunctionsSchrödinger’s equation

Solutions give wave functions and energy levels of electrons

Wave function Wave that corresponds to electron Called orbitals for electrons in atoms

Amplitude of wave function Can be related to probability of finding

electron at that given point

Nodes Regions of wire where electrons will not be

found45


Orbitals Characterized by Three Quantum Numbers:

Quantum Numbers: Shorthand Describes characteristics of electron’s position Predicts its behavior

n = principal quantum number All orbitals with same n are in same shell

ℓ = secondary quantum number Divides shells into smaller groups called

subshells

mℓ = magnetic quantum number Divides subshells into individual orbitals 46


n = Principal Quantum Number Allowed values: positive integers from 1 to

n = 1, 2, 3, 4, 5, …

Determines: Size of orbital

Total energy of orbital

Number of nodes (points where * = 0)

RHhc = 2.18 x 1018 J/atom

For given atom, Lower n = Lower (more negative) E

= More stable

2H

2

n

hcRZE

47


ℓ = Orbital Angular Momentum QN Allowed values: 0, 1, 2, 3, 4, 5…(n – 1)

Letters: s, p, d, f, g, h

Orbital designation number nℓ letter

Possible values of ℓ depend on n n different values of ℓ for given n

Specifies orbital angular momentum of e Determines Shape of orbital Angular variation of e path Kinetic energy of orbital

48


mℓ = Magnetic Quantum Number Allowed values:

mℓ = ℓ, ℓ+1, ℓ+2, …, 0 , …, ℓ2, ℓ1, ℓ

Possible values of mℓ depend on ℓ

There are 2ℓ +1 different values of mℓ for given ℓ

z axis component of orbital angular momentum

Determines orientation of orbital in space To designate specific orbital, you need

three quantum numbers

(n, ℓ, mℓ)49


Table 8.1 Summary of Relationships Among the Quantum Numbers n, ℓ,

and m

50


Energy Level Diagram for H Atom or Other 1 e– Ion

4s

3s

2s

1s

Ene

rgy

4p

3p

2p

3d

4d

All orbital subshells with same n value have same Energy

2lo

2hi

2

n

2

n

1ZE hcRHE = Elo – Ehi

2H

2

n

hcRZE

51


Orbital Energies in Many Electron Atoms

1. Each orbital represented by circle or line

2. Now different subshells ( values) have different E

3. All orbitals of same subshell = same E

4. As you go up in energy, spacing between successive shells (n values) decreases as number of subshells increases

Leads to overlapping of several subshells 4s/3d 5s/4d 6s/4f/5d

52


Orbitals of Many Electrons

53

Orbital Designation

Based on first 2 quantum numbers

number for n and letter for

How many e can go in each orbital? Need another

quantum number


Spin Quantum Number, ms Arises out of behavior

of e in magnetic field e acts like a top Spinning charge is like

a magnet e behave like tiny

magnets Leads to 2 possible

directions of e spin up and down north and south

54

Possible Values:

+½ ½


Pauli Exclusion Principle No two e in same atom can have same set of

all four quantum numbers (n, , m, ms)

Can only have 2 e per orbital 2 e s in same orbital must have opposite spin

e s are paired

Odd number of es Not all spins paired Have unpaired es

Even number of es Depends on number of orbitals

55


Consequences of Pauli Exclusion Principle

56


Know from Magnetic Properties

Two e s in same orbital with different spin Spins paired—diamagnetic Sample not attracted to magnetic field Magnetic effects tend to cancel each other

Two e s in different orbital with same spin Spins unpaired—paramagnetic Sample pulled into magnetic field Magnetic effects add

Measure extent of attraction Gives number of unpaired spins

57


Your Turn!Which of the following is a valid set of four quantum numbers (n, ℓ, mℓ, ms)?

A. 3, 2, 3, +½

B. 3, 2, 1, 0

C. 3, 0, 0, -½

D. 3, 3, 0, +½

E. 0, -1, 0, -½

58


Your Turn!What is the maximum number of electrons allowed in a set of 4p orbitals?

A.14

B.6

C.0

D.2

E.10

59


Ground State e ArrangementsElectron Configurations

Distribution of es among orbitals of atom 1. List subshells that contain electrons2. Indicate their electron population with

superscriptEx. N is 1s2 2s2 2p3

Orbital Diagrams Way to represent es in orbitals

1. Represent each orbital with circle (or line)2. Use arrows to indicate spin of each electron Ex. N is

60

1s 2s 2p


Energy Level Diagram for Multi e Atom/Ion

4s

3s

2s

1s

Ene

rgy

4p

3p

2p

3d

4d5s

5p

4f6s

How to put e– into a diagram?

Need some rules

61


Aufbau Principle Building-up principle

Fill lowest energy subshell before going to next highest energy subshell

Fill lowest n first Within given n, fill lowest first Within given , fill highest m first

Within given m, fill highest ms (+ ½ , ) first

Pauli Exclusion Principle 2 e per orbital Spins opposite

62


Hund’s Rule If you have more than 1 orbital all at the

same energy Put 1e into each orbital with spins

parallel (all up) until all are half filled

Before pair up es in same orbital

Why? Repulsion of e in same region of space Empirical observation based on magnetic

properties

63


Orbital Diagram and e Configurations: N Z = 7

4s

3s

2s

1s

Ene

rgy

4p

3p

2p

3d

Each arrow represents electron

1s2 2s2 2p3

64


4s

3s

2s

1s

Ene

rgy

4p

3p

2p

3d

Orbital Diagram and e- Configurations: V Z = 23

Each arrow represents an electron1s2 2s2 2p6 3s2 3p6 4s2 3d3

65


Learning Check

4s

3s

2s

1s

Ene

rgy

4p

3p

2p

3d

4d5s

5p6s

Give electron configurations and orbital diagrams for Na and As

Na Z = 11

As Z = 33

66

1s22s22p63s1

1s22s22p63s23p64s23d104p3


Your Turn!The ground state electron configuration for Ca is:

A.[Ar] 3s1

B.1s2 2s2 2p6 3s2 3p5 4s2

C.[Ar] 4s2

D.[Kr] 4s1

E.[Kr] 4s2

67


Periodic Table Divided into regions of 2, 6, 10, and 14

columns = maximum number of electrons in s, p, d,

and f sublevels

68


Each row (period) represents different energy level

Each region of chart represents different type of sublevel

69

Sublevels and the Periodic Table


Now Ready to Put es into Atoms Electron configurations must be consistent

with:

Pauli Exclusion principle 2 e per orbital, spins opposite

Aufbau principle Start at lowest energy orbital

Fill, then move up

Hund’s rule 1 e in each orbital of same, spins parallel

Only pair up if have to70


Where Are The Electrons?

n= 1 1H

2He

n= 2 3Li

4Be

5B

6C

7N

8O

9F

10Ne

n= 3 11Na

12Mg

13Al

14Si

15P

16S

17Cl

18Ar

n= 4 19K

20Ca

21Sc

22Ti

23V

24Cr

25Mn

26Fe

27Co

28Ni

29Cu

30Zn

31Ga

32Ge

33As

34Se

35Br

36Kr

n= 5 37Rb

38Sr

39Y

40Zr

41Nb

42Mo

43Tc

44Ru

45Rh

46Pd

47Ag

48Cd

49In

50Sn

51Sb

52Te

53I

54Xe

n= 6 55Cs

56Ba

57La

72Hf

73Ta

74W

75Re

76Os

77Ir

78Pt

79Au

80Hg

81Tl

82Pb

83Bi

84Po

85At

86Rn

n= 7 87Fr

88Ra

89Ac

104Rf

105Db

106Sg

107Bh

108Hs

109Mt

110Ds

111Rg

58Ce

59Pr

60Nd

61Pm

62Sm

63Eu

64Gd

65Tb

66Dy

67Ho

68Er

69Tm

70Yb

71Lu

90Th

91Pa

92U

93Np

94Pu

95Am

96Cm

97Bk

98Cf

99Es

100Fm

101Md

102No

103Lr

71

Each box represents room for electron. Read from left to right

“ns” orbital being filled “np” orbital being filled “(n – 1)d” orbital being filled “( n – 2)f” orbital being filled


Read Periodic Table to Determine e- Configuration –

He Read from left to right

1st e– goes into period 1

1st type of sublevel to fill = “1s”

He has 2 e–

e– configuration for He is: 1s2

72

n= 1 1H

2He

n= 2 3Li

4Be

n= 3 11Na

12Mg

n= 4 19K

20Ca

21Sc

22Ti

23V

24Cr

25Mn

26Fe

27Co

28Ni

n= 5 37Rb

38Sr

39Y

40Zr

41Nb

42Mo

43Tc

44Ru

45Rh

46Pd

n= 6 55Cs

56Ba

57La

72Hf

73Ta

74W

75Re

76Os

77Ir

78Pt

n= 7 87Fr

88Ra

89Ac

104Rf

105Db

106Sg

107Bh

108Hs

109Mt

110Ds



Electron Configuration of Boron (B)

73

n= 1 1H

2He

n= 2 3Li

4Be

5B

6C

7N

8O

9F

10Ne

n= 3 11Na

12Mg

13Al

14Si

15P

16S

17Cl

18Ar

n= 4 19K

20Ca

21Sc

22Ti

23V

24Cr

25Mn

26Fe

27Co

28Ni

29Cu

30Zn

31Ga

32Ge

33As

34Se

35Br

36Kr

n= 5 37Rb

38Sr

39Y

40Zr

41Nb

42Mo

43Tc

44Ru

45Rh

46Pd

47Ag

48Cd

49In

50Sn

51Sb

52Te

53I

54Xe

n= 6 55Cs

56Ba

57La

72Hf

73Ta

74W

75Re

76Os

77Ir

78Pt

79Au

80Hg

81Tl

82Pb

83Bi

84Po

85At

86Rn

n= 7 87Fr

88Ra

89Ac

104Rf

105Db

106Sg

107Bh

108Hs

109Mt

110Ds

111Rg

B has 5 e–

Fill first shell… Fill two subshells in 2nd shell, in order of

increasing E Electron Configuration B = 1s22s22p1


Noble Gas Core Notation for Mn

n= 1 1H

2He

n= 2 3Li

4Be

5B

6C

7N

8O

9F

10Ne

n= 3 11Na

12Mg

13Al

14Si

15P

16S

17Cl

18Ar

n= 4 19K

20Ca

21Sc

22Ti

23V

24Cr

25Mn

26Fe

27Co

28Ni

29Cu

30Zn

31Ga

32Ge

33As

34Se

35Br

36Kr

n= 5 37Rb

38Sr

39Y

40Zr

41Nb

42Mo

43Tc

44Ru

45Rh

46Pd

47Ag

48Cd

49In

50Sn

51Sb

52Te

53I

54Xe

n= 6 55Cs

56Ba

57La

72Hf

73Ta

74W

75Re

76Os

77Ir

78Pt

79Au

80Hg

81Tl

82Pb

83Bi

84Po

85At

86Rn

n= 7 87Fr

88Ra

89Ac

104Rf

105Db

106Sg

107Bh

108Hs

109Mt

110Ds

111Rg

58Ce

59Pr

60Nd

61Pm

62Sm

63Eu

64Gd

65Tb

66Dy

67Ho

68Er

69Tm

70Yb

71Lu

90Th

91Pa

92U

93Np

94Pu

95Am

96Cm

97Bk

98Cf

99Es

100Fm

101Md

102No

103Lr

74


Find last noble gas that is filled before Mn Next fill sublevels that follow [Ar] 4s 3d2 5


Writing Electron Configurations Fill pattern across row in Periodic Table is:

ns [(n2)f ] [(n1)d ]np where n = row number in periodic table

Must rearrange e configuration List in order of increasing n Within n level, list in order of increasing Why?

Once 3d and 4f are filled, they become part of core e

Harder to remove e in core or with lower n75


Learning CheckWrite the correct ground state electron configuration for each of the following elements. List in order of increasing n and within each shell, increasing ℓ.

1. K Z = 19

= 1s2 2s2 2p6 3s2 3p6 4s1

2. Ni Z = 28

= 1s2 2s2 2p6 3s2 3p6 4s2 3d8 = 1s2 2s2 2p6 3s2 3p6 3d8 4s2

3. Pb Z = 82

= 1s22s22p63s23p64s23d104p65s24d10

5p66s24f145d106p2

= 1s22s22p63s23p63d104s24p64d104f145s25p65d106s26p2

76


Your Turn!What is the correct ground state electron configuration for Si?

A.1s2 2s2 2p6 3s2 3p6, no unpaired spins

B.1s2 2s2 2p6 3s2 3p4, no unpaired spins

C.1s2 2s2 2p6 2d4, 4 unpaired spins

D.1s2 2s2 2p6 3s2 3p2, 2 unpaired spins

E.1s2 2s2 2p6 3s1 3p3, 4 unpaired spins

77


Your Turn!An element with the electron configuration [Xe]4f145d76s2 would belong to which class on the periodic table?

A.transition elements

B.alkaline earth elements

C.halogens

D.rare earth elements

E.alkali metals

78


Chemical Reactivity Periodic Table arranged by chemical

reactivity Depends on outer shell e’s (highest n)

Arranged by n

Each row is different n

Core electrons Inner e’s = those with n < nmax

Buried deep in atom

Don’t normally play role in chemical bond formation

79


Abbreviated Electron Configurations = Noble Gas

Notation [noble gas of previous row] + es filled in nth row

Represents core + outer shell es Use to emphasize that only outer shell

electrons react

Ex. Ba = [Xe] 6s2 see above

Ru = [Kr] 4d6 5s2

S = [Ne] 3s2 3p4

80


Look at Group IIA

Z Electron Configuration Abbrev

Be 4 1s2 2s2 [He] 2s2

Mg 12 1s2 2s2 2p6 3s2 [Ne] 3s2

Ca 20 1s2 2s2 2p6 3s2 3p6 4s2 [Ar] 4s2

Sr 38 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s2 [Kr] 5s2

Ba 56 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2

[Xe] 6s2

Ra 88 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p6 7s2

[Rn] 7s2

81

All have ns2 outer shell electrons

Only difference is value of n


Shorthand Orbital Diagrams

82

S [Ne]

3s 3p

Write out lines for orbital beyond Noble gas

Higher energy orbital to right Fill from left to rightAbbreviated Orbital Diagrams

Ru [Kr]

4d 5s


Valence Shell Electron Configurations One last type of electron configuration

Use with representative elements (s and p block elements)—longer columns

Here only e’s in outer shell important for bonding

Only e’s in s and p subshells

Valence Shell = outer shell

= occupied shell with highest n

Result - use even more abbreviated notation for e configurations

Sn = 5s2 5p2 83


Electronic Configurations: A few exceptions to rules

Element Expected Experimental

Cr

Cu

Ag

Au

[Ar] 3d4 4s2

[Ar] 3d9 4s2

[Kr] 4d9 5s2

[Xe] 5d9 6s2

84

[Ar] 3d5 4s1

[Ar] 3d10 4s1

[Kr] 4d10 5s1

[Xe] 5d10 6s1

Exactly filled and exactly half-filled subshells have extra stability

Means that you can promote 1 electron into next higher energy orbital to gain this extra stability


Your Turn!The orbital diagram corresponding to the ground state electron configuration for N is:

A.

B.

C.

D.

E.

85

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p


Your Turn!Which of the following choices is the correct electron configuration for a cobalt atom?

4s 3d

A.[Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑

B.[Ar] ↑ ↑↓ ↑↓ ↑↓ ↑↓

C.[Ar] ↑ ↑↓ ↑↓ ↑↓ ↑ ↑

D.[Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑

E.[Ar] ↑↓ ↑↓ ↑↓ ↑ ↑ ↑

86


Heisenberg’s Uncertainty Principle Can’t know both exact position and exact

speed of subatomic particle simultaneously Such measurements always have certain

minimum uncertainty associated with them

4h

mvx

87

x = particle position

mv = particle momentum = mass × velocity of particle

h = Planck’s constant = 6.626 × 10–

34 J∙s


Heisenberg’s Uncertainty PrincipleMacroscopic scale

Errors in measurements << value

Subatomic scale Errors in measurements or > value

If you know position exactly, know nothing about velocity

If you know velocity exactly, know nothing about position

88


Consequence of HUP Can’t talk about absolute position Can only talk about e probabilities

Where is e likely to be?

= wavefunction Amplitude of e wave

2 = probability of finding e at given location

Probability of finding e in given region of space = square of amplitude of wave at that point

89


Electron Cloude dot picture = snapshots

Lots of dots = large amplitude of wave

= High probability of finding e

e density How much of es charge packed into given

volume

High Probability High e charge or Large e density

Low Probability Low e charge or Small e density

90


1s Orbital Representations

a. Dot-density diagram

b. Probability of finding electron around given point, ψ2, with respect to distance from nucleus

c. Radial probability distribution = probability of finding electron between r and r + x from nucleus

rmax = Bohr radius91


e Density Distribution Determined by

e density No sharp boundary Gradually fades away

“Shape” Imaginary surface enclosing 90% of e density

of orbital Probability of finding e is same everywhere on

surface

Shape Size nOrientatio

nm

92


Effect of n on s Orbital

In any given direction probability of finding e same

All s orbitals are spherically shaped

Size as n

93


Spherical Nodes At higher n, now have spherical nodes

Spherical regions of zero probability, inside orbital

Node for e wave Imaginary surface where e density = 0

2s, one spherical node, size larger

3s, two spherical nodes, size larger yet

In general: Number of spherical nodes

= n ℓ 194


p Orbitals

Possess one (1) nodal plane through nucleus e density only on 2 sides of nucleus 2 lobes of e density

All p orbitals have same overall shape Size as n For 3p get spherical node

95


Representations of p Orbitals Constant probability surface for

2p orbital

Simplified p orbital emphasizing directional nature of orbital

All 3 p orbitals in p sub shell One points along each axis

96

x

y

z2px

x

y

z2py 2pz

x

y

z


There Are Five Different d Orbitals

Four with four lobes of e density

One with two lobes and ring of e density

Result of two nodal planes though nucleus

Number of nodal planes through nucleus = ℓ

97


Your Turn!Which sketch represents a pz orbital?

98

x

y

x

z

y

z

x

xy

z

y

z

x

A. B.

D. E.

C.


Periodic Properties: Consequences of e Configuration

Chemical and physical properties of elements Vary systematically with position in periodic

table

i.e. with element's e configuration

To explain, must first consider amount of + charge felt by outer e s (valence e s) Core e s spend most of their time closer to

nucleus than valence (outer shell) e s

Shield or cancel out (screen out, neutralize) some of + charge of nucleus

99


Learning check: Li 1s2 2s1

3 p+ in nucleus 2 e in close (1s) Net + charge felt

at outer e

~ 1 p+

Effective Nuclear Charge (Zeff)

Net positive (+) charge outer e feels

Core e s shield valence e s from full nuclear charge

100


Shielding Electrons in same subshell don't shield each

other Same average distance from nucleus

Trying to stay away from each other

Spend very little time one below another

Zeff determined primarily by

Difference between charge on nucleus (Z) and

Charge on core (number of inner e s)

101


Penetration Within same n shell

e s in 2s orbital are closer to nucleus than e s in 2p orbital

2s feel more of nuclear charge and partially shield 2p from nucleus

Likewise, 3s penetrates inside 3p which penetrates inside 3d

Gives rise to ordering of energy levels Ens < Enp < End < Enf

102


Your Turn!What value is the closest estimate of Zeff for a valence electron of the calcium atom?

A.1

B.2

C.6

D.20

E.40

103


Atomic Size Theory suggests sizes of atoms and ions

indistinct Experiment shows atoms/ions behave as if

they have definite size C and H have ~ same distance between

nuclei in large number of compounds

Atomic Radius (r) Half of distance between two like atoms

H—H C—C etc. Usually use units of picometer 1 pm = 1 x 1012 m Range 37 – 270 pm for atoms 104


Trends in Atomic Radius (r) Down Column (group)

Zeff essentially constant n, outer e s farther away from nucleus and

radius Across row (period)

n constant Zeff, outer e s feel larger Zeff and radius

Transition Metals and Inner Transition Metals Size variations less pronounced as filling core n same (outer e s) across row Zeff and r more gradually

105


Atomic and Ionic Radii (in pm)

106


Ionic Radii down column (group)

across row (period)

Cations smaller than parent atom

Same Zeff, less e s,

Radius contracts

Anions larger than parent atom

Same Zeff, more e s

Radius expands

107


Your Turn!Which of the following has the smallest radius?

A.Ar

B.K+

C.Cl–

D.Ca2+

E.S2–

108


Ionization Energy Energy required to remove e from gas

phase atom Corresponds to taking e from n to n = 1st IE M (g) M+ (g) + e

IE = E

Trends: IE down column (group) as n IE across row (period) as Zeff

109

2

2effH

n

hcZRIE


Comparing 1st IE’s

110

Largest 1st IEs are in upper right

Smallest 1st IEs are in lower left


Successive Ionization Energies

slowly as remove each successive e

See big in IE When break

into exactly filled or half filled subshell

111


Table 8.2: Successive Ionization Energies in kJ/mol for H through Mg

112


Your Turn!Place the elements C, N, and O in order of increasing ionization energy.

A.C, N, O

B.O, N, C

C.C, O, N

D.N, O, C

E.N, C, O

113


Electron Affinity (EA) Potential energy change associated with

addition of 1e to gas phase atom or ion in the ground state

X (g) + e X (g)

O and F very favorable to add electrons 1st EA usually negative (exothermic) Larger negative value means more

favorable to add e

114


Table 8.3 Electron Affinities of Representative Elements

115


Trends in Electron Affinity (EA) EA becomes less exothermic down column

(group) as n e harder to add as orbital farther from nucleus

and feels less + charge

EA becomes more exothermic across row (period) as Zeff Easier to attract e as + charge

116


Successive EAs Addition of 1st e often exothermic Addition of more than 1 e s requires

energy Consider addition of electrons to oxygen:

117

Change: EA(kJ/mol)

O(g) + e– O–(g) –141

O–(g) + e– O2–(g) +844

Net:

O(g) + 2e– O2–(g) +703


Your Turn!Which of the following has the largest electron affinity?

A.O

B.F

C.As

D.Cs

E.Ba

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Chapter 8: The Quantum Mechanical Atom

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Transcript of Chapter 8: The Quantum Mechanical Atom