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Chapter 6: Electronic Structure of Atoms

Light is a form of electromagnetic radiation (EMR):

- an oscillating charge, such as an electron, gives rise to electromagnetic radiation:

Electric Field

Magnetic Field

Chapter 6: Electronic Structure of Atoms

- Both the Electric and the Magnetic field propagate through
- space

- In vacuum, both move at the speed of light(3 x 108 m/s)

Chapter 6: Electronic Structure of Atoms

- Electromagnetic radiation is characterized by
- wavelength (), or frequency () and
- amplitude (A)

l

A = intensity

l

l

Chapter 6: Electronic Structure of Atoms

Frequency (n) measures how many wavelengths pass a point per second:

1 s

Chapter 6: Electronic Structure of Atoms

Electromagnetic radiation travels at the speed of light:

c = 3 x 108 m s-1

Relation between wavelength, frequency, and amplitude:

c =l n

Chapter 6: Electronic Structure of Atoms

RedOrangeYellowGreenBlueUltraviolet

Chapter 6: Electronic Structure of Atoms

What is the wavelength, in m, of radiowaves transmitted by

the local radio station WHQR 91.3 MHz?

Chapter 6: Electronic Structure of Atoms

A certain type of laser emits green light of 532 nm. What frequency does this wavelength correspond to?

Chapter 6: Electronic Structure of Atoms

Classically, electromagnetic radiation (EMR) was thought to have only wave-like properties.

Two experimental observations challenged this view:

Blackbody radiation

Photoelectric Effect

Chapter 6: Electronic Structure of Atoms

Blackbody radiation

- Hot objects emit light

- The higher T, the higher
- the emitted frequency

Chapter 6: Electronic Structure of Atoms

Blackbody radiation

prediction of classical theory

= there would be NO DARKNESS

Brightness

“ultraviolet catastrophe”

T2

T1

wavelength (l)

visible region

Chapter 6: Electronic Structure of Atoms

Blackbody radiation

Max Planck (1858 - 1947)

- light is emitted by oscillators

- high energy oscillators require a minimum amount of energy to be excited:
- E = h
- energy is not provided by temperature in “black body”

Chapter 6: Electronic Structure of Atoms

Blackbody radiation

frequency of oscillator

E = h

Planck’s constant = 6.63 x 10-34 J s

Energy of radiation is related to frequency, not intensity

Chapter 6: Electronic Structure of Atoms

What is the energy of a photon of electromagnetic radiation

that has a frequency of 400 kHz?

= 2.65 x 10-28 J

Chapter 6: Electronic Structure of Atoms

Photoelectric Effect

Albert Einstein (1879-1955)

e-

e-

e-

e-

- Light of a certain minimum frequency is required to dislodge electrons from metals

Chapter 6: Electronic Structure of Atoms

Photoelectric Effect

- Ability of light to dislodge electrons from metals is related to its frequency, not intensity

E = h

- This means that light comes in “units” of h

- Intensity is related only to the number of “units”

- The h “unit” is called a quantum of energy

- A quantum of light (EMR) energy = photon

Chapter 6: Electronic Structure of Atoms

Relationship between Energy, Wavelength, and Frequency:

Chapter 6: Electronic Structure of Atoms

What is the energy of a photon of light of 532 nm?

= 3.74 x 10-19 J

Chapter 6: Electronic Structure of Atoms

Electromagnetic Radiation

stream of particles

(photons)

wave

or

E = h n

Whether light behaves as a wave or as a stream of photons depends on themethod used to investigate it !

Chapter 6: Electronic Structure of Atoms

Understanding light in terms of photons helped understand atomic structure

many light sources produce a continuous spectrum

Chapter 6: Electronic Structure of Atoms

Thermally excited atoms in the gas phase emit line spectra

continuous spectrum (all wavelengths together: white light)

line spectrum (only some wavelengths: emission will have a color)

1.097 x 107 m-1

positive integers

(e.g. 1,2,3, etc)

Chapter 6: Electronic Structure of Atoms

Photograph of the H2 line spectrum (Balmer series) in the visible region

(1825-1898)

Johann Balmer (1825-1898)

Chapter 6: Electronic Structure of Atoms

Niels Bohr was the first to offer an explanation for line spectra

Bohr Model of the Hydrogen Atom

- Only orbits of defined energy and radii are permitted in the hydrogen atom

- An electron in a permitted orbit has a specific energy and will not radiate energy and will not spiral into the nucleus

- Energy is absorbed or emitted by the electron as the electron moves from one allowed orbit into another. Energy is absorbed or emitted as a photon of E = hn

Chapter 6: Electronic Structure of Atoms

Niels Bohr was the first to offer an explanation for line spectra

electron orbits

n = 1

n = 2

n = 3

n = 4

n = 5

n = 6

nucleus

Bohr’s Model of the Hydrogen Atom

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

Bohr’s Model of the Hydrogen Atom

Energy

absorption of a photon

e

Ground State

nucleus

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

Bohr’s Model of the Hydrogen Atom

Energy

e

Ground State

nucleus

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

Bohr’s Model of the Hydrogen Atom

Energy

“excited state”

e

Ground State

nucleus

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

Bohr’s Model of the Hydrogen Atom

Energy

e

Ground State

nucleus

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

Bohr’s Model of the Hydrogen Atom

Energy

e

Ground State

emission of a photon

nucleus

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

Which of these transitions represents

an absorption process?

(a)

(b)

(c)

Energy

Which of these transitions involves the

largest change in energy?

Which of these transitions leads to the

emission of the longest wavelength photon?

Ground State

Does this wavelength correspond to a high or low frequency?

nucleus

Energy of electron in a given orbit:

n = 6

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

n = Principal Quantum Number (main energy levels)

h=Planck’s constant, c=speed of light, RH = Rydberg constant

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

For an electron moving from n = 4 to n = 2:

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

For an electron moving from n = 4 to n = 2:

DE = - 4.09 x 10-19 J

n = 5

n = 4

n = 3

n = 2

n = 1

Chapter 6: Electronic Structure of Atoms

The energy of the photon emitted is:

E = 4.09 x 10-19 J

What wavelength (in nm) does this

energy correspond to?

l = 486 x 10-9 m

= 486 nm

Chapter 6: Electronic Structure of Atoms

The Wave Behavior of Matter

If light can behave like a stream of particles (photons)…

… then (small) particles should be able to behave like waves, too

For a particle of mass m, moving at a velocity v:

De Broglie Wavelength

e.g electrons have a wavelength (electron microscope!)

Chapter 6: Electronic Structure of Atoms

The Uncertainty Principle

Werner Heisenberg (1901-1976)

and Niels Bohr

Chapter 6: Electronic Structure of Atoms

The Uncertainty Principle

It is impossible to know both the exact position and the exact

momentum of a subatomic particle

uncertainty in momentum, mv

uncertainty in position, x

Chapter 6: Electronic Structure of Atoms

Quantum Mechanics and Atomic Orbitals

Erwin Schrödinger (1887-1961)

Chapter 6: Electronic Structure of Atoms

Quantum Mechanics and Atomic Orbitals

- Schrödinger proposed wave mechanical model of the atom

- Electrons are described by a wave function, ψ

- The square of the wave function, ψ2, provides information on
- the location of an electron (probability density or electron density)

Chapter 6: Electronic Structure of Atoms

Quantum Mechanics and Atomic Orbitals

- the denser the stippling, the
- higher the probability of finding
- the electron

- shape of electron density
- regions depends on energy of
- electron

y

x

Chapter 6: Electronic Structure of Atoms

Bohr’s model:

n = 1

orbit

electron circles around nucleus

Schrödinger’s model:

orbital

n = 1

or

electron is somewhere

within that spherical region

Chapter 6: Electronic Structure of Atoms

Bohr’s model:

- requires only a single quantum number (n) to describe an orbit

Schrödinger’s model:

- requires three quantum numbers (n, l, and m) to describe an orbital

n: principal quantum number

l : second or azimuthal quantum number

ml: magnetic quantum number

- energy of electron in a given orbital:

Chapter 6: Electronic Structure of Atoms

Schrödinger’s model:

(1) n = principal quantum number (analogous to Bohr model)

- the higher n, the higher the energy of the electron

- is always a positive integer: 1, 2, 3, 4 ….

- lis normally listed as a letter:

Value of l: 0 1 2 3

letter: spdf

Chapter 6: Electronic Structure of Atoms

Schrödinger’s model:

(2)l = azimuthal quantum number

- takes integral values from 0 to n-1

e.g.

n = 3

- ldefines the shape of an electron orbital

(1 of 3)

d-orbital

(1 of 5)

f-orbital

(1 of 7)

Chapter 6: Electronic Structure of Atoms

Schrödinger’s model:

z

y

x

s-orbital

Chapter 6: Electronic Structure of Atoms

Schrödinger’s model:

(3) ml = magnetic quantum number

- takes integral values from -lto +l, including 0

e.g.

l = 2

- mldescribes the orientation of an electron orbital in space

Chapter 6: Electronic Structure of Atoms

Shells:

- are sets of orbitals with the same quantum number, n

- a shell of quantum number n has n subshells

Subshells:

- are orbitals of one type within the same shell

- total number of orbitals in a shell is n2

4f subshell

Chapter 6: Electronic Structure of Atoms

3

n =

1

2

4

l =

0

0, 1

0, 1, 2

0, 1, 2, 3

1s

2s, 2p

3s, 3p, 3d

4s, 4p, 4d, 4f

ml =

0

0, -1,0,1

0; -1,0,1; -2,-1,0,1,2

0; -1,0,1; -2,-1,0,1,2; -3,-2,-1,0,1,2,3

# orbitals

in subshell

1

1

3

3

5

1

3

5

1

7

Total # of

orbitals

in shell

1

4

9

16

Chapter 6: Electronic Structure of Atoms

3s-room

3p-room

3deluxe-room

3rd floor

2s-room

2promotion-room

2nd floor

standard-room

1st floor

Chapter 6: Electronic Structure of Atoms

What is the designation for the n=3, l=2 subshell ?

How many orbitals are in this subshell ?

What are the possible values for ml for each of these orbitals ?

Chapter 6: Electronic Structure of Atoms

Which of the following combinations of quantum numbers

is possible?

n=1, l=1, ml= -1

n=3, l=0, ml= -1

n=3, l=2, ml= 1

n=2, l=1, ml= -2

Chapter 6: Electronic Structure of Atoms

Which combination of quantum numbers is possible for the

orbital shown below?

(a) n=1, l=0, ml= 0

(c) n=3, l=3, ml= -2

(d) n=3, l=2, ml= -1

(b) n=2, l=-1, ml= 1

Chapter 6: Electronic Structure of Atoms

There is a fourth quantum number that characterizes electrons:

spin magnetic quantum number, ms

ms can only take two values, +1/2 or -1/2

Chapter 6: Electronic Structure of Atoms

Pauli’s Exclusion Principle:

No two electrons in an atom can have the same set of 4 quantum numbers, n, l, ml, and ms

For a given orbital, e.g. 2s, n, l, ml are fixed:

n=2, l=0, ml =0

=> an orbital can only contain two electron if they differ in ms

Chapter 6: Electronic Structure of Atoms

A maximum of 2 electron can occupy one orbital, IF these two electrons have opposite spin:

n=2, l=0, ml =0, ms = +1/2

n=2, l=0, ml =0, ms = -1/2

2s

2p

arrows pointing up/down indicate electron spin

Chapter 6: Electronic Structure of Atoms

Energy levels in the hydrogen atom:

all subshells of a given shell

have the same energy

Chapter 6: Electronic Structure of Atoms

Energy levels in many-electron atoms:

- In many-electron atoms, the energy of an orbital increases with l, for a given n

- In many-electron atoms, the lower energy orbitals get filled first

- orbitals with the same energy are said to be degenerate

Chapter 6: Electronic Structure of Atoms

Electron Configurations:

Line Notation:

1H

1s1

2He

1s2

1s22s1

3Li

1s22s2

4Be

1s22s22p2

6C

1s22s22p3

7N

10Ne

1s22s22p6

1s22s22p63s1

11Na

Chapter 6: Electronic Structure of Atoms

Electron Configurations:

Hund’s Rule:

For degenerate orbitals, the energy is minimized when the number of electrons with the same spin is maximized

=> degenerate orbitals (p, d, etc)

get filled with one electron each first (same spin).

1s22s22p3

7N

Chapter 6: Electronic Structure of Atoms

the Aufbau Principle helps you to remember the order in which orbitals get filled:

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d 6f

7s 7p 7d 7f

Chapter 6: Electronic Structure of Atoms

Line notation

1s22s22p63s23p2

14Si

[Ne]

3s23p2

Condensed line notation

orbital diagram

(no energy info)

3

d

2

p

1

“coreelectrons”

s

“valence (outer shell) electrons”

Chapter 6: Electronic Structure of Atoms

Line notation

1s22s22p63s23p2

14Si

[Ne]

3s23p2

Condensed line notation

orbital diagram

(no energy info)

3

d

2

p

1

s

Valence electrons take part in bonding

Chapter 6: Electronic Structure of Atoms

What is the electronic structure of Cl?

3s23p5

[Ne]

17Cl :

valence electrons (7)

3

d

2

p

1

coreelectrons

=

electron configuration

of the preceding noble gas

s

coreelectrons

=

electron configuration

of the preceding noble gas

Chapter 6: Electronic Structure of Atoms

What is the electronic structure of Ca?

[Ar]

4s2

20Cl :

(4s orbital is filled before 3d !)

4

f

3

d

2

p

1

s

=

electron configuration

of the preceding noble gas

Chapter 6: Electronic Structure of Atoms

What is the electronic structure of Br?

[Ar]

3d104s24p5

35Br :

(4s orbital is filled before 3d !)

valence electrons (7)

4

f

3

For main group elements,

electrons in a filled d-shell

(or f-shell) are not valence

electrons

d

2

p

1

s

Chapter 6: Electronic Structure of Atoms

Does it matter in which order the electron configuration is written ?

1s22s22p63s23p63d104s24p5

ordered by orbital number

35Br :

or:

1s22s22p63s23p64s23d104p5

ordered by energy

4

f

3

d

2

p

1

NO, both are correct!

s

Chapter 6: Electronic Structure of Atoms

What is the electron configuration of vanadium (V)?

[Ar]

3d34s2

23V:

(4s orbital is filled before 3d !)

4

f

3

d

2

valence electrons (5)

p

1

coreelectrons

=

electron configuration

of the preceding noble gas

s

3d44s2

is less stable than

[Ar]

3d54s1

Chapter 6: Electronic Structure of Atoms

What is the electron configuration of chromium (Cr)?

[Ar]

3d54s1

24Cr:

4

f

3

d

2

p

1

s

A half-filled or completely filled d-shell is a preferred configuration

Chapter 6: Electronic Structure of Atoms

What is the electronic structure of the Ca ion?

[Ar]

4s2

20Ca :

[Ar]

20Ca2+ :

4

f

3

d

2

p

1

s

Chapter 6: Electronic Structure of Atoms

- Metals tend to lose electrons to form cations
- Nonmetals tend to gain electrons to form anions

- Atoms tend to gain or lose the number of electrons

needed to achieve the

electron configuration of the closest noble gas

Chapter 6: Electronic Structure of Atoms

What is the electronic structure of the ion formed by Se?

[Ar]

3d104s24p4

34Se :

[Ar]

3d104s24p6

= [Kr]

34Se2- :

4

f

3

d

2

p

1

s

Chapter 6: Electronic Structure of Atoms

What is the electronic structure of the ion formed by Br?

[Ar]

3d104s24p5

35Br :

[Ar]

3d104s24p6

= [Kr]

35Br- :

4

f

3

d

2

p

1

s

Chapter 6: Electronic Structure of Atoms

What is the electronic structure of the ion formed by Rb?

[Kr]

5s1

37Rb :

[Kr]

37Rb+ :

5

4

f

3

d

2

p

1

s

have the same electron configuration:

37Rb+

,

35Br-

,

34Se2-

, and 36Kr

Chapter 6: Electronic Structure of Atoms

37Rb+ :

[Ar]

3d104s24p6

= [Kr]

35Br- :

[Ar]

3d104s24p6

= [Kr]

34Se2- :

[Ar]

3d104s24p6

= [Kr]

they are isoelectronic

b.

c.

d.

Chapter 6: Electronic Structure of Atoms

Which of the four orbital diagrams written below for nitrogen violates the Pauli Exclusion Principle?

violates Hund’s rule

(all spins must point in the same direction)

violates Hund’s rule

(degenerate orbitals get one electron each, first)

doesn’t violate anything

violates Pauli’s Exclusion Principle

there are two same spin electrons in one orbital, i.e. all 4 quantum numbers are the same – which is impossible

1s

2s

2p

Chapter 6: Electronic Structure of Atoms

What is the total number of orbitals in the fourth shell (n=4) ?

a. 16 b. 12 c. 4 d. 3

what is the total number of different s,p, d and f orbitals?

n=4

l = 0 1 2 3

s p d f

0

-1,0,1

-3,-2,-1,0,1,2,3

ml =

-2,-1,0,1,2

one s + three p + five d + 7 f orbitals

=

16 orbitals

(n2)

Chapter 6: Electronic Structure of Atoms

What is the number of subshells in the third shell (n=3) ?

a. 18 b. 9 c. 3 d. 1

How many different types of orbitals are there?

n=3

l = 0 1 2

s p d

Chapter 6: Electronic Structure of Atoms

What is the electron configuration of the sodium cation, Na+ ?

a. 1s22s22p63s1 b. 1s22s22p6

c. 1s22s22p63s2 d. 1s22s22p7

11Na+

= 11 electrons -1 = 10 electrons

1s2

2s2

2p6

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