Zumdahl Ch 07 Atomic structure powerpoint

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Chapter 7

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Chapter 7

Table of Contents

7.1 Electromagnetic Radiation 7.2 The Nature of Matter

7.3 The Atomic Spectrum of Hydrogen 7.4 The Bohr Model

7.5 The Quantum Mechanical Model of the Atom 7.6Quantum Numbers

7.7Orbital Shapes and Energies

7.8Electron Spin and the Pauli Principle 7.9Polyelectronic Atoms

7.10The History of the Periodic Table

7.11The Aufbau Principle and the Periodic Table 7.12Periodic Trends in Atomic Properties

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Chapter 7

Table of Contents

• Objective: Students will be able to answer questions about electromagnetic radiation.

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Section 7.1

Electromagnetic Radiation

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Section 7.1

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Questions to Consider

• Why do we get colors?

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Section 7.1

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Electromagnetic radiation

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Section 7.1

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Three characteristics:

• wavelength ( ) – distance between two peaks or troughs in a wave.

• frequency ( ) – number of waves (cycles) per second that pass a given point in space

• speed (c) – speed of light (2.9979×108 m/s)





=



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Section 7.1

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Section 7.1

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Classification of Electromagnetic Radiation (see p. 276)Lower _energy

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Section 7.1

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Ex. 1: Microwave ovens use a frequency of 2.45 GHz

(2.45 x 109 cycles/s). What is the wavelength of this

radiation?

• Answer:

 = c = 2.9979 x 108 m______________

 s 2.45 x 109 s-1

=



c



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Section 7.2

The Nature of Matter

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Section 7.2

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• Max Planck (1858-1947) discovered that energy

can be gained or lost only in integer multiples of h:

E = nh

• where n is an integer (1, 2, 3 …)

• h is Planck’s constant = 6.626 x 10-34 Js

  = frequency

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Section 7.2

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• A system can transfer energy only in whole quanta (or photons or “packets”).

• Energy seems to have particulate properties too.

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Section 7.2

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• Electromagnetic radiation is a stream of “particles” called photons.

photon = = _

hc

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Section 7.1

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Ex. 2: Find the energy of 1 mole of x-ray photons ( = 2.36 nm).

• Answer: E = h or E = hc



• E = hc = 6.626 x 10-34 J s 2.9979 x 108 m 6.022 x 1023

 s 2.36 x 10-9_{m mol}

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Section 7.2

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The Photoelectric Effect

• Einstein, early 20th century

• Light strikes a metal surface; e-s are

ejected; current flows.

• _{A minimum frequency of light is needed}

for this effect.

• Above that minimum, greater light

intensity leads to higher current (more e-s

ejected).

• _{So light has particle-like properties!}

• Think of light as a high energy photon. If

its energy is high enough, it collides with

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Section 7.2

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• Energy and mass can be interconverted: E = mc2

• Dual nature of light:

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Section 7.2

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DeBroglie equation:  = h/mv

• Ex: Find the deBroglie wavelength for an electron (m = 9.110 x 10-31 kg) traveling 1% the speed of

light.

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Section 7.2

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DeBroglie equation:  = h/mv

• Ex: Find the deBroglie wavelength for an

electron (m = 9.110 x 10-31 kg) traveling 1% the

speed of light.

 = h = 6.626 x 10-34 kg m2 s

mv s 9.110 x 10-31 kg 2.9979 x 106 m

= 2.425 x 10-10 m

(similar to x-rays)

Answer: = h mv

Note: Unit was J•s, now it’s kg m2/s

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Section 7.2

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End of lesson

• See homework on board.

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Section 7.3

The Atomic Spectrum of Hydrogen

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• Continuous spectrum (when white light is passed through a prism) contains all the wavelengths of visible light

• _{Hydrogen emission spectrum (p. 284) has only a}

few lines, at distinct wavelengths: 410 nm, 434 nm, 656 nm. Why?

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Section 7.3

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• This shows that only certain

energies are allowed for the e- in the

H atom.

• _{Bohr (1885 – 1962) developed the}

quantum model for H, the idea that the energy of the e- in the H atom is

“quantized.” (i.e. The e- moves only

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Section 7.3

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•Why is it significant that the color emitted from the hydrogen emission spectrum is not white?

•_{How does the emission spectrum support the}

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Section 7.4

The Bohr Model

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• Bohr’s model gave H atom energy levels consistent with the H emission spectrum.

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Section 7.4

The Bohr Model

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a) An energy-level diagram for electron transitions

b) An orbit-transition diagram

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Section 7.4

The Bohr Model

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Section 7.4

The Bohr Model

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• The Bohr model leads to:

• For a single e- transition from one energy level to

another:

ΔE = change in energy of the atom (energy of the emitted photon)

n_final = integer; final energy level n_initial = integer; initial energy level … and  = hc/E

18

2 2

final initial

1

= 2.178 10 J























E

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Section 7.4

The Bohr Model

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levels of the H atom and proposes only certain allowed circular orbits for the e-.

• _{As the e}- becomes more tightly bound, its energy

becomes more negative relative to the zero-energy reference state (free electron). As the

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Section 7.4

The Bohr Model

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Note:

• Bohr’s model is incorrect. This model only works for H.

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Section 7.4

The Bohr Model

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Ex: Find the amount of energy (and the wavelength of light) needed to bump the H electron from n = 2 to n = 4.

18

2 2

final initial

1

= 2.178 10 J























E

n

We set n_final= 4; n_initial = 2.

E = +4.084 x 10-19 J

18

2 2

final initial

1

= 2.178 10 J























E

n

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Section 7.4

The Bohr Model

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Ex: Find the amount of energy (and the wavelength of light) needed to bump the H electron from n = 2 to n = 4.

 = hc =

E

_{= 4.864 x 10}-7 m = 486 nm (cyan part of visible

spectrum)

6.626 x 10-34 Js 2.9979 x 108 m

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Section 7.4

The Bohr Model

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Exercise

What color of light is emitted when an

excited electron in the hydrogen atom falls from:

a) n = 5 to n = 2 b) n = 4 to n = 2 c) n = 3 to n = 2

Which transition results in the

longest wavelength of light?

blue, λ = 434 nm

green, λ = 486 nm

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Section 7.4

The Bohr Model

Return to TOC strontium sodium lithium potassium copper

Many elements give off characteristic light which can help identify them.

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Section 7.4

The Bohr Model

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End of lesson

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Section 7.5

The Quantum Mechanical Model of the Atom

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Objective: Students will be able to answer

questions about the quantum mechanical model of the atom (although quantum numbers

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Section 7.5

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• _{The Bohr model was unsatisfactory because:} • _e-s do not follow a circular orbit

• did not work for elements past hydrogen

• Schrodinger (1887 – 1961) treated the e- as a

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Section 7.5

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 has no physical meaning , but 2 relates to e

-probability within an orbital.

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Section 7.5

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 You can’t know both the position and

momentum of a particle at the same time.

Δx = uncertainty in a particle’s position

Δ(mν) = uncertainty in a particle’s momentum h = Planck’s constant



m



4

  

 



h

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Section 7.5

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Section 7.5

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Physical meaning of a wave function

• The square of the function indicates the probability of finding an electron near a particular point in space.

• _{Probability distribution – intensity of color is}

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Section 7.5

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Orbitals

• region in space where e-s are found

• difficult to define precisely

• a wave function

• _{a 3-D electron density map of a particular size}

and shape

• _{Shapes are defined by a boundary that encloses}

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Section 7.5

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Section 7.5

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Section 7.6

Quantum Numbers

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•

Principal quantum number (

n = 1, 2, 3…

) –

energy level; orbital size and energy.

•

Angular momentum quantum number (

l

=

0,1, … n-1

) – sublevel; orbital shape

•

Magnetic quantum number (

m

_l

= -

l

, … 0, …

+

l

) – orbital orientation in space relative to

the other orbitals in the atom.

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Section 7.6

Quantum Numbers

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n l Orbital

designation

ml Number

of

orbitals

shapes

1 0 1s 0 1

2 0 2s 0 1

1 2p -1, 0, +1 3

3 0 3s 0 1

1 3p -1, 0, +1 3

2 3d -2 … +2 5

4 0 4s 0 1

1 4p -1, 0, +1 3

2 4d -2 … +2 5

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Section 7.6

Quantum Numbers

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Section 7.6

Quantum Numbers

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Exercise

For principal quantum level n = 3, determine the number of allowed sublevels (different

values of l), and give the designation of each.

# of allowed sublevels = 3

l

= 0, 3s

l

= 1, 3p

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Section 7.6

Quantum Numbers

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Exercise

For l = 2, determine the magnetic quantum numbers (m_l) and the number of orbitals.

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Section 7.7

Orbital Shapes and Energies

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Section 7.7

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Section 7.7

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Section 7.7

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Section 7.7

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Section 7.7

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4f orbitals

Who wants to be in

charge of making a

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Section 7.7

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Do these sets of quantum numbers exist? Why or why not?

(a) n = 2, l = 0, ml = 1, ms = + ½

(b) n = 3, l = 3, ml = -3, ms = - ½

(c) n = 2, l = 0, ml = 0, ms = + ½

(d) n = 3, l = -1, ml = -1, ms = - ½

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Section 7.7

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End of lesson

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Section 7.8

Electron Spin and the Pauli Principle

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Copyright © Cengage Learning. All rights reserved 57 Section 7.8 - 10

• Objectives: Students will understand and apply the Pauli exclusion principle.

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Section 7.8

Electron Spin and the Pauli Principle

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• Remember that electron spin quantum number (m_s) – can be +½ or -½.

• Pauli exclusion principle - in a given atom, no two electrons can have the same set of four quantum numbers.

• _{An orbital can hold only two electrons, and}

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Section 7.9

Polyelectronic Atoms

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• Electron correlation problem:

 Since the electron pathways are unknown, the electron repulsions cannot be calculated

exactly.

• When electrons in a polyelectronic atom are

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Section 7.9

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Hydrogen vs. polyelectronic atom

1s

2s 2p

3s 3p 3d

1s

2s 2p

3s 3p

3d

All orbitals of the same value of n are

degenerate (i.e. have the same energy).

Due to penetration (p. 299),

s e-s can be closer to the

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Section 7.9

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• A 2s electron penetrates to the nucleus more than one in the 2p orbital.

• This causes an electron in a 2s orbital to be attracted to the nucleus more strongly than an electron in a 2p orbital.

• _{Thus, the 2}_s_{orbital is lower in energy than}

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Section 7.9

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Section 7.9

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Section 7.9

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Comparing the radial probability

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Section 7.10

The History of the Periodic Table

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• Newlands proposed octaves.

•

Mendeleev

is given credit for our current periodic table…

• _{… which represents the patterns in the}

chemical properties of the elements.

• _{Mendeleev predicted existence of Ge and}

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Section 7.10

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How many orbitals exist for each set of quantum numbers?

a. n = 2, l = 2 b. n = 4, l = 3

c. n = 3, l = 2, m_l = -1 d. n = 2, m_l = +1

e. n = 5, l = 3

a. 0 b. 7

c. 1 d. 1

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Section 7.10

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End of lesson.

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Section 7.1

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EMR Quiz 2017

Random helpful equations:

E = hc/  = h/mv h = 6.626 x 10-34 J·s

c = 2.9979 x 108 m/s

E = -2.178 x 10-18 J (1/n

final2 – 1/ninitial2)

mass of e- = 9.109 x 10-28 g

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Section 7.11

The Aufbau Principle and the Periodic Table

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Objective:

• Students will be able to draw orbital diagrams and write electron configurations for any

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Section 7.11

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Aufbau principle

• As protons are added one by one to the

nucleus to build up the elements, electrons are similarly added to hydrogen–like orbitals.

1s 2s 2p

• H • He

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Section 7.11

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Hund’s rule

• The lowest energy configuration for an atom is the one having the

maximum number of unpaired electrons allowed by the Pauli principle in a particular set of

degenerate (same energy) orbitals.

• _{Or: Electrons “prefer” to be in}

separate orbitals of equal energy

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Section 7.11

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1s 2s 2p

• Li • … • C • …

1s 2s 2p 3s 3p

• P

Orbital diagrams

Done. Notice the total number of arrows = ___? = the atomic number.

… due to

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Section 7.11

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Electron configurations

• An oxygen atom has two electrons in the 1s sublevel, two electrons in the 2s sublevel, and four electrons in the 2p sublevel.

• _{We write:}

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Section 7.11

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Orbital diagram

• shows how many e-s an atom has in each of

its occupied orbitals.

O: 1s22s22p4

O: 1s 2s 2p

Electron configuration

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Section 7.11

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P: 1s22s22p63s23p3

or: [Ne] 3s23p3

core valence

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Section 7.11

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Ch. 7.2 EMR Quiz

1. What is the characteristic wavelength (in nm) of an electron with a velocity of 4.97 x 106 m/s?

electron mass = 9.11 x 10-28 g E = hc/

E = -2.178 x 10-18 J (1/n

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Section 7.11

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Ch. 7.2 EMR Quiz

1. Calculate the energy in kJ that corresponds to the transition of 1.00 mole of hydrogen e-s from the

n = 1 to the n = 4 state. Is light absorbed or emitted?

electron mass = 9.11 x 10-28 g E = hc/

E = -2.178 x 10-18 J (1/n

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Section 7.11

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• the electrons in the outermost principal quantum level of an atom.

Ne: 1s22s22p6 (valence electrons = 8)

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Section 7.11

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Section 7.11

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The orbitals being filled for elements in various parts of the periodic table

s

d

f

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Section 7.11

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81

Note 1 – 18 numbering system.

Note Cr, Cu exceptions – apparent stability of filled and half-filled sublevels (see middle of p. 305).

Try:

orbital diagram for K

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Section 7.11

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Determine the expected electron configurations for each of the following.

a) S

1s22s22p63s23p4 or [Ne]3s23p4

b) Ba

[Xe]6s2

c) Eu

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Section 7.12

Periodic Trends in Atomic Properties

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Periodic trends

• Ionization energy

• Electron affinity (tomorrow)

• _{Atomic radius}

Objective: Students will be able to explain periodic table trends in atomic radius and ionization

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Section 7.12

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Ionization energy

• Energy required to remove an electron from a gaseous atom or ion.

 X(g) → X+(g) + e–

Mg → Mg+ + e– I

1 = 735 kJ/mol (1st IE)

Mg+ → Mg2+ + e– I

2 = 1445 kJ/mol (2nd IE)

Mg2+ → Mg3+ + e– I

3 = 7730 kJ/mol *(3rd IE)

*Core electrons are bound much more tightly

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Section 7.12

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Ionization energy

IE generally increases to the right due to … IE g en er al ly i n cr e as es u p w ar d d u e to …

greater nuclear charge

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Section 7.12

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Ionization energy

Look at IE₁ values on Table 7.5 (p. 310). Explain the trend.

•(Greater nuclear charge  greater IE.)

Look at IE₁ values on Table 7.6 (p. 310). Explain the trend.

•_{(More shielding moving down the P.T.}__{lower IE)}

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Section 7.12

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Section 7.12

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Concept check

Explain why the graph of ionization energy

versus atomic number (across a period) is not linear.

electron repulsions

Where are the exceptions?

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Section 7.12

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Why the decrease

from N to O? … because O has

more e-/e

-repulsion…

… so it’s easier to remove the

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Section 7.12

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Concept check

Which atom would require more energy to remove an electron? Why?

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Section 7.12

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Concept check

Which atom would require more energy to remove an electron? Why?

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Section 7.12

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Concept check

Which has the larger second ionization energy? Why?

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Section 7.12

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Section 7.12

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Atomic radius (defined on p. 313)

decreases due to…

In c re as es d u e t o …

greater nuclear charge

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Section 7.12

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Section 7.12

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Concept check

Which is the larger atom? Why?

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Section 7.12

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Exercise

Arrange the elements oxygen, fluorine, and sulfur according to increasing:

 Ionization energy

S, O, F

 Atomic size (radius)

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Section 7.12

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Exercise

Which is largest? Why? a. Al, Ga, In

b. P, S, Cl c. F+, F,

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Section 7.12

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End of lesson

• Homework – see board.

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Section 7.12

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Objective

• Students will understand the pattern of electron affinity on the periodic table.

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Section 7.12

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Electron affinity

• energy change associated with the addition of an electron to a gaseous atom.

 X(g) + e– → X–(g) H = - ...

generally negative because nonmetal atoms do attract an extra e-.

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Section 7.12

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Electron affinity



EA



generally increases to the right

due to greater nuclear charge

| E

A

|

 gen

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Section 7.13

The Properties of a Group: The Alkali Metals

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The periodic table – Final thoughts

1. The number and type of valence electrons primarily determine chemical properties.

2. Electron configuration relates to position on the periodic table.

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Section 7.13

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Section 7.13

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The periodic table – Final thoughts

4. Basic division of the elements is into metals and nonmetals.

metals nonmetals

low IE high IE; most

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Section 7.13

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Metals Versus Nonmetals

Most

activ

e meta ls

Most

activ

e non

meta ls

(not

the n

oble

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Section 7.13

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The alkali metals (see p. 316)

• Li, Na, K, Rb, Cs, and Fr

 _{Most chemically reactive of the metals}

 _{React with nonmetals to form ionic solids}

 _{All have a similar e}- configuration (… ns1)

 _{Going down group:}

 _{Ionization energy decreases}

 _{Atomic radius increases}

 _{Density increases}

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Section 7.13

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End of lesson

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Section 7.13

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Zumdahl Ch 07 Atomic structure powerpoint

Chapter 7

Chapter 7

Questions to Consider

Section 7.1

Ex. 1: Microwave ovens use a frequency of 2.45 GHz

Section 7.2

Section 7.1

ejected; current flows.

Section 7.2

light.

Answer: = h mv

few lines, at distinct wavelengths: 410 nm, 434 nm, 656 nm. Why?

“quantized.” (i.e. The e- moves only

Section 7.4

a) An energy-level diagram for electron transitions

another:

levels of the H atom and proposes only certain allowed circular orbits for the e-.

Section 7.4

spectrum)

Section 7.4

Section 7.5

Section 7.5

momentum of a particle at the same time.

Section 7.5

Section 7.5

Principal quantum number (

Section 7.6

Exercise

Section 7.7

Section 7.7

4f orbitals

Who wants to be in

Section 7.7

Copyright © Cengage Learning. All rights reserved 57 Section 7.8 - 10

Section 7.9

Section 7.9

s e-s can be closer to the

Section 7.9

Comparing the radial probability

Section 7.10

Mendeleev

chemical properties of the elements.

End of lesson.

Objective:

Section 7.11

Hund’s rule

maximum number of unpaired electrons allowed by the Pauli principle in a particular set of

… due to

Electron configurations

Section 7.11

Copyright © Cengage Learning. All rights reserved 75 Electron configuration shortcut

Ch. 7.2 EMR Quiz

1. What is the characteristic wavelength (in nm) of an electron with a velocity of 4.97 x 106 m/s?

Ch. 7.2 EMR Quiz

1. Calculate the energy in kJ that corresponds to the transition of 1.00 mole of hydrogen e-s from the

Section 7.11

Section 7.11

Section 7.11

Section 7.12

Ionization energy

Section 7.12

Section 7.12

Which atom would require more energy to remove an electron? Why?

Section 7.12

Section 7.12

Section 7.12

Section 7.12

Electron affinity

2. Electron configuration relates to position on the periodic table.

4. Basic division of the elements is into metals and nonmetals.

Section 7.13

Section 7.13