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Unit 01 (Chp 6,7):

Atoms and

Periodic Properties

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

John D. BookstaverSt. Charles Community College

St. Peters, MO 2006, Prentice Hall, Inc.

Development of the Atomic Model

• Indivisible

• Identical

• React in fixed ratios

+• + stuff

• – electrons

• empty space

• nucleus

NaLi Cu

Rutherford’s atomic model didn’t explain

properties of matter (color, reactivity, …)

Development of the Atomic Model

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prism

helium

(He)lamp

prism

white light � continuous spectrum

elements � discrete lines of E & f

Atomic Emission Spectra

(only specific colors of energy & frequency)

A mystery for

Niels Bohr.

Hydrogen Emission Spectrum

e–’s emit

(–) energy, move back

to inner

levels(n=5 to n=2)

e–’s absorb

(+) energy, move to

outer

levels(n=2 to n=5)

EXCITED

state

GROUND

state

∆E

5

242

Which transition shows a light wave

of the greatest energy?

32

n=5 to n=2

(1913–Niels Bohr)Bohr’s Shell Model

+

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Photon Energy as Light Waves

• Distance between same point on adjacent

waves is the _______________. (m)

• Number of Waves passing a given point

per unit time is the ______________. (Hz)(s–1)

λλλλ and νννν are

_________ proportional

wavelength (λλλλ)

frequency (νννν)

inversely

All light waves

move at the same speed, so

which color has

more energy?

All EM waves travel the same speed:

the speed of light (c), 2.998 ×××× 108 m/s.c = λλλλνννν

R O Y G B I V

(higher E) (higher νννν) (shorter λλλλ)

Low Frequency High Frequency

Electromagnetic (EM) Spectrum

Low

Energy

High

Energy

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Photon (Light) Calculations

Given wavelength (λλλλ) of light, one can

calculate the energy (E) of 1 photon of that light:

E = hννννc = λλλλνννν (given on Exam)

HW p. 253

#14,25ab,26,34

λ , ν (inverse) E , ν (direct)

λλλλ↔ νννν νννν ↔ E

2.998 × 108 m/s 6.626 × 10–34 J•s(constants)

Schrödinger Wave Equation:

Heisenberg Uncertainty Principle: The more precisely a particle’s

motion is known,…

…the less precisely itsposition is known.(particle)

(wave)

3-D regions of probability (ORBITALS) in sublevelsin each fixed energy level which better explains reactivity.

(1926–Schrodinger, Plank, de Broglie, etc. )Quantum Mechanical Model

s , p , d , f

(E , λ , ν)

(probable locations)

“quantized” into specific multiples

ofwavelengths,

electrons occupy only specific levels (shells)of “quantized” energy

(& wavelength & frequency)

Electrons as Waves (instead of particles)

but none in between.

Quantum Mechanical Model

nucleus

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1803 Dalton

Atomic Theory

1904 Thomson

Plum Pudding

1911 Rutherford

Nuclear Model

1913 Bohr

Shell Model

1926 Quantum Mechanical

Model

Development of Atomic Models

+

+

–

–

––

–

–

1. (Shell) principle energy level (n) (1,2,3,4 …)

2. (Sub-shell) shape

3. (Orbital) 3-D arranged

4. (Electron) spin up/down

Where are the electrons really?

x z

s (1) p (3)(not rings)

d (5) f (7)

y

HW p. 255

#57ac, 60

1s2 2s2 2p4

energy

level(shell, n)

1s2 2s2 2p4

Orbital Notation

+8

Oxygen (O)

sublevel

shape(s,p,d,f)

1s2 2s2 2p4

Electron Configuration (arrangement)

# of e–’s in each sublevel

1s2 2s2 2p4

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Electron Configuration (arrangement)

1s2 2s2 2p41s2 2s2 2p4

+8

1s2 2s2 2p6 3s1Na

1s2 2s2 2p6 3s2 3p1

[Ne] 3s2 3p5 (noble gas core configuration)

Al

Cl

Oxygen (O)

How many valence e–’s?

6

(outer level)

E-Config? Element?

__

__

____________

1s2 2p3 3p6 4p23d10

(3d fills after 4s)

Aufbau: Fill lowest energyorbitals first.+

nucleus

Pauli Exclusion:no e–’s same props

(opp. spin) (↑↓)

Hund:1 e– in equal orbitals before pairing

(↑↑↑↓↓↓)

?

4s22p62s2 3s2

d orbital e–’s are core e–’s …NOT valence e–’s

Electron Configuration of IonsIon E-Con

(i) F–

(ii) Ca2+

(iii) S2–

(iv) Na+

(v) Al3+

1s2 2s2 2p6 [Ne]

1s2 2s2 2p6 3s2 3p6 [Ar]

1s2 2s2 2p6 3s2 3p6 [Ar]

1s2 2s2 2p6 [Ne]

1s2 2s2 2p6 [Ne]

Which ions are isoelectronic?F– , Na+ , Al3+ Ca2+ , S2–

List 3 species isoelectronic with Ca2+ & S2–.

P3– , Cl– , Ar, K+ , Sc3+ , Ti4+, V5+, Cr6+, Mn7+

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• Paramagnetic:

species attracted by a magnet

(caused by unpaired electrons).

Fe: [Ar] ↑↓ ↑↓ ↑ ↑ ↑ ↑4s 3d

• Diamagnetic:

species repelled by magnets

(caused by all paired electrons)

Zn: [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓4s 3d

Other Aspects of Electron Configs

(“di-” is 2)

Other Aspects of Electron Configs

• d block metals lose their outer s electrons

before any core d electrons to form ions.

Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6

Fe2+ 1s2 2s2 2p6 3s2 3p6 3d6

Fe3+ 1s2 2s2 2p6 3s2 3p6 3d5

• d block (trans. metals) have colored ions

b/c light excites e– transitions in d orbitals

HW p.255

#74

Spectroscopy

SPECTROSCOPIC TECHNIQUE EM REGION APPLICATION

WATCH this 6 min Video Explanation of PES at HOME.

Microwave MicrowaveMolecular Structure by

molecular Rotation

IR Infrared Types of bonds by

bond Vibration

Vis/UV Atomic Emission Spectra

(lines of frequencies/colors)

Visible & Ultraviolet

Transition of e–’sbetween energy levels

PES (Photoelectron Spectroscopy) X-rayIonization of e–’sshows e– configuration

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higher peak = more e–’s

Photoelectron Spectroscopy (PES)

Rela

tive # o

f e

–’s

Binding Energy

Which peak is H and which is He?

He

H

...or Ionization Energy(required to remove e–’s)

1s2

1s1

6 5 4 3 2 1 0

���� further left = more energy required

(stronger attractiondue to more protons)

(MJ/mol)

higher peak = more e–’s

Photoelectron Spectroscopy (PES)

Rela

tive # of e

–’s

Binding Energy

Which peak is H and which is He?

He

H

...or Ionization Energy(required to remove e–’s)

1s2

1s1

6 5 4 3 2 1 0

���� further left = more energy required

(stronger attractiondue to more protons)

(MJ/mol)

1s22s2

2p6?

Identify the

element& e-config

Ne

PES (A)

PES (B)

1s2 2s2

2p6

3s2

3p6

3d10

4s2 4p2

?4s1

n = 1 n = 2 n = 3 n = 4

Identify element

(A)

Identify element

(B)

Ge

K

1s2 2s2

2p6

3s2

3p6

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PES (X)

Write the complete electron configuration of

element (X), and identify the element.

1s2 2s2

2p6

3s2

3p6

3d10

4s24p1

1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1 Ga1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 WS 3a

Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

John D. Bookstaver

St. Charles Community College

St. Peters, MO

2006, Prentice Hall, Inc.

Unit 1 (Chp 7):

Periodic Properties…or…

Periodicity of Trends in

Atomic Properties

Periodic Trends

• We will explain observed trends in

�Atomic (and Ionic) Radius

�Ionization energy

�Electronegativity

size

lose e–

attract e–

Zeff & shielding(explains ALL periodic trends and properties)

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• effective nuclear charge, (Zeff):

Zeff & Shielding

• shielding, (S):

(core e–’s)

Na atom

Z = +11

Zeff = +1

inner core e–’s shield valence

e–’s from nuclear attraction.

Z = nuclear charge (+proton’s)S = shielding

+11

Zeff = Z − S

↑↑↑↑attraction

↓↓↓↓shielding

↑↑↑↑Zeff

Atomic Radius

-due toincreasing

shielding

(more energy levels)

incre

ases d

ow

n a

gro

up

-due to increasing Zeff

(more protons)

decreases across a period

↑↑↑↑att.

=shield

↑↑↑↑Zeff

↓↓↓↓att.

↑↑↑↑shield

=Zeff

Ionic Radius

•Cations are smaller than

neutral atoms.

�outermost electron(s) are removed and loses a shell

�core shell closer to nucleus

�inner e–’s ↓↓↓↓shielded (↑↑↑↑Zeff)

e–e–

Na+

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Ionic Radius

• Anions are larger than

their parent atoms.

�electrons are added and repulsions are increased

(=Zeff & =shielding)

Arrange the following species by

increasing size: Ar, K+, Ca2+, S2–, Cl–

Ca2+ < K+ < Ar < Cl– < S2–

e–e–

e–

HW

p. 292#13,28

Ionization Energy (IE)

• more energy to remove next electron

• IE1 < IE2 < IE3, …

• once all valence e–’s are removed, the next e– is on an inner level with ↑↑↑↑attraction (↓↓↓↓shielding & ↑↑↑↑Zeff).

• energy required to remove an electron

look for a huge jump in IE

huge jump in IE4 b/c 4th e– on inner level(must have 3 valence e–’s)

increases across a period

decre

ases d

ow

n a

gro

up

Trends in First IE

-due to increasing Zeff

(more protons)

↑↑↑↑att.

=shield

↑↑↑↑Zeff-due toincreasing

shielding

(more energy levels)

↓↓↓↓att.

↑↑↑↑shield

=Zeff

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5B & 8O exceptions to trend.

Does this graph support your

understanding of IE1 and the Periodic Table?

Why?

increase across period

(↑↑↑↑Zeff , =shielding)

1st IE tends to…

Exceptions to 1st IE Trend

1st IE of B < Be b/c…

The e– in 2p orbital of B is

higher energy than the e– in 2s orbital of Be ; less energy

needed to remove 1st e– in B.

↑↓

2s

↑↓

2s

↑

2pBeB

↑ ↑

increase across period

(↑↑↑↑Zeff , =shielding)

1st IE tends to…

Exceptions to 1st IE Trend

1st IE of O < N b/c…

The paired e– in 2p orbital

of O experiences e–---e–

repulsion requiring less

energy to remove 1st e– in O.

↑↓

2s

↑↓

2s

↑↓

2pNO ↑ ↑↑

2p

HW p. 292 #38,46

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Trends in Electronegativity (EN)-ability of an atom to attract electrons when bonded(sharing e–’s) with another atom.

decre

ases d

ow

n a

gro

up

increases across a period-due to increasing Zeff

(more protons)

↑↑↑↑att.

=shield

↑↑↑↑Zeff

-due toincreasing

shielding

↓↓↓↓att.

↑↑↑↑shield

=Zeff

Periodic Table

Elements arranged by…

atomic #

Periodic Table

Metals on the left

(80% of all elements)

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Nonmetals on the right

(except H)

Periodic Table

Metalloids border the stair-step

(Al is metal)

Periodic Table

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