lecture outline: chapter 7 periodic...

Lecture outline: Chapter 7Periodic properties

1. Electrostatic effects2 Atomic size2. Atomic size3. Ionization energy4 El t ffi it4. Electron affinity5. Summarize some periodic properties

1S. Ensign, periodic properties

Some important termsp• Electron configuration: the order of filling of

orbitals; tells us which orbitals are filled or partially filled

• Core electrons: those in filled shellsCore electrons: those in filled shells“inner shell electrons”

• Valence electrons: those beyond the filled• Valence electrons: those beyond the filled shells“o ter shell electrons”“outer shell electrons”

• Effective nuclear chargeZeff = Z - S


Electron configurations of the elements,color coded by subshell typey yp


How far are the electrons from the nucleus for three 1s e-

2s & 2p e-

He: 1s2

y

Noble gases (He, Ne, Ar)?2s & 2p e

Ne: 1s2 2s22p6

Ar: 1s2 2s22p6 3s23p6.

ron

dens

ityad

ial e

lect

r

1s e-

2 & 2Ra 2s & 2p e-

3s & 3p e-

1s e-

Distance from nucleus ( ) 1 Å 4S. Ensign, periodic properties

Electrostatic interactions are responsible for the differences in orbital sizes for different atoms

•particles with like charges repel one another (electrostatic repulsion)

•particles with opposite charges are attracted to each other (electrostatic attraction)

•The strength of an electrostatic attraction is directly related to g ythe relative magnitudes of the charges on the particles

•The strength of an electrostatic attraction is inversely relatedto the distance between the interacting particles


The potential energy of two interacting charged particles

QQkE 21 d

dQQkE 21= Q1 Q2

d

dQQdQQE 21∝d

S. Ensign, periodic properties6

Electrostatic interactions are responsible for the differences in orbital si es for different atomsdifferences in orbital sizes for different atoms

Zeff = Z - Sn = 4

eff

Nuclear charge (atomic #)Effective nuclear charge

Z p+ 1

n = 2

n = 3 Nuclear charge (atomic #)# of shielding electrons

•A simplified model for electronicZ p n = 1 A simplified model for electronic structure: treat all electrons of a given n value as if they were identical (they’re really not, of course)

•Assume that electrons in the same shell do not shield each

th ff ti lother effectively

7S. Ensign, periodic propertiesd

QQE 21∝


He: 1s2 Zeff = Z - S

n = 4

eff


2

n = 3


2 p+ n = 1

n = 2 •Treat all electrons of a given shell (n value) as identical

•Assume that electrons in the same shell do not shield each other effectively


QQE 21∝


Ne: 1s2 2s22p6 Zeff = Z - S

n = 4

p eff


2

n = 3


n = 1

n = 210 p+ •Treat all electrons of a given

shell (n value) as identical



QQE 21∝


Ar: 1s2 2s22p6 3s23p6 Zeff = Z - S

n = 4

p p eff


2

n = 3


n = 1

n = 218 p+ •Treat all electrons of a given

shell (n value) as identical



QQE 21∝

How far are the electrons from the nucleus for three

1s e-: Zeff = +18

He: 1s2y

from the nucleus for three Noble gases (He, Ne, Ar)?2s & 2p e-

He: 1sNe: 1s2 2s22p6

Ar: 1s2 2s22p6 3s23p6.ron

dens

ity

p p

adia

l ele

ctr

1s e-: Zeff = +10

2 & 2Ra 2s & 2p e-

3s & 3p e-

1s e-: Zeff = +2

Distance from nucleus ( ) 11S. Ensign, periodic properties

The concepts of nuclear charge, inner l t hi ldi d l t t tielectron shielding and electrostatic

attraction/repulsion are important in predicting three important chemical properties of atoms:p p

• Atomic size• Ionization energy• Electron affinityy


Atomic size (radius)( )

n = 3

18 p+ n = 1

n = 2

n = 3

n 1

Ar: 1s2 2s22p6 3s23p613

S. Ensign, periodic properties


Ar: 1s2 2s22p6 3s23p6p p

n = 2

n = 3

18 p+ n = 1

n = 2is really:



18 p+ n = 1

n = 2

n = 3

r = 0.97 Å = 97 pmp

Golf ball nucleus


Atomic radius: the typical distance from the nucleus to a boundary of the surroundingnucleus to a boundary of the surrounding

electron cloud• What defines the boundary?What defines the boundary?• Van der Waals radius (unbonded touching)

C l t di ( l l t d f b d l th• Covalent radius (calculated from bond lengths measured for molecules)

• Calculated radii from theoretical models• Calculated radii from theoretical models


Atomic radii of elements 1-57

300

pm)

200

250

c ra

dius

(p

150

atom

ic

100

0

50

atomic number of element0 10 20 30 40 50 60

0


Calculated atomic radii of s and p block elements

IIncreasinng atomic Increasing atomic radius from R to L in a rowradius fr

Increasing atomic radius from R to L in a row

rom T to BB

in a per

18


riod

Two opposing effects from Left Right and Top Bottom:

increase n increase orbital sizeincrease n, increase orbital sizeIncrease Zeff, decrease orbital size

Increassing atomic Increasing atomic radius from R to L in a rowradius from

Increasing atomic radius from R to L in a row

m T to B

in a


17

a period

Li vs. Be (L to R)

Li: 1s2 2s1 Be: 1s2 2s2Li: 1s2 2s1 Be: 1s2 2s2

n = 2

n = 3

n = 2

n = 3

n = 1

n = 2

n = 1

n = 23 p+ 4 p+


QQE 21∝

Li vs. Na (Top to Bottom)

Li: 1s2 2s1 Na: 1s2 2s22p6 3s1Li: 1s2 2s1 Na: 1s2 2s22p6 3s1

n = 2

n = 3

n = 2

n = 3

n = 1

n = 2

n = 1

n = 23 p+ 11p+


QQE 21∝

Chapter 7 materialChapter 7 material

1. Electrostatic effects2. Atomic size3. Ionization energy4 Electron affinity4. Electron affinity


Ionization energy: The minimum energy required to remove an gy gy qelectron from the ground state of the isolated gaseous atom

X X+X(g) X+(g) + e-

When reported with units of kJ/mol, it is the energy required to remove one mol of electrons from one mol ofrequired to remove one mol of electrons from one mol of gaseous atoms


Ionization of the hydrogen atom:

H(g) H+(g) + e-

n = ∞“Zero point” E = 0

6n = 7

n = 8

Increasing E ΔE (+) for removing ann = 4

n = 5n = 6

removing an electron from the atom’s ground state

n = 3

ground state orbitn = 2

24

n = 1 “Ground state” E = (-)24


Ionization of the sodium atom:

Na(g) Na+(g) + e-Na: 1s2 2s22p6 3s1

n = ∞“Zero point” E = 0

6n = 7

n = 8

Increasing EΔE (+)n = 4

n = 5n = 6

( )

n = 3 “Ground state” E = (-)

n = 2

25

n = 125


Successive ionizations for Silicon

I. # Elec. Config. before and after ionization

Ion formed

Zefffelt by departing

Energy req. (kJ) to remove

Si: 1s2 2s22p6 3s23p2

e- e-

I1 [Ne]3s23p2

[Ne]3s23p1

Si+ +4 786

I [N ]3 23 1 Si2+ 4 1 577n = 3

n = 4

I2 [Ne]3s23p1

[Ne]3s2

Si2+ +4 1,577

I3 [Ne]3s2

[Ne]3s1

Si3+ +4 3,228n = 1

n = 2

n = 3

14 p+[Ne]3s1

I4 [Ne]3s1

[He}2s22p6

Si4+ +4 4,354

n 1

I5 [He}2s22p6

[He]2s22p5

Si5+ +12 16,100


QQE 21∝

2500

Periodic trends for the first ionization energy: X(g) X+(g) + e-

2000

ener

gy

1500

2000

niza

tion

1000

1500

first

ion

500

1000

0

500

atomic number of element0 10 20 30 40 50 60

0


What determines the energy required to remove an electron from an outer shell?remove an electron from an outer shell?

• Zeff felt by the valence electron• Distance of electron from the nucleus

(reflected by atomic radius values)

-Z p+-

-

-


QQE 21∝

What happens to Zeff and atomic radius when f l ft t i ht i i d?we move from left to right in a period?

•Increase Zeff, increase I1eff, 1

•Decrease size, increase I1


QQE 21∝

What happens to Zeff and atomic radius when d i ?we move down in a group?

•Zeff stays same for a given groupeff y g g p

•Increase size, decrease I1

S. Ensign, periodic properties30d

QQE 21∝

2500

Some “irregularities” in the trend for first ionization energy

2000

2500 He

Ne

ener

gy

1500

2000

NF

Ar

niza

tion

1000

1500

HBe C

O PCl

first

io

500

1000

BMg

Al

SiS

0

Li NaAl

atomic number of element0 5 10 15 20


Explanation of trend irregularities: Be to B and Mg to Al

Zeff = Z - S3s

3p

2p

3p

eff2s

2p

n = 3 2s

2p3s

n = 31s

n = 110p+

n = 210p+

n = 2

n = 1

Si li ti “ i h ll” d l I lit bit l l i

Ne: 1s2 2s22p6Ne: 1s2 2s22p6


Simplistic “onion shell” model treats s and p orbitals of a given “n” value as having equal energies

In reality s orbitals are lower in energy than p orbitals for a given “n” value

y2000

2500

Ne

He

Explanation of trend irregularities”: Be to Bfir

st io

niza

tion

ener

gy

1000

1500

BeMg

B

C

N

O

F

Si

PS

ClAr

H


f

0

500 AlLi Na

3s

3pn = 3

2s2p

2s

2p4 p+

n = 2

n = 1Be: 1s2 2s2

1s

Be


Be

y2000

2500

Ne

He

Explanation of trend irregularities”: Be to Bfir

st io

niza

tion

ener

gy

1000

1500

BeMg

B

C

N

O

F

Si

PS

ClAr

H


f

0

500 AlLi Na

3s

3pn = 3

2s2p

2s

2p5 p+

n = 2

n = 1B: 1s2 2s22p1

1s

B


BA filled s orbital provides some shielding of the nuclear charge felt by an electron in a p orbital of the same “n” value

y2000

2500

Ne

He

Explanation of trend irregularities”: Mg to Alfir

st io

niza

tion

ener

gy

1000

1500

BeMg

B

C

N

O

F

Si

PS

ClAr

H


f

0

500 AlLi Na

2p3s

3p

3s

3p 2s

n = 2

3s

n = 3

2s

2p 10p+ n = 1

Mg: 1s2 2s22p63s2

1s

Mg

Mg: 1s 2s 2p 3s


MgA filled s orbital provides some shielding of the nuclear charge felt by an electron in a p orbital of the same “n” value

y2000

2500

Ne

He

Explanation of trend irregularities”: Mg to Alfir

st io

niza

tion

ener

gy

1000

1500

BeMg

B

C

N

O

F

Si

PS

ClAr

H


f

0

500 AlLi Na

2p3s

3p

3s

3p 2s

n = 2

3s

n = 3

2s

2p 10p+ n = 1

Al: 1s2 2s22p63s23p1

1s

Al

Al: 1s 2s 2p 3s 3p


AlA filled s orbital provides some shielding of the nuclear charge felt by an electron in a p orbital of the same “n” value

Explanation of trend irregularities: N to O and P to S

2s

2p

2s

2p

2s 2s

1s 1s

O: 1s2 2s22p4N: 1s2 2s22p3


The e- configuration np3 (a “spin party”) is good

3p

Explanation of trend irregularities: N to O and P to S

3p

3s

p

3s

p

2s

2p

2s

2p

2s 2s

1s 1s

S: 1s2 2s22p63s23p4P: 1s2 2s22p63s23p3



Electron affinity (EA)y ( )• The energy change associated with the

addition of an electron to a gaseous atom gor ion

X+( )+ e- X( )

X(g) + e- X-(g)

• (-) E.A. : E is released (usually, but not always the case)

X (g)+ e X(g)

always the case)• (+) E.A. : E is added

When reported with units of kJ/mol, it is the energy change associated with adding one mol of electrons to one mol of gaseous atoms or ions

39

one mol of gaseous atoms or ions


Electron affinity values/trendsMore negative EA (in general) ex. n = 2, 5 and 8

nera

l)E

A (in

gen

hang

e in

Eot

muc

h ch

40

No


Why the trends in electron affinity values?

More negative EA (in general)

gen

eral

)

Z p+--

e in

EA

(in-

ch c

hang

eN

ot m

u


Explanation for why group 2A elements have positive or small negative electron affinitiesg

2p3s

3p

2s

n = 2

3s

n = 3

10p+ n = 1

Add e-..

S. Ensign, periodic properties42Mg: 1s2 2s22p63s2

Explanation for why group 6A elements have positive or “out of pattern” negative electron affinities

2p 2p

2s

p

2s

p

e-

1s

N-: 1s2 2s22p4N: 1s2 2s22p3

1s


pp


lecture outline: chapter 7 periodic...

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