General Chemistry Students of the Faculty of Pharmacy Prof

Prentice Hall © 2005General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Chapter Eight

1

General Chemistry

Students of the Faculty of Pharmacy

Prof. Dr. M. Hassan Al-Samman


Chapter Eight

2

Chapter Eight

Electron Configurations, Atomic

Properties, and the Periodic Table


Chapter Eight

3

Orbital Energy Diagrams

Subshells

within a shell

are at the same

energy level in

hydrogen:

2s = 2p.

Subshells are split

in a multielectron

atom:

2s < 2p.

Orbital energies are

lower in a

multielectron atom …

…than in the

hydrogen atom.


Chapter Eight

4

•Quantum Numbers

Schrödinger proposed a wave equation for the

electron , the four solutions of this equation called

the quantum numbers.

Electron Configurations

The principal quantum number (n)-1

This number assigns the shell which the electron

belongs to , and n takes the positive integer values :

1,2,3,4,5,6,7 but not zero.

Sometimes we use the letters instead of the numbers to

assign the shells K, L, M, N ….

.


Chapter Eight

5


The orbital angular momentum quantum number (l)-2

This number assigns the type of the orbital which the

electron belongs to, and l takes the positive integer

values from zero to n-1, usually we use the letters

instead of the numbers to assign the type of the orbital:

s, p, d, and f.

Example:

When n=4, the angular momentum quantum number

takes the values: 0, 1, 2, and 3

0 1 2 3

s p d f

So in the fourth shell 4 we have four types of orbitals.

.


Chapter Eight

6


The magnetic quantum number (m)-3

This number assigns the orientation of the orbital which

the electron belongs to in space, and m takes the values

from –l to +l including zero.

Example:

When l=0 (s orbitals), m=0, there is only one orientation

(one orbital).

When l=1 (p orbitals), m= -1, 0, +1, there are three

orientations (3 orbitals).

When l=2 (d orbitals), m= -2, -1, 0, 1, 2, there are five

orientations (5 orbitals).

When l=3 (f orbital) m= -3, -2, -1, 0, 1, 2, 3, there are

seven orientations (7 orbitals)


Chapter Eight

7


The spin quantum number (s)-4

This number assigns the direction of the spinning of the

electron, clock wise or counter clock wise , and takes the

values +1/2 and -1/2.


Chapter Eight

8

The shapes of the orbitals

• When l=0 ( the case of s orbitals ) , then

• m=0


Chapter Eight

9

• When l =1 ( the case of p orbitals )

• m= -1 , 0 , +1 , there are three orbitals:



Chapter Eight

10

• When l=2 , (the case of d orbitals) , m will take the values:

• m= -2 , -1 , 0 , +1 , +2 :



Chapter Eight

11

• An electron configuration describes the distribution of electrons among the various orbitals in the atom.

• Electron configuration is represented in two ways.


The spdf notation uses

numbers to designate a

principal shell and letters

(s, p, d, f) to identify a

subshell; a superscript

indicates the number of

electrons in a designated

subshell.


Chapter Eight

12

In an orbital (box) diagram a box represents each

orbital within subshells, and arrows represent

electrons. The arrows’ directions represent electron

spins; opposing spins are paired.


N:


Chapter Eight

13

• Electrons ordinarily occupy orbitals of the lowest energy available.

• No two electrons in the same atom may have all four quantum numbers alike.

• Pauli exclusion principle: one atomic orbital can accommodate no more than two electrons, and these electrons must have opposing spins.

• Of a group of orbitals of identical energy, electrons enter

empty orbitals whenever possible (Hund’s rule).[ in the

orbitals with identical energy , the electrons tend to be

single as possiple.]

• Electrons in half-filled orbitals have parallel spins (same

direction).

Rules for Electron Configurations


Chapter Eight

14

Order of Subshell Energies

• The electrons fill the orbitals according to Muller's rule.

• Follow the arrows from the top: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc.

• Subshells that are far from the nucleus may exhibit exceptions to the filling order.


Chapter Eight

15

Order of Subshell Energies


Chapter Eight

16

• The Aufbau principle describes a hypothetical “building-up” of an atom from the one that precedes it in atomic number.

(Z = 1) H 1s1

(Z = 2) He 1s2

(Z = 3) Li 1s2 2s1

• Noble-gas-core abbreviation: we can replace the portion that

corresponds to the electron configuration of a noble gas with

a bracketed chemical symbol. It’s easier to write …

(Z = 3) Li [He]2s1

(Z = 22) Ti [Ar]4s2 3d2

The Aufbau Principle

To get He, add one

electron to H.

To get Li, add one

electron to He.


Chapter Eight

17

Example 8.1

Write electron configurations for sulfur, using both the spdf

notation and an orbital diagram, then:

1- refer to the outer shell orbitals and core-electrons.

2-use the noble gas abbreviation to write the electron

configuration

A-

a) S= 1s², 2s², 2p⁶, 3s², 3p⁴

b) The orbitals diagram

1s² 2s² 3s²2p⁶ 3p⁴


Chapter Eight

18

-In the noble gas core abbreviated :

S = [ Ne ] 3s², 3p⁴

S = [ Ne ]

3s² 3p⁴


Chapter Eight

19

Example 8.2

Give the complete ground-state electron configuration of a

strontium atom (a) in the spdf notation and (b) in the

noble-gas-core abbreviated notation.

Z Sr = 38

Sr = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹º, 4p⁶, 5s²

Sr = [ Kr ] 5s²


Chapter Eight

20

Exceptions to the Aufbau Principle

Half-filled d subshell plus

half-filled s subshell has

slightly lower in energy

than s2 d4.

Filled d subshell plus

half-filled s subshell has

slightly lower in energy

than s2 d9.

More exceptions occur

farther down the periodic

table. They aren’t always

predictable, because energy

levels get closer together.


Chapter Eight

21

• The valence shell is the outermost occupied principal shell.

The valence shell contains the valence electrons.

• For main group elements, the number of valence shell

electrons is the same as the periodic table group number

(2A elements: two valence electrons, etc.)

The period number is the same as the principal quantum

number n of the electrons in the valence shell.

• Electrons in inner shells are called core electrons.

Example: As [Ar] 4s2 3d104p3

Valence Electrons and Core Electrons

Five valence electrons, for which n = 4

28 core electrons


Chapter Eight

22

• To obtain the electron configuration of an anion

by the aufbau process, we simply add the

additional electrons to the valence shell of the

neutral nonmetal atom.

• The number added usually completes the shell.

• A nonmetal monatomic ion usually attains the

electron configuration of a noble gas atom.

O2– : [Ne]

Br– : [Kr]

Electron Configurations of Ions


Chapter Eight

23

Atom Ion (or)

F 1s2 2s22p5 F– 1s2 2s22p6 [Ne]

S [Ne] 3s2 3p4 S2– [Ne] 3s2 3p6 [Ar]

Fe [Ar] 4s2 3d6 Fe2+ [Ar] 4s2 3d6 [Ar] 3d6

Ti [Ar] 4s2 3d2 Ti4+ [Ar] 4s2 3d2 [Ar]

Sr [Kr] 5s2 Sr2+ [Kr] 5s2 [Kr]

What would be

the configuration

of Fe3+? Of Sn2+?

Valence electrons

are lost first.

Electron Configurations of Ions (cont’d)


Chapter Eight

24


Chapter Eight

25

Example 8.3

Write the electron configuration of the Co3+ ion using:

a) spdf notation

b) Orbital diagram

c) noble-gas-core abbreviated spdf notation.

A-

Co = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d⁷

Co³⁺= 1s², 2s², 2p⁶, 3s², 3p⁶, 4s⁰, 3d⁶ Co³⁺= [Ar] 4s⁰, 3d⁶

Co³⁺=

Co³⁺= [Ar] 4s⁰, 3d⁶


Chapter Eight

26

Important Rule:

In the transition elements, the d orbitals tend to

be : empty , half filled or fully filled.

1- Chromium Cr , Molybdenum Mo and

Tungsten W, suppose to be half filled.

2- Copper Cu silver Ag and gold Au, suppose

to be fully filled


Chapter Eight

27

Example

Write the electron configuration of chromium atom Cr (

Z=24 ) using:

a) spdf notation

b) Orbital diagram


Solution:

₂₄Cr = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d⁵

₂₄Cr = = [Ar] 4s¹, 3d⁵

₂₄Cr =


Chapter Eight

28

Example

Write the electron configuration of chromium atom Cu (

Z=29 ) using:

a) spdf notation

b) Orbital diagram


Solution:

₂₂₂₂₉Cu = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d¹⁰

₂₂₂₂₉Cu = [Ar] 4s¹, 3d¹⁰

₂₂₂₂₉Cu=


Chapter Eight

29

Types of the bond:

There are two types of the bonds:

σ bond:

Each linear combination gives a σ bond.

Molecular Orbitals


Chapter Eight

30

σ bond:

Molecular Orbitals


Chapter Eight

31

σ bond:

Molecular Orbitals


Chapter Eight

32

π bond :

Each parallel combination gives a π bond

Molecular Orbitals


Chapter Eight

33

π bond :

Notice:

If there is more than one bond between two atoms, one of them

suppose to be a σ bond , so in the case of the presence of double

covalent bond between two atoms , one of them suppose to be a σ

bond while the other bond suppose to be a π bond, and in the

case of the presence of triple covalent bond between two atoms ,

one of them suppose to be a σ bond, while the other two bonds

Molecular Orbitals


Chapter Eight

34

suppose to be π bonds .

- π bond is a weak bond comparing with σ bond, so in the

unsaturated organic compounds having double bonds , can easily

break the π bond .

Molecular Orbitals


Chapter Eight

35

Molecular Structures:

There are many theories in describing the molecular structures of

the molecules and compounds , but we well deal only with the

most common theories:

1- the molecular orbital theory

2- the valence bond theory ( hybridization theory)

The molecular bond theory:

The molecular bond theory consists that in the atoms ,

there are atomic orbitals ( s, p, d and f) and in the

molecules , there are molecular orbitals.

Molecular Orbitals


Chapter Eight

36

We use the same rules of the electron arrangement in

the molecules that are used in the atoms , such as :

- The electron fills the lowest energy level as possible.

- The maximum number of electrons in the orbital is

two , an my be one or empty .

- In the case of degenerate orbitals ( orbitals with the

same energy level) , the electrons tend to be single as

possible.

- The molecular orbitals form by the linear

combination of the atomic orbitals.

Molecular Orbitals


Chapter Eight

37

So by the combination of two atomic orbitals , two

molecular orbitals will be formed , one of them with a

low energy level and we call it : bonding orbital , and

the second with high energy level and we call it : anti-

bonding orbital.

H₂ molecule:

In H₂ molecule , 1S orbital of the first hydrogen atom

combines with 1S orbital of the second hydrogen atom to

form two molecular orbitals , one of them is a bonding

orbital with a low energy level and the second is anti-

bonding orbital with high energy level :

Molecular Orbitals


Chapter Eight

38

Molecular Orbitals


Chapter Eight

39

The valence bond theory:

The valence bond theory differentiate between the

central atom and side atoms , such as in methane

molecule CH₄ , the carbon atom is the central atom

while the hydrogen atoms are side atoms.

How to use the valence bond theory in describing the

molecular structure of a compound?

1- we write the electron configuration for each atom in

the compound.

2- we chick if the number of single electrons of the

Molecular Orbitals


Chapter Eight

40

central atom is enough to combine with the number of

side atoms, if not we should excite the central atom by

moving one electron from low energy orbital to a higher

energy orbital .

3- we overlap ( hybridize) the outer shell orbital of

central atom ( that with single electrons or pair of

electrons) to form the suitable hybrid orbitals which

show the geometry of the compound:

- The overlapping between S orbital and one P orbital

gives two hybrid orbitals of the type SP , according to

the repulsion the shape suppose to be a linear structure

.

Molecular Orbitals


Chapter Eight

41

- The overlapping between S orbital and two P orbitals

gives three hybrid orbitals of the type SP² , according

to the repulsion the shape suppose to be a tri angular

in plane.

- - The overlapping between S orbital and three P

orbital gives four hybrid orbitals of the type SPᶟ ,

according to the repulsion the shape suppose to be a

pyramid structure( tetrahedral structure).

4- then each hybrid orbital combine with an orbital of

the side atoms to form a σ bond.

Molecular Orbitals


Chapter Eight

42

Example:

Describe the molecular structure of beryllium hydride

BeH₂ .

Solution:

1- we write the electron configuration of each atom in

the molecule:

H : 1S¹

Be: 1S² , 2S²

2- we notice that the number of the single electrons in

the central atom is not enough to combine with two

Molecular Orbitals


Chapter Eight

43

hydrogen atoms, so we excite the central atom , so the

electron configuration will become:

Be*: 1s² , 2s¹ , 2 pᵪ¹

3- by overlapping s orbital with p orbital we get two

hybrid orbitals of the type : sp , each of them with single

electron , giving a linear shape.

Molecular Orbitals


Chapter Eight

44

4- each sp orbital combines with 1s orbital of hydrogen

atom forming a σ bond.

Molecular Orbitals


Chapter Eight

45

Example:

Describe the molecular structure of boron hydride BH₃

Solution:


the molecule:

H : 1S¹

B: 1S² , 2S² , 2pᵪ¹


the central atom is not enough to combine with three

Molecular Orbitals


Chapter Eight

46

Hydrogen atoms, so we excite the central atom , and the


B*: 1s² , 2s¹ , 2 pᵪ¹ , 2py¹

3- by overlapping s orbital with two p orbitals we get

three hybrid orbitals of the type : sp² , each of them with

single electron , giving a triangular structure in plane.

4- each sp² orbital combines with 1s orbital of hydrogen


Molecular Orbitals


Chapter Eight

47

Molecular Orbitals


Chapter Eight

48

Example:

Describe the molecular structure of methane CH₄

Solution:


the molecule:

H : 1S¹

C: 1S² , 2S² , 2pᵪ¹ , 2py¹


the central atom is not enough to combine with four

Molecular Orbitals


Chapter Eight

49

Hydrogen atoms, so we excite the central atom , and the


C*: 1s² , 2s¹ , 2 pᵪ¹ , 2py¹ , 2pz¹

3- by overlapping s orbital with the three p orbitals we

get four hybrid orbitals of the type : spᶟ , each of them

with single electron , giving a tetrahedral structure

(Pyramid).

4- each spᶟ orbital combines with 1s orbital of hydrogen


Molecular Orbitals


Chapter Eight

50

Molecular Orbitals


Chapter Eight

51

• Diamagnetism is the weak repulsion associated

with paired electrons.

• Paramagnetism is the attraction associated with

unpaired electrons.

– This produces a much stronger effect than the

weak diamagnetism of paired electrons.

• Ferromagnetism is the exceptionally strong

attractions of a magnetic field for iron and a few

other substances.

Magnetic Properties


Chapter Eight

52

• The magnetic properties of a substance can be determined

by weighing the substance in the absence and in the

presence of a magnetic field.

Magnetic Properties (cont’d)

The mass appears to have increased, so this

substance must be _______increased_____

and must have (unpaired) electrons.


Chapter Eight

53

Example 8.4

A sample of chlorine gas is found to be diamagnetic. Can

this gaseous sample be composed of individual Cl atoms?


Chapter Eight

54

Certain physical and chemical properties recur at regular intervals, and/or vary in regular fashion, when the elements are arranged according to increasing atomic number.

Melting point, boiling point, hardness, density, physical state, and chemical reactivity are periodic properties.

We will examine several periodic properties that are readily explained using electron configurations.

Periodic Properties


Chapter Eight

55

Half the distance between the

nuclei of two atoms is the

atomic radius.

Covalent radius: half the

distance between the nuclei

of two identical atoms joined

in a molecule.

Metallic radius: half the

distance between the nuclei

of adjacent atoms in a solid

metal.

Periodic Properties: Atomic Radius


Chapter Eight

56

• Atomic radius increases from top to bottom

within a group.

• The value of n increases, moving down the

periodic table.

• The value of n relates to the distance of an

electron from the nucleus.



Chapter Eight

57

• Atomic radius decreases from left to right within a period.

• Why? The effective nuclear charge increases from left to

right, increasing the attraction of the nucleus for the valence

electrons, and making the atom smaller.


Mg has a greater

effective nuclear

charge than Na, and

is smaller than Na.


Chapter Eight

58

Atomic Radii of the Elements


Chapter Eight

59

Example 8.5

With reference only to a periodic table, arrange each set of

elements in order of increasing atomic radius:

(a) Mg, S, Si (b) As, N, P (c) As, Sb, Se

A-

(a) S , Si , Mg

(b) N , P , As

(c) Se , As , Sb


Chapter Eight

60

The ionic radius of

each ion is the

portion of the

distance between

the nuclei occupied

by that ion.

Ionic Radii


Chapter Eight

61

• Cations are smaller than the atoms from which

they are formed; the value of n usually decreases.

Also, there is less electron–electron repulsion.

Ionic Radii


Chapter Eight

62

• Anions are larger than the

atoms from which they are

formed.

• Effective nuclear charge is

unchanged, but additional

electron(s) increase electron–

electron repulsion.

• Isoelectronic species have the

same electron configuration;

size decreases with effective

nuclear charge.

Ionic Radii


Chapter Eight

63

Some

Atomic

and

Ionic

Radii


Chapter Eight

64

Example 8.6

Refer to a periodic table but not to Figure 8.14, and

arrange the following species in the expected order of

increasing radius:

Ca2+, Fe3+, K+, S2–, Se2–

A-

S2–, Fe3+, Ca2+, K+, Se2–


Chapter Eight

65

• Ionization energy (I) is the energy required to

remove an electron from a ground-state gaseous

atom.

• I is usually expressed in kJ per mole of atoms.

M(g) M+(g) + e– ΔH = I1

M+(g) M2+(g) + e– ΔH = I2

M2+(g) M3+(g) + e– ΔH = I3

Removing one elcI is usually expressed in kJ per

mole of atoms

Ionization Energy


Chapter Eight

66

• I1 < I2 < I3

– Removing an electron from a positive ion is more difficult than removing it from a neutral atom.

• A large jump in I occurs after valence electrons are completely removed (why?).

• I1 decreases from top to bottom on the periodic table.

– n increases; valence electron is farther from nucleus.

• I1 generally increases from left to right, with exceptions.

– Greater effective nuclear charge from left to right holds electrons more tightly.

Ionization Energy Trends


Chapter Eight

67

Selected

Ionization

Energies

Compare I2 to I1 for a 2A

element, then for the

corresponding 1A element.

Why is I2 for each 1A element

so much greater than I1?

Why don’t we see the same trend

for each 2A element? I2 > I1 … but

only about twice as great …


Chapter Eight

68

Selected Ionization Energies

General trend in I1: An increase

from left to right, but …

…I1 drops, moving

from 2A to 3A.

The electron being

removed is now a p electron

(higher energy, easier to

remove than an s).

I1 drops again

between 5A and 6A.

Repulsion of the

paired electron in 6A

makes that electron

easier to remove.


Chapter Eight

69

First Ionization Energies

Change in trend

occurs at 2A-3A

and at 5A-6A for

each period …

… but the change

becomes smaller at

higher energy levels.


Chapter Eight

70

Example 8.7

Without reference to Figure 8.15, arrange each set of

elements in the expected order of increasing first

ionization energy.

(a) Mg, S, Si (b) As, N, P (c) As, Ge, P


Chapter Eight

71

Electron affinity (EA) is the energy change that occurs

when an electron is added to a gaseous atom:

M(g) + e– M–(g) ΔH = EA1

• A negative electron affinity means that the process is

exothermic.

• Nonmetals generally have more affinity for electrons than

metals do. (Nonmetals like to form anions!)

• Electron affinity generally is more negative or less positive

on the right and toward the top of the periodic table.

Electron Affinity


Chapter Eight

7272

Properties from the Periodic Table

Is the tendency of the atom in a compound towards

the electrons.

The electro negativity increase when we move in

the periodic table from left to right and from

bottom to the top.

The highest figures are at the upper left side corner

of the periodic table.

EOS

The Electro negativity:


Chapter Eight

73

Selected Electron Affinities

The halogens have a

greater affinity for

electrons than do the alkali

metals, as expected.


Chapter Eight

74

Example 8.8 A Conceptual Example

Which of the values given is a reasonable estimate of the

second electron affinity (EA2) for sulfur?

S–(g) + e– S2–(g) EA2 = ?

–200 kJ/mol +450 kJ/mol

+800 kJ/mol +1200 kJ/mol


Chapter Eight

75

A Summary of Trends


Chapter Eight

76

Example 8.9

In each set, indicate which is the more metallic element.

(a) Ba, Ca (b) Sb, Sn (c) Ge, S

Example 8.10 A Conceptual Example

Using only a blank periodic table such as the one in Figure

8.17, state the atomic number of (a) the element that has

the electron configuration 4s2 4p6 4d5 5s1 for its fourth and

fifth principal shells and (b) the most metallic of the fifth-

period p-block elements.


Chapter Eight

77

• The noble gases are on the far right of the periodic table

between the highly active nonmetals of Group 7A and the

very reactive alkali metals.

• The noble gases rarely enter into chemical reactions

because of their stable electron configurations.

• However, a few compounds of noble gases (except for He

and Ne) have been made.

The Noble Gases


Chapter Eight

78

Atoms emit energy when

electrons drop from higher to

lower energy states (Ch.7).

Elements with low first

ionization energies can be

excited in a Bunsen burner

flame, and often emit in the

visible region of the

spectrum.

Elements with high values

of IE1 usually require

higher temperatures for

emission, and the emitted

light is in the UV region of

the spectrum.

BaSrCa

KNaLi

Flame Colors


Chapter Eight

79

• An acidic oxide produces an acid when the oxide reacts with water.

• Acidic oxides are molecular substances and are generally the oxides of nonmetals.

• Basic oxides produce bases by reacting with water.

• Often, basic oxides are metal oxides.

• An amphoteric oxide can react with either an acid or a base.

Acidic, Basic, and Amphoteric

Oxides


Chapter Eight

80

Properties of the Oxides of the

Main-group Elements

The metalloids and some

of the heavier metals

form amphoteric oxides.


Chapter Eight

81

Cumulative Example

Given that the density of solid sodium is 0.968 g/cm3,

estimate the atomic (metallic) radius of a Na atom. Assess

the value obtained, indicating why the result is only an

estimate and whether the actual radius should be larger or

smaller than the estimate.

Solution:

d= mass/volume V=mass/d

Volume of one gram of sodium atoms = 1/0.968 cm3

Volume of 23 g (1 mol ) of Na = 23 x 1.03305 cm3Volume

of Na atom =

23.76015 / 6.022 x 10²³ = 3.94 x 10⁻²³ cm3


Chapter Eight

82

Cumulative Example cont.

-Volume of Na atom =

23.76015 / 6.022 x 10²³ = 3.94 x 10⁻²³ cm3

-By remembering : V = 4/3 r³

-We can calculate the radius of Na atom:

r = 2.11 x 10⁻⁸ or

r = 211 pm

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General Chemistry Students of the Faculty of Pharmacy Prof