Atomic Structure Note

2.1 Bohr’s atomic model

2.2 Quantum mechanical model

2.3 Electronic configuration

1

At the end of this topic students should be able to:-

a) Describe the Bohr’s atomic model.

b) Explain the existence of electron energy levels in an atom.

c) Calculate the energy of electron using:

d) Describe the formation of line spectrum of hydrogen atom.

e) Calculate the energy change of an electron during transition.

f) Calculate the photon of energy emitted by an electron that produces a particular wavelength during transition.

2

2n

1

3

g)perform calculations involving the rydberg

equation for lyman, balmer, paschen, brackett

and pfund series.

h)calculate the ionisation energy of hydrogen

atom from lyman series.

i) state the weaknesses of bohr's atomic model.

j) state the dual nature of electron using de

broglie's postulate and heisenberg's uncertainty

principle

In 1913, a young Dutch physicist, Niels Böhr proposed a theory of atom that shock the scientific world.

The atomic model he described had electrons circling a central nucleus that contains positively charged protons.

4

Böhr also proposed that these orbits can only

occur at specifically “permitted” levels only

according to the energy levels of the electron

and explain successfully the lines in the

hydrogen spectrum.

BOHR’S ATOMIC MODELS

5

1. Electron moves in circular orbits about the nucleus. In

moving in the orbit, the electron does not radiate any

energy and does not absorb any energy.

Postulates

H

Nucleus

(proton) H 1

1

BOHR’S ATOMIC MODELS

ii) The energy of an electron in a hydrogen atom is quantised, that is, the electron has only a fixed set of allowed orbits, called stationary states. ( e can only exist on specific orbit,not between orbit)

6

n=1

n=2

n=3

H Nucleus

(proton)

[ orbit = stationary state = energy level = shell ]

BOHR’S ATOMIC MODELS

Postulates

7

3. At ordinary conditions the electron is at the ground state

(lowest level). If energy is supplied, electron absorbed

the energy and is promoted from a lower energy level to

a higher ones. (Electron is excited)

4. Electron at its excited states is unstable. It will fall back

to lower energy level and released a specific amount of

energy in the form of light. The energy of the photon

equals to the energy difference between the levels.

BOHR’S ATOMIC MODELS

Postulates

Ground state

the state in which the electrons have their lowest energy

Excited state

the state in which the electrons have shifted from a lower

energy level to a higher energy level

Energy level

energy associated with a specific orbit or state

Points to Remember

8

• The energy of an electron in its level is given by:

RH (Rydberg constant) or A = 2.18 10-18J.

n (principal quantum number) = 1, 2, 3 …. (integer)

Note:

• n identifies the orbit of electron

• Energy is zero if electron is located infinitely far from nucleus

• Energy associated with forces of attraction are taken to be negative (thus, negative sign)

2Hnn

1RE

9

THE BOHR ATOM

• Radiant energy emitted when the electron

moves from higher-energy state to lower-

energy state is given by the difference in

energy between energy levels:

2

i

Hin

1RE

10

THE BOHR ATOM

E = Ef - Ei

2

f

Hfn

1RE2

i

H2

f

Hn

1R

n

1RE

2

f

2

i

Hn

1

n

1RE

where

Thus, ni = initial orbit

nf = final orbit

• The amount of energy released by the electron is called a photon of energy.

• A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength.

where; h (Planck’s constant) =6.63 10-34 J s = frequency

c

11

THE BOHR ATOM

E = h

Where; c (speed of light) = 3.00 108 ms-1

Thus, hcΔE

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

Electron is excited from lower to higher

energy level. A specific amount of energy

is absorbed

E = h = E1-E3 (+ve)

Electron falls from higher to lower energy level .

A photon of energy is released.

E = h = E3-E1 (-ve)

12

Energy level diagram for the hydrogen atom

Pote

ntial en

erg

y

n = 1

n = 2

n = 3

n = 4

n =

Energy

released

Energy

absorbed

13

14

14

Example

Calculate:

i) The E of an e- has when it occupies

when it was at n=3 & n=4.

ii) The E of the photon emitted when

one mole e- of drops from the fourth E

level to third E level.

iii) The frequency & wavelength of this

photon.

15

15

Solution :

2Hnn

1RE

i) at n=3, E3 = -RH 1 = - 2.18 x 10-18 J

32 9

= -2.422 x 10-19 J

at n=4, E4 = -RH 1 = - 2.18 x 10-18 J

42 16

= -1.363 x 10-19J

16

n = 4

n = 3

ΔE = Ef – Ei = E3 – E4

= -2.422x 10-19J- (-1.363 x 10-19J)

= -1.06 x 10-19J

-ve sign indicates that E is released

when e- falls.

ii)

E released by 1 mol of e-,

ΔE = -1.06 x 10-19J x 6.023 x 1023 mol-1

= - 63 843.8 Jmol-1 = -63.8438 kJmol-1

E

16

17

iii) Frequency, v = ΔE = 1.06 x 10-19J

h 6.63 x 10-34 Js

= 1.599 x 1014 s-1

= 1.599 x 1014Hz

# -ve sign of ΔE is ignored because frequency is

always +ve

Wavelength, λ = c = 3.0 x 108 ms-1

v 1.599 x 1014 s-1

= 1.876 x 10-6 m

= 1876 nm

# 1 m = 109 nm

Exercises: 1) Calculate the energy of an electron in the second

energy level of a hydrogen atom. (-5.448 x 10-19 J) 2) Calculate the energy of an electron in the energy

level n = 6 of a hydrogen atom. 3) Calculate the energy change (J), that occurs when

an electron falls from n = 5 to n = 3 energy level in a hydrogen atom.

(answer: 1.55 x 10-19J) 4) Calculate the frequency and wavelength (nm) of

the radiation emitted in question 3.

18

19

Emission Spectra

Continuous

Spectra

Line

Spectra

Continuous Spectrum A spectrum consists all wavelength components

(containing an unbroken sequence of frequencies) of the visible portion of the electromagnetic spectrum are present.

It is produced by incandescent solids, liquids, and compressed gases.

20

Regions of the Electromagnetic Spectrum

21

• When white light from incandescent lamp is passed through a slit then a prism, it separates into a spectrum.

• The white light spread out into a rainbow of colours produces a continuous spectrum.

• The spectrum is continuous in that all wavelengths are present and each colour merges into the next without a break.

22

FORMATION OF CONTINUOUS SPECTRUM

23

Line Spectrum (atomic spectrum)

A spectrum consists of discontinuous & discrete lines

produced by excited atoms and ions as the electrons

fall back to a lower energy level. The radiation

emitted is only at a specific wavelength or frequency.

It means each line corresponds to a specific

wavelength or frequency.

•Line spectrum are composed of only a few

wavelengths giving a series of discrete line

separated by blank areas

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prism film

The emitted light (photons) is then separated into its components by

a prism. Each component is focused at a definite position, according

to its wavelength and forms as an image on the photographic plate.

The images are called spectral lines.

FORMATION OF ATOMIC / LINE SPECTRUM

25

26

Line Emission Spectrum of Hydrogen Atoms

26

FORMATION OF ATOMIC / LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5 n =

En

erg

y

When an electrical discharge is passed through a sample of

hydrogen gas at low pressure, hydrogen molecules decompose to

form hydrogen atoms.

Radiant energy (a

quantum of energy)

absorbed by the atom (or

electron) causes the

electron to move from a

lower-energy state to a

higher-energy state.

Hydrogen atom is said to

be in an excited state

(very unstable).

27

FORMATION OF ATOMIC / LINE SPECTRUM

Emission of photon

n = 2

n = 3

n = 4

n = 5

n = 6 n =

En

erg

y When the electrons fall

back to lower energy

levels, radiant energies

(photons) are emitted in

the form of light

(electromagnetic radiation

of a particular frequency or

wavelength)

28

FORMATION OF ATOMIC / LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5 n =

Lyman Series

Emission of photon

Line

spectrum E

Energy

29

FORMATION OF ATOMIC / LINE SPECTRUM

n = 1

n = 2

n = 3

n = 4

n = 5 n =

Lyman Series

Emission of photon

Line

spectrum

Balmer Series

E

Energy

30

Emission series of hydrogen atom

n = 1

n = 2

n = 3

n = 4

n =

222

AE

23

AE3 24

4

AE

211

AE

31

Lyman series

Balmer series

Brackett series

Paschen series

Pfund series

Exercise: Complete the following table

Series n1 n2 Spectru

m region

Lyman 2,3,4,…

2 3,4,5,…

Paschen 4,5,6,… Infrared

4 5,6,7,… Infrared

5 6,7,8,… Infrared 32

ultraviolet

Visible Balmer

Brackett

Pfund

1

3

33

33

The following diagram shows the line spectrum of hydrogen atom. Line A is the first line of the Lyman series.

Example 1

Specify the increasing order of the radiant energy,

frequency and wavelength of the emitted photon.

Which of the line that corresponds to :

i) The shortest wavelength?

ii) The lowest frequency?

Draw the energy level diagram for corresponding line

spectrum above.

Line

spectrum E

A B C D E

Line E

Line A

34

A B C D E n = 1

n = 2

n = 3

n = 4

n = 5

n = 6

E

iii)

34

35

35

Describe the transitions of electrons that

lead to the lines W, and Y, respectively.

Example 2

Solution

Line

spectrum

W Y

Balmer series

For W: transition of electron is from n=4 to n=2

For Y: electron shifts from n=7 to n=2

Exercise The following diagram depicts the line spectrum of hydrogen atom. Line A is the first line of the Lyman series.

36

Line

spectrum E

Specify the increasing order of the radiant energy,

frequency and wavelength of the emitted photon.

Which of the line that corresponds to

i) the shortest wavelength?

ii) the lowest frequency?

A B C D E

Line E

Line A

Homework

Calculate En for n = 1, 2, 3, and 4. Draw a

diagram showing how the energy, at

different values of n, increases vertically

and indicate by vertical arrows the electron

transitions that lead to lines in:

a) Lyman series

b) Paschen series

37

Significance of Atomic Spectra

• In Lyman series, the frequency of the

convergence of spectral lines can be used to

find the ionisation energy of hydrogen atom:

IE = h

• The frequency of the first line of the Lyman

series > the frequency of the first line of the

Balmer series.

38 Lyman Series

Line

spectrum E

Balmer Series

Line

spectrum

A B C D E

Exercise

39

Paschen series

Solution

Which of the line in the Paschen series corresponds to the

longest wavelength of photon?

Describe the transition that gives rise to the line.

Line A.

The electron moves from n=4 to n=3.

40

40

EXERCISES

Calculate En for n = 1, 2, 3, and 4.

Draw an energy level diagram and line

spectrum for the transition of electron that

lead to the formation of 4 lines in :

a) Pfund series

b) Paschen series

• Wavelength emitted by the transition of

electron between two energy levels is

calculated using Rydberg equation:

41

Rydberg Equation

RH = 1.097 107 m-1

= wavelength

Since should have a positive value thus n1 < n2

where

2

2

2

1

H n

1

n

1 R

λ

1

Exercise

Calculate

a) the wavelength in nm

b) the frequency

c) the energy

that associated with the second line in the Balmer series of the hydrogen spectrum.

42

Solution (a)

Second line of

Balmer series:

the transition of

electron is from

n2=4 to n1=2

= RH

1 1

n12

n22

1

= (1.097x107 m 1)

1 1

22

42

1

= x 1

109 m

nm

= 486 nm

4.86x10 7 m

43

43

Frequency, v = c

= 3.0 x 10 8 ms-1

4.86 x 10-7 m

= 6.173 x 1014 s-1

Energy, ΔE = hv

= 6.63 x 10 -34Js x 6.173 x 1014s-1

= 4.093 x 10 -19J

Solution (c)

Solution (b)

Example

Calculate the wavelength, in nanometers of

the spectrum of hydrogen corresponding to

ni = 2 and nf = 4 in the Rydberg equation.

44

Solution:

Rydberg equation:

1/λ = RH (1/ni2 – 1/nf2)

ni = 2 nf = 4

RH = 1.097 x 107m

1/λ = RH (1/22 – 1/42)

= RH(1/4-1/16)

λ = 4.86m x 102 m

= 486nm

Example

Use the Rydberg equation to calculate the

wavelength of the spectral line of hydrogen atom

that would result when an electron drops from the

fourth orbit to the second orbit. Name the series of

line formed.

45

Solution:

1/λ = RH (1/n1 2 – 1/n2 2)

n1 = 2 n2 = 4

1/λ = 1.097 x 107 (1/22 – 1/42)

λ = 4.86 x 10-7 m

= 486 nm

*e dropped to the second orbit (n=2),

>>> Balmer series

EXAMPLE 3

Calculate the wavelengths of the fourth line in

the Balmer series of hydrogen.

46

n1 = 2 n2 = 6

RH = 1.097 x 107m-1

λ = 4.10 x 10-7 m

RH 22 62

1 1 1 =

λ

Different values of RH and its usage

1. RH = 1.097 107 m-1

2

f

2

i

Hn

1

n

1RE

47

RH n2

1 n2

2

1 1 1 =

λ

RH = 2.18 x 10-18 J

n1 < n2

ni – initial orbit

nf - final orbit

EXAMPLE 4

Calculate the energy liberated when an electron from the fifth energy level falls to the second energy level in the hydrogen atom.

48

RH n2

1 n2

2

1 1 1 =

λ

52 1.097 x 107

22

1 1 1 =

λ

1

λ = 0.2303 X 107 m-1

ΔE = 4.58 x 10-19 J

hcΔE

49

ΔE = (6.63 10-34Js)X(3.00 108 ms-1) X (0.2303 X 107 m-1)

For Lyman series n1=1 and n2 = ∞ calculate;

i ) Wavelength

ii ) Frequency

iii ) Wave number of the last line of hydrogen spectrum in Lyman series

Wave number = 1/wavelength

50

EXERCISE 5

Ans: i) 9.116 x 10-8 m ii) 3.29 x 1015 s-1 iii) 1.0970 x 107 m-1

The weakness of Bohr’s Theory

• Bohr was successful in introducing the idea of

quantum energy and in explaining the lines of

hydrogen spectrum.

• His theory could not be extended to predict the

energy levels and spectra of atoms and ions

with more than one electron.

• His theory can only explain the hydrogen

spectrum or ions contain one electron eg He+,

Li2+.

• Modern quantum mechanics retain Bohr’s

concept of discrete energy states and energy

involved during transition of electrons but totally

reject the circular orbits he introduced. 51

At the end of this topic students

should be able to:-

1) Define the term orbital.

2) State all the four quantum numbers of

an electron in an orbital.

3) Sketch the shape of s, p and d orbitals

with the correct orientations.

52

2n

1

Atomic Orbital

An orbital is a three-dimensional region in space around the nucleus where there is a high probability of finding an electron.

53

Definition

ATOMIC ORBITAL

•Orbital - region in space around the

nucleus where there is a high

probability of finding an electron.

•Where electron can be expected to be

found.

•Orbit – the path of an e- as it travels

round the nucleus of an atom.

54

QUANTUM NUMBERS

• Quanta - discrete amounts of E that an e-

absorbs as it moves up an E level or releases

when it moves down to lower E level.

• The positions and orbits of e- referred as E state

and described by 4 quantum numbers :

i. Principal quantum number (n)

ii. Angular momentum/Azimuthal quantum

number (l)

iii. Magnetic quantum number (m)

iv. Spin quantum number (s)

55

PRINCIPAL QUANTUM NUMBER, n

- Determines the energy & size of an orbital.

- n is large – size of atom is larger & distance e- from nucleus is greater.

- The principal quantum number may have only integral values: n =1, 2, 3, …,

.

56

n 1 2 3 4

shell K L M N

Orbital size

Energy increases

PRINCIPAL QUANTUM NUMBER, n

57

ANGULAR MOMENTUM QUANTUM NUMBER, l

58

Alternative name: Subsidiary / Azimuthal / Orbital

Quantum Number

- Indicates the shape of the atomic orbital, the

types of orbitals, and the angular momentum

of the electron.

- The allowed values of l are 0, 1, 2,…, (n 1).

- Depends on value of n.

ANGULAR MOMENTUM QUANTUM NUMBER, l

59

Symbol Orbital shape

0 s spherical

1 p dumbbell

2 d cloverleaf

3 f ~

Numerical

value of l

Example

60

Shell, n Sub-shell, l Called as

2

4

0

1

2s

2p

0

1

2

3

4s

4p

4d

4f

MAGNETIC QUANTUM NUMBER, m

•Determines the orientation of orbital in

space.

•Permitted value for m depend on the value of l.

•It has integer value ranging from –l to +l .

•Indicates the maximum number of orbitals for a particular value of l.

•Example :

l = 1, m = -1, 0, 1

61

ELECTRON SPIN QUANTUM NUMBER, s

•Determines the

direction of spinning

motions of an

electron on its own

axis.

•Clockwise or counter

clockwise.

•The electron spin

quantum number

has a value of : 62

+ 1

2 -

1

2 or

The relationship between the values of n, , and m

n

(<n)

Orbital

notati

on

m

(- m +)

No. of

degenerat

ed orbitals

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 2,1,0,-1,-2 5 63

Example

64

n l shap

e

m s No. of

orbital

No. of

e-

1

2

3

4

1s 0

0

1

0

1

2

0

1

2

3

0

2s

3s

2p

3p

3d

4s

4p

4d

4f

0

0

0

-1, 0, 1

-1, 0, 1

-1, 0, 1

-2, -1, 0, 1, 2

-2, -1, 0, 1, 2

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

2, 3

1/2

1/2

1/2

1/2

1/2

1/2

1/2

1/2

1/2

1/2

1 2

1

3 8

18

32

1

3

3

1

5

7

5

Exercise

State whether or not each of the

following symbols is an acceptable

designation for an atomic orbital.

Explain what is wrong with the

unacceptable symbols.

65

b) 6g

a) 2d

c) 7s

d) 5i

n=2 l=2

(l < n)

n=6

n=7

n=5

l=4

l=0

l=6

unacceptable

acceptable

unacceptable

acceptable

SHAPE OF ATOMIC ORBITALS

a) s orbitals

When l = 0

Spherical shape.

As n increases, s orbital gets larger

66

Shape of s orbital with different n

SHAPE OF ATOMIC ORBITALS

b) p orbitals

When l = 1

Dumbbell shaped

Consists of 3 p orbital, each with the

same size, shape and energy; they differ

from one another only in orientation - px,

py, and pz.

Correspond m of -1, 0, and +1.

67

SHAPE OF ATOMIC ORBITALS

68

c) d - orbitals

When l = 2, m = -2, -1, 0, 1, 2

The orbitals are: dyz, dxz, dxy, dx2-y2, dz2

SHAPE OF ATOMIC ORBITALS

69

70

2.3 Electronic Configuration

At the end of this topic students should

be able to:-

a. State Aufbou principle, Hund’s rule and Pauli ‘s exclusion principle.

b. Apply the rules in (a) to fill electrons into atomic orbital.

c. Write the electronic configuration of atoms and monoatomic ions using spdf notation.

d) Explain the anomalous electronic configurations of chromium and copper.

71

2n

1

Representing Electronic Configuration

Method 1: Orbital diagram

72

O: 8

1s 2s 2p

Method 2: spdf notation

O: 8 1s 2s 2p 2 2 4

platform

Concentric circle

box

Rules for Assigning Electrons to Orbitals

73

ii) Aufbau Principle

Electrons fill the lowest energy orbitals first and other

orbitals in order of ascending energy.

The order of filling orbitals is:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s

1s 2s 2p 1s

2s

3s

4s

5s

2p

3p

4p

5p

3d

4d

5d

4f

5f

Relative Energy Level of Atomic Orbitals

74

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d

en

erg

y

n=1

n=2

n=3

n=4

1s

2s 2p

3s

4s

3p

4p

3d

4d 5s

Orbital energy levels

in the H atom

Orbital energy levels

in a many-electron atom

ii) Pauli Exclusion Principle

75

Rules for Assigning Electrons to Orbitals

No two electrons in an atom can have the same four

quantum numbers (n, , m, s)

1s

a b c

e(a)

e(b)

e(c)

n m s

1 0

0 1

0

0 1

0

0

1 2

1 2

1 2

( ) , , ,

) ( , , ,

+

+

iii) Hund’s Rule

76

Rules for Assigning Electrons to Orbitals

Only when all the degenerate orbitals (a group of

orbitals of identical energy e.g. three p-orbitals and five d-

orbitals) contain an electron do the electrons begin to

occupy these orbitals in pairs. The electrons in half-filled

orbitals have the same spins, that is, parallel spins.

2p

Indicate which of the following orbital diagrams are

acceptable or unacceptable for an atom in ground state.

Explain what mistakes have been made in each and draw

the correct orbital diagram:

77

Exercise

1s 2s 2p

1s 2s 2p

1s 2s 2p

?

?

1s 2s 2p

?

1s 2s 2p

?

1s 2s 2p

?

Draw ‘electrons-in-boxes’ diagram of the electronic

configuration of titanium, Ti (Z = 22). Also, write the ground-

state electronic configurations for Ti and Ti2+ ion.

78

Exercise

1s 2s 2p

Ti:

3s 3p 4s

3d

1s 2 2s 2 2p 6 Ti: 3s 2 3p 6 3d 2 4s 2

1s 2 2s 2 2p 6 Ti2+: 3s 2 3p 6 3d 2

Points to remember

• The electronic configuration of atom or monatomic ion at

ground state

Distribution of electrons obeys Aufbau principle, Pauli

exclusion principle and Hund’s rule

• Each atomic orbital can only accommodate a maximum of 2

electrons

• Atomic orbital is a 3-D region in space around the nucleus

where there is a high probability of finding an electron.

• Assigning electrons to subshells

s-orbital a max of 2 electrons (ns2)

p-orbitals a max of 6 electrons (np6)

d-orbitals a max of 10 electrons (nd10) 79

The Anomalous Electronic Configurations of

Cr and Cu

• Cr and Cu have electron configurations which are

inconsistent with the Aufbau principle. The anomalous

are explained on the basis that a completely filled or half-

filled orbital is more stable.

Element Expected Observed/actual

Cr (Z=24) [Ar] 3d4 4s2 [Ar] 3d5 4s1

Cu (Z=29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

80

For Chromium : 24 e-

Expected :

24Cr = 1s2 2s2 2p6 3s2 3p6 4s2 3d4

Orbital diagram:

24Cr = [ Ar ]

Observed :

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

Orbital diagram:

24Cr = [ Ar ]

81

3d 4s

3d 4s

For Copper : 29 e-

Expected :

29Cu = 1s2 2s2 2p6 3s2 3p6 4s2 3d9

Orbital diagram:

29Cu = [ Ar ]

Observed :

29Cu = 1s2 2s2 2p6 3s2 3p6 4s1 3d10

Orbital diagram:

29Cu = [ Ar ]

82

3d 4s

3d 4s

• Orbitals that are half–filled (s1, p3 or d5)

or completely filled (s2, p6 or d10) have

extra stability due to the equal

symmetrical distribution of charge

around an atom.

83

84

z = 21

z = 30

1. How many 2p orbitals are there in an atom?

2. How many electrons can be placed in the 3d sub-shell?

Thinking question

85

3.What is the electron configuration of Mg?

4.What are the possible quantum numbers for the last

(outermost) electron in Cl?

86

THE END

87