4.1 Atomic Spectra.

� Spectrum ~ display of component of a beam of radiation.

� When white ray passes through a prism it forms rainbow colour.

� A continuous spectrum is spectrum composed of visible light of all

wavelength while a discontinuous spectrum which the line

spectrum represent the atomic emissions

� Productions of line spectrum depend on what types of element

used and the visible light emitted also depend on it.

� Spectrophotometer is the instrument that is used to split and

dispersed into individual wavelength.

� The study of line spectra can be used to

� Identify an element

� Determine the ionisation energy of an element

� Determine the arrangement of electrons in an atom of an

element.

The main components of a typical spectrophotometer

Monochromator

(wavelength selector)

disperses incoming

radiation into continuum of

component wavelengths

that are scanned or

individually selected.

Sample in compartment

absorbs characteristic

amount of each incoming

wavelength.

Computer

converts

signal into

displayed

data.

Source produces radiation in

region of interest. Must be

stable and reproducible. In

most cases, the source emits

many wavelengths.

Lenses/slits/collimaters

narrow and align beam.

Detector converts

transmitted radiation

into amplified electrical

signal.

7.3

4.2 The Atomic Emission Spectrum of Hydrogen

� In hydrogen, the atomic emission spectrum of hydrogen shows a

few distinct series of line. These are the

� Lyman series – in …………………………… region

� Balmer series – in …………………………. region.

� Paschen series – in ………………………… region

� The diagram below shows the example of the line emission

produced through the spectrophotometer.

ultraviolet

visible

infrared

4.2.1 Formation of atomic spectra.

� The absorption spectrum of a substance is a continuous spectrum of

radiation produced by examining the substance through

spectrophotometer.

� Hence, there is a absorption and emission process occur in the

formation of spectra where

Absorption spectroscopy Emission spectroscopy

↑ Occur when electron which is occupying at

lower energy level (ground level / state)

absorb light at a certain frequency.

↑ When enough energy is absorbed, electron is

able to ‘excite’ to a higher energy level.

↓ Occur when electron which is located at a high

energy level (after absorbed energy) released

the energy in the form of radiation

↓ When energy is released, electron falls to the

original ground state, emitting the radiation

and hence give the spectrum

n = 3

n = 2

n = 1

n = 3

n = 2

n = 1

� The following characteristics of any series should be noted :

� The spectrum consists of discrete lines, each having its own

discrete frequency.

� The wavelengths represented by broken lines are called spectrum

(plural : spectra) Each spectrum represents the wavelength (or

frequency) in a series. The frequency emitted can be related to

the spectral line by the Rydberg’s Equation

� The value of n1 and n2 for the atomic spectrum of hydrogen are

given in Table below

λ = wavelength (in m)

n1 = series of the spectra line

n2 = line where the emission of energy begin.

RH = Rydberg constant = 1.097 x 107 m-1

−=

2

2

2

1

111

nnRHλ

Series n1 n2

Lyman 1 2, 3, 4 …….

Balmer 2 3, 4, 5 …….

Paschen 3 4, 5, 6 …….

Example 1 : Calculate the wavelength of the first

line of Lyman series

Example 2 : Calculate the wavelength of the third

line of Balmer series

Example 3 : Calculate the wavelength of the forth

line of Paschen series

Example 4 : Calculate the wavelength of the last

line of Balmer series

In Lyman series, n1 = 1

First line in series, so n2 = 1 + 1 = 2

−=

2

2

2

1

111

nnRHλ

−×=

λ 22

7

2

1

1

110097.1

1

λ = 122 nm

In Balmer series, n1 = 2

First line in series, so n2 = 2 + 3 = 5

−=

2

2

2

1

111

nnRHλ

−×=

λ 22

7

5

1

2

110097.1

1

λ = 434 nm

In Passchen series, n1 = 3

First line in series, so n2 = 3 + 4 = 7

−=

2

2

2

1

111

nnRHλ

−×=

λ 22

7

7

1

3

110097.1

1

λ = 1005 nm

In Balmer series, n1 = 2

Last line in series, so n2 = 2 + ∞ = ∞

−=

2

2

2

1

111

nnRHλ

∞−×=

λ 22

7 1

2

110097.1

1

λ = 365 nm

Example 5 : Calculate the frequency of the second

line of Paschen series

Example 6 : Calculate the frequency of the last line

of Lyman series

m

sm100.3f;

)wavelength(

)lightofspeed(c f frequency, Since

18

λ

×=

λ=

−

• In 1900, Max Planck proposed the quantum theory. He mentioned that

radiant energy was emitted in discrete packets or “quantum” Each

quantum of energy emitted, E, is proportional to the frequency, f, of the

radiation

• Where energy, E = Planck constant (h) x frequency (f)

• h = 6.63 x 10-34 J s

)wavelength(

)lightofspeed(c)tstanconPlanck(hE,Energy

λ×=

In Passchen series, n1 = 3

Second line in series, n2 = 3 + 2 = 5

−=

2

2

2

1

111

nnRHλ

−×=

λ 22

7

5

1

3

110097.1

1

1 / λ = 7.80 x 105 m-1

f = c x (1 / λ) = (3.0 x 108)(7.80 x 105)

f = 2.34 x 1014 s-1

In Lyman series, n1 = 1

last line in series, so n2 = 1 + ∞ = ∞

−=

2

2

2

1

111

nnRHλ

∞−×=

λ 22

7 1

1

110097.1

1

1 / λ = 1.097 x 107 m-1

f =c x (1 / λ) =(3.0 x 108)(1.097 x 107)

f = 3.29 x 1015 s-1

Example 7 : Calculate the energy required to

excite one electron to the fourth line in

Passchen series

Example 8 : Calculate the energy required to

excite one electron to the last line in Balmer

series.

In Passchen series, n1 = 3

Fourth line in series, n2 = 3 + 4 = 7

−=

2

2

2

1

111

nnRHλ

−×=

λ 22

7

7

1

3

110097.1

1

1 / λ = 9.95 x 105 m-1

f = c x (1 / λ) = (3.0 x 108)(9.95 x 105)

f = 2.99 x 1014 s-1

E = h f

E = 6.63 x 10-34 x 2.99 x 1014

E = 1.98 x 10-19 J / e

In Balmer series, n1 = 2

Last line in series, n2 = 2 + ∞ = ∞

−=

2

2

2

1

111

nnRHλ

∞−×=

λ 22

7 1

2

110097.1

1

1 / λ = 2.74 x 106 m-1

f = c x (1 / λ) = (3.0 x 108)(2.74 x 106)

f = 8.23 x 1014 s-1

E = h f

E = 6.63 x 10-34 x 8.23 x 1014

E = 5.45 x 10-19 J / e

4.3 Electronic Energy Levels

� Hydrogen emission spectrums consist of several series of discrete

lines in different limited wavelength.

� This implies that hydrogen atoms do not emit light at of all possible

energies but only emit light with certain amount of energy.

� Bohr made these assumptions to explain line in spectrum.

� Electron move in orbit around nucleus (like planet and sun)

� Energy of electron is quantised (only have certain energy). Electron

closer to nucleus has lower energy and vice versa.

� Electron of atom will fill up the lowest energy level first. But when

energy is given to the electron, it moves from a lower level to a

higher level. This process is called excitation.

� If the energy loss from the excited level to original level, electron

will emit electromagnetic radiation. This is what caused the

emission of spectral lines. The convergence of spectral line show

the difference between successive energy level become ↓ with

increasing distance of the energy levels from the nucleus

� Electron are not randomly distributed but are arranged

in a series of shells or orbits which are situated at

various distance from the molecules corresponding to

differ E-level.

� Though Bohr theory proves its unsatisfactory when

come to spectra of more complicated atoms. Thus

Rydberg equation can only used for atoms or ions

contain only 1 electron.

� For atom or ion with many electrons, the analysis of

atomic spectra become more difficult.

Wavelength increase Frequency increase

4.4 Calculate the Ionisation energy of Hydrogen from its

Line Spectrum

� When energy level of the hydrogen spectrum increased, there’s a

possibility where it reaches the convergence limit of the energy

level where n =

� When this occur, electron are not able to return to ………………..,

and eventually the hydrogen atom is …………….., according to the

following equation

� Equation :

� For this to occur, a certain amount of energy is required. The

energy required to remove one electron is called

……………………………… Here, a positive charged ion is formed.

� There are 2 methods of calculating ionisation energy of hydrogen

through the line spectrum

ground level ionised

H (g) � H+ (g) + e-

ionisation energy

Using Rydberg and Planck equations

� Supposed that electron at energy level n1 = 1 absorbed energy and it is

excited. When reaches the convergence limit, n2 = , the wavelength can

be calculated

� Using Rydberg equation n1 = n2 =

� For each electron, the energy absorbed can be determined using the

Planck equation

� Since this much energy is required to excite one electron, so for 1 mole of

electron, the energy required

� Hence, the ionisation energy of hydrogen is …………. kJ mol-1

−=

2

2

2

1

111

nnRHλ

α−=

λ 22

7 1

1

110x097.1

1

∞

1 ∞710x097.1

1=

λ

E = h f E = h c 1/λE = (6.63 x 10-34)(3.0 x 108)(1.097 x 107)

E = 2.182 x 10-18 J / e

Since in 1 mol contain 6.02 x 1023 e, so

Ionisation energy, ∆H = E x NA

= 2.182 x 10-18 J / e x 6.02 x 1023

= 1.313 x 106 J / mol+ 1313

4.5 Atomic Orbital

� Heinsberg uncertainty principle = position & momentum of an

electron cannot be known with great precision but probability of

finding an electron in a certain position can be calculated.

� Orbital - the space within which there’s a ………………… probability of

finding an ……………………. The electron in the orbital is describe as

occupying 3-D space around the nucleus. The nucleus is described as

being surrounded by

Orbital – boundary within which there

is 95% (high) chances to find an

electron

high

electron

Type of

orbitalShape of orbital

Number of

electron

filled

S

P

2

6

Type of

orbitalShape of orbital

Number of

electron

filled

f 7 orbitals

10

14

� Filling of orbitals in shell

� Electrons filled in the shell of an atom through occupation of

orbitals in each shell

� The formula used to calculate the number of orbitals that can

occupy each shell is

No of orbitals in shell = n2 ; (n = the number of shell)

Shell,

n

No. of

orbitalsType of orbitals Orbital in shell

1

2

3

4

n2 = 12 = 1 s 1s

n2 = 22 = 4 s , p 2s 2p

n2 = 32 = 9 s , p , d 3s 3p 3d

n2 = 42 = 16 s , p , d , f 4s 4p 4d 4f

4.6 Electronic configuration

� Electronic configuration of an element is the expression on how

the electrons inside the element filled in its orbitals accordingly

� There are 3 rules for assigning electrons to the orbitals of an

atom in the ground state.

1 . ………………………………………………………………….

2 . ………………………………………………………………………….

3 . ………………………………………………………………………….

1.The Aufbau Principle ~ e- occupy orbitals in the order of the

energy levels of the orbitals. Orbital with the lowest energy are

always occupied first.

� The arrangement of energy level from the lowest to the highest

is shown in the diagram below

Aufbau Principle

Pauli’s Exclusion Principle

Hund’s Rule

� Energy level : 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p

2. The Pauli exclusion principle ~ 2 electron may occupy the same orbital but

these 2 electrons must have opposite spin.

Same spin (cannot occur) Opposite spin

(Pauli exclusion principle)

3. The Hund’s Rule ~ when electrons are placed in a set of orbitals with equal

or degenerated energies, the electrons must occupy them singly with parallel

spins before they occupy the orbital in pair In other words, an atom tend to

have as many unpaired electrons as possible. Example : when filling in 3

electrons in p orbital

ElementNo of

e- Orbital diagramElectronic

configuration

Hydrogen

H____

1s

Helium

He____

1s

Lithium

Li ____ ____

1s 2s

Beryllium

Be ____ ____

1s 2s

Boron

B____ ____ ____ ____ ____

1s 2s 2 p

Carbon

C____ ____ ____ ____ ____

1s 2s 2 p

Nitrogen

N____ ____ ____ ____ ____

1s 2s 2 p

1 1s1

2 1s2

3 1s22s1

4 1s22s2

5 1s22s22p1

6 1s22s22p2

7 1s22s22p3

ElementNo of

e- Orbital diagramElectronic

configuration

Oxygen

O____ ____ ____ ____ ____

1s 2s 2 p

Fluorine

F____ ____ ____ ____ ____

1s 2s 2 p

Neon

Ne____ ____ ____ ____ ____

1s 2s 2 p

Sodium

Na____ ____ ____ ____ ____ ____

1s 2s 2 p 3s

Magnesium

Mg____ ____ ____ ____ ____ ____

1s 2s 2 p 3s

Aluminium

Al____ ____ ____ ____ ____ ____ ___ ___ ___

1s 2s 2 p 3s 3 p

Silicon

Si____ ____ ____ ____ ____ ____ ____ ____ ____

1s 2s 2 p 3s 3 p

8 1s22s22p4

9 1s22s22p5

10 1s22s22p6

11 1s22s22p63s1

12 1s22s22p63s2

13 1s22s22p63s23p1

14 1s22s22p63s23p2

ElementNo of

e- Orbital diagram Electronic configuration

Phospho-

rous, P____ ____ ____ ____ ____ ____ ____ ____ ____

1s 2s 2 p 3s 3 p

Sulphur

S____ ____ ____ ____ ____ ____ ____ ____ ____

1s 2s 2 p 3s 3 p

Chlorine

Cl____ ____ ____ ____ ____ ____ ____ ____ ____

1s 2s 2 p 3s 3 p

Argon

Ar____ ____ ____ ____ ____ ____ ____ ____ ____

1s 2s 2 p 3s 3 p

Potassium

K____ ____ ____ ____ ____ ____ ____ ____ ____ ___

1s 2s 2 p 3s 3 p 4s

Calcium

Ca____ ____ ____ ____ ____ ____ ____ ____ ____ ___

1s 2s 2 p 3s 3 p 4s

15 1s22s22p63s23p3

16 1s22s22p63s23p4

17 1s22s22p63s23p5

18 1s22s22p63s23p6

19 1s22s22p63s23p64s1

20 1s22s22p63s23p64s2

1s22s22p63s23p63d14s2

1s22s22p63s23p63d24s2

1s22s22p63s23p63d34s2

1s22s22p63s23p63d54s1

1s22s22p63s23p63d54s2

1s22s22p63s23p63d64s2

1s22s22p63s23p63d74s2

1s22s22p63s23p63d84s2

1s22s22p63s23p63d104s1

1s22s22p63s23p63d104s2

� According to Hund’s rule : ………………………………………………………..…

………………………………………

� For the case of chromium, it has the valence electronic as

………………………………... instead of …………………………….

� This is due to the extra stability achieved due to the equal /

symmetrical distribution of charge around the atom within the half-filled

(p3 and d5) orbital and full-filled orbital (p6 and d10)

� What happened before and after the electron filling in the 4s orbital can

be explained by the diagram below

Position of 3 d and 4 s orbital before

electron filling in

Position of 3d and 4s orbital after

electron filling in

____ ____ ____ ____ ____

3 d

____

4s

____

4s

____ ____ ____ ____ ____

3 d

Electron filling in orbitals at the same

energy level with single electron before pairing

3d54s1 3d44s2

� When this occurs, the 3rd electron is expected to be filled in 3d orbital

(Aufbau principle). Until the 6th electron is filling in, a degeneration of

orbital occur where now the energy level between 3d and 4s orbital are

the same, thus one of the electron from the 4s “jumped” to 3d orbital.

When this happened, an extra stability occur at the half-filled 3 dorbital has achieved. When the 7th electron is filled, it will be placed in

4s orbital

Position of 3 d and 4 s orbital before

the

6th electron filling in

Position of 3d and 4s orbital after the

6th electron filling in

____

4s

____ ____ ____ ____ ____

3 d

____

4s

____ ____ ____ ____ ____

3 d

� Same thing happened when comes to the case of copper. An

extra stability is achieved when the orbital in 3 d is full-filled

with electrons.

Position of 3 d and 4 s orbital before

the

11th electron filling in

Position of 3d and 4s orbital after the

11th electron filling in

____

4s

____ ____ ____ ____ ____

3 d

____

4s

____ ____ ____ ____ ____

3 d

� 3Li : ……………………………. Li+ : ………………………………

� 5B : …………………………….. B3+ : ……………………………….

� 7N : …………………………….. N3- : ……………………………….

� 9F : ……………………………… F- : ……………………………….

� 12Mg : …………………………. Mg2+ : …………………………….

� 14Si : ………………………….. Si4+ : ……………………………….

� 16S : ……………………………. S2- : ………………………………

� 19K : …………………………………… K+ : …………………………………….

� 21Sc : ………………………………….. Sc2+ : ………………………………….

� 23V : ………….………………………. V2+ : …………………………………..

� 28Ni : ………………………………….. Ni+ : …………………………………..

� 30Zn : …………………………………. Zn2+ : …………………………………..

� 35Br : …………………………………. Br- : ……………………..............

1s22s1 1s2

1s22s22p1 1s2

1s22s22p3 1s22s22p6

1s22s22p5 1s22s22p6

1s22s22p63s2 1s22s22p6

1s22s22p63s23p2 1s22s22p6

1s22s22p63s23p4 1s22s22p63s23p6

1s22s22p63s23p64s1 1s22s22p63s23p6

1s22s22p63s23p63d14s2 1s22s22p63s23p63d1

1s22s22p63s23p63d34s2 1s22s22p63s23p63d3

1s22s22p63s23p63d84s2 1s22s22p63s23p63d84s1

1s22s22p63s23p63d104s2 1s22s22p63s23p63d10

1s22s22p63s23p63d104s24p5 1s22s22p63s23p63d104s24p6

Block of Periodic

TableGroup of elements

s-block Group 1 and 2 (Those with valence electron ns1 and ns2)

p-blockGroup 13 , 14 , 15 , 16 , 17 , 18

(Those with valence electrons ns2 np1, 2, 3, 4, 5, 6)

d-blockGroup 3 – 12

(Transition element with valence electron (n–1)d1-10 ns2 )

f-block Lanthanides and Actinides Series

•From the Periodic Table, elements can be classified accordingly to the block.

� Based on the valence electron, the Group of an element can be

determined.

Valence electron Group

ns1 1

ns2 2

ns2np1 13

ns2np2 14

ns2np3 15

ns2np4 16

ns2np5 17

ns2np6 18