Topic2_AtomicStructure

CHAPTER 2

ATOMIC STRUCTURE

2.1 Bohr’s Atomic Model

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 at its level

(orbit) using.

Learning Outcomes

E n=−RH1n 2 , RH=2.18x10−18 J

Learning Outcomes

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.

E=RH 1n1

2 −1n2

2 where RH=2.18×10−18 J

E=h where =c /

Learning Outcomes

g) Perform calculation involving the Rydberg equation for Lyman, Balmer, Paschen, Brackett and Pfund series:

g) Calculate the ionisation energy of hydrogen atom from the Lyman series.

1

=RH 1n1

2 −1n2

2 where RH=1.097×107 m−1

n1n2

Learning Outcomes

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.

Bohr’s Atomic Model In 1913, a young

Dutch physicist, Niels Böhr proposed a theory of atom that shook the scientific world.

The atomic model he described had electrons circling a electrons circling a central nucleus central nucleus that contains positively change proton.

Bohr’s Atomic Model

Böhr also proposed that these orbits can only occur at specifically specifically “permitted” levels “permitted” levels according to the energy levels of the electron and explain successfully the lines in the hydrogen spectrum.


First Postulates Electron moves in

circular orbits about the nucleus.

While moving in the orbit, the electron does not radiate or absorb any energy.

[orbit = energy level=shell]

Orbit is a pathway where the electron is move around the nucleus.

Orbit


Second Postulate The moving electron

has a specific amount of energy; its energy is quantised.

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

RH (Rydberg constant) = 2.18 x 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

E n=−R H1n2


Third postulate 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)


Fourth Postulate 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 (photon).

The energy of the photon equals the energy difference between levels.

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

Electron is excited from lower to higherenergy level.A specific am ount of energy is absorbedE = h = E 3-E 1

Electron falls from higher to lower energy leve l .A photon is em itted .E = h = E 1-E 3

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

Electron is excited from lower to higherenergy level.A specific am ount of energy is absorbedE = h = E 3-E 1

Electron falls from higher to lower energy leve l .A photon is em itted .E = h = E 1-E 3

Radiant energy emitted when the electron moves from higher-energy state to lower-energy state is given by:

Where: E=Ef−E i

2

iH2

fH n

1Rn1RE

2

f2

iH n

1n1RE

E f=−RH 1n f

2 E i=−R H 1n i

2 Thus,

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.

=cv

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

Thus : hc

ΔE

Where : h (Planck's constant) = 6.63 x 10-34 J s

v = frequency (s-1)

E =hv

Rydberg Equation Wavelength emitted by the transition of

electron between two energy levels is calculated using Rydberg equation:

1

=RH 1n1

2 −1n2

2 where RH=1.097×107 m−1

n1n2

Example 1 Calculate the wavelength, in nanometers of the

spectrum of hydrogen corresponding to n = 2 and n = 4 in the Rydberg equation.

Exercises:1. Calculate the energy of hydrogen electron in the:

(a) 1st orbit

(b) 3rd orbit

(c) 8th orbit

1. Calculate the energy change (J), that occurs when an

electron falls from n = 5 to n = 3 energy level in a

hydrogen atom.

2. Calculate the frequency and wavelength (nm) of the

radiation emitted in question 2.

Emission Spectra

Emission Spectra

Continuous Spectra

LineSpectra

Continuous Spectrum

A spectrum consists of radiation distributed over all wavelength without any blank spot.

Example : electromagnetic spectrum, rainbow It is produced by white light (sunlight or

incandescent lamp) that passed through a prism

Formation of Continuous Spectrum

Regions of the Electromagnetic Spectrum

Line Spectrum (atomic spectrum) A spectrum consists of discontinuous & discrete

lines with specific wavelength.

It is composed when the light from a gas

discharge tube containing a particular element

is passed through a prism.

Formation of Atomic / Line Spectrum

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.



Example : The line emission spectrum of hydrogen atom

Line spectrum are composed a few wavelengths giving a series of discrete line separated by blank areas

It means each line corresponds to a specific each line corresponds to a specific wavelength or frequency.wavelength or frequency.

Formation of Line Spectrum When electron absorbed radiant energy, they

will move from lower energy level to higher energy level (excited state).

This excited electron is unstable and it will fall back to lower energy level.

During the transition, electron will release energy in the form of light with specific wavelength and can detected as a line spectrum.

Differences Between Line & Continuous Spectra

Continuous Spectrum A spectrum that

contains all wavelength without any blank spots.

Example: Rainbow.

Line Spectrum A spectrum that

contain only specific wavelengths.

A spectrum of discrete lines with certain wavelengths.

Example: Emission spectrum an element.

Formation of Line Spectrum (Lyman Series)

n = 1

n = 2

n = 3

n = 4n = 5

Lyman Series

Emission of photonLinespectrum λ

E

Energy

n = ∞

Formation of Line Spectrum(Balmer Series)

n = 1

n = 2

n = 3

n = 4n = 5n = ∞

Lyman Series

Emission of photonLinespectrum

Balmer SeriesλE

Energy

Energy Level in Hydrogen Atom

Example

Series nf niSpectrum

region

Lyman 2,3,4,…

2 3,4,5,…

Paschen 4,5,6,… Infrared

4 5,6,7,… Infrared

5 6,7,8,… Infrared

ultraviolet

Visible/uvBalmer

Brackett

Pfund

1

3

Complete the following table

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

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?

Linespectrum

E λ

A B C D E

v

Example

Example

The line spectrum of Balmer is given as below:

Describe the transitions of electrons that lead to the lines

W, and Y, respectively.Solution

W Y

For W: transition of electron is from n =4 to n = 2For Y: electron shifts from n = 7 to n = 2

Example

Linespectrum

ABCDE

Paschen series

(a) Which of the line in the Paschen series

corresponds to the longest wavelength of photon?

(b) Describe the transition that gives rise to the line.Solution

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

Example

With refer to the second line in the Balmer series of the hydrogen spectrum, Calculate;

a)the wavelength in nm

b)the frequency

c)the energy

ExampleRefer to last line of hydrogen spectrum in Lyman

series, Calculate:

a) Wavelength

b) Frequency

c) Wave number; where wave number = 1

For Lyman series; n1 = 1 & n2 = ∞Ans:i. 9.116 x10-8 m ii. 3.29 x1015 s-1 iii. 1.0970 X 107 m-1

Ionization Energy

Defination : Ionization energy is the minimum energy required to remove an electron in its ground state from an atom (or an ion) in gaseous state.

M (g) → M+ (g) + e ∆H = +ve

Ionization Energy The hydrogen atom is ionised when electron is

removed from its ground state (n = 1) to n = ∞.

At n = ∞, the potential energy of electron is

zero, here the nucleus attractive force has no

effect on the electron (electron is free from

nucleus)

Examplen1 = 1, n2 = ∞

∆E = RH (1/n12 – 1/n2

2)

= 2.18 x 10-18 (1/12 – 1/∞2) = 2.18 x 10-18 (1 – 0)

= 2.18 x 10-18 JIonisation energy = 2.18 x 10-18 x 6.02 x 1023 J mol-1

= 1.312 x 106 J mol-1

= 1312 kJ mol-1

Example

Calculate the energy to ionized :

(a) a hydrogen atom.

(b) 1 mol of hydrogen atom.

Solution

E=−R H 1n1

2 −1n2

2=−2.18×10−18 1

12 −1∞2

=2.18×10−18 J

(a)

Solution

(b) 1 H atoms need 2.18 x 1018 J

1 mol H atom

= 2.18 x 1018 x 6.02 x 1023

= 1.31 x 106 J

The energy to ionized 1 mol of hydrogen atom is

1.31 X 106 J

Example 10.97 10.66 10.52 10.27 9.74 8.22

The Lyman series of the spectrum of hydrogen is

shown above. Calculate the ionisation energy of

hydrogen from the spectrum.

Solution

ΔE = h X c/λ = h x c x wave no.= (6.626x10-34 Js)(3x108 ms-1)(10.97x106 m-1)= 218.06 x 10-20 J= 2.18 x 10-18J

Ionisation energy = (2.18 x 10-18) (6.02x1023 J mol-1)= 1.312 x 106 J mol-1

= 1312 kJ mol-1

The weaknesses of Bohr’s Theory It can only explain the hydrogen spectrum or

any spectrum of ions contain one electron. example: He+, Li2+.Therefore, it did not account for the emission spectrum of atom containing more than 1 electron.

Electron are wavelike, we can’t define the

precise location of a wave because a wave

extends in space.

de Broglie’s Postulate In 1924 Louis de Broglie proposed that not only

light but all matter has a dual nature and possesses both wave and particle properties.

Electron is both particle and wave.

Tiny particle such as electron does have wave

properties.

De Broglie deduced that the particle and wave

properties are related by the expression:

Example

Electron has dual nature properties. Why don't we see the wave properties of a Baseball?

de Broglie’s Postulate

h = Planck constant (J s)m = particle mass (kg)μ = velocity (m/s)λ = wavelength of a matter wave

=h

mμ

Heisenberg’s Uncertainty Principle It is impossible to know simultaneously both the momentum

p (defined as mass times velocity) and the position of a particle with certain.

Stated mathematically,

where Δx = uncertainty in measuring the position

Δp = uncertainty in measuring the momentum= Δmv

h = Planck constant

x p≥h4π

2.2 Quantum Mechanical Model

Learning Outcomes

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

Define the term orbital. State the four quantum numbers in an orbitals. sketch the 3-D shape of s, p and d orbitals.

Atomic Orbital

Definition

An orbital is a three-

dimensional region in

space around the

nucleus where there

is a high probability of

finding an electron.

Quantum Numbers

Each of the electrons in an atom is described and characterised by a set of four quantum numbers, namely

principal quantum number, n

angular momentum quantum number, l

magnetic quantum number, m

electron spin quantum number, s.

Principal Quantum Number, n n determines the energy level (electron shell)

and size of an orbital.

The principal quantum number n, may have +ve

value starting from n =1, 2, 3, …, ∞.

As n increase :

i) the orbital become larger

ii) electron has higher energy

n 1 2 3 4

Orbital size

Energy increases

Principal Quantum Number, n

Angular Momentum Quantum Number, l

Alternative name:- Subsidiary Quantum Number- Azimuthal Quantum Number- Orbital Quantum Number

The value of l indicates the shape of the

atomic orbital.

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


Letters are assigned to different numerical values of

Value of l Symbol012

3

Orbital shapespd

f

sphericaldumbbellcloverleaf


value is depend on n. (i.e., 0 ≤ l < n). If n = 1, l = 0 (s-orbital)

If n = 2, l = 0 (s-orbital)

= 1 (p-orbital)

If n = 3, l = 0 (s-orbital)

= 1 (p-orbital)

= 2 (d-orbital)

two subshells (s and p orbitals)

three subshells (s, p, and dorbitals)

One subshell (s orbital)

Magnetic Quantum Number, m Describe the orientation of orbitals in space. Possible values of m depend on the value of l.

For a given l, m can be : -l, …, 0, …, +l Example:

If l = 0, m = 0 » 1 orientation of s orbital If l = 1, m = -1,0,+1 » 3 orientation of p orbital

(px, py, pz)

If l = 2, m = -2,-1, 0,+1,+2 » 5 orientation of d orbital ( dxy,dxz,dyz,dx2

-

y2,dz2)

Electron Spin Quantum Number, s The value of s represent the

direction of an electron rotation on its own axis.

either clockwise or

anticlockwise

It has 2 value : +½ and -½

Shape of Atomic Orbital

s orbitals Spherical shape with the nucleus at the centre.

When l = 0 , m = 0 , only 1 orientation of s

orbital.

The larger value of n, the size of s orbital gets

larger.

Shape of s orbital with different n

1s 2s 3s

Z

Y

X

Z

Y

X

Z

Y

X

Shape of Atomic Orbitals

p orbitals

Can be represent as a pair of dumb-bell shaped

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

3 orientation of p-orbitals px, py, and pz.

As n increases, the p orbitals get larger.

Shape of p orbital

px py pz

Shape of Atomic Orbitals

d orbitals

All the d orbitals do not look alike.

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

There are five orientation of d orbitals.

Shape of d orbital

dx2-y2 dz2 dxy

dxz dyz

Set of Four Quantum Numbers 4 quantum number n,l,m and s enable us to

label completely an electron in any orbital of an atom. Example:

4 quantum numbers of 2s orbital electron are

n = 2 , l = 0 , m = 0 and s = +½ and -½

Can be simplified as (2,0,0,+½) or (2,0,0,-½)n, l , m, s n, l , m, s

Exercise

Shell n l Orbital notation m No. of orbitals

1 1 0 1s 0 1

2 20 2s 0 1

1 2p -1,0,+1 3

Exercise Predict the following quantum numbers whether

they are allowed or not

(a) (1,0,0,-½)

(b) (2,0,1,1)

(c) (0,1,1,+½)

(d) (4,1,0,-½)

2.3 Electronic Configuration

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

State and apply Aufbau principle, Hund's rule and

Pauli exclusion principle in filling of electrons in

orbitals of an atom.

Write the electronic configuration of atoms and

monoatomic ions.

(a) Orbital diagram

(b) spdf notation

Learning Outcomes Explain the anomalous electronic configurations

of chromium and copper.

Introduction The electronic configuration of an atom show

how electron are filled in the orbital.

Electronic configuration describes the

arrangement of electron in an atom.

Electronic Configuration

Box

1s 2s 2p

Method 1: Orbital diagram

Example: 8O


Method 2: s,p,d,f notation

Example: 8O

1s2 2s2 2p4

Principal quantum number, n

Azimuthal quantum number, l

Number of electrons in the subshells


To enable us to do electronic configuration, we have to obey the following rules:

a) The Aufbau Principle

b) The Pauli Exclusion Principle

c) The Hund rule

Aufbau Principle State that electrons are filled in the orbitals in

order of increasing energy.

Electrons should occupy the orbital with the

lowest energy first before enters the one with

higher energy.

Relative Energy Level of Atomic Orbitals

ener

gy

Orbital energy levels in a many-electron atom

2s2p

3s

4s

3p

4p

3d

4d5s

1s

n=2

n=3

n=4

n=1

n=5

Ene

rgy

Order of orbitals (filling) in multi-electron atom

Order of orbitals (filling) in multi-electron atom

The order of filling orbitals is:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s Start with the 1s orbital and move downward,

following the arrows. Example:

(a) 4Be (b) 10Ne

Electronic configuration 4Be : 1s2 2s2

Electronic configuration 10Ne : 1s2 2s2 2p6

Pauli Exclusion Principle No two electrons in an atom can have the same

four quantum numbers (n, l, m, s).

Eg : Li (3 electrons)

Hund’s Rule States that when electrons are added to the

orbital of equivalent energy (degenerate orbitals), each orbital are filled singly with electron of the same spin first before it is paired.

The electron in half-filled orbitals have the same

spins, that is, parallel spins.

Example

1) Carbon (6 electron)

2) Oxygen (8 electron)

1s 2s 2p

1s 2s 2p

ExerciseWrite the electronic configuration of the following

atom or ion:

(a) C

(b) Ne

(c) Al

(d) Al3+

(e) Cl

(f) Cl-

The Anomalous Electronic Configurations of Cr and Cu

Element Expected (Aufbau Principle) Observed/actual

Cr (Z=24) 1s22s22p63s23p6 4s2 3d4 1s22s22p63s23p6 4s1 3d5

Cu (Z=29) 1s22s22p63s23p6 3d9 4s2 1s22s22p63s23p6 3d10 4s1

Cr and Cu have electron configurations which are inconsistent with the Aufbau principle. The anomalous are explained on the basis that a filled or half-filled orbital is more stable.

Orbital diagram

24Cr : 18[Ar]

24Cr : 18[Ar]

3d orbital with a half filled orbital arrangement are more stable.

Actual

*Half filled orbital arrangement increase stability of Cr atom

3d 4s

3d 4s

18Ar : 1s2 2s2 2p6 3s2 3p6

Copper expected orbital notation (Aufbau Principle)

Cu : 18[Ar]

3d orbital with fully filled orbital arrangement is more stable.

Copper actual orbital notation

Cu : 18[Ar]

3d 4s

3d 4s

18Ar : 1s2 2s2 2p6 3s2 3p6

Topic2_AtomicStructure

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