topic 1 introduction to atomic structure 2012-1

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Course Title: General and Inorganic

Chemistry Course Code: FME 151/FBE 105/FCE 181

Lecturer and contacts

Mr. Vincent Madadi

Department of Chemistry, University of Nairobi

P. O. Box 30197-00100,

Nairobi, Kenya

Chemistry Dept. Rm 114

Tel: 4446138 ext 2185

Email: [email protected], [email protected]

Website: http://www.uonbi.ac.ke/staff/vmadadi

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Course content

1) Atomic structure:

– Atomic structure & electromagnetic radiation,

– Borh’s atomic model & wave mechanical model of atomic

structure;

– electron configuration & effective nuclear charge.

– Periodic table & atomic properties

2) Chemical bonding:

– ionic bond, covalent bond, metallic bond, hydrogen bond.

3) Chemical kinetics:

– Rate of reactions and rate equations, factors that influence rate of

reaction, order of reaction, molecularity of reaction, activation

energy,

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Course content cont.

4) Chemical equilibrium:

– Equilibrium law, Equilibrium constant, homogeneous

equilibrium, heterogeneous equilibrium, Lechatelier’s principle.

5) Solutions:

– types of solutions, Henry’s law, solubility and solubility curves,

solubility product, ionic equilibrium.

6) Adsorption:

– types of adsorption, adsorption isotherms, application of

adsorption.

7) Environmental significance of some elements and salts:

– Environmental segments, types of pollutants, air pollution, acid

rain, smog formation, green house effect, ozone layer, water

treatment.02/10/2012 3

Examination

• Two CATs = 15%

• 7 Practicals = 15%

• Final Exam = 70%

• Tutorial questions (Contribute to CAT marks)

• CAT 1 (2nd November 2012)

• CAT 2 (30th November 2012)02/10/2012 4

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1. ATOMIC STRUCTURE

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1.1 Composition of atomParticle Symbol Charge Charge (C) Mass (Atomic

mass units)Mass (kg)

Proton p+ + +1.6 × 10-19 C 1 1.673 x 10-27

Electron e- - -1.6 × 10-19 C 1/1837 9.108 x 10-31

Neutron N 0 0 1 1.675 × 10-27

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Name Symbol Calculations Representation of subatomic particles in elements

Hydrogen

H

p = 1; n = 0; e = 1 A = p + n = 1+0 = 1; Z = e = 1

1

H 1

Helium

He p = 2; n = 2; e = 2 A = p + n + 2+2 =4; Z = e = 2

2

He 4

Neon

Ne

p = 10; n = 10; e =10 A = p + n = 10+10 = 20 Z = e = 10

10

Ne 20

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http://periodictable.com/Posters/Poster3.2000.JPG

1.2 Spectroscopya) Definition:

• Spectroscopy: The study of interaction of electromagnetic radiation

with matter.

• Electromagnetic radiation: This is a stream of energy (called quanta

or photon) moving in the direction of propagation and perpendicular

to electric and magnetic field.

• Characteristics of electromagnetic radiation (emr) : All forms of emr:

1) Do not require a medium to travel

2) Travel with velocity of light = 3 x 108 m/s

3) Have dual nature – exhibit both wave and particle nature

4) It has electric and magnetic components oscillating in plane

perpendicular to each other and perpendicular to the direction

of propagation.02/10/2012 8

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Characteristics of electromagnetic

radiation

• Example of an electromagnetic wave with the magnetic field

oscillating parallel to the z-axis, the electric field oscillating

parallel to the y-axis, and the wave moving along the x-axis.

• The x, y, and z-axes are perpendicular to each other.

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Direction of propagation

b) Properties of electromagnetic radiation

waves1. Wavelength (symbol = λ pronounced as “lambda”): The

distance between two consecutive peaks in the wave.

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1 m = 109 nm = 1010 Å = 1012 pm

nm = nanometre; = angstrom pm = Pico-metre

niki

At least this is easy

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Properties of electromagnetic radiation

waves ...

2) Frequency ( symbol = ν , pronounced as “nu”): This is the

number of waves or cycles that pass a point on the x-axis

each second.

3) Velocity (symbol = c): All forms of electromagnetic radiation

travel at the speed of light, c = 3 x 108 m/s

• Relationship between wavelength, frequency and velocity

• These are related by the expression:

C = λ ν ............................Eq. 1

� λ = c/ ν and ν = c/ λ .....Eq. 202/10/2012 11

Properties of electromagnetic radiation

waves ...• Wave number ( symbol ν , pronounced as nu bar): This is the

reciprocal of the wavelength.

ν = 1/ λ = ν/c .............................Eq. 3

• Examples:

• Example 1. The limits of microwave region are approximately 1 GHz and 100

GHz, and those of the millimetre wave region are approximately 100 GHz and 300

GHz. Calculate the wavelengths associated with these frequencies .

• Ans: From equation 2, λ = c/ ν

� For 1 GHz, λ = 3 x 108 m/s /1 x 109 s-1 = 0.3 m

100 GHz, λ = 3 x 108 m/s /100 x 109 s-1 = 3 mm

300 GHz, λ = 3 x 108 m/s /300 x 109 s-1 = 1 mm

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c) Calculating energy associated with

electromagnetic radiation- Planck’s theory

of radiation.

1) Emission or absorption of energy does not take place

continuously but discontinuously in form of small packets or

bundles called quanta ( or photons in case of light).

• A quantum is the smallest packet of energy wave that can exist

independently.

2) Each quantum is associated with a definite amount of energy

which is proportional to the frequency of radiation i.e.

E α ν or E = hν ...........Eq. 4

Quantum or photon

Where h = Planck’s constant = 6.626 x 10-34 Js

02/10/2012 13

Calculating energy associated with

electromagnetic radiation ...3) The amount of energy emitted or absorbed by a body is in whole

number multiple of hν (or quantum)

E = nhν .........................Eq. 5

Where n = 1, 2, 3 etc but not 0.1, 0.8 etc.

• This is called quantisation of energy implying that energy occurs

in whole number multiples of hν

• Example 2

• Calculate the energy of one photon of radiation of:

a) Frequency 4.6 GHz

b) Wave numbers 37,000 cm-1

c) What is the energy of one mole of these photons?02/10/2012 14

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Calculating energy associated with

electromagnetic radiation ...

Ans:

a) E = hv = 6.626 x 10-34 Js x 4.6 x 109 s-1 = 3 x 10-24 J

1 mol = 6.022 x 1023 particles = Avogadro’s number of particles

� For 1 mole of photons, E = NAhv = 6.022 x 1023 mol-1 x 3 x 10-24 J = 1.8 J

b) From equation 3, ν = 1/ λ = ν/c ; Hence E = hcv = 6.626 x 10-34 Js x 3 x 108

m/s x 37,000 x 102 m-1 = 7.3 x 10-19 J

• For 1 mol, E = NAhv = N

Ahcv

= 6.022 x 1023 mol-1 x 7.3 x 10-19 J = 440,000 Jmol-1 = 440

kJmol-1

[Note: the other unit for energy is electron volt (eV).

1 eV = 1.60218 x 10-19 J = 96.485 kJmol-1

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d) Electromagnetic radiation spectrum• Definition: This is the entire range of electromagnetic

radiations separated into different wavelengths or frequencies.

• Example:

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Spectrum• Definition: This is the separation or analysis of a composite

radiation into different wavelengths or frequencies .

• Example: when white light is passed through a prism a

spectrum of seven colours (VIBGYOR) is obtained.

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Interaction of electromagnetic radiation

with matter /origin of spectrum• When an atom is bombarded by emr, it absorbs energy and

move to excited state with higher energy.

• Two possible transitions occur giving rise to two different types

of spectra:

1) Absorption spectrum

2) Emission spectrum

i.e.

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hvhv

E0

E1

E1

E0

Energy is absorbed, absorption spectrum produced∆E = E1 - E0 = hv

Energy is emitted, emission spectrum produced∆E = E1 - E0 = hv

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Interaction of electromagnetic radiation

with matter cont.

• The spectrum produced is characteristic for each element,

hence can be used to identify the element.

1) Emission spectrum:

• Emission spectrum is produced by heating a substance directly

in a flame or electrically and passing the emitted radiation

through a prism or a grating.

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Sample inexcited state Grating Spectrometer Frequency

Inte

nsity

Types of emission spectra

i) Continuous emission spectrum

� It consists of a wide band of continuous wavelengths which

appear as continuous band of light.

� It is produced by incandescent solids e.g. Hot filament , hot

iron etc.

� The intensity of spectrum is not uniform over the entire

range i.e. is maximum at particular wavelengths

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Types of emission spectra cont.

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ii) Line or atomic emission spectrum

� It consists of discrete bright lines produced by gases or vapours

of a substance in atomic state.

� The lines are regularly spaced but differ in their intensities.

� Examples:

a) hydrogen line spectrum

b) sodium vapour spectrum

c) mercury vapour spectrum

� Illustration:

• Hydrogen line

spectrum

Examples of atomic emission spectra of

elements

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Types of emission spectra cont.

iii) Band emission spectrum

� It consists of luminous bands separated by dark spaces.

� It is produced by substances in molecular state.

� At high resolution, each band is observed to be comprised of

very fine lines.

� Examples: a) vacuum tubes

b) Carbon arc with metallic salt in its core

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Luminous bands

2) Absorption spectrum

• Absorption spectrum is produced when light from a source

that emits a continuous range of wavelengths is passed

through a substance (solid, liquid or gas) at lower temperature

and observed through a spectrometer.

• The substance absorbs some of the wavelengths leaving dark

lines (or dark bands or continuous dark region) at their places.

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Source of Continuous radiation Grating Sample Spectrometer Frequency

Inte

nsity

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Types of absorption spectrum

• Depending on the type of substance absorbing the radiation,

the spectrum can be:

i) Continuous absorption spectrum

� Occurs when a continuous range of wavelengths are absorbed

by the sample producing continuous dark regions e.g. When

the glass absorbs all wavelengths except red

ii) Line absorption spectrum

� It consists of discrete dark lines produced when absorbing

materials are in vapour or gaseous phase.

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Types of absorption spectrum

iii) Band absorption spectrum

� It consists of dark bands produced in absorption spectrum of

aqueous solution e.g. KMnO4 gives five dark absorption

bands.

• Difference between spectra

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1.3) Photoelectric effect

• When a photon of light of sufficient frequency (v) and energy (hv)

strikes a metal surface, electrons are ejected.

• Part of the photon energy is consumed to separate electrons from

the metal surface (threshold energy or hvo),

• The remaining energy is imparted to eject electron to a certain

velocity (v), thus the emitted electron gains kinetic energy (KE or

½ mv2). This can be represented as:

hv = hvo + ½ mv2 .....................Eq. 6

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Photon energy or incident energy Work function of

A metal

Kinetic energy of an emitted electron

Photoelectric effect ...

• Where , h = Planck’s constant = 6.626 x 10-34 Js

V = frequency of incident photon

Vo = threshold frequency

V = velocity of electron

me = mass of electron = 9.108 x 10-31 kg

� Illustration of photoelectric

effect

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e-

e-

e-

e-

e-

UV light (hv)

Stream ofelectrons

Flow of electrons

Ammeter to measure current

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Photoelectric effect ...

• When EM radiation with frequency above threshold frequency

(vo) is shined on the metal surface, an electric current registers

on the meter .

• For frequency below the cut-off frequency (vo ), no current is

obtained, even at very high intensities.

• As frequency increases above vo , the kinetic energy of electron

increases as illustrated below.

02/10/2012 29

KineticEnergy

vo Frequency (v)

Calculations involving photoelectric effect• Example 3:

When sodium metal surface is exposed to radiation of 300 nm, electrons with kinetic

energy KE = 1.68 x 105 Jmol-1 are emitted.

a) Calculate the maximum energy needed to remove an electron from the metal

surface.

b) What is the maximum wavelength that will cause photoelectric effect?

Ans:

Step 1

Energy of photon = hv = hvo + KE = h x c/λ + KE

= (6.626 x 10-34 Js) x (3 x 108 m/s) = 6.626 x 10-19 J

300 x 10-9 m

Step 2

KE of 1 electron = KE of 1 e- in Jmol-1/Avogadro’s number02/10/2012 30

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Calculations involving photoelectric effect ...

• KE of 1 electron = 1.68 x 105 Jmol-1 = 2.789 x 10-19 J

6.022 x 1023 mol-1

hvo = Emin = hv - KE

= (6.626 x 10-19 – 2.789 x 10-19)J = 3.837 x 10-19 J

b) From Emin = hvo or Emin = hc/λ we can determine λmax, i.e.

� 3.837 x 10-19 J = (6.626 x 10-34 Js) x (3 x 108 m/s)

λ

� λ = (6.626 x 10-34 Js) x (3 x 108 m/s) /3.837 x 10-19 J

= 5.181 x 10-7 m

02/10/2012 31

1.4 Bohr’s Atomic Model• Bohr’s atomic model explains the

origin of the spectrum of

hydrogen and hydrogen-like

species e.g. He+, Li2+ etc.

• Three main postulates of Bohr’s

atomic model

1) Electron in an atom behaves like

a material particle and revolves

round the nucleus in a fixed

circular orbits or shells called

energy shells or energy levels

� Electron in a particular shell is

associated with a certain

amount of energy.

� Energy increases with increasing

distance from nucleus.

� E1 < E2 < E3 etc.

� The values of n (the principal

quantum number ) are n = 1 =

k, n= 2= L etc.

2) As long as the electron is

revolving in a particular orbit, it

does not absorb nor emit

energy. Hence, these energy

levels are also referred to as

stationary states.

� When electron jumps from

lower energy n=1 with energy

E1 to higher energy n=2 with

energy E2, it absorbs energy in

form of photon.02/10/2012 32

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Bohr’s Atomic Model ...• Energy change ΔE = E2 – E1 = hv = hc/λ

........................................................ Eq. 7

� When electron falls from higher energy

level n=2 with energy E2 to lower energy

level n=1 with energy E1, energy is

emitted.

• ΔE = E2 – E1 = hv = hc/λ .............. Eq. 8

• The released energy appears as a

spectral line in the emission spectrum.

3) Electron can only move in the orbits in

which the angular momentum (mvr ) of

the revolving electron is an integral

multiple of h/2π

� mvr = nh/2 π ............................. Eq. 9

• Where,

m = mass of electron = 9.108 x 10-31

kg

V = velocity of electron

r = radius of the orbit in which

electron is revolving

h = Planck’s constant = 6.626 x 10-34

Js

n = an integral number which

denotes the number of orbit.

• The equation mvr = nh/2 π means

that the angular momentum of the

revolving electron is quantised which

implies that the magnitude of the

angular momentum is always a whole

number not fraction.02/10/2012 33

Bohr’s Atomic Model ...• By applying the postulates and classical laws of physics, Bohr

was able to derive expressions for: 1) Energy of revolving electron

2) Velocity of revolving electrons

3) The radius of the nth orbit.

4) Frequency and wavelengths of spectral lines emitted.

1. Derivation of the expression for the radius of nth orbit of hydrogen

atom and hydrogen-like species.

Step1:

– Considering hydrogen or hydrogen-like species with atomic number =

z,

– Assuming electron of charge e- is revolving round the nucleus.

– The charge on the nucleus will be = ze.

– Assume r = the distance between the revolving electron and the

nucleus = radius of the orbit, m = mass of electron and V = tangential

velocity of electron

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Bohr’s Atomic Model ...

• At any time, the electron experiences two types of forces:

1) Electrostatic force of attraction (centripetal force or CP)-the force of

attraction between the nucleus and the revolving electron.

� CP = 1 x Ze2 .......................................................... Eq. 10

4πεο r2

Where, εο = permittivity constant of the media = 8.854 x 10-12 J-1C2m-1

2) Force of repulsion (centrifugal force or CF) – the force acts outwards

from the nucleus hence keeps electron revolving away from the orbit.

� CF = mv2/r ..........................................................................Eq. 11,

• At any moment, Centrifugal force equals centripetal force

02/10/2012 35


� 1 x Ze2 = mv2

4πεο r2 r

� V2 = Ze2 x 1 ................................................Eq. 12

4πεο mr

From quantisation of angular momentum,

mvr = nh/2π v2 = n2h2/4π2m2r2 ......Eq 13

Equating equation 12 and 13,

� = Ze2 x 1 = n2h2 .............................Eq. 14

4πεο mr 4π2m2r2

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• Making r the subject,

� r = n2h2 x 4π εο m

4π2m2 Ze2

� r = n2h2 εο

πm e2 Z

� r = εο h2 n2 .............................................................................Eq. 15

πm e2 Z

� Substituting the values of π = 3.14; h = 6.626x10-34 Js; εο = 8.85

x 10-12J-1C2m-1; m = 9.108 x 10-31 kg; e- = 1.602 x 10-19 C

02/10/2012 37


• Noting that J = kgm2s-2, the value of the radius of nth orbit

can be calculated as:

• rn = n2 x (6.626x 10-34 Js)2 x (8.85 x 10-12J-1C2m-1)

3.14 x (9.108 x 10-31 kg) x (1.602 x 10-19C)2 x Z

= n2 x 5.29 x 10-11 x J2S2j-1C2m-1 = n2 x 5.29 x 10-11 x Js2m-1

kgC2 x Z kg x Z

= 5.29 X 10-11 X n2 X kgm2s-2s2m-1

kg x Z

rn = 0.529 x 10-10 (n2/z) m ...............................................Eq. 16

02/10/2012 38

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Relationship between rn and (r1)H

• From equation 15,

� rn = εο h2 x n2 .....................................Eq. 17

πm e2 Z

� (r1)H = εο h2 x 12 .........................................Eq. 18

πm e2 1

• Dividing equation 17 by Eq.18,

� Rn/ (r1)H = εο h2 x n2 x πme2 = n2

πm e2 Z εο h2 z

� rn = (r1)H x n2 or rn = 0.529 x 10-10 x n2 m ..........Eq. 19

Z Z

02/10/2012 39

Relationship between rn and (r1)H ...

Example 5: Calculate the radius of 3rd orbit of hydrogen atom.

Ans. From equation 15,

� rn = εο h2 x n2 = 0.529 x 10-10 x n2 m

πm e2 Z Z

For n = 3 and Z = 1; we obtain the radius of the 3rd

orbit of H- atom given by,

� (R3)H = 0.529 x 10-10 x 32 m = 4.761 x 10-10 m

1

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Expression for velocity of electron

revolving in the nth orbit

• Velocity of electron revolving in hydrogen atom and hydrogen-

like species:

� Noting that centripetal (1 x Ze2 )

4π εο r2 and centrifugal (mv2/r)

forces are equal;

� Ze2 = mv2/r or v2 = Ze2 ..........................Eq. 20

4π εο r2 4π εο rm

From quantisation of angular momentum,

mvr = nh/2π or v =nh/2πmr .......................................Eq. 21

02/10/2012 41


revolving in the nth orbit cont.

• Dividing Eq. 20 by Eq.21 ;

� V2 = Ze2 x 2πmr = Ze2

v 4π εοrm nh 2 εοnh

� Vn = Ze2 .....................................................Eq. 22

2 εοnh

Substituting the values of constants,

e- = 1.602 x 10-19C, εο = 8.85 x 10-12 J-1C2m-1 and h = 6.626 x

10-34 Js

Vn = Z x (1.602 x 10-19C)2 .

2 x (8.85x10-12 J-1C2m-1) x n x (6.626 x 10-34 Js)

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revolving in the nth orbit cont.Vn = 0.2185 x 107 Z ms-1 ...................................................Eq. 23

n

• Number of revolutions made by electron moving in the nth orbit

• Number of revolutions = Vn ........................................Eq.24

2πrn

From Eq. 22 and 23 Velocity of e- = Vn = Ze2 = 0.2185 x 107 Z ms-1

2 εοnh n

From Eq. 15 and 16,

radius of nth orbit = rn = h2 εο n2 = 0.529 x 10-10 x n2 m

πm e2 Z Z02/10/2012 43


revolving in the nth orbit cont.

Assuming that an atom is circular (Bohr’s Theory),

No. of revolutions = (0.2185 x 107 Z ms-1)x ( z .)

n 2π x 0.529 x 10-10 x n2 m

� No. of revolutions = 65.711 x 1014 x Z2 s-1 ...........................Eq. 25

n3

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Calculating Energy of electron in nth Orbit

of H-atom and hydrogen-like atoms• The sum energy of electron (En) = Potential Energy (PE) +

Kinetic Energy (KE)

� En = ½ mv2 – Ze2/4π εοr .............................................. Eq. 26

� For e- revolving in nth orbit, centripetal and centrifugal forces are

equal (from Eq. 20),

� Ze2 = mv2/r or mv2 = Ze2 or ½ mv2 = Ze2 .....Eq. 27

4π εοr2 4π εοr 8π εοr

� Substituting Eq. 27 into Eq. 26 for kinetic Energy,

� En = Ze2 – Ze2 = - Ze2 .................................Eq. 28

8π εοr 4π εοr 8π εοr02/10/2012 45


of H-atom and hydrogen-like atoms cont.

� Substituting r with h2 εο n2 from equation 15,

πm e2 Z

� En = - Ze2 x πm e2 Z = - m e4 Z2 ............................Eq. 29

8π εο h2 εο n2 8 εο

2 h2 n2

Where, m = 9.108 x 10-31 kg, e- = 1.602 x 10-19 C,

εο = 8.85 x 10-12 J-1C2m-1, h = 6.626 x 10-34 Js, J = kgm2s-2

Substituting the values of constants,

En = -(9.108x 10-31kg) x (1.602 x 10-19 C)4 x Z2

8 x (8.85 x 10-12 J-1C2m-1)2 x (6.626 x 10-34 Js)2 n2

02/10/2012 46

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24


of H-atom and hydrogen-like atoms cont.

En

= -2.18 x 10-18 x Z2 x kgC4

n2 (J-1C2m-1)2 (Js)2

En = -2.18 x 10-18 x Z2 x kgC4

n2 J-2C4m-2J2s2

En = -2.18 x 10-18 x Z2 x kgm2s-2

n2

En = -2.18 x 10-18 x Z2 J atom-1 ...................................................Eq. 30

n2

02/10/2012 47

Energy of electron increases with increasing energy level

Relationship between En and (E1)H

• From Eq. 29,

En = - m e4 Z2 (E1)H = - m e4 x 12

8 εο2 h2 n2 8 εο

2 h2 12

Dividing En by (E1)H ,

� En = - m e4 Z2 x - 8 εο2 h2 12 = Z2

(E1)H 8 εο2 h2 n2 m e4 12 n2

� En = (E1)H x Z2 .......................................................Eq. 31

n2

� Substituting the values (E1)H from equation 30,

En = - 2.18 x 10-18 x Z2/n2 J atom-1

� The negative energy implies as e- approaches the nucleus, E is

released hence energy of e- decreases.

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25

Calculating ionisation (IE) energy of H-

atom hydrogen like species• IE is defined as the energy involved in exciting an e- from the

lowest energy level to infinity (n = ∞).

• E1 = - me4 Z2 = -2.18 x 10-18 J atom-1

8 εο2 h2 n2

• E∞ = - me4 Z2 = -2.18 x 10-18 J atom-1 = 0

8 εο2 h2 (N∞)2 (N∞)2

• IE = ΔE = E∞ - E1

= 0 - (- 2.18 x 10-18 Z2 J atom-1

n2

IE = 2.18 x 10-18 Z2 J atom-1 .........................................Eq. 32

n202/10/2012 49

Ionisation energy increases across the period and decreases down the group.

Calculating wavelength(λ) and frequency

(v) of spectral lines cont.

• When electron falls from higher orbit (n = 2) to a lower orbit (n =

1) H emission spectrum is produced.

• The spectrum consists of a large number of spectral lines

grouped into 5 series i.e. Lyman, Balmer, Paschen, Bracket &

Pfund series

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26


(v) of spectral lines cont.Step1: Determine the energy

change.

• Assuming electron from n=2 to n=2;

• For n1, E1 = - me4 x Z2

8 εο2 h2 n1

2

• For n2, E2 = - me4 x Z2

8 εο2 h2 n2

2

• ΔE = E2-E1 = - me4 x Z2 - (-me4 x Z2 )

8εο2 h2 n2

2 8εο2 h2 n1

2

• ΔE = E2-E1 = me4 x Z2 x ( 1 - 1 )

8 εο2 h2 n1

2 n22

.................................Q33

Step 2: Determine frequency

• But ΔE = hv

• hv = me4 Z2 x( 1 - 1)

8 εο2 h2 n1

2 n22

� v = me4 x Z2 x( 1 - 1 )

8 εο2 h3 n1

2 n22

..............................................................Q34

Substituting the values of constants:

Note, m = 9.108 x 10-31 kg, e- = 1.602 x

10-19 C, εο = 8.85 x 10-12 J-1C2m-1,

h = 6.626 x 10-34 Js, J = kgm2s-202/10/2012 51


(v) of spectral lines cont.� Substituting the constants,

� me4 = 3.2906 x 1015 s-1

8 εο2h3

� V = 3.2906 x 1015 Z2 x (1 - 1 ) s-1 or

n12 n2

2

Hz .....................................Eq. 35

Step 3: Determine wave numbers

From Equation 3 and Eq. 34,

� v = 1/λ = v/c = me4 x Z2 x (1 - 1 )

8 εο2 h3c n1

2 n22

But the expression me4

8 εο2 h3c

is called Rydberg constant of hydrogen

atom = 1.0976 x 107 m-1

Thus, v = 1.0976 x 107 Z2 ( 1 - 1 )

n12 n2

2

= RH Z2 (1 - 1 )

n12 n2

2

But from Eq. 34 , V = me4 x Z2 x (1 - 1 )

8 εο2 h3 n1

2 n22

� V = me4 x c x Z2 x ( 1 - 1 )

8 εο2 h3 c n1

2 n22

02/10/2012 52

02/10/2012

27


(v) of spectral lines cont.

Rydberg constant, RH = me4 =

8 εο2 h3 c

Thus, V = RHc Z2 ( 1 - 1 ) s-1 or

Hz ...Eq. 37

n12 n2

2

Step 4: Determine wavelength (λ)

From 1/λ = v = v/c , the value of λ

can be calculated as,

� λ = c/v = 1/ v

• Calculate the frequency, energy and wavelength

of the radiation corresponding to the spectral

line of the lowest frequency in Lyman series in

the spectrum of H-atom.

• Ans.

• From 1/λ = v = RH

(1 - 1 )

n12 n2

2

• The lowest line in the Lyman series

corresponds to e- moving from n = 2

to n = 1.

• From Eq. 36,

� v = 1/λ = v/c = me4 x Z2 x (1 - 1 )

8 εο2 h3c n1

2 n22

.........................................................Eq. 36

02/10/2012 53

Example cont.

= (1.09678 x 107 m-1) [1/12 – 1/22] = 1.09678 x 107 m-1 x ¾

� λ = 4 = 1.215 x 10-7 m

3 x 1.09678 x 107 m-1

Frequency, v = c/λ = 3 x 108 m/s = 2.46 x 1015 s-1

1.215 x 10-7 m

� Change in energy = ΔE = hv = me4 x c x h x Z2 x ( 1 - 1 )

8 εο2 h3 c n1

2 n22

= RH x c x h x Z2 x ( 1 - 1 )

n12 n2

2

= 1.09678 x 107 m-1 x (6.626 x 10-34 Js)x (3 x 108 ms-1) x ¾

= 16.34 x 10-19 J02/10/2012 54

02/10/2012

28

Bohr’s Atomic model

1) Explained the

stability of atom

2) Explained how

emission spectrum is

produced

3) Explained how

absorption spectrum

is produced

4) Explained the origin

of spectral lines in

hydrogen spectrum

eg. Lyman, Balmer,

Paschen, Bracket

and Pfund series

1) It does not explain the origin of spectra

given by multi-electron species.

2) It assumes the circular orbits in which

the electrons revolve are planar instead

of 3 dimensional.

3) It does not explain the cause of Zeeman

and Stark effects:

1) Zeeman effects- splitting of spectral lines in

magnetic field

2) Stark effect - splitting of spectral lines in

electric field

4) It does not account for uncertainty

principle and dual nature of electrons.

5) It does not explain the origin of fine

structures observed in the spectral lines

using high resolution microscope.02/10/2012 55

A : Achievements B: Limitations

Wave Mechanical Concepts of Atomic

structure• The theory assumes that all small particles such as electrons,

protons, neutrons and atoms when in motion possess wave

properties such as wavelength (λ), amplitude (A), and frequency

(v ).

• De Broglie derived expression describing the relationship

between the wavelength associated with the mass m of a body

moving with velocity v.

� λ = h/mv ..................................................................Eq. 37

� Where λ = wavelength of the particle, h = Planck’s constant, v =

velocity of the particle and m = mass of the particle.

02/10/2012 56

02/10/2012

29

Wave Mechanical Concepts of Atomic

structure cont.• Derivation of De Broglie’s equation

• From Planck’s E = hv,

• From Einstein’s equations of energy,

E = mc2 .......................................................Eq. 38

Equating the two expressions,

� mc2 = hv = hc/λ

� mc = h/ λ

� λ = h/mc = h/p

� Where p = momentum of the particle.

• Implication of De Broglie’s Equation

1) Everything in nature possesses both wave and particle properties.

2) Properties of large particles are well characterised by particle properties

whereas for small objects wave properties are suitable.02/10/2012 57

Examples of calculations involving De

Broglie's equation

• Calculate the wavelength of a ball of mass 100 g moving with velocity of 1000 cm/s

and that of an electron moving at velocity of 2.188 x 10-8 cm/s.

• Ans:

i) λ = h/mv = 6.626 x 10-34 Js = 6.626 x 10-34 m

0.1 kg x 10 m/s

ii) For an electron, λ = h/mv = 6.626 x 10-34 m

9.108 x 10-31 kg x 2.188 x 10-10 ms-1

= 3.32 x 10-10 m

• The wave length of the ball is too small to be measured by

spectroscopic techniques, whereas that of electron is comparable

with that of X-ray and hence can be measured.02/10/2012 58

02/10/2012

30

Heisenberg’s Uncertainty Principle

• It states that both position and momentum of the particle

cannot be determined with absolute exactness or certainty.

� If momentum or velocity of the particle is accurately

determined, the measurement of position will be less precise.

02/10/2012 59

e-

Photon

+

Momentum of e- changes whena photon of light strikes it.

e- changes momentum at the instant of collusion

Nucleus

Expected path of the electron

Heisenberg’s Uncertainty Principle cont.

• The uncertainty in measurement of position (Δx) and

uncertainty in determining momentum (Δp) or velocity (Δmv)

is given by Heisenberg’s expression,

Δx x Δp ≥ h/4π ...........................................................Eq. 39

or Δx x m Δv ≥ h/4π ...................................................Eq. 40

Example:

• A cricket ball of mass 200g has uncertainty in position of 5 pm. Calculate

uncertainty in the velocity of the ball.

Ans: Δx x Δp = h/4π or Δx x m Δv ≥ h/4 π ; Thus Δv = h/4π m Δx

02/10/2012 60

02/10/2012

31

Heisenberg’s Uncertainty Principle cont.

� Δv x 5 x10-12 m x 0.2 kg = 6.626 x 10-34 Js /4 x 3.14

� Δv = 6.626 x 10-34 Js . = 0.5275 x 10-22 x Js

5 x 10-12 m x 0.2 kg x 4 x 3.14 m x kg

� Δv = 5.275 x 10-23 x kgm2s-2s = 5.275 x 10-23 m/s

m x kg

� Example: According to Bohr’s theory of H-atom, the velocity of an electron in

the first orbit is 2.183 x 106 m/s. If uncertainty in position of electron is 5 pm,

calculate the uncertainty in velocity.

02/10/2012 61

Heisenberg’s Uncertainty Principle cont.• Δx x Δp ≥ h/4π or Δx x m Δv ≥ h/4π

� 5 x 10-12 m x 9.108 x 10-31 kg x Δv = 6.626 x 10-34 Js

4 x 3.14

� Δv = 6.626 x 10-34 Js = 1.158 x 107 ms-1

5 x 10-12 m x 9.108 x 10-31 kg

Explanation:

• Uncertainty in velocity of the cricket ball is very small compared to

the actual velocity that an object of this size can have. Hence for

macroscopic objects, both velocity and position can be determined

with accuracy.

• The uncertainty in velocity of electron in the first orbit is greater than

the velocity of electron in the orbit.

� For microscopic objects, it is not possible to simultaneously

determine both position and velocity of the particle with accuracy.02/10/2012 62

02/10/2012

32

Schr’o’ndinger’s Wave Equation

• Is based on the assumption that if electron behaves like a

wave, there must be a wave equation describing its motion.

• Schr’o’dinger’s equation describes the position of electron in

three dimensional space around the nucleus.

• If we represent a wave by the sine function,

• then, Ψ = A sin (2πx/λ) ..................................................Eq.41

• Where A = amplitude of the wave, x = displacement from the

origin, λ = wavelength of the wave.

• Differentiating the equation 41 twice with respect to (wrt) x,

02/10/2012 63

Schr’o’ndinger’s Wave Equation cont.

• δΨ/δx = A(2π/λ) cos (2πx/λ)

• δ2Ψ/δx2 = (-4π2/λ2)A sin(2πx/λ) = - (4π2/λ2) Ψ ...........................Eq. 42

• Noting that total Energy E = PE + KE = V + ½mv2

� 2E = 2V +mv2 .......................................................................Eq. 43

• Making v2 the subject,

� V2 = 2(E – V)/m ......................................................................Eq. 44

� V = 2(E-V)/m

� From De Broglie’s equation,

� λ = h/mv = λ2 = h2/m2v2 ........................................................Eq. 45

02/10/2012 64

02/10/2012

33


• Substituting for v2,

� λ2 = h2/m2 x m/2(E-V) = h2/2m(E-V) .............................Eq. 46

� Substituting Eq. 46 into Eq. 42

� δ2Ψ/δx2 = - (4π2/λ2) Ψ = (4π2 2m(E-V)/h2) Ψ

� δ2Ψ/δx2 + 8π2 m(E-V)/h2) Ψ = 0 ....................................Eq. 47

� This is the wave equation for a particle moving along the x -

axis.

02/10/2012 65


• For 3 dimensional space,

� δ2Ψ/δx2 + δ2Ψ/δy2 + δ2Ψ/δz2 + 8π2 m(E-V)/h2)Ψ = 0 ...........Eq. 48

� This is the Schr’o’dinger’s equation

� The equation can be further reduced to,

� 2 Ψ + 8π2m (E-V) Ψ = 0 ............................................Eq. 49

h2

Where, 2 = δ2/δx2 + δ2/δy2 + δ2/δz2 and is called the Laplacian

operator02/10/2012 66

∇

∇

02/10/2012

34


• The equation 49 can be rearranged into,

(V- 2h2) Ψ = E Ψ

8π2m

• Or H Ψ = E Ψ ........................................................................Eq. 50

• Where H = V - 2h2/8π2m is called the quantum mechanical

Hamiltonian Operator, and E is the Eigen value.

• The equation is called Schrodinger equation in operator form, and

the corresponding wave function Ψ is called Eigen function.

02/10/2012 67

∇

∇

Significance of the wave function (Ψ)

• The wave function Ψ (psi) can be negative or positive, thus it has

no practical significance since the probability of finding an

electron can only be positive or zero.

• The wave function squared (Ψ2) is always positive, hence gives the

probability of finding an electron around the nucleus.

• Solutions to Schr’o’dinger equation

• Solving the schr’o’dinger equation for each electron gives the

shape of atomic orbital, which is defined as the three dimensional

space in which the probability of finding electron is maximum.

02/10/2012 68

02/10/2012

35

Differences between an orbit and atomic orbital

Orbit Atomic orbital

1. Is a definite circular path at a fixed

distance from the nucleus in which electron

revolves

1. A 3 dimensional region or space around

the nucleus within which the probability of

finding electron with a definite energy is

maximum.

2. It indicates the exact position of an

electron in an atom

2. Does not specify definite position of

electron in the tom since electron due to

wave nature cannot be at fixed distance

from the nucleus

3. There is certainty about movement of

electron in an orbit

3. No certainty about movement of an

electron in an orbital

4. It represents planar motion of electron It represents 3 dimensional motion of

electron

5. Maximum number of electrons in orbit =

2n2 where n = number of orbit

5. Maximum number of electrons in an

orbital = 2, the two must have opposite

spins

6. Orbits are circular in shape 6. Orbitals have different shapes e.g. S-

orbitals are spherical, p-orbitals are egg

shaped etc.02/10/2012 69

Quantum numbers

• These are integral numbers that describe:

1) the energy of electron in an orbit,

2) The position of electron from the nucleus

3) Shape and the number of orientations of orbital round its own

axis

4) Orientation of the spinning of electron round its own axis

• There are four types of quantum number i.e.

1) The principal quantum number (n)

2) Azumuthal quantum number (l)

3) Magnetic quantum number (m)

4) Spin quantum number (s)

02/10/2012 70

02/10/2012

36

The principal quantum number (n)

1) It represents the number of shells (orbit) or main energy level in

which the electron revolves round the nucleus e.g. n = 1; 2; 3 e.tc.

2) It gives the distance of the electron from the nucleus (radius of the

orbit) e.g. r = εο h2 n2 .

πm e2 Z

3) It gives the energy (E) of an electron in an orbit i.e.

En = - m e4 Z2

8 εο2 h2 n2

4) It gives the maximum number of electrons that can be

accommodated in a given shell, i.e. maximum number of electrons

= 2n2

02/10/2012 71

Azimuthal Quantum Number (l)

• It determines the magnitude of the orbital angular momentum

hence also called orbital angular momentum quantum number.

• It explains the appearance of the group of closely spaced lines in

the hydrogen spectrum.

• It is disgnated by l

• It assumes values from 0, 1, 2 ... (n-1)

• e.g. For n = 3, l = 0, 1, 2.

• The values of l represent a particular sub-shell within the principal

shell i.e.02/10/2012 72

02/10/2012

37

Azimuthal Quantum Number (l) cont.• Examples:

• For l = 0 s-sub-shell (or s- orbital)

l = 1 p- sub-shell (or p-orbital)

l = 2 d- sub-shell (or d-orbital)

l = 3 f- sub-shell (or f-orbital)

• The letters s, p, d and f are obtained from the description of the

spectral lines i.e. Sharp, principal, diffused and fundamental

respectively.

• The total number of sub-shells in a given shell equals the number

of the main shell e.g. for n = 1, the only sub-shell is s; for n = 2,

two sub-shells present are s and p etc.

02/10/2012 73

Azimuthal Quantum Number (l) cont.• The maximum number of electrons that can be accommodated

in a given subshell = 2(2l +1)

• Example:

02/10/2012 74

Orbital Value of l 2(2l +1)

s 0 2

p 1 6

d 2 10

f 3 14

02/10/2012

38

Magnetic Quantum Number or

orientation quantum number (m)

• It explains the presence of additional spectral lines seen when

magnetic field is applied to the source of spectrum

• It gives the total number of orbital orientations within a given sub-

shell.

• It takes values from –l, 0, +l

• Example:

l = 0; m = 0

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

l = 2; m = -2, -1, 0, +1, +2

l = 3; m = -3, -2, -1; 0; +1; +2; +302/10/2012 75

Spin Quantum Number (s)

• Arises due to the fact that electron while moving round the

nucleus also rotates around its own axis.

• The spinning can be in clockwise or anti-clockwise direction.

02/10/2012 76

• Representation of electron spinning:1) Clockwise (+ ½ ) or ( ) or (α spin)2) Anticlockwise (- ½) or ( ) or (β spin)

02/10/2012

39

Relationship between quantum numbers

02/10/2012 77

Principal

Quantum

number (n)

Values of l

and number

of sub-shells

Total

Number of

electrons in

a sub-shell =

2(2l+1)

Different

values of m

Max

number of

electrons in

orbital = 2 x

number of

orbitals

Max

number of

electrons in

main shell =

2n2

1 0

(s -sub-shell)

2 (2x0 +1) =

2

0 2 x 1 = 2 2 x 12 = 2

2 0

(2s orbital)

1 (2 p-

orbital)

2 (2x0 +1) =

2

2(2 x 1+1) =

6

0

0; +1; -1

2 x 1= 2

2 x 3 = 6

2 x 22 = 8

3 0 (3 s-

orbital)

1 (3 p-

orbital)

2 (3 d-

orbital)

2(2x0+1) = 2

2(2x1+1) = 6

2(2x2+1) =

10

0

0; -1;+1

0; -2; -1; +1;

+2

2x1=2

2 x 3 = 6

2 x 5 = 10

2 x 32 + 18

Types of atomic orbitals• Atomic orbitals have different shapes

1) s-orbitals

• They are symmetrical in shape and non-directional and occur

when l = 0; m = 0; n ≥ 1

• Accommodate maximum of 2 electrons

• For larger values of n, the size of s-orbital increases i.e. The

number of radial nodes increases

• A node is a region in space where the probability of finding an

electron is zero. At a node, Ψ2 = 0

• For an s-orbital, the number of nodes is (n - 1).02/10/2012 78

02/10/2012

40

79

ss-- orbitalsorbitals

P- Orbitals• Occur when l = 1; m = -1; 0; +1; n ≥ 2

• There are three p-orbitals px, py, and pz. (The three p-orbitals lie

along the x-, y- and z- axes of a Cartesian system. The letters

correspond to allowed values of ml are -1, 0, and +1.)

• Each has two egg shaped lobes on each side of the nucleus

• The probability of finding electron in each lobe is the same e.g.

Degenerate orbitals (same energy level)

• Both lobes are separated by nodal plane which passes through the

nucleus. The electron density on the plane and the nucleus is zero.

02/10/2012 80

02/10/2012

41

d-Orbitals• Occur when n ≥ 3; l = 2; m = -2; -

1; 0; +1; +2

• They are dumb-bell shaped

• All the five orbitals are

degenerate

• Three of the d-orbitals lie in a

plane bisecting the x-, y- and z-

axes, the remaining two lie in a

plane aligned along the x-, y-

and z-axes.

• Four of the d-orbitals have four

lobes each, but one has two

lobes and a collar.

02/10/2012 81

f-Orbitals• They occur when n≥ 4; l = 3; m = -3; -2; -1; 0; +1; +2; +3

• There 7 types of f-orbitals

• Almost no covalent bonding because metal orbitals are so

contracted

• Radial functions same for all nf orbitals

• Three angular nodes (nodal planes) orthogonal to s, p and d

orbitals

02/10/2012 82

02/10/2012

42

Electron configuration

• This is the distribution of electrons into different atomic orbitals in

order of increasing energy levels. It shows in which orbitals the

electrons for an element are located.

• The filling of atomic orbitals follow 3 principles@

1) Pauli’s exclusion principle

2) Aufbau’s principle

3) Hund’s rule of maximum multiplicity

1) Paul’s exclusion principle

� It states that no two electrons in an atom can have the same set

of the four quantum numbers e.g.

� For two electrons in the same orbital, the spin quantum number

must be different i.e. Spinning in different directions02/10/2012 83

84

Electron Spin and the Pauli Electron Spin and the Pauli Exclusion PrincipleExclusion Principle

Since electron spin is quantized, we

define ms = spin quantum number = ±±±±½.

Two electrons in the same orbital

must have opposite spins.

S = -1/2 S = +1/2

02/10/2012

43

Paul’s exclusion principle cont.Main

shell (n)

Combin

ation

number

Different Values of n, l, m,

and s

Electron

with upward

and

downward

arrows

Total

number of

electrons in

orbital

Total

number of

electrons in

main shell

1 1

2

N = 1; l= 0; m= 0; s= +1/2

N = 1; l = 0; m = 0; s = -1/2

↑

↓

2 (1s2) (↑↓) 2 x 12 = 2

2 1 N = 2; l= 0; m= 0; s= +1/2 ↑ 2 (2s2) (↑↓) 2x22 = 8

2 N = 2; l = 0; m = 0; s = -1/2 ↓

3 N = 2; l = 1; m = 0; s = +1/2 ↑ 2 (2pz2)

(↑↓)

4 N = 2; l = 1; m = 0; s = -1/2 ↓

5 N = 2; l = 1; m = +1; s = +1/2 ↑ 2 (2py2)

(↑↓)

6 N = 2; l = 1; m = +1; s = -1/2 ↓

7 N = 2; l = 1; m = -1; s = +1/2 ↑ 2 (2px2)

(↑↓)

8 N = 2; l = 1; m = -1; s = -1/2 ↓02/10/2012 85

Aufbau Principle• Also referred to as the building up principle

• Each added occupies higher energy sub-shells only after the

lower sub-shells have been filled completely

• The energy of orbital is determined by the n+l rule

1) Orbital having lower value of n+l will have lower energy e.g.

2s< sp

2) Orbitals having similar (n+l) values, the orbital with lower

value of n will have lower energy eg. 3p<4s

02/10/2012 86

02/10/2012

44

OrbitalsOrbitals in Many Electron Atomsin Many Electron Atoms

• Energy levels of orbitals

87

Order of filling atomic orbitals

Hund’s Rule of Maximum Multiplicity

• single electrons with same spin occupy each equal-energy

orbital before additional electrons with opposite spins can

occupy the same orbitals

02/10/2012 88

02/10/2012

45

• For shorthand electron

configurations

– Write the core electrons

corresponding to the filled

Noble gas in square brackets.

– Write the valence electrons

explicitly.

• Example,

• P: 1s22s22p63s23p3 but Ne is

1s22s22p6

• Therefore, P: [Ne]3s23p3.

• For atoms having having p, d

and f orbitals, the most stable

electron configuration is when

the d and f orbitals are half or

fully filled.

• Example:

• Cu (Z = 29)

Stable ec: = 1s2, 2s2, 2p6, 3s2,

3p6, 4s1, 3d10

• Cr (Z = 24)

Stable ec: = 1s2, 2s2, 2p6, 3s2,

3p6, 4s1 3d5

89

ShortShort formform ofof electronelectron configurationsconfigurations

Definitions

• Outermost shell: is the shell with maximum number of n value

that is fully or partially filled. Also called valence shell, and

electrons in this shell are called valence electrons

• Inner shells: are the shells other than the valence shell. The total

number of electrons in the inner shells are called core electrons

or kernel electrons

• Penultimate shell: this is the shell just before the valence shell

i.e. (n-1)nth shell, where n represents the valence shell.

• Antepenultimate shell: this is the (n-2)nth shell where n refers to

the valence shell.02/10/2012 90

02/10/2012

46

91

Effective Nuclear Charge

• Effective nuclear charge: is the net charge experienced by an electron in a many-electron atom.

• It is not the same as the charge on the nucleus because of the effect of the inner electrons.

• It results from the degeneracy of the orbitals with the same value of n.

• In a many-electron atom, for a given value of n, the effective nuclear charge decreases with increasing values of l.

• Z eff = Nuclear Charge (Z) – Screening Constant (σ)

� Z eff = Z - σ ................................................Eq. 51

Period Table 92

Screening and Penetration

The effects of other electrons in an atom on an electron is screening the positive charge. Since electrons are waves, they penetrate into space “occupied by other electrons”. No assumption can be made so that we can treat many-electron atoms as H-like atoms.

Thus, we assume the charge experienced by an electron as Zeff, the effective atomic charge (or number).

Thus, the energy of many electrons is

En = – RH ---------; RH = – 13.6 eV, the Rydberg costant

for H

Energies of sub-shells are also affected by the quantum number l, as we have pointed out before, but quantum numbers l and m also affect sizes.

Zeff2

n 2

02/10/2012

47

Calculation of effective nuclear charge

• Calculation of the effective nuclear charge is based on the Slates

rules:

1) Determination of the nuclear charge for electron in s and p

orbitals:

• The value of σ for electron in s or p orbital of the nth shell of an

atom or ion is the sum of the following:

1) Each electron in nth shell contributes 0.35 to the value of σ

2) Each electron in the (n-1)th shell contributes 0.85 to σ

3) Each electron in the (n-2)th shell and inner shells contributes 1

to the value of σ02/10/2012 93

Calculation of effective nuclear charge

4) No contribution to the value

of σ by electron in (n+1)th

shell and the shells above

5) If σ being calculated is for

electron in 1s orbital, the

remaining electron

contributes 0.3 to the value

of σ

Example:

1. Calculate the values of σ and

Zeff for 4s and 3d electrons in

a) Mn; Cu ; Cr

For Cu,

• EC: 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d10

• Rearranging: (1s2, 2s2, 2p6), (3s2,

3p6, 3d10) ,(4s1)

• σ = (1x 10)+(0.85x 18) + (0.35x 0)

= 25.30

• Z eff = Z - σ

= 29 – 25.30 = 3.70

02/10/2012 94

02/10/2012

48

Calculation of effective nuclear charge for

electron in d-0rbital• The value of σ is the sum of the

following:

a) No contribution to σ by

electron in the nth orbital

b) Each electron in the (n-1)d

orbital contribute 0.35 to the

value of σ

c) Each electron in the (n-1)s, (n-

1)p orbitals and inner shells

contribute 1 to the value of σ

Examples:

• Calculate the value of σ and

zeff for electron in 3d orbital

of: Cu; Mn ;Cr

For Cu,

• EC: 1s2, 2s2, 2p6, 3s2, 3p6, 4s1,

3d10

• Rearranging: (1s2, 2s2, 2p6, 3s2,

3p6 ) (3d10) ,(4s1)

• σ = (1x 18)+(0.35x 9) + (0x 1)

= 21.15

• Z eff = Z - σ

= 29 – 21.15 = 7.85

95

96

ElectronElectron ConfigurationsConfigurations andand thethePeriodicPeriodic TableTable

02/10/2012

49

Characteristics of s-block elements

• These are elements in which the last electron fills the s-orbital

• They are located at the extreme left of the periodic table

(Group IA and IIA)

• They consist of high electropositive metals

• The valence shell configuration is ns1-2

• The outer shell is partially filled

02/10/2012 97

Characteristics of p-block elements• The occur on the extreme right hand of the Periodic Table

• The nth shell of the noble gases is completely filled

• The (n-1)th shell of p-block elements of the 3rd period has 8

electrons, whereas that of the 4th, 5th and 6th periods has 18

electrons

• The (n-2)th shell of p-block elements of the 5th has 18 electrons

whereas that of the 6th has 32 (18+14) due to the inclusion of 14

Lanthanides

• The 2nd shell of the p-block elements has 3 electrons whereas the

3rd other periods has 8 electrons. The 1st shell of this block has 2

electrons02/10/2012 98

02/10/2012

50

Characteristics of d-block elements• They occur between s and p block elements

• The general electron configuration is ns p(n-1)d1-10

• General characteristics d block elements

1) They often form coloured compounds

2) They can have a variety of different oxidation states

3) They are often good catalysts e.g. Mn, Fe, Co, Cr

4) They are silvery-blue at room temperature (except copper and

gold)

5) They are solids at room temperature (except mercury)

6) They form complexes

7) They are often paramagnetic.

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Characteristics of f-block elements

1) Lanthanides

1) Silvery metals.

2) High melting points.

3) Found mixed in nature and

hard to separate.

4) Used in:

• Movie projectors

• Welder’s goggles

• TV and Computer monitors

2) Actinides

1) Radioactive elements.

2) Only 3 exist in nature.

3) Remaining are synthetic

(transuranium elements)

– greater atomic number

than uranium.

4) Decay quickly.

5) Used in:

• Home smoke detectors,

nuclear power plants.

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Atomic properties

• These are properties that change from one

element to the other, hence used to

characterise elements

1) Atomic radius

2) Ionization energy

3) Electronegativity

4) Electron affinity

5) Magnetic properties

6) Acidity and alkalinity of oxides

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102

Ionization EnergyIonization Energy

•The first ionization energy, I1, is the amount of energy required to remove the first outermost electron from a gaseous atom:

Na(g) → Na+(g) + e-.

•The second ionization energy, I2, is the energy required to remove the second outermost electron from a gaseous ion:

Na+(g) → Na2+

(g) + e-.

•The larger the ionization energy, the more difficult it is to remove the electron.

•There is a sharp increase in ionization energy when a core (non-valence) electron is removed.

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103

Ionization EnergyIonization EnergyPeriodic Trends in Ionization EnergyPeriodic Trends in Ionization Energy

•Ionization energy decreases down a group. �This means that the outermost electron is more readily removed as we go down a group.

�As the atom gets bigger, it becomes easier to remove an electron from the most spatially extended orbital.

•Ionization energy generally increases across a period.As we move across a period, Zeff increases.

�Therefore, it becomes more difficult to remove an electron.

104

Trend in Ionization Energy

Ionization energy, I, is the energy required to convert a gaseous atom or ion into a gaseous ion, in eV per ion or in J or kJ per mole. For example,

Mg (g) → Mg+ (g) + e–; I1 = 738 kJ / mol = 7.65 eV/atom

Mg+ (g) → Mg2+ (g) + e–; I2 = 1451 kJ / mol = 15.0 eV/atom

The effective atomic number Zeff, may be estimated using,

Eeff = --------- a positive value

But ionization energy is not Eeff.

Zeff2

n 2

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Period Table 105

Variation of I1 as Z Varies

How does first ionization energy I1vary in a group and in a period, why?

Decreases and increases respectively

Period Table 106

The In of Group n

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Trends in periodic table – atomic radius

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108

ElectronElectron ShellsShells andand thethe SizesSizes ofof AtomsAtoms

Electron Shells in AtomsElectron Shells in AtomsConsider a simple diatomic molecule.

The distance between the two nuclei is called the bond distance .

If the two atoms which make up the molecule are the same, then half the bond distance is called the covalent radius of the atom.

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Period Table 109

Atomic radii as Z increases

Period Table 110

From Atoms to Ions

arrange the following lists by increasing atomic radius.Na+, Li+, K+, Cs+, Xe, I-

Br– , Cl –, I –, F –

Be2+, Li+, B, C, O–2, F–, N

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Period Table 111

Trends of Ionic radii

Ions usually have the same electronic configuration as an inert gas.

He

Li+ Be2+ N3– O2– F – Ne

Na+ Mg2+ P3– S2– Cl – Ar

K+ Ca2+ Se2– Br – Kr

Rb+ Sr2+ Te2– I – Xe

Cs+ Ba2+

Iso-electrons: these are ions of different elements with same electron configuration but different ionic sizes

112

Electron AffinityElectron Affinity•Electron affinity is the opposite of ionization energy.

•Electron affinity is the energy change when a gaseous atom gains an electron to form a gaseous anion:

Cl(g) + e- → Cl-(g)

Electron affinity of different in eV per ion or in J or kJ per mole.

For example,

F (g) + e– → F – (g); EA = – 328 kJ / mol = – 3.4 eV/atom

Li (g) + e– → Li– (g); EA = – 59.6 kJ / mol = – 0.62 eV/atom

O (g) + e– → O– (g); EA1= – 141 kJ / mol = – 1.46 eV/atom

O– (g) + e– → O2– (g); EA2 = 744 kJ / mol = 7.71 eV/atom

The variation of EA is very irregular as Z increases. There is no particular trend in groups and in periods.

Note the relationship of EAs and Is.

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Electronegativity

δδδδ+ δδδδ– δδδδ0 δδδδ0

H Cl H H

This is the ability of an atom to attract the shared electrons towards itself.

This creates a partial negative charge on a more electronegative atom and a partial positive charge on a less elctronegative atom, hence the bond is polarised

Electronegativity cont.

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Period Table 115

Magnetic Properties• Materials can be divided into three types according to

their magnetic properties. Be able to explain these terms:

• Diamagnetic material: substance slightly repelled by a magnetic field. There are no unpaired electrons.

• Paramagnetic material: substance slightly attracted by a magnetic field. There are some unpaired electrons, (single e– in an orbital)

• Ferromagnetic material: substances strongly attracted by magnetic field. Magnetic domains line up in these type, Fe, Fe2O3 etc.

• Which of these are paramagnetic, H, Na, Mg, Cl, Cl –, Ag, Fe

116

Oxides and Strong AcidsAside from HCl, other strong acids are derived from oxides of N, S, Cl:

HNO3 H2SO4 HClO4HClO3

Some oxides also form weak acids:

HCO3 HNO2 H2SO3 HClO2

In contrast, look at some strong bases

NaOHKOH Ca(OH)2RbOH Sr(OH)2CsOH Ba(OH)2

topic 1 introduction to atomic structure 2012-1

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