Post on 11-Jul-2016
description
02/10/2012
1
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: vmadadi@uonbi.ac.ke, madadivin2002@yahoo.com
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
02/10/2012 5
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
02/10/2012 6
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.
02/10/2012 9
Direction of propagation
b) Properties of electromagnetic radiation
waves1. Wavelength (symbol = λ pronounced as “lambda”): The
distance between two consecutive peaks in the wave.
02/10/2012 10
1 m = 109 nm = 1010 Å = 1012 pm
nm = nanometre; = angstrom pm = Pico-metre
02/10/2012
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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
02/10/2012 12
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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
02/10/2012 15
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.
02/10/2012 17
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.
02/10/2012 18
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
02/10/2012
10
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.
02/10/2012 19
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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11
Types of emission spectra cont.
02/10/2012 21
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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12
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
02/10/2012 23
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.
02/10/2012 24
Source of Continuous radiation Grating Sample Spectrometer Frequency
Inte
nsity
02/10/2012
13
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.
02/10/2012 25
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
02/10/2012 26
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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
02/10/2012 27
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
02/10/2012 28
e-
e-
e-
e-
e-
UV light (hv)
Stream ofelectrons
Flow of electrons
Ammeter to measure current
02/10/2012
15
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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17
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
02/10/2012 34
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18
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
Bohr’s Atomic Model ...
� 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
02/10/2012 36
02/10/2012
19
Bohr’s Atomic Model ...
• 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
Bohr’s Atomic Model ...
• 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
02/10/2012
20
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
02/10/2012 40
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21
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
Expression for velocity of electron
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)
02/10/2012 42
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22
Expression for velocity of electron
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
Expression for velocity of electron
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
02/10/2012 44
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23
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
Calculating Energy of electron in nth Orbit
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
Calculating Energy of electron in nth Orbit
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.
02/10/2012 48
02/10/2012
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
02/10/2012 50
02/10/2012
26
Calculating wavelength(λ) and frequency
(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
Calculating wavelength(λ) and frequency
(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
Calculating wavelength(λ) and frequency
(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
Schr’o’ndinger’s Wave Equation cont.
• 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
Schr’o’ndinger’s Wave Equation cont.
• 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
Schr’o’ndinger’s Wave Equation cont.
• 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
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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