CHAPTER 2
ATOMIC STRUCTURE
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lLECTURE 1
2.1 Bohr’s Atomic Model
lBohr’s
Atomic Model
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lAt the end of this topic students should be able to:-
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b)Explain the existence of electron
energy levels in an atom.
c)Calculate the energy of electron using :
l En = - RH (1/n2) ,
l RH = 2.18 x
10-18 J
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In 1913, a young Dutch
physicist, Niels Bohr proposed a theory of atom
that shook the scientific world.
The atomic model he
described had electrons circling a
central nucleus that
contains positively charged protons.
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l10:61 (Yunus)
lKAMU TIDAK BERADA DALAM
SUATU KEADAAN DAN TIDAK MEMBACA SUATU AYAT DA AL QURAN DAN KAMU TDAK MENGERJAKAN SESUATU
PEKERJAAN , MELAINKAN KAMI MENJADI SAKSI ATAS MU DI WAKTU KAMU MELAKUKAN NYA.
TIDAK LUPUT DARI PENGETAHUAN TUHANMU BIARPUN SEBESAR ZARAH (ATOM) DI BUMI ATAU
PUN LANGIT. TIDAK ADA YANG LEBIH KECIL DAN TIDAK PULA YANG LEBIH BESAR DARI
ITU, MELAINKAN (SEMUA TERCATAT) KITAB YANG NYATA (LAUH MAHFUZ)
l4:40 (AN NISAA)
lSESUNGGUHNYA ALLAH TIDAK
MENGANIAYA SESEORANG WALAUPUN SEBESAR ZARAH, NESCAYA ALLAH AKAN MELIPAT
GANDAKAN DAN MEMBERIKAN DARI SISINYA PAHALA YANG BESAR.
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2. The energy of an electron in a
hydrogen atom is quantised, that
is, the electron has only a fixed set of allowed orbits, called stationary states.
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lThe energy of an electron in its level is
given by:
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RH (Rydberg constant)
or A = 2.18´10-18J.
n (principal quantum number) = 1, 2, 3
….¥ (integer)
Note:
ln identifies the orbit of electron
lEnergy is zero if electron is located
infinitely far from nucleus
lEnergy associated with forces of attraction
are taken to be negative (thus, negative sign)
lLECTURE 2
At the end of this topic
students should be able to:-
d) Describe the formation of
line spectrum of hydrogen atom
e) Calculate the energy
change of an electron during transition.
DE = RH (1/n12 - 1/n22) ,
where RH = 2.18 x
10-18 J
f) Calculate the photon of energy emitted by an electron that produces
a particular wavelength during transition
DE = hn where n= c/λ
lEmission
Spectra
lContinuous Spectrum
A spectrum consists all
wavelength components (containing an unbroken sequence of frequencies) of the visible
portion of the electromagnetic spectrum are present.
It is produced by incandescent solids, liquids, and compressed
gases.
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lWhen white light from
incandescent lamp is passed through a slit then
a prism, it separates into a spectrum.
lThe white light spread out
into a rainbow of colours produces a continuous
spectrum.
lThe spectrum is continuous
in that all wavelengths are presents and each colour merges
into the next without a break.
lLine Spectrum (atomic spectrum)
A spectrum consists of discontinuous & discrete lines produced by excited atoms and ions as the
electrons fall back to a lower energy level. The radiation emitted is only at a
specific wavelength or frequency. It means each line corresponds
to a specific wavelength or frequency.
lLine
spectrum are composed of only a few wavelengths giving a series of discrete
line separated by blank areas
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lExercise: Complete the following table
lExercise
The following diagram depicts the line
spectrum of hydrogen atom. Line A is the first line of the Lyman series.
lExercise
Describe the transitions of electrons that
lead to the lines W, and Y, respectively.
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lHomework
Calculate En for n = 1, 2, 3, and 4. Make a
one-dimensional graph showing energy, at different values of n, increasing vertically. On
this graph, indicate by vertical arrows transitions that lead to lines in
a) Lyman series
b) Paschen series
lSignificance
of Atomic Spectra
lIn Lyman series, the
frequency of the convergence of spectral lines can be used to find the
ionisation energy of hydrogen atom:
IE = hn¥
lThe frequency of the first
line of the Lyman series > the frequency of the first line of the Balmer
series.
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lExercise
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lRadiant energy emitted when the electron
moves from higher-energy state to lower-energy state is given by the difference
in energy between energy levels:
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lThe amount of energy released by the electron is called a
photon of energy.
lA photon of energy is emitted in the form of radiation with
appropriate frequency and wavelength.
where;
h (Planck’s constant) =6.63 ´ 10-34 J s
n = frequency
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lExercises:
1)Calculate
the energy of an electron in the second energy level of a hydrogen atom.
(-5.448 x 10-19 J)
2)Calculate
the energy of an electron in the energy level n = 6 of an hydrogen atom.
3) Calculate the energy change (J), that
occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen
atom.
(answer: 1.55 x 10-19J)
4)Calculate
the frequency and wavelength (nm) of the radiation emitted in question 3.
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lLECTURE 3
At the end of this topic
students should be able to:-
g) Perform calculations involving the Rydberg equation for Lyman, Balmer, Paschen, Brackett and Pfund series .
1/ λ = RH (1/n12 - 1/n22) ,
where RH = 1.097 x
107 m-1 and n1<n2
h) Calculate
the ionisation energy of hydrogen atom from Lyman series.
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i)State the
weaknesses of Bohr’s atomic model.
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j)State the
dual nature of electron using the
Broglie’s postulate and Hesseinberg’s uncertainty principle.
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lWavelength emitted by the
transition of electron between two energy levels is calculated using Rydberg
equation:
lExample
Calculate the wavelength, in nanometers of the spectrum of hydrogen
corresponding to ni = 2 and nf = 4 in the Rydberg equation.
lExample
Use
the Rydberg
equation to calculate the wavelength of the spectral line of hydrogen atom that
would result when an electron drops from the fourth orbit to the second orbit,
then identified the series the line would be found.
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Calculate the wavelengths of the fourth line
in the Balmer series of hydrogen.
l Different values of RH and its
usage
1.RH = 1.097 ´ 107 m-1
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Calculate the energy liberated when an
electron from the fifth energy level falls to the second energy level in
the hydrogen atom.
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Calculate what is;
i )
Wavelength
ii ) Frequency
iii ) Wave number of the
last line of hydrogen spectrum
in Lyman series
Wave number = 1/wavelength
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lDefinition
: Ionization energy is the minimum energy required to remove one mole of
electron from one mole of gaseous atom/ion.
M (g) ® M+ (g) +
e DH = +ve
lThe hydrogen atom is said to
be ionised when electron is removed
from its ground state (n = 1) to n = ¥.
lAt n = ¥, the potential energy of electron is zero, here the nucleus
attractive force has no effect on the electron
(electron is free from nucleus).
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ln1 = 1, n2 = ∞
∆E = RH (1/n12 – 1/n22)
= 2.18 X 10 -18 (1/12 – 1/ ∞ 2)
= 2.18 X 10 -18 (1 – 0)
= 2.18 X 10 -18 J
Ionisation energy
= 2.18 X 10 -18x 6.02 X
1023J mol-1 =1.312 x 106 J mol-1
= 1312 kJ mol-1
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10.97
10.66 10.52 10.27 9.74 8.22
wave number (x106 m-1)
The Lyman series of the
spectrum of hydrogen is shown above. Calculate the ionisation energy of
hydrogen from the spectrum.
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ΔE = hc/λ
=h x c / λ = h x c x wave no.
= 6.626 x 10-34 J s x 3 x 108 m s-1 x 10.97x 106 m-1 = 218.06x 10-20 J
= 2.18 x 10-18J
Ionisation energy
= 2.18 X 10 -18x 6.02 X 1023 J mol-1
=1.312 x 106 J mol-1
= 1312 kJ mol-1
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Compute the ionisation energy of hydrogen
atom in kJ mol-1.
lThe weakness of Bohr’s
Theory
1.His theory could not be
extended to predict the energy levels and spectra of atoms and ions with more
than one electron. It only can explain the hydrogen spectrum or ions contain
one electron eg He+, Li2+.
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2.Electrons are restricted to
orbit the nucleus at certain fixed distances
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3.It cannot explain for the
dual nature of electron
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4.It cannot explain for the
extra lines formed in the hydrogen spectrum.
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Davisson & Germer observed the diffraction of
electrons when a beam of electrons was directed at a nickel crystal.
Diffraction patterns produced by scattering electrons from crystals are very
similar to those produced by scattering X-rays from crystals. This experiment
demonstrated that electrons do indeed possess
wavelike properties.
Thus, can the ‘position’ of a wave be specified???
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Broglie’s Postulate
In 1924 Louis de Broglie proposed that not only light but all matter has a dual nature and possesses
both wave and corpuscular properties. De
Broglie deduced that the particle and wave properties are related by the
expression:
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lHeisenberg’s
Uncertainty Principle
It is impossible to know simultaneously both
the momentum p (defined as mass times velocity) and the position of a particle with certain.
lLECTURE 4
l2.2 QUANTUM MECHANICAL MODEL
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l At the end of this topic students should be
able to:-
b)State the four quantum numbers of
an electron in an orbital (n, l,m,s).
l i) principal quantum number, n
l ii) angular momentum quantum
number, l
l iii) magnetic quantum number, m
l iv) electron spin quantum number, s.
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lAtomic
Orbital
An orbital is a
three-dimensional region in space around the nucleus where there is a high
probability of finding an electron.
lQuantum
Numbers
Each of the electrons in an
atom is described and characterised by a set of four quantum numbers, namely
a) principal quantum number, n
b) angular momentum quantum number, l
c) magnetic quantum number, m
d) electron spin quantum number, s.
lPrincipal
Quantum Number, n
lThe
value of n
determines the energy of an orbital and thereby the energy of the electron in
that particular orbital.
lThe
principal quantum number may have only integral values: n =1, 2, 3, …, ¥.
lAngular
Momentum Quantum Number, l
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lMagnetic
Quantum Number, m
lThe direction
or orientation of the magnetic field is
determined by the value of m.
lPossible values of m
depend on the value of l. For a given l, m can
be : -l, …, 0, …, +l
(-l £ m £ +l)
If l = 0, m can only be 0 Þ one orbital in s-subshell
If l = 1, m can be -1, 0, +1 Þ three orbitals in p-subshell
If l = 2, m can be -2, -1, 0, +1, +2 Þ five orbitals in d-subshell
lMagnetic Quantum Number, m
The number of m values indicates the number of orbitals in a
subshell with a particular l value.
The values of n = 2 and l = 1
indicate that we have a 2p-subshell, and in this subshell we have three 2p-orbitals (because there are three values of m,
given by -1, 0 and +1)
lElectron Spin Quantum
Number, s
lThe value of s determines the direction of spinning motions of an electron (either clockwise or counter clockwise)
which is spinning on its own axes, as Earth does.
lThe electron spin quantum number has a value
of
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lExercise: Complete the following table
lExercise
State whether or not each of
the following symbols is an acceptable designation for an atomic orbital.
Explain what is wrong with the unacceptable symbols.
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LECTURE 5
At the end of this topic students should be
able to:-
c) Sketch the shapes of s,p and d orbitals with the correct orientations.
lShape of Atomic Orbitals
a)s orbitals
üSpherical shape with the
nucleus at the centre.
üThe probability of finding
electrons at the distance r from the nucleus is the same from all direction.
üWhen l = 0
üAs n increases s orbital
ü gets larger
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b) p orbitals
l When l = 1
ldumbbell
shaped
l three p-orbitals px, py, and pz.
lcorrespond
m
of -1, 0, and +1.
lAs n increases, the p-orbitals get larger.
lAll p-orbitals have a node at the nucleus.
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lLECTURE 6
& 7
l2.3 ELECTRONIC CONFIGURATION
lElectronic
Configuration
l At the end of this
topic students should
l be able to:-
a)
State
and use Aufbau principle, Hund’s rule and Pauli exclusion
principle
b)Apply
rules in (a) to fill electrons into atomic orbitals.
c)
Write the electronic
configuration of an atom and monostomic ions by using spdf notation .
ld) Explain the anomalous electronic
configurations of chromium and copper.
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lRepresenting Electronic
Configuration
Method 1: Orbital diagram
lRules for Assigning
Electrons to Orbitals
lRelative Energy Level of
Atomic Orbitals
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ii) Pauli Exclusion Principle
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iii) Hund’s Rule
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lHund’s rule states that when
electrons are added to the orbital of equivalent energy (for degenerate
orbital), each orbital are filled singly with electron of the parallel spins
(same) first before it is paired.
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Indicate which of the
following orbital diagrams are acceptable or unacceptable for an atom in ground
state. Explain what mistakes have been made in each and draw the correct
orbital diagram:
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Draw
‘electrons-in-boxes’ diagram of the electronic configuration of titanium, Ti (Z
= 22). Also,
write the ground-state electronic configurations for Ti and Ti2+ ion.
lIMPORTANT!
lIn an empty atom, the 4s
orbital has a lower energy compared to that of the 3d orbital. That is why
electrons fill the 4s orbital first before filling the 3d orbital.
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lHowever, once electron/s is/are added to the 3d orbital, the 4s
electrons are repelled
to a higher energy level. The 3d orbitals now have lower energy than 4s.
lPoints to remember
lThe electronic configuration of atom or monatomic ion at ground state
Þ Distribution of electrons
obeys Aufbau
principle, Pauli exclusion principle and Hund’s rule
lEach atomic orbital can only accommodate a maximum of 2 electrons
lAtomic
orbital is a 3-D region in space around the nucleus where
there is a high probability of finding an electron.
lAssigning electrons to subshells
s-orbital Þ a max of 2 electrons (ns2)
p-orbitals Þ a max of 6 electrons (np6)
d-orbitals Þ a max of 10 electrons (nd10)
lThe Anomalous Electronic Configurations of Cr
and Cu
lCr and Cu have electron
configurations which are inconsistent with the Aufbau principle. The anomalous
are explained on the basis that a filled or half-filled
orbital is more stable.
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Ø
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24Cr : 18[Ar]
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The actual orbital notation
24Cr : 18[Ar]
Half filled orbital is more
stable
(possesses an extra, added stability)
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Copper predicted orbital notation
ØCu : [Ar]
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The actual orbital notation
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ØCu : [Ar]
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Write the ground-state electronic configuration
and explain the anomalous case for Cr (Z=24) and Cu ( Z=29)
lWriting Electronic Configuration for Negative
Ion
Add electron according to Aufbau
Principle.
Example:
i. Cl-
ii. O2-
lWriting Electronic Configuration for Positive
Ions
Remove
electron from the outermost orbital (largest value of n)
Example:
i. Mg2+
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ii. K+
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iii. Fe2+
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