1. Quantum Theory and the Electronic Structure of Atoms
Chapter7
Quantum Theory and the Electronic Structure of Atoms
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2. 7.1 From classical physics to quantum theory.
Chapter7
7.1 From classical physics to quantum theory.
7.3 Bohr’s theory of the hydrogen atom.
7.6 Quantum numbers .
7.7 Atomic orbitals .
7.8 Electron configurations .
( ), ( ), ( )

3. 7.1 From classical physics to quantum theory.

4. Quantum Theory نظرية الكم
1-how many electrons are present in a particular atom?
كم عدد الإلكترونات الموجودة في كل ذرة.
2-What energies do individual electron possess?
ما هي الطاقة التي يحملها كل إلكترون من الإلكترونات الموجودة في الذرة.
3- where in the atom can electrons be founds?
أين توجد الإلكترونات في الذرة.

5. Wave is a form of energy transfer
Properties of Waves
Wavelength (l) is the distance between identical points on successive waves. (distance/waves)
Wavelength is expressed in units of m, cm, nm
Amplitude is the vertical distance from the midline of a wave to the peak or trough.

6. The speed of the wave (u) = l x n
Properties of Waves
Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). (waves/time)
1 Hz= s-1
The speed of the wave (u) = l x n
Distance/time= (distance/waves) × (waves/time)
7.1

8. Electromagnetic radiation
Maxwell (1873), proposed that visible light consists of electromagnetic waves.
Electromagnetic waves
has an electric field component and a magnetic field component.
These two components have the same wavelength and frequency, and hence the same speed, but they travel in mutually perpendicular planes.
Speed of light (c) in vacuum = 3.00 x 108 m/s
All electromagnetic radiation
l x n = c
7.1

10. Electromagnetic Radiation is the emission and transmission of energy in the form of electromagnetic waves.
The wavelength of electromagnetic waves is given in nm

11. c = l × n l = c/n l = 3.00 × 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m
A photon has a frequency of 6.0 x 104 Hz. Convert
this frequency into wavelength (nm).
Does this frequency fall in the visible region? l n
c = l × n
l = c/n
l = 3.00 × 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm
Radio wave

12. Photon is a “particle” of light
Plank’s Quantum theory
Energy (light) is emitted or absorbed in discrete units (quantum).
E = h x n = h c / l
= n h n
Planck’s constant (h)
h = 6.63 x J•s
n = 1, 2,3, ……..
Energy is always emitted or absorb in multiple hn
‘’Photoelectric Effect” Solved by Einstein
Photon is a “particle” of light
energy of photon E = hn
7.1

13. E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
When copper is bombarded with high-energy electrons, X-ray is emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X-ray is nm.
E = h x n
E = h x c / l
E = 6.63 x (J•s) x 3.00 x 10 8 (m/s) / x 10-9 (m)
E = 1.29 x J
7.2

14. Example 7.2 E = hν v = C/λ E = h C/λ
Calculate the energy (in joules) of:
A photon with a wavelength of 5x104 nm (IR region)
A photon with a wavelength of 5x10-2 nm (X-ray region)
E = hν
v = C/λ
E = h C/λ
1 nm=10-9 m
λa= 5x104 nm , λb= 5x10-2 nm , speed of light (C)= 3x108 m/s
Plank’s Constant (h) = 6.63x10-34 J.s
Ea = (6.63x10-34 J.s) (3x108 m/s)
5x104x10-9m
Ea = 3.98x10-21 J
Eb = (6.63x10-34 J.s) (3x108 m/s)
5x10-2x10-9m
Eb = 3.98x10-15 J

15. 7.3 Bohr’s theory of the hydrogen atom.

16. n (principal quantum number) = 1,2,3,…
Bohr’s Model of the Atom (1913)
1- e- can only have specific (quantized) energy values
2- light is emitted as e- moves from one energy level to a lower energy level
E = nhn
n= 1,2,3,.... hn
2hn
En = -RH
( ) 1 n2
n (principal quantum number) = 1,2,3,…
2hn < hn
RH (Rydberg constant) = 2.18 x 10-18J
7.3

17. E = nhv E = nhv 6 5 4 3 2 Ground State 1
Ground state or ground level is the lowest energy state of a system (in this case is the atom).
Excited state or excited level is the higher level in energy than the ground state.
E = nhv 6 5 4 3
E = nhv 2
Ground State 1

18. ( ) ( ) ( ) Ephoton = ∆E = Ef - Ei 1 Ef = -RH n2 1 Ei = -RH n2 1
( ) 1 n2 f
nf = 4
Ei = -RH
( ) 1 n2 i
nf = 3 i f
∆ E = RH
( ) 1 n2
nf = 2
ni > nf
emission
∆ E negative
ni < nf
absorbed
∆ E positive
nf = 1
7.3

19. ( ) ∆E = RH 1 n2 Ephoton = Ephoton = 2.18 × 10-18 J × (1/25 - 1/9)
Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. i f
∆E = RH
( ) 1 n2
Ephoton =
Ephoton = 2.18 × J × (1/25 - 1/9)
Ephoton = ∆E = × J
Ephoton = h C / λ
λ = h × C / Ephoton
λ = 6.63 × (J•s) × 3.00 × 108 (m/s)/1.55 × 10-19J
λ = 1.28 × 10-6 m ×109 = 1280 nm
7.3

20. Worked Example 7.4

21. 7.6 Quantum numbers .

22. Summary for Quantum Numbers 1- Principal Quantum Number (n)
Specifies the distance of e- from the nucleus (shell)
Values of (n) = 1, 2, 3, ….., ∞
Increase of (n) energy n 1 2 3 4 5 6 7
shell Name K L M N O P Q
For certain value of (n),
there are (n2) No. of orbitals
and (2n2) No. of electrons
No. of electron
2n2
No. of orbital n2 n
2= 2(1)2
1=12 1
8= 2(2)2
4=22 2
18= 2(3)2
9=32 3

23. 2- Angular Momentum (secondary) Quantum Number (l)
Specifies the shape of the orbital in space (subshell)
Values of (l) = 0, 1, 2, 3,….(n-1)
s p d f
Increase of (l) energy l 1 2 3
subshell Name s p d f
Subshell
l (0……..(n-1)) n s 1
s, p
0,1 2
s, p, d
0,1,2 3
s, p, d, f
0,1,2,3 4

24. 3- Magnetic Quantum Number ml
Specifies the orientation of the orbital in space.
for a given value of l
ml = -l, (-l+1), …., 0, ….,(+l-1) +l
For each l there are (2l+1) orbital
(orbital ) ml
(2l+1)
subshell
(0……..(n-1)) l n 1 s 4
-1,0,+1 3 p
-2,-1,0,+1,+2 5 d 2
-3,-2,-1,0,+1,+2,+3 7 f
The number of ml values indicates the number of orbitals in subshell

25. Shell n (1,2,3……)
Subshell l (s, p, d, f)
Orbital ml (s has one orbital)
(p has three orbitals px , py, pz )
(d has five orbitals)
( f has seven orbitals)
The maximum number of e- for one orbital = 2e-
The maximum number of e- for subshell= ( No. subshell)×2e-
s = 1×2=2 e-, p =3×2=6 e- , d =5×2=10 e- , f =7×2=14 e-

26. 4- Spin Quantum Number ms
Specifies the orientation of the spin axis of an electron.
Clockwise مع عقارب الساعة
Counterclockwise ضد عقارب الساعة
Values of (ms)= +½ or -½

27. 2px = 2py = 2pz (the same size and shape) f >d >p >s
3s >2s>1s
2px = 2py = 2pz (the same size and shape)
f >d >p >s
(n2)
(2l+1)
For certain value of (n),
there are (n2) No. of orbitals and (2n2) No. of electrons
For each l there are (2l+1) orbital
7.6

28. Schrödinger Wave Equation
Y = fn(n, l, ml, ms)
Existence (and energy) of electron in atom is described by its unique wave function Y.
Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers.
Each seat is uniquely identified
Each seat can hold only one individual at the same time
7.6

29. Schrödinger Wave Equation
Y = fn(n, l, ml, ms)
Shell – electrons with the same value of n (2s and 2p)
Subshell – electrons with the same values of n and l (2p)
Orbital – electrons with the same values of n, l, and ml (2px , 2py or 2pz )
How many electrons can an orbital hold?
If n, l, and ml are fixed, then ms = ½ or - ½
Y = (n, l, ml, +½)
or Y = (n, l, ml, -½)
An orbital can hold 2 electrons
7.6

30. How many 2p orbitals are there in an atom?
If l = 1, then (2l+1)=3 orbitals
ml = -1, 0, +1 2p
l = 1
How many electrons can be placed in the 3d subshell?
n=3
If l = 2, then (2l+1)=5 orbitals
ml = -2, -1, 0, +1, +2 3d
5 orbitals contain 5×2=10 e-
l = 2
7.6

31. Which of the orbital diagrams is not follow the pauli exclusion principle?

32. Worked Example 7.6
n=4
If l=2,
so the ml values are = +2, +1, 0, -1, -2 4d
l = 2

33. Worked Example 7.7
If n=3,
so the total number of orbitals are given by:
n2 = 32 = 9 orbitals

34. n=3 3p
l = 1
No. of orbitals = 2l+1 = 3 orbitals
No. of electrons = 2× 3 = 6 electrons

35. ( ) Energy of orbitals in a single electron atom 1 En = -RH n2
Energy only depends on principal quantum number n
n=3
excited state
En = -RH
( ) 1 n2
n=2
1s <2s =2p <3s =3p =3d <4s =4p =4d =4f
Energies of orbitals increase as follows:
Ground state
n=1

36. Energy of orbitals in a multi-electron atom
Energy depends on n and l
n=3, l = 2
n=3, l = 1
n=3, l = 0
n=2 l = 1
n=2, l = 0
n=1,l = 0
7.7

37. “Fill up” electrons in lowest energy orbitals (Aufbau principle)
H 1 electron
He 2 electrons
Li 3 electrons
Be 4 electrons
B 5 electrons
H 1s1
He 1s2
Li 1s22s1
Be 1s22s2
B 1s22s22p1

38. The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule).
C 6 electrons
N 7 electrons
O 8 electrons
F 9 electrons
Ne 10 electrons
C 1s22s22p2
N 1s22s22p3
O 1s22s22p4
F 1s22s22p5
Ne 1s22s22p6
7.7

39. Which of the orbital diagrams is not follow Hund's rule?

40. Order of orbitals (filling) in multi-electron atom

41. 7.8 Electron Configuration.

42. in the orbital or subshell
Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom.
1-The electron configuration of the electron in a ground-state hydrogen atom:
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
2-The electron configuration can also be represented by:
Orbital diagram
This way is useful to determine the four quantum No. especially
ml and ms for each electron in an atom.
The spin of electron
1s1 H
7.8

43. 1s < 2s < 2p < 3s < 3p < 4s
What is the electron configuration of 12Mg?
1s < 2s < 2p < 3s < 3p < 4s
12Mg 1s22s22p63s2
Electron configuration using noble gas core
= 12 electrons
[10Ne] 1s22s22p6
Abbreviated as 12Mg [Ne]3s2
Valence electrons
(Determine the No. of group in periodic table)
Determine the period in which the element is located
What are the possible quantum numbers for the last (outermost) electrons in 17Cl ?
1s < 2s < 2p < 3s < 3p < 4s
17Cl 1s22s22p63s23p5
= 17 electrons
Abbreviated as 17Cl [Ne]3s2 3p5 +1 -1 3p
Last electrons added to 3p orbital
n = 3
l = 1
ml = -1, 0, +1
ms = +½ or -½
7.8

44. Outermost subshell being filled with electrons
Classification of groups of elements in the periodic table according to
the type of subshell being filled with electron (blocks)
1A, 2A
Representative elements
3A, 4A, 5A,6A,7A
Transition metals
Lanthanides (rare earth) and actinides series

45. Ground State Electron Configurations of the Elements (IUPAC)
ns2np6
Ground State Electron Configurations of the Elements (IUPAC)
ns1
ns2np1
ns2np2
ns2np3
ns2np4
ns2np5
ns2
Two irregularities
d10 d1 d5 4f 5f
8.2

46. Noble gas core(page307)
Transition metals(page 309)
Lanthanides or rare earth series(page309)
Actinide series(page309)
What is the electron configuration of Cr24?
What is the electron configuration of Cu29?

47. 24Cr
[18Ar]
1s2 2s2 2p6 3s2 3p6 4s2 3d4
less stable
24Cr
1s2 2s2 2p6 3s2 3p6 4s1 3d5
more stable
ns2 (n-1)d4 Less stable
ns1 (n-1)d5 more stable
29Cu
[18Ar]
1s2 2s2 2p6 3s2 3p6 4s2 3d9
Less stable
29Cu
1s2 2s2 2p6 3s2 3p6 4s1 3d10
more stable
ns2 (n-1)d9 less stable
ns1 (n-1)d10 more stable

48. The same chemical properties
The same group
3s2 3p1
[10Ne] Al
4s2 3d10 4p5
[18Ar] Br
4s2 3d10 4p1
[18Ar] Ga

50. كل عنصر بعدد ذري فردي بارا وليس كل زوجي دايا
Any atom with an odd number of electron will always paramagnetic
كل عنصر بعدد ذري فردي بارا وليس كل زوجي دايا 2p 2p
Paramagnetic
المغناطيسية المتوازية
Diamagnetic
المغناطيسية المزدوجة
Substances are those contain
unpaired electrons and attracted
by a magnet.
Substances are those do not contain
unpaired electrons and slightly
repelled by a magnet.

51. Which of the following elements would be expected to be paramagnetic
Which of the following elements would be expected to be paramagnetic? Diamagnetic? He, Be, Li, N
All of the electrons are spin-paired in diamagnetic elements so their subshells are completed, causing them to be unaffected by magnetic fields. Paramagnetic elements are strongly affected by magnetic fields because their subshells are not completely filled with electrons. So, to determine whether the elements are paramagnetic or diamagnetic, write out the electron configuration for each element. 1s
He: 1s2 subshell is filled
Be: 1s22s2 subshell is filled
Li: 1s22s1 subshell is not filled
N: 1s22s22p3 subshell is not filled
diamagnetic 2s
diamagnetic 2s
Paramagnetic +1 -1 2p
Paramagnetic

52. -1

54. General Rules for Assigning Electrons to Atomic orbitals
مما سبق يمكننا وضع القواعد التالية لكيفية عمل التوزيع الإلكتروني على مختلف الأغلفة الفرعية والأفلاك الذرية. هذه القواعد هي:
لكل غلاف رئيس (shell) قيمته n هناك عدد n أيضا من الأغلفة الفرعية ((subshell أي للغلاف n =2 هناك غلافان فرعيان هما 2s و 2p.
كل غلاف فرعي له القيمة l يحتوي على عدد (2l+1) من الأفلاك(orbitals) مثلا الغلاف الفرعي p يحتوي على 3 أفلاك.
لا يمكن أن يوجد أكثر من إلكترونين في كل فلك ذري (atomic orbital) وبالتالي فإن أقصى عدد للإلكترونات في كل غلاف فرعي (subshell) هي ضعف عدد أفلاكه.
يمكن تعيين العدد الأقصى من الإلكترونات في كل غلاف رئيس(shell) حسب العلاقة 2n 2.

55. Problems
7.3 – 7.8 – 7.16 – 7.18
7.32 – 7.34 – 7.120
7.56 – 7.58 – 7.62 – 7.66 – 7.70
7.76 – 7.78 – 7.79 – 7.84 – 7.88 – 7.90 – 7.124