1. Atomic Structure and Periodicity Chapter 7
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2. The speed (u) of the wave = l x n
Properties of Waves
Wavelength (l) is the distance between identical points on successive waves.
Amplitude is the vertical distance from the midline of a wave to the peak or trough.
Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s).
The speed (u) of the wave = l x n

3. Speed of light (c) in vacuum = 3.00 x 108 m/s
Maxwell (1873), proposed that visible light consists of electromagnetic waves.
Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves.
Speed of light (c) in vacuum = 3.00 x 108 m/s
All electromagnetic radiation
l x n = c

5. l x n = c l = c/n l = 3.00 x 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
l x n = c
l = c/n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
l = 5.0 x 1012 nm

6. Mystery #1, “Heated Solids Problem” Solved by Planck in 1900
When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths.
Radiant energy emitted by an object at a certain temperature depends on its wavelength.
Energy (light) is emitted or absorbed in discrete units (quantum).
E = h x n
Planck’s constant (h)
h = 6.63 x J•s

7. Photon is a “particle” of light
Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 hn
Light has both:
wave nature
particle nature
KE e-
Photon is a “particle” of light
hn = KE + W
KE = hn - W
where W is the work function and
depends how strongly electrons
are held in the metal

8. 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 rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays 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

9. Line Emission Spectrum of Hydrogen Atoms

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

12. E = hn
E = hn

13. ( ) ( ) ( ) Ephoton = DE = Ef - Ei 1 Ef = -RH n2 1 Ei = -RH n2 1
nf = 2
ni = 3
nf = 1
ni = 3
Ef = -RH
( ) 1 n2 f
Ei = -RH
( ) 1 n2 i
nf = 1
ni = 2 i f
DE = RH
( ) 1 n2

15. 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
DE = RH
( ) 1 n2
Ephoton =
Ephoton = 2.18 x J x (1/25 - 1/9)
Ephoton = DE = x J
Ephoton = h x c / l
l = h x c / Ephoton
l = 6.63 x (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
l = 1280 nm

16. Why is e- energy quantized?
De Broglie (1924) reasoned that e- is both particle and wave.
2pr = nl l = h mu
u = velocity of e-
m = mass of e-

17. What is the de Broglie wavelength (in nm) associated with a 2
What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?
l = h/mu
h in J•s
m in kg
u in (m/s)
l = 6.63 x / (2.5 x 10-3 x 15.6)
l = 1.7 x m = 1.7 x nm

18. Chemistry in Action: Laser – The Splendid Light
Laser light is (1) intense, (2) monoenergetic, and (3) coherent

19. Chemistry in Action: Electron Microscopy
le = nm
STM image of iron atoms
on copper surface
Electron micrograph of a normal red blood cell and a sickled red
blood cell from the same person

20. Schrodinger Wave Equation
In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e-
Wave function (y) describes:
. energy of e- with a given y
. probability of finding e- in a volume of space
Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems.

21. Schrodinger Wave Equation
y is a function of four numbers called
quantum numbers (n, l, ml, ms)
principal quantum number n
n = 1, 2, 3, 4, ….
distance of e- from the nucleus
n=1
n=2
n=3

22. Where 90% of the
e- density is found
for the 1s orbital
Radial probability max. at 52.9 pm from nucleus, equal to radius of innermost orbit

23. Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
angular momentum quantum number l
for a given value of n, l = 0, 1, 2, 3, … n-1
l = s orbital
l = p orbital
l = d orbital
l = f orbital
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
Shape of the “volume” of space that the e- occupies

24. l = 0 (s orbitals)
l = 1 (p orbitals)

25. l = 2 (d orbitals)

26. Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
magnetic quantum number ml
for a given value of l
ml = -l, …., 0, …. +l
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
orientation of the orbital in space

27. ml = -1, 0, or 1
3 orientations is space

28. ml = -2, -1, 0, 1, or 2
5 orientations is space

29. Schrodinger Wave Equation
(n, l, ml, ms)
spin quantum number ms
ms = +½ or -½
ms = +½
ms = -½

30. Schrodinger Wave Equation
quantum numbers: (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 (E, R12, S8)
Each seat can hold only one individual at a time

32. Schrodinger Wave Equation
quantum numbers: (n, l, ml, ms)
Shell – electrons with the same value of n
Subshell – electrons with the same values of n and l
Orbital – electrons with the same values of n, l, and ml
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

33. How many 2p orbitals are there in an atom?
If l = 1, then ml = -1, 0, or +1 2p
3 orbitals
l = 1
How many electrons can be placed in the 3d subshell?
n=3
If l = 2, then ml = -2, -1, 0, +1, or +2 3d
5 orbitals which can hold a total of 10 e-
l = 2

34. ( ) Energy of orbitals in a single electron atom 1 En = -RH n2
Energy only depends on principal quantum number n
n=3
n=2
En = -RH
( ) 1 n2
n=1

35. 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

36. “Fill up” electrons in lowest energy orbitals (Aufbau principle)
C 6 electrons ? ?
B 5 electrons
B 1s22s22p1
Be 4 electrons
Be 1s22s2
Li 1s22s1
Li 3 electrons
He 1s2
He 2 electrons
H 1 electron
H 1s1

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

38. Order of orbitals (filling) in multi-electron atom
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

39. in the orbital or subshell
Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom.
number of electrons
in the orbital or subshell
1s1
principal quantum
number n
angular momentum
quantum number l
Orbital diagram
1s1 H

40. What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2
= 12 electrons
Abbreviated as [Ne]3s2
[Ne] 1s22s22p6
What are the possible quantum numbers for the last (outermost) electron in Cl?
Cl 17 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s23p5
= 17 electrons
Last electron added to 3p orbital
n = 3
l = 1
ml = -1, 0, or +1
ms = ½ or -½

41. Outermost subshell being filled with electrons

43. Paramagnetic
Diamagnetic
unpaired electrons
all electrons paired 2p 2p

44. When the Elements Were Discovered

45. Ground State Electron Configurations of the Elements
ns2np6
Ground State Electron Configurations of the Elements
ns1
ns2np1
ns2np2
ns2np3
ns2np4
ns2np5
ns2
d10 d1 d5 4f 5f

46. Classification of the Elements

47. Electron Configurations of Cations and Anions
Of Representative Elements
Na [Ne]3s1
Na+ [Ne]
Atoms lose electrons so that cation has a noble-gas outer electron configuration.
Ca [Ar]4s2
Ca2+ [Ar]
Al [Ne]3s23p1
Al3+ [Ne]
H 1s1
H- 1s2 or [He]
Atoms gain electrons so that anion has a noble-gas outer electron configuration.
F 1s22s22p5
F- 1s22s22p6 or [Ne]
O 1s22s22p4
O2- 1s22s22p6 or [Ne]
N 1s22s22p3
N3- 1s22s22p6 or [Ne]

48. Cations and Anions Of Representative Elements+1 +2 +3 -3 -2 -1

49. Isoelectronic: have the same number of electrons, and hence the same ground-state electron configuration
Na+: [Ne]
Al3+: [Ne]
F-: 1s22s22p6 or [Ne]
O2-: 1s22s22p6 or [Ne]
N3-: 1s22s22p6 or [Ne]
Na+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne
What neutral atom is isoelectronic with H- ?
H-: 1s2
same electron configuration as He

50. Electron Configurations of Cations of Transition Metals
When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals.
Fe: [Ar]4s23d6
Mn: [Ar]4s23d5
Fe2+: [Ar]4s03d6 or [Ar]3d6
Mn2+: [Ar]4s03d5 or [Ar]3d5
Fe3+: [Ar]4s03d5 or [Ar]3d5

51. Effective nuclear charge (Zeff) is the “positive charge” felt by an electron.
Zeff = Z - s
0 < s < Z (s = shielding constant)
Zeff  Z – number of inner or core electrons
Zeff
Core Z
Radius (pm) Na Mg Al Si 11 12 13 14 10 1 2 3 4
186
160
143
132

52. Effective Nuclear Charge (Zeff)
increasing Zeff
increasing Zeff

53. Atomic Radii
covalent radius
metallic radius

55. Trends in Atomic Radii

56. Comparison of Atomic Radii with Ionic Radii

57. Cation is always smaller than atom from which it is formed.
Anion is always larger than atom from which it is formed.

58. The Radii (in pm) of Ions of Familiar Elements

59. Chemistry in Action: The 3rd Liquid Element?
117 elements, 2 are liquids at 250C – Br2 and Hg
223Fr, t1/2 = 21 minutes
Liquid?

60. Ionization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state.
I1 + X (g) X+(g) + e-
I1 first ionization energy
I2 + X+(g) X2+(g) + e-
I2 second ionization energy
I3 + X2+(g) X3+(g) + e-
I3 third ionization energy
I1 < I2 < I3

62. Variation of the First Ionization Energy with Atomic Number
Filled n=1 shell
Filled n=2 shell
Filled n=3 shell
Filled n=4 shell
Filled n=5 shell

63. General Trends in First Ionization Energies
Increasing First Ionization Energy
Increasing First Ionization Energy

64. Electron affinity is the negative of the energy change that occurs when an electron is accepted by an atom in the gaseous state to form an anion.
X (g) + e X-(g)
F (g) + e X-(g)
DH = -328 kJ/mol
EA = +328 kJ/mol
O (g) + e O-(g)
DH = -141 kJ/mol
EA = +141 kJ/mol

66. Variation of Electron Affinity With Atomic Number (H – Ba)