1. Autumn 1999
2. The best available balances can weigh amounts as small
as 10-5 g. If you were to count out water molecules at the
rate of one per second, how long would it take to count a
pile of molecules large enough to weigh 10-5 g?

2. Autumn 1999
2. The best available balances can weigh amounts as small
as 10-5 g. If you were to count out water molecules at the
rate of one per second, how long would it take to count a
pile of molecules large enough to weigh 10-5 g?
1 molecule H2O has mass of = 18 amu
1 mole H2O has mass of 18 g  x 1023 molecules

3. Autumn 1999
2. The best available balances can weigh amounts as small
as 10-5 g. If you were to count out water molecules at the
rate of one per second, how long would it take to count a
pile of molecules large enough to weigh 10-5 g?
1 molecule H2O has mass of = 18 amu
1 mole H2O has mass of 18 g  x 1023 molecules
10-5 g  10-5/18 moles = 5.6 x 10-7 moles

4. Autumn 1999
2. The best available balances can weigh amounts as small
as 10-5 g. If you were to count out water molecules at the
rate of one per second, how long would it take to count a
pile of molecules large enough to weigh 10-5 g?
1 molecule H2O has mass of = 18 amu
1 mole H2O has mass of 18 g  x 1023 molecules
10-5 g  10-5/18 moles = 5.6 x 10-7 moles
 5.6 x 10-7 x x 1023 molecules = 3.35 x 1017 molecules

5. Autumn 1999
2. The best available balances can weigh amounts as small
as 10-5 g. If you were to count out water molecules at the
rate of one per second, how long would it take to count a
pile of molecules large enough to weigh 10-5 g?
1 molecule H2O has mass of = 18 amu
1 mole H2O has mass of 18 g  x 1023 molecules
10-5 g  10-5/18 moles = 5.6 x 10-7 moles
 5.6 x 10-7 x x 1023 molecules = 3.35 x 1017 molecules
3.35 x 1017 s

6. Autumn 2000
13. Hemoglobin absorbs light of wavelength 407 nm.
Calculate the energy (in J) of one millimole of photons of
this light.

7. Autumn 2000
13. Hemoglobin absorbs light of wavelength 407 nm.
Calculate the energy (in J) of one millimole of photons of
this light.
E = hn = hc/l
= x x 3 x 108 /407 x 10-9 = J s m s-1 m-1

8. Autumn 2000
13. Hemoglobin absorbs light of wavelength 407 nm.
Calculate the energy (in J) of one millimole of photons of
this light.
E = hn = hc/l
= x x 3 x 108 /407 x 10-9 = J s m s-1 m-1
= 4.88 x J

9. Autumn 2000
13. Hemoglobin absorbs light of wavelength 407 nm.
Calculate the energy (in J) of one millimole of photons of
this light.
E = hn = hc/l
= x x 3 x 108 /407 x 10-9 = J s m s-1 m-1
= 4.88 x J
1 millimole = 10-3 mole = x 1020 photons

10. Autumn 2000
13. Hemoglobin absorbs light of wavelength 407 nm.
Calculate the energy (in J) of one millimole of photons of
this light.
E = hn = hc/l
= x x 3 x 108 /407 x 10-9 = J s m s-1 m-1
= 4.88 x J
1 millimole = 10-3 mole = x 1020 photons
energy of 1 millimole of photons  x 1020x 4.88 x J

11. Autumn 2000
13. Hemoglobin absorbs light of wavelength 407 nm.
Calculate the energy (in J) of one millimole of photons of
this light.
E = hn = hc/l
= x x 3 x 108 /407 x 10-9 = J s m s-1 m-1
= 4.88 x J
1 millimole = 10-3 mole = x 1020 photons
energy of 1 millimole of photons  x 1020x 4.88 x J
= 294 J

13. Quantization of e- energy levels by standing
wave assumption
Derived a mathematical description for e-s as
3-d standing waves in atoms:
H = E
, the wave function, is a function spatial
position of e-;
H, is a mathematical function which allows the
calculation of the e- energy (Hamiltonian
operator); and
E, is the total energy of the electron

14. What is 2? " Amplitude probability function "
e.g. for a linear standing wave: Amplitude = sin x ,
where x is distance along the wave

15. What is 2? " Amplitude probability function "
e.g. for a linear standing wave: Amplitude = sin x ,
where x is distance along the wave
Why amplitude probability function ?
Heisenberg Uncertainty Principle (1927):

16. What is 2? " Amplitude probability function "
e.g. for a linear standing wave: Amplitude = sin x ,
where x is distance along the wave
Why amplitude probability function ?
Heisenberg Uncertainty Principle (1927):
"There is a fundamental limit to how precisely we can
simultaneously determine both the position (x) and
the momentum (m.v) of a particle"

17. What is 2? " Amplitude probability function "
e.g. for a linear standing wave: Amplitude = sin x ,
where x is distance along the wave
Why amplitude probability function ?
Heisenberg Uncertainty Principle (1927):
"There is a fundamental limit to how precisely we can
simultaneously determine both the position (x) and
the momentum (m.v) of a particle"
x.mv > h/4
i.e. x.mv > (10-35) m2 kg s-1

18. Example: For a v of 0.1m/s, calculate x for (a) an
electron (m=9.11 x 10-31kg) and (b) a football
(m=0.4kg).

19. Example: For a v of 0.1m/s, calculate x for (a) an
electron (m=9.11 x 10-31kg) and (b) a football
(m=0.4kg).
(a) x > (10-35)/9.11(10-31)(0.1) = 5.79(10-4) m

20. Example: For a v of 0.1m/s, calculate x for (a) an
electron (m=9.11 x 10-31kg) and (b) a football
(m=0.4kg).
(a) x > (10-35)/9.11(10-31)(0.1) = 5.79(10-4) m
x of e- is big c.f. size of an atom (2(10-10m))!

21. Example: For a v of 0.1m/s, calculate x for (a) an
electron (m=9.11 x 10-31kg) and (b) a football
(m=0.4kg).
(a) x > (10-35)/9.11(10-31)(0.1) = 5.79(10-4) m
x of e- is big c.f. size of an atom (2(10-10m))!
(b) x > (10-35)/0.4(0.1) = 1.31(10-33) m

22. Example: For a v of 0.1m/s, calculate x for (a) an
electron (m=9.11 x 10-31kg) and (b) a football
(m=0.4kg).
(a) x > (10-35)/9.11(10-31)(0.1) = 5.79(10-4) m
x of e- is big c.f. size of an atom (2(10-10m))!
(b) x > (10-35)/0.4(0.1) = 1.31(10-33) m
x of ball is v. small c.f. its size (0.3m)

23. Example: For a v of 0.1m/s, calculate x for (a) an
electron (m=9.11 x 10-31kg) and (b) a football
(m=0.4kg).
(a) x > (10-35)/9.11(10-31)(0.1) = 5.79(10-4) m
x of e- is big c.f. size of an atom (2(10-10m))!
(b) x > (10-35)/0.4(0.1) = 1.31(10-33) m
x of ball is v. small c.f. its size (0.3m)
Conclusion
H.U.P. only important for v. small particles such as e-

24. Major implication of H.U.P.: It is not possible to find
out the exact position & velocity of an e- in an atom

25. Major implication of H.U.P.: It is not possible to find
out the exact position & velocity of an e- in an atom
Can only define the shapes of electron "clouds" in
terms of probability of finding electron at a given spot

26. Major implication of H.U.P.: It is not possible to find
out the exact position & velocity of an e- in an atom
Can only define the shapes of electron "clouds" in
terms of probability of finding electron at a given spot
Probability or e- density is given by the term 2

27. Major implication of H.U.P.: It is not possible to find
out the exact position & velocity of an e- in an atom
Can only define the shapes of electron "clouds" in
terms of probability of finding electron at a given spot
Probability or e- density is given by the term 2
Probability maps for a particular wavefunction are
called electron orbitals

28. Major implication of H.U.P.: It is not possible to find
out the exact position & velocity of an e- in an atom
Can only define the shapes of electron "clouds" in
terms of probability of finding electron at a given spot
Probability or e- density is given by the term 2
Probability maps for a particular wavefunction are
called electron orbitals
Orbitals usually drawn as shape which encloses 90%
of the total e- density for that wavefunction

29. Characterisation of Electron Orbitals
Many different solutions to the Schroedinger
Equation for an electron in an atom.

30. Characterisation of Electron Orbitals
Many different solutions to the Schroedinger
Equation for an electron in an atom.
Each solution represented by an orbital

31. Characterisation of Electron Orbitals
Many different solutions to the Schroedinger
Equation for an electron in an atom.
Each solution represented by an orbital
A series of quantum numbers are used to
describe the various properties of an orbital:

32. Characterisation of Electron Orbitals
Many different solutions to the Schroedinger
Equation for an electron in an atom.
Each solution represented by an orbital
A series of quantum numbers are used to
describe the various properties of an orbital:
Principle quantum number (n) has integral values
(1,2,3, etc.) & describes orbital size and energy

33. Characterisation of Electron Orbitals
Many different solutions to the Schroedinger
Equation for an electron in an atom.
Each solution represented by an orbital
A series of quantum numbers are used to
describe the various properties of an orbital:
Principle quantum number (n) has integral values
(1,2,3, etc.) & describes orbital size and energy
Angular momentum quantum number (l) has
integral values from 0 to n-1 for each value of n &
describes the orbital shape

34. Magnetic quantum number (ml) has integral values
between l and -l, including zero & describes the
orientation in space of the orbital relative to the
other orbitals in the atom

35. Magnetic quantum number (ml) has integral values
between l and -l, including zero & describes the
orientation in space of the orbital relative to the
other orbitals in the atom
Spin quantum number (ms) is either +1/2 or -1/2 for
a given e- & describes the direction of spin of the
e- on its axis

36. Magnetic quantum number (ml) has integral values
between l and -l, including zero & describes the
orientation in space of the orbital relative to the
other orbitals in the atom
Spin quantum number (ms) is either +1/2 or -1/2 for
a given e- & describes the direction of spin of the
e- on its axis
Pauli Exclusion Principle: "no two electrons in an
atom can have the same set of quantum numbers",
or, only two electrons (of opposite spin) per orbital.

37. Some Quantum Nos. & Orbitals in the H Atom
Sub-shell No. of
n l Designation ml Orbitals No. of e-

38. Some Quantum Nos. & Orbitals in the H Atom
Sub-shell No. of
n l Designation ml Orbitals No. of e- s

39. Some Quantum Nos. & Orbitals in the H Atom
Sub-shell No. of
n l Designation ml Orbitals No. of e- s s
1 2p , 0,

40. Some Quantum Nos. & Orbitals in the H Atom
Sub-shell No. of
n l Designation ml Orbitals No. of e- s s
1 2p , 0, s
1 3p , 0,
2 3d -2, -1, 0, +1,
4...etc...

42. Which of the following are valid sets of quantum numbers?
(a) n =1, l =1, ml = 0, ms = 1/2

43. Which of the following are valid sets of quantum numbers?
(a) n =1, l =1, ml = 0, ms = 1/2
invalid l = 0 to n-1
(b) n =2, l =1, ml = 0, ms = 1/2
valid

44. Which of the following are valid sets of quantum numbers?
(a) n =1, l =1, ml = 0, ms = 1/2
invalid l = 0 to n-1
(b) n =2, l =1, ml = 0, ms = 1/2
valid
(c) n =2, l =0, ml = -1, ms = 1/2
invalid ml = -l to l

45. Which of the following are valid sets of quantum numbers?
(a) n =1, l =1, ml = 0, ms = 1/2
invalid l = 0 to n-1
(b) n =2, l =1, ml = 0, ms = 1/2
valid
(c) n =2, l =0, ml = -1, ms = 1/2
invalid ml = -l to l
(d) n =3, l =1, ml = 0, ms = 0
invalid ms = -1/2 or +1/2

46. Which of the following are valid sets of quantum numbers?
(a) n =1, l =1, ml = 0, ms = 1/2
invalid l = 0 to n-1
(b) n =2, l =1, ml = 0, ms = 1/2
valid
(c) n =2, l =0, ml = -1, ms = 1/2
invalid ml = -l to l
(d) n =3, l =1, ml = 0, ms = 0
invalid ms = -1/2 or +1/2
Write a valid set of quantum numbers for each of the
following sub-shells:
(a) 2s n = 2, l = 0, ml = 0, ms = 1/2

47. Which of the following are valid sets of quantum numbers?
(a) n =1, l =1, ml = 0, ms = 1/2
invalid l = 0 to n-1
(b) n =2, l =1, ml = 0, ms = 1/2
valid
(c) n =2, l =0, ml = -1, ms = 1/2
invalid ml = -l to l
(d) n =3, l =1, ml = 0, ms = 0
invalid ms = -1/2 or +1/2
Write a valid set of quantum numbers for each of the
following sub-shells:
(a) 2s n = 2, l = 0, ml = 0, ms = 1/2
(b) 2p n = 2, l = 1, ml = -1, ms = 1/2

48. Which of the following are valid sets of quantum numbers?
(a) n =1, l =1, ml = 0, ms = 1/2
invalid l = 0 to n-1
(b) n =2, l =1, ml = 0, ms = 1/2
valid
(c) n =2, l =0, ml = -1, ms = 1/2
invalid ml = -l to l
(d) n =3, l =1, ml = 0, ms = 0
invalid ms = -1/2 or +1/2
Write a valid set of quantum numbers for each of the
following sub-shells:
(a) 2s n = 2, l = 0, ml = 0, ms = 1/2
(b) 2p n = 2, l = 1, ml = -1, ms = 1/2
(c) 3d n = 3, l = 2, ml = -2, ms = 1/2

49. Principal quantum number specifies energy of e-

50. Principal quantum number specifies energy of e-
For hydrogen, Schroedinger showed that
E = - hR/n2
where R is the Rydberg constant = 3.29 X 1015 Hz

51. Principal quantum number specifies energy of e-
For hydrogen, Schroedinger showed that
E = - hR/n2
where R is the Rydberg constant = 3.29 X 1015 Hz
Can calculate from theory the hydrogen spectrum

52. Principal quantum number specifies energy of e-
For hydrogen, Schroedinger showed that
E = - hR/n2
where R is the Rydberg constant = 3.29 X 1015 Hz
Can calculate from theory the hydrogen spectrum
For transition from n = 3 to n = 2 state

53. Principal quantum number specifies energy of e-
For hydrogen, Schroedinger showed that
E = - hR/n2
where R is the Rydberg constant = 3.29 X 1015 Hz
Can calculate from theory the hydrogen spectrum
For transition from n = 3 to n = 2 state
DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4]

54. Principal quantum number specifies energy of e-
For hydrogen, Schroedinger showed that
E = - hR/n2
where R is the Rydberg constant = 3.29 X 1015 Hz
Can calculate from theory the hydrogen spectrum
For transition from n = 3 to n = 2 state
DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4]
= h x 4.57 x 1014 Hz = hn and n = 4.57 x 1014 Hz

55. Principal quantum number specifies energy of e-
For hydrogen, Schroedinger showed that
E = - hR/n2
where R is the Rydberg constant = 3.29 X 1015 Hz
Can calculate from theory the hydrogen spectrum
for transition from n = 3 to n = 2 state
DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4]
= h x 4.57 x 1014 Hz = hn and n = 4.57 x 1014 Hz
l = c/n = 3 x 108 ms-1/ 4.57 x 1014 s-1 = 656 nm

56. Line spectrum of hydrogennm

57. For hydrogen, E = - hR/n2
If H atom acquires enough energy e- can be promoted from
one level to the next

58. For hydrogen, E = - hR/n2
If H atom acquires enough energy e- can be promoted from
one level to the next
To promote an e- from n =1 to n = 2 level
DE = -hR[1/22 - 1/12] = - h x 3.29 x 1015[1/4 - 1] = 1.63 x J

59. For hydrogen, E = - hR/n2
If H atom acquires enough energy e- can be promoted from
one level to the next
To promote an e- from n =1 to n = 2 level
DE = -hR[1/22 - 1/12] = - h x 3.29 x 1015[1/4 - 1] = 1.63 x J
To promote an e- from n = 2 to n = 3 level
DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4] = 3.03 x J

60. For hydrogen, E = - hR/n2
If H atom acquires enough energy e- can be promoted from
one level to the next
To promote an e- from n =1 to n = 2 level
DE = -hR[1/22 - 1/12] = - h x 3.29 x 1015[1/4 - 1] = 1.63 x J
To promote an e- from n = 2 to n = 3 level
DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4] = 3.03 x J
To remove e- (ionise) from the atom
DE = -hR[1/2 - 1/12] = hR = 2.18 x J

61. For hydrogen, E = - hR/n2
If H atom acquires enough energy e- can be promoted from
one level to the next
To promote an e- from n =1 to n = 2 level
DE = -hR[1/22 - 1/12] = - h x 3.29 x 1015[1/4 - 1] = 1.63 x J
To promote an e- from n = 2 to n = 3 level
DE = -hR[1/32 - 1/22] = - h x 3.29 x 1015[1/9 - 1/4] = 3.03 x J
To remove e- (ionise) from the atom
DE = -hR[1/2 - 1/12] = hR = 2.18 x J
To ionise one mole of hydrogen atoms
DE = NAhR = 1.31 x 103 kJ H H+ + e-

62. Principle quantum number
n = 1, 2, 3,… describes orbital size and energy
Angular momentum quantum number
l = 0 to n describes orbital shape
Magnetic quantum number
ml = l, l-1…-l describes orientation in space
of the orbital relative to the other
orbitals in the atom
Spin quantum number
ms = +1/2 or -1/ describes the direction of spin
of the e- on its axis
Pauli Exclusion Principle: "no two electrons in an
atom can have the same set of quantum numbers",
or, only two electrons (of opposite spin) per orbital.

63. Write a valid set of quantum numbers for each of the
following sub-shells:
(a) 2 s n = 2, l = 0, ml = 0, ms = - 1/2
n = 2, l = 0, ml = 0, ms = ± 1/2
2 combinations

64. Write a valid set of quantum numbers for each of the
following sub-shells:
(a) 2 s n = 2, l = 0, ml = 0, ms = - 1/2
n = 2, l = 0, ml = 0, ms = ± 1/2
2 combinations
(b) 2 p n = 2, l = 1, ml = -1, ms = - 1/2
n = 2, l = 1, ml = -1, 0 or 1, ms = ± 1/2
6 combinations

65. Write a valid set of quantum numbers for each of the
following sub-shells:
(a) 2 s n = 2, l = 0, ml = 0, ms = - 1/2
n = 2, l = 0, ml = 0, ms = ± 1/2
2 combinations
(b) 2 p n = 2, l = 1, ml = -1, ms = - 1/2
n = 2, l = 1, ml = -1, 0 or 1, ms = ± 1/2
6 combinations
(c) 3 d n = 3, l = 2, ml = -2, ms = - 1/2
n = 3, l = 2, ml = -2, -1, 0, 1, or 2, ms = ± 1/2
10 combinations

66. How many orbitals in a subshell?
l = 1, px, py, pz
l = 2, dxy,, dxz,, dyz ,, dx2-y2, dz2 5

67. How many orbitals in a subshell?
l = 1, px, py, pz
l = 2, dxy,, dxz,, dyz ,, dx2-y2, dz2 5
2 l + 1 orbitals per subshell

68. How many orbitals in a subshell?
l = 1, px, py, pz
l = 2, dxy,, dxz,, dyz ,, dx2-y2, dz2 5
2 l + 1 orbitals per subshell
How many orbitals in a shell?
n = 1, 1s
n = 2, 2s, 2px, 2py, 2pz
n = 3, 3s, 3px, 3py, 3pz, 3dxy,, 3dxz,, 3dyz ,, 3dx2-y2, 3dz2 9

69. How many orbitals in a subshell?
l = 1, px, py, pz
l = 2, dxy,, dxz,, dyz ,, dx2-y2, dz2 5
2 l + 1 orbitals per subshell
How many orbitals in a shell?
n = 1, 1s
n = 2, 2s, 2px, 2py, 2pz
n = 3, 3s, 3px, 3py, 3pz, 3dxy,, 3dxz,, 3dyz ,, 3dx2-y2, 3dz2 9
n2 orbitals per principal quantum level

70. Hydrogen atom-
all orbitals within a shell have the same energy
electrostatic interaction between e- and proton

71. Hydrogen atom-
all orbitals within a shell have the same energy
electrostatic interaction between e- and proton
Multi-electron atoms-
the energy level of an orbital depends not only on the
shell but also on the subshell
electrostatic interactions between e- and proton and other e-

72. Quantum Mechanical Model for Multi-electron Atoms
electron repulsions
He He+ + e- E = 2372 kJ mol-1
He has two electron which repel each other

73. Quantum Mechanical Model for Multi-electron Atoms
electron repulsions
He He+ + e- E = 2372 kJ mol-1
He has two electron which repel each other
He+ He2+ + e- E = 5248 kJ mol-1
He+ has one electron, no electrostatic repulsion

74. Quantum Mechanical Model for Multi-electron Atoms
electron repulsions
He He+ + e- E = 2372 kJ mol-1
He has two electron which repel each other
He+ He2+ + e- E = 5248 kJ mol-1
He+ has one electron, no electrostatic repulsion
Less energy required to remove e- from He than from He+
Shielding of outer orbital electrons from +ve nuclear
charge by inner orbital electrons
=> outer orbital electrons have higher energies

75. Quantum Mechanical Model for Multi-electron Atoms
Penetration effect of outer orbitals within inner orbitals:
ns > np > nd
For a given n, energy of s < energy of p < energy of d

76. Quantum Mechanical Model for Multi-electron Atoms
Penetration effect of outer orbitals within inner orbitals:
ns > np > nd
For a given n, energy of s < energy of p < energy of d
Effective nuclear charge (Zeff) experienced by an electron
is used to quantify these additional effects.

77. Quantum Mechanical Model for Multi-electron Atoms
Penetration effect of outer orbitals within inner orbitals:
ns > np > nd
For a given n, energy of s < energy of p < energy of d
Effective nuclear charge (Zeff) experienced by an electron
is used to quantify these additional effects.
Example: Sodium, Na, Z = 11
Na 1s e- : Zeff = shielding effect is small
Na 3s e- : Zeff = large shielding effect by inner e-’s
penetration effect counteracts
this to a small extent

78. Orbital Energies
3dxy
3dxz
3dyz
3dx2-y2
3dz2
3px
3py
3pz 3s
Energy
2px
2py
2pz 2s 1s

79. Electronic Configuration: Filling-in of Atomic Orbitals
Rules: 1. Pauli Principle

80. Electronic Configuration: Filling-in of Atomic Orbitals
Rules: 1. Pauli Principle
2. Fill in e-'s from lowest energy orbital
upwards (Aufbau Principle)

81. Electronic Configuration: Filling-in of Atomic Orbitals
Rules: 1. Pauli Principle
2. Fill in e-'s from lowest energy orbital
upwards (Aufbau Principle)
3. Try to attain maximum number of unpaired
e- spins in a given sub-shell (Hund's Rule)

82. Electronic Configuration: Filling-in of Atomic Orbitals
Rules: 1. Pauli Principle
2. Fill in e-'s from lowest energy orbital
upwards (Aufbau Principle)
3. Try to attain maximum number of unpaired
e- spins in a given sub-shell (Hund's Rule)
H (Z = 1) 1s1
2s 2p
Energy 1s

83. Electronic Configuration: Filling-in of Atomic Orbitals
Rules: 1. Pauli Principle
2. Fill in e-'s from lowest energy orbital
upwards (Aufbau Principle)
3. Try to attain maximum number of unpaired
e- spins in a given sub-shell (Hund's Rule)
N (Z = 7) 1s2, 2s2, 2p3, 2p 2s
Energy 1s

84. Electronic Configuration: Filling-in of Atomic Orbitals
Rules: 1. Pauli Principle
2. Fill in e-'s from lowest energy orbital
upwards (Aufbau Principle)
3. Try to attain maximum number of unpaired
e- spins in a given sub-shell (Hund's Rule)
B (Z = 5) 1s2, 2s2, 2p1 2p 2s
Energy 1s

85. Electronic Configuration: Filling-in of Atomic Orbitals
Rules: 1. Pauli Principle
2. Fill in e-'s from lowest energy orbital
upwards (Aufbau Principle)
3. Try to attain maximum number of unpaired
e- spins in a given sub-shell (Hund's Rule)
F (Z = 9) 1s2, 2s2, 2p5 2p 2s
Energy 1s

86. Hydrogen
2s 3s 4s
1s 2p 3p 4p
3d 4d 4f
Multi-electron atoms
1s s 3s 4s 5 s
2p 3p 4p

87. 1s s px 2py 2pz
H 1s1
He 1s2
Li 1s2, 2s1
Be 1s2, 2s2
B 1s2, 2s2, 2px1
C 1s2, 2s2, 2px1, 2py1
N 1s2, 2s2, 2px1, 2py1, 2pz1
O 1s2, 2s2, 2px2, 2py1, 2pz1
F 1s2, 2s2, 2px2, 2py2, 2pz1
Ne 1s2, 2s2, 2px2, 2py2, 2pz2

88. H 1s1 He 1s2
Li [He], 2s1 Be [He], 2s2

89. H 1s1 He 1s2
Li [He], 2s1 Be [He], 2s2
B [He], 2s2, 2p1 Ne [He], 2s2, 2p6
Na [He], 2s2, 2p6, 3s1  [Ne], 3s1

90. H 1s1 He 1s2
Li [He], 2s1 Be [He], 2s2
B [He], 2s2, 2p1 Ne [He], 2s2, 2p6
Na [He], 2s2, 2p6, 3s1  [Ne], 3s1
Mg [He], 2s2, 2p6, 3s2  [Ne], 3s2
Al [Ne], 3s2, 3p1 Si [Ne], 3s2, 3p2

91. H 1s1 He 1s2
Li [He], 2s1 Be [He], 2s2
B [He], 2s2, 2p1 Ne [He], 2s2, 2p6
Na [He], 2s2, 2p6, 3s1  [Ne], 3s1
Mg [He], 2s2, 2p6, 3s2  [Ne], 3s2
Al [Ne], 3s2, 3p1 Si [Ne], 3s2, 3p2
P [Ne], 3s2, 3p3 S [Ne], 3s2, 3p4
Cl [Ne], 3s2, 3p5 Ar [Ne], 3s2, 3p6

92. H 1s1 He 1s2
Li [He], 2s1 Be [He], 2s2
B [He], 2s2, 2p1 Ne [He], 2s2, 2p6
Na [He], 2s2, 2p6, 3s1  [Ne], 3s1
Mg [He], 2s2, 2p6, 3s2  [Ne], 3s2
Al [Ne], 3s2, 3p1 Si [Ne], 3s2, 3p2
P [Ne], 3s2, 3p3 S [Ne], 3s2, 3p4
Cl [Ne], 3s2, 3p5 Ar [Ne], 3s2, 3p6
outermost shell - valence shell
most loosely held electron and are the most important
in determining an element’s properties

93. K [Ar], 4s1 Ca [Ar], 4s2
Sc [Ar], 4s2, 3d1 Ca [Ar], 4s2, 3d2

94. K [Ar], 4s1 Ca [Ar], 4s2
Sc [Ar], 4s2, 3d1 Ca [Ar], 4s2, 3d2
Zn [Ar], 4s2, 3d10 Ga [Ar], 4s2, 3d10, 3p1
Kr [Ar], 4s2, 3d10, 3p6

95. K [Ar], 4s1 Ca [Ar], 4s2
Sc [Ar], 4s2, 3d1 Ca [Ar], 4s2, 3d2
Zn [Ar], 4s2, 3d10 Ga [Ar], 4s2, 3d10, 3p1
Kr [Ar], 4s2, 3d10, 3p6
Anomalous electron configurations
d5 and d10 are lower in energy than expected
Cr [Ar], 4s1, 3d not [Ar], 4s2, 3d4
Cu [Ar], 4s1, 3d10 not [Ar], 4s2, 3d9

96. Electron Configuration of Ions
Electrons lost from the highest energy occupied orbital
of the donor and placed into the lowest unoccupied orbital
of the acceptor (placed according to the Aufbau principle)

97. Electron Configuration of Ions
Electrons lost from the highest energy occupied orbital
of the donor and placed into the lowest unoccupied orbital
of the acceptor (placed according to the Aufbau principle)
Examples:
Na [Ne], 3s1 Na+ [Ne] + e-
Cl [Ne], 3s2, 3p5 + e- Cl- [Ne], 3s2, 3p6
Mg [Ne], 3s2 Mg2+ [Ne]
O [He], 2s2, 2p4 O2- [He], 2s2, 2p6

98. Modern Theories of the Atom - Summary
Wave-particle duality of light and matter
Bohr theory
Quantum (wave) mechanical model
Orbital shapes and energies
Quantum numbers
Electronic configuration in atoms