Principle quantum number n = 1, 2, 3,….. describes orbital size and energy

Angular momentum quantum numberl = 0 to n-1 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 -1/2 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.

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 = ± 1/22 combinations

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 = ± 1/22 combinations

(b) 2 p n = 2, l = 1, ml = -1, ms = - 1/2

n = 2, l = 1, ml = -1, 0 or 1, ms = ± 1/26 combinations

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 = ± 1/22 combinations

(b) 2 p n = 2, l = 1, ml = -1, ms = - 1/2

n = 2, l = 1, ml = -1, 0 or 1, ms = ± 1/26 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 = ± 1/210 combinations

How many orbitals in a subshell?l = 0, 1s 1

l = 1, px, py, pz 3

l = 2, dxy,, dxz,, dyz ,, dx2

-y2, dz

2 5

How many orbitals in a subshell?l = 0, 1s 1

l = 1, px, py, pz 3

l = 2, dxy,, dxz,, dyz ,, dx2

-y2, dz

2 5

2 l + 1 orbitals per subshell

How many orbitals in a subshell?l = 0, 1s 1

l = 1, px, py, pz 3

l = 2, dxy,, dxz,, dyz ,, dx2

-y2, dz

2 5

2 l + 1 orbitals per subshell

How many orbitals in a shell?n = 1, 1s 1

n = 2, 2s, 2px, 2py, 2pz 4

n = 3, 3s, 3px, 3py, 3pz, 3dxy,, 3dxz,, 3dyz ,, 3dx2

-y2, 3dz

2 9

How many orbitals in a subshell?l = 0, 1s 1

l = 1, px, py, pz 3

l = 2, dxy,, dxz,, dyz ,, dx2

-y2, dz

2 5

2 l + 1 orbitals per subshell

How many orbitals in a shell?n = 1, 1s 1

n = 2, 2s, 2px, 2py, 2pz 4

n = 3, 3s, 3px, 3py, 3pz, 3dxy,, 3dxz,, 3dyz ,, 3dx2

-y2, 3dz

2 9

n2 orbitals per principal quantum level

Hydrogen atom-

all orbitals within a shell have the same energy

electrostatic interaction between e- and proton

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 theshell but also on the subshell

electrostatic interactions between e- and proton and other e-

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

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

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

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

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.

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 = 10.3 shielding effect is small

Na 3s e- : Zeff = 1.84 large shielding effect by inner e-’s

penetration effect counteracts this to a small extent

Z

Z

0

1

2

3

4

5

6

7

8

0 5 10 15 20

Z /atomic number

Zef

f/ e

ffe

cti

ve

nu

cle

ar

ch

arg

e

Orbital Energies

Energy

1s

2s

2px 2py 2pz

3s

3px 3py 3pz

3dxy 3dxz 3dyz 3dx2-y2 3dz2

Electronic Configuration: Filling-in of Atomic Orbitals

Rules: 1. Pauli Principle

Electronic Configuration: Filling-in of Atomic Orbitals

Rules: 1. Pauli Principle

2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle)

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)

Energy

1s

2s 2p

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

Energy

1s

2s

2p

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,

Energy

1s

2s

2p

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

Energy

1s

2s

2p

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

Hydrogen

2s 3s 4s1s 2p 3p 4p

3d 4d4f

Multi-electron atoms

1s 2s 3s 4s 5 s

2p 3p 4p

3d 4d

H 1s1

He 1s2

Li 1s2, 2s1

Be 1s2, 2s2

B 1s2, 2s2, 2px1

C 1s2, 2s2, 2px1, 2py

1

N 1s2, 2s2, 2px1, 2py

1, 2pz1

O 1s2, 2s2, 2px2, 2py

1, 2pz1

F 1s2, 2s2, 2px2, 2py

2, 2pz1

Ne 1s2, 2s2, 2px2, 2py

2, 2pz2

1s 2s 2px 2py 2pz

H 1s1 He 1s2

Li [He], 2s1 Be [He], 2s2

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

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, 3p1Si [Ne], 3s2, 3p2

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, 3p1Si [Ne], 3s2, 3p2

P [Ne], 3s2, 3p3S [Ne], 3s2, 3p4

Cl [Ne], 3s2, 3p5 Ar [Ne], 3s2, 3p6

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, 3p1Si [Ne], 3s2, 3p2

P [Ne], 3s2, 3p3S [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

K [Ar], 4s1 Ca [Ar], 4s2

Sc [Ar], 4s2, 3d1 Ti [Ar], 4s2, 3d2

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

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, 3d5 not [Ar], 4s2, 3d4

Cu [Ar], 4s1, 3d10 not [Ar], 4s2, 3d9

Electron Configuration of Ions

Electrons lost from the highest energy occupied orbitalof the donor and placed into the lowest unoccupied orbitalof the acceptor (placed according to the Aufbau principle)

Electron Configuration of Ions

Electrons lost from the highest energy occupied orbitalof the donor and placed into the lowest unoccupied orbitalof 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

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

Compare the energies of photons emitted by tworadio stations, operating at 92 MHz (FM) and 1500 kHz (MW)?

Compare the energies of photons emitted by tworadio stations, operating at 92 MHz (FM) and 1500 kHz (MW)?

E = h

92 MHz = 92 x 106 Hz => E = 6.626 x 10-34 x 92 x 106 = 6.1 x 10-26J

Compare the energies of photons emitted by tworadio stations, operating at 92 MHz (FM) and 1500 kHz (MW)?

E = h

92 MHz = 92 x 106 Hz => E = 6.626 x 10-34 x 2 x 106 = 1.33 x 10-27J

1500 kHzE = 6.626 x 10-34 x 1.5 x 106 = 9.94 x 10-28J

The energy from radiation can be used to break chemical bonds. Energy of at least 495 kJ mol-1 is required to break the oxygen-oxygen bond. What is the wavelength of this radiation?

The energy from radiation can be used to break chemical bonds. Energy of at least 495 kJ mol-1 is required to break the oxygen-oxygen bond. What is the wavelength of this radiation?

E = hc/

495 x 103 J mol-1 495 x 103 J mol-1/NA

= 8.22 x 10-19 J per molecule

The energy from radiation can be used to break chemical bonds. Energy of at least 495 kJ mol-1 is required to break the oxygen-oxygen bond. What is the wavelength of this radiation?

E = hc/

495 x 103 J mol-1 495 x 103 J mol-1/NA

= 8.22 x 10-19 J per molecule

= 6.626 x 10-34 x 3 x 108/ 8.22 x 10-19

= 242 x 10-9 m = 242 nm.

Autumn 1999

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

Autumn 1999

2. The best available balances can weigh amounts as smallas 10-5 g. If you were to count out water molecules at therate 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 16 + 2 = 18 amu

1 mole H2O has mass of 18 g 6.022 x 1023 molecules

Autumn 1999

2. The best available balances can weigh amounts as smallas 10-5 g. If you were to count out water molecules at therate 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 16 + 2 = 18 amu

1 mole H2O has mass of 18 g 6.022 x 1023 molecules

10-5 g 10-5/18 moles = 5.6 x 10-7 moles

Autumn 1999

2. The best available balances can weigh amounts as smallas 10-5 g. If you were to count out water molecules at therate 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 16 + 2 = 18 amu

1 mole H2O has mass of 18 g 6.022 x 1023 molecules

10-5 g 10-5/18 moles = 5.6 x 10-7 moles

5.6 x 10-7 x 6.022 x 1023 molecules = 3.35 x 1017 molecules

Autumn 1999

2. The best available balances can weigh amounts as smallas 10-5 g. If you were to count out water molecules at therate 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 16 + 2 = 18 amu

1 mole H2O has mass of 18 g 6.022 x 1023 molecules

10-5 g 10-5/18 moles = 5.6 x 10-7 moles

5.6 x 10-7 x 6.022 x 1023 molecules = 3.35 x 1017 molecules

3.35 x 1017 s

Autumn 2000

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

Autumn 2000

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

E = h = hc/

= 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1

Autumn 2000

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

E = h = hc/

= 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1

= 4.88 x 10-19 J

Autumn 2000

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

E = h = hc/

= 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1

= 4.88 x 10-19 J

1 millimole = 10-3 mole = 6.022 x 1020 photons

Autumn 2000

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

E = h = hc/

= 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1

= 4.88 x 10-19 J

1 millimole = 10-3 mole = 6.022 x 1020 photons

energy of 1 millimole of photons 6.022 x 1020x 4.88 x 10-19 J

Autumn 2000

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

E = h = hc/

= 6.626 x 10-34 x 3 x 108 /407 x 10-9 = J s m s-1 m-1

= 4.88 x 10-19 J

1 millimole = 10-3 mole = 6.022 x 1020 photons

energy of 1 millimole of photons 6.022 x 1020x 4.88 x 10-19 J

= 294 J