Post on 13-Dec-2015
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
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
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
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 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