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What’s coming up???• Oct 25 The atmosphere, part 1 Ch. 8• Oct 27 Midterm … No lecture• Oct 29 The atmosphere, part 2 Ch. 8• Nov 1 Light, blackbodies, Bohr Ch. 9• Nov 3,5 Postulates of QM, p-in-a-box Ch. 9• Nov 8,10Hydrogen and multi – e atoms Ch. 9• Nov 12 Multi-electron atoms Ch.9,10• Nov 15 Periodic properties Ch. 10• Nov 17 Periodic properties Ch. 10• Nov 19 Valence-bond; Lewis structures Ch. 11• Nov 22 VSEPR Ch. 11• Nov 24 Hybrid orbitals; VSEPR Ch. 11, 12• Nov 26 Hybrid orbitals; MO theory Ch. 12• Nov 29 MO theory Ch. 12• Dec 1 bonding wrapup Ch. 11,12• Dec 2 Review for exam
The Final Exam
• December 13 (Monday) • 9:00 – 12:00
• Cumulative (covers everything!!)• Worth 50% of total mark
• Multiple choice
The Final Exam
• From my portion, you are responsible for:
– Chapter 8 … material from my lecture notes– Chapter 9 … everything– Chapter 10 … everything– Chapter 11 … everything– Chapter 12 … everything except 12.7
The Final Exam
• You will need to remember
– Relationship between photon energy and frequency / wavelength
– De Broglie AND Heisenberg relationships– Equations for energies of a particle-in-a-box
AND of the hydrogen atom– VSEPR shapes AND hybribizations which
give them
COMBINATION OF ORBITALS
1sA + 1sB = MO1
builds up electron density between nuclei.
Remember, when we take linear combinations of orbitals we get out as many as we put in.
Here, the sum of the 2 orbitals
90% probability
COMBINATION OF ORBITALS
1sA + 1sB = MO1
builds up electron density between nuclei.
1sA – 1sB = MO2
results in low electron density between nuclei
EEnergy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A B
ADDITION gives an….
Energy more negative than average of original orbitals
Energy more positive than average of original orbitalsSUBTRACTION gives an….
1s
1s*
E1s
1s*
1s
1s
He2: (1s)2(1s*)2
He HeHe2
The bonding effect of the (1s)2 is cancelled by the
antibonding effect of (1s*)2
The He2 molecule is not
a stable species.
BOND ORDER
= {
A high bond order indicates high bond energy and short bond length.
# of bonding electrons(nb)
# of antibonding electrons (na)
– 1/2 }
A measure of bond strength and molecular stability.
If # of bonding electrons > # of antibonding electrons
Bondorder
the molecule is predicted to be stable
Consider H2+,H2,He2
+,He2……….
= 1/2 (nb - na)
1s*
1s
Magnetism
Bond order
Bond energy (kJ/mol)
Bond length (pm)
First row diatomic molecules and ions
H2+
Para-
½
225
106
E
He2+
Para-
½
251
108
He2
—
0
—
—
H2
Dia-
1
436
74
E
1s 1s
1s
Electron configuration for DILITHIUM
2s
2s*
2s
2s
(1s)2(1s*)2(2s)2
Li2
Bond Order = 1/2 (nb - na)
= 1/2(4 - 2) =1
A single bond.
E2s
2s*
2s
2s
Li2 (2s)2
Li LiLi2
(1s)2(1s*)2 assumed
Only valence orbitals contribute to molecular bonding
E2s
2s*
2s
2s
Be2Be BeBe2
Electron configuration for DIBERYLLIUM
Configuration: (2s)2(2s*)2 Bond order?
B2
The Boron atomic configuration is
1s22s22p1
form molecular orbitals.
So we expect B to use 2p orbitals to
How do we do that???
Combine them by addition and subtraction
BUT … remember there are 3 sets of p-orbitals to combine
The M.O.’s formed by p orbitals
2p*
2p
2p
2p*
The do not split as much as the because of weaker overlap.
E2p 2p
Combine this with the s-orbitals…..
E
Expected orbital splitting:
2s
2s*
2s
2s
2p
2p*
2p
2p
2p
2p*
The do not split as much because of weaker overlap.
But the s and p along the internuclear axis DO interact
This pushes the 2p up..
E
MODIFIED ENERGY LEVEL DIAGRAM
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p* interaction
Notice that the 2p and 2p
have changed places!!!!
Now look at B2...
E
Electron configuration for B2:
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*
(2s)2(2s*)2(2p)2
Abbreviated configuration
(1s)2(1s*)2(2s)2(2s*)2(2p)2
E
2s
2s*
2s
2s
Bond order
2p
2p*
2p2p
2p
2p*Molecule is predicted to be stable and paramagnetic.
1/2(nb - na)
= 1/2(4 - 2) =1
SECOND ROW DIATOMICS
B2 C2 N2 O2 F2
E
2s
2s*
2s
2s
2p
2p*
2p2p
2p
2p*
2s
2s*
2s
2s
2p*
2p
2p
2p
2p*
2p
Li2
O O
Back to Oxygen
2p*
2p*
2p
2p
2s*
2s
E
12 valence electrons
BO = 2 but PARAMAGNETIC
BUT REMEMBER …THE LEWIS STRUCTURE WAS DIAMAGNETIC
2p*
2p*
2p or 2p
2por 2p
2s*
2s
Magnetism
Bond order
Bond E. (kJ/mol)
Bond length(pm)
Second row diatomic molecules
B2
Para-
1
290
159
C2
Dia-
2
620
131
N2
Dia-
3
942
110
O2
Para-
2
495
121
F2
Dia-
1
154
143
E
NOTE SWITCH OF LABELS
Example: Give the electron configuration and bond order
for O2, O2+ , O2
- & O22-. Place them in order of bond
strength and describe their magnetic properties.
Step 1:Determine the number of valence electrons in each:
O2+ : 6 + 6 - 1 = 11
O2– : 6 + 6 + 1 = 13
O22- : 6 + 6 + 2 = 14
O2 : 6 + 6 = 12
Step 2: Determine the valence electrons configurations:
O2 : (2s)2(2s*)2 (2p)2(2p)4 (2p*)2
O2+ :
O2– :
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
O2 : (2s)2(2s*)2 (2p)2 (2p)4(2p*)2
O2+ : (2s)2(2s*)2 (2p)2 (2p)4(2p*)1
O2– : (2s)2(2s*)2 (2p)2 (2p)4(2p*)3
O22- : (2s)2(2s*)2 (2p)2 (2p)4(2p*)4
2p*
2p*
2p
2p
2s*
2s
E
O2 O2+ O2
– O22-
O2 : B.O. = (8 - 4)/2 = 2
O2+ : B.O. = (8 - 3)/2 = 2.5
O2– : B.O. = (8 - 5)/2 = 1.5
O22- : B.O. = (8 - 6)/2 =
1
Step 3: Determine the bond orders of each species:
2p*
2p*
2p
2p
2s*
2s
E
NITRIC OXIDE (NO)Number of valence electrons: 5 + 6 = 11
USE THE MO DIAGRAM FOR HOMONUCLEAR DIATOMIC MOLECULES WITH s-p INTERACTION AS AN APPROXIMATION FOR < 12 ELECTRONS
Put the electrons in…..
Molecule is stable and paramagnetic.
NITRIC OXIDE (NO)
2p*
2p*
2p
2p
2s*
2s
E
Bond order 522
38.
Experimental data agrees.
NO+ and CN-
Ions are both stable and diamagnetic.
NO+: Number of valence electrons: 5 + 6 - 1 = 10
CN–: Number of valence electrons: 4 + 5 + 1 = 10
ISOELECTRONIC
2p*
2p*
2p
2p
2s*
2s
EBond order 03
2
28.
Experimental data agrees.
TRIPLE BOND