Lecture 8 and 9

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131

Ten Important Types of Overlap in General HomoDinuclear M.O.’s

Are there any other possibilities that have gone unmentioned?

132

Five s+px s+py px+py px+pz py+pz

Five s-px s-py px-py px-pz py-pz

Q. What about these? A. They are zero overlap

133

Due to the three different types of atomic orbitals depicted below, we also have three different types of M.O.’s*

*These three types of a.o.’s can combine with one another to give m.o.’s that have zero ( σ ), one (π), or two (δ) nodal planes σ, π, δ bonding σ, π*, δ* antibonding

134

Examples of Diatomic M.O. Treatment (1) F2 Molecule: F is 22s22p5 valence electrons Buried, next to nucleus does not participate in bonding Remember effective nuclear charge increases left → right in periodic table (adding protons to atoms whose electrons are going into the same shell) F- effective Nuclear Charge is high → 2s/2p orbital energies are therefore quite different (p orbitals are more shielded than s orbitals) 1s – very low in energy 2s – still very low in energy 2p – higher energy due to being more shielded from nuclear charge so their relative I.E. is less

135

σ4* π2* 2p

2p' π1 σ3

E σ2* 2s

2s' F atomic orbitals

σ1 m.o.’s

F atomic orbitals

Fill the diagram with the 14 valence electrons (7 from each F)

F2 Electronic Configuration is : σ12 σ2*2 σ32 π14 π2*4 (σ levels are non degenerate) (π levels are doubly degenerate) Net bonding is: σ12 σ2*2 σ32 π14 π2*4 one σ bond based on the 2pz….2pz overlap

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Bond order in MO theory is: (#of bonding electrons - # of antibonding electrons)/2 2e- per bond Example 2 Li2 Main difference between Li2 and F2 is that the 2s and 2p separation is much lesser in Li2 Li 1s22s12p0 Need to understand why σ3 goes up

The electronic configuration is σ12 based only on s…s overlap. It is a weak bond because s…s overlap is poor (compared say to s-p σ overlap)

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Li2 (σ) Weak s…s bond looks like this in terms of the electron density contour. Each new contour line as you go in from perimeter is a double of e- density

Note: this bond doesn’t depend on any of the higher energy M.O.’s

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The difference in the separation of 2s and 2p lead to different energy orderings for Li2 → F2 *Crossing of π, and σ3 occurs at O2 when mixing becomes unimportant.

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What are the bond order for the series? Li2 σ12 B.O. = 2/2 = 1 Be2 σ12 σ2*2 B.O. = 0 B2 σ12 σ2*2π12 B.O. = 1 C2 σ12 σ2*2π14 B.O. = 2 N2 σ12 σ2*2π14 σ32 B.O. = 3 O2 σ12 σ2*2 σ32 π14 π2*2 B.O. = 2 F2 σ12 σ2*2 σ32 π14 π2*4 B.O. = 1 - N2 has highest bond order (:N≡N:), the shortest, and the strongest bond - O2 is a double bond and a paramagnetic molecule because the last two electrons go in the π set unpaired * O2 Lewis Structure is correct, but it does not predict two unpaired electrons.

140

Q. What about Ne2? A. This is an unstable molecule: σ12 σ2*2 σ32 π14 π2*4 π4*2 16 valence electrons (B.O. =0) Heteronuclear Diatomic Molecules AB rather than A2 means that the atomic orbitals no longer begin at the same energies.

Contrast M.O. Diagram for CO with N2:

141

N1 a.o.’s

M.O.’s N2

10 valence electrons: σ12 σ2*2 π14 σ32 (B.O. =3.0)

N2 a.o.’s

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M.O. Diagrams for the Isoelectronic N2 and CO molecules are very different. They help explain the different reactivites of N2 vs CO in a way that Lewis Diagrams never could (or formal charges!) CO N2 (1) Highest e- are in a (1) Highest e- are in a strongly bonding slightly antibonding orbital σ12σ1*2π14σ32, orbital σ12σ1*2π14σ3*2 (higher in energy pz-pz σ orbital than the starting (2) As a consequence N2 a.o.’s) is very stable, + -eN2 →N 2 weakens the (2) As a consequence CO is not as stable, N-N bond. I.E. is CO →CO+ actually very high leads to a strong C≡O bond!

143

What does ionization of N2 versus CO have to do with their activation? (meaning destroying the molecule or even attaching it as a “ligand” to metals through lone pair) A. Lewis basicity of CO is much higher than N2 :C≡O: This end binds to Lewis acids such as M+ ions very easily. The highest energy l.p. is σ*3 which is primarily C-based.

:N≡N: Very difficult to get this lone pair to donate (buried in energy!)

144

Extrapolation of M.O. Theory Diatomic to Polyatomic Linear Triatonics like BeH2 which can form only σ orbitals

1) In each bonding orbitals, B, The e- density is continuous over adjacent atoms, in antibonding orbitals, A, there is a node. 2) In each bonding orbital, the electron pair is spread out (delocalized) over entire molecule.

145

BeH2 M.O. Diagram

Main Features

Be atom

Two H atoms

Two equal energy One 2s and three 1s orbitals are 2p orbitals lie at placed in the much higher diagram. energy than H 1s They are low in and they are close energy due to together due to higher effective lower effective nuclear charge. nuclear charge Bonding in H-Be-H σ12σ22 = 4 Bonding electrons Distributed over two Be-H bonds. So two single bonds.

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Trigonal Planar Molecules AB3 BF3 , CO32- , NO3Recall, we invoked π- bonding in CO32- and NO3- as part of “resonance” structures like these:

Q. But how does M.O. theory account for πbonding? More to the point: Can M.O. theory explain the one π bond in the above structure needing to be in 3 places at the same time? A. YES!!!

147

M.O. Treatment of AB3 planar molecules. For example BF3, NO3-, CO32Requires two different groups of atomic orbitals to be considered. 1) Hybrid Orbitals on central atom, A, and on B that will be used to make σ- bonds (A-B bonds) and used to house lone pairs (in plane) 2) Group Orbitals on outer atoms, B, that are made of pz orbitals (out-of-plane) that can overlap with the pz orbital on central atom A. BF3

3e- + 3(7e-) = 24e-

CO32-

4e- + 3(6e-) + 2e- = 24e-

NO3-

5e- + 3(6e-) + 1e- = 24e-

The xy plane is the plane of molecule The z axis comes out of paper

148

149

150

151

152

Diagrams showing how π- bonding and π- antibonding M.O.’s arise from overlap of G.O.3 with the pz Orbital of A.

π1*

BF3 Only 6 valence CO32- electrons are pz NO3 available for π- orbital on A bonding. The 18ebefore this are involved in σ interactions (24enon-bonding systems).

G1

π1

G2

G3

pz orbitals on B make up these three

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The part of the M.O. Diagram that depicts the π- bonding is:

if BF3 then: each of these is a single bond based on sp2 overlap

and

there is 1/3 of a π bond (1 π bond over 3 atoms)

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Note, we did not draw a complete M.O. diagram here with all of the orbitals and interactions. It is too complicated to try and get the relative energies of the starting orbitals and M.O.’s correct. Nevertheless, we succeeded in developing a qualitative picture of the bonding that holds true for these types of molecules.

155

Multi-Center Bonding in Electron Deficient Molecules This happens when you don’t have enough electrons to have a two-electron bond between all adjacent atoms. Examples (classic ones) *

There are eight adjacent pairs of atoms in these molecules but count electrons… 2B 6e- (3 each) H 6e12e-

You need 16e- to make 8 bonds but you have only 12e- which is only enough for 6 bonds

*These are not planar (note that BH3 and Al(CH3)3 are not really the way the formulae indicate)

Consider these two resonance forms (canonical forms)

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This implies that in each bridge, that one electron pair is shared between (or distributed over) two B•••H bonds. This would lead to a bond order of ½ for each B•••H bridge bond but still result in the other two

bonds being normal 2e- bonds.

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Q. Doesn’t this seem a little artificial to you? A. Yes. There is a better way to think about this with M.O. theory B has sp3 hybridization for tetrahedral Boron - BH2 has two ordinary bonds made from two of the four sp3 hybrids and the H 1s orbitals.

These BH2 fragments are coplanar - the remaining two sp3 hybrids overlap in a perpendicular orientation with the bridging H atoms

158

Formation of three-center two-electron Bonds in B2H6

159

Combinations of B sp3 hybrids and H1s orbital

160

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Lecture 8 and 9

131 Ten Important Types of Overlap in General HomoDinuclear M.O.’s Are there any other possibilities that have gone unmentioned? 132 Five s+px s+...

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