MO Theory Extra Review

8/10/2019 MO Theory Extra Review

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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?


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132

Five Five

s+px s-pxs+py s-py

px+py px-pypx+pz

px-pz

py+pz py-pz

Q.

What aboutthese?

A.

They are zero

overlap


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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 (

), one (

), ortwo (!) nodal planes

", , ! bonding

", *, !* antibonding


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134

Examples of Diatomic M.O. Treatment

(1)

F2Molecule:

F is22s

22p

5 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 fromnuclear charge so their relative I.E. is less


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135

F2

Electronic Configuration is :

"12"2

*2"3

21

42

*4 ("levels are non degenerate)

(levels are doubly degenerate)

Net bonding is:

"12"2

*2"3

21

42

*4

one "bond based on the 2pz.2pzoverlap

F

atomic

orbitalsm.o.s

2 2 '

2p'2p

"1

"2*

"3

"4*

1

2*Fill the

diagram

with the 14valence

electrons

(7 from

each F)

E

F

atomic

orbitals


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136

Bond order in MO theory is:

(#of bonding electrons - # of antibonding

electrons)/2 2e-per bond

Example 2 Li2

Main difference between Li2and F2is that the 2s and

2p separation is much lesser in Li2

Li 1s22s12p0

Need to understand why "3goes up

The electronic configuration is "12based only on ss

overlap. It is a weak bond because ss overlap is poor

(compared say to s-p "overlap)


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137

Li2 (")

Weak ss bond looks like this in terms of the electron

density contour. Each new contour line as you go infrom perimeter is a double of e-density

Note: this bond doesnt 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

, and"

3occurs at O2when mixingbecomes unimportant.


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What are the bond order for the series?

Li2"

1

2

B.O. = 2/2 = 1Be2 "12

"2*2

B.O. = 0

B2"12

"2*21

2 B.O. = 1

C2"12

"2*21

4B.O. = 2

N2 "12

"2*21

4"3

2B.O. = 3

O2"12

"2*2

"321

42

*2B.O. = 2

F2"12"2*2"32142*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 set

unpaired

* O2Lewis Structure is correct, but it does not predict

two unpaired electrons.


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Q. What about Ne2?

A. This is an unstable molecule:"12

"2*2

"321

42

*44

*2

16 valence electrons (B.O. =0)

Heteronuclear Diatomic Molecules

AB rather than A2means that the atomic orbitals no

longer begin at the same energies.

Contrast M.O. Diagram for CO with N2:


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N1 M.O.s N2

a.o.s N2 a.o.s

10 valence electrons:"1

2"2

*21

4"3

2(B.O. =3.0)


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142

M.O. Diagrams for the Isoelectronic N2and CO

molecules are very different. They help explain the

different reactivites of N2vs CO in a way that LewisDiagrams never could (or formal charges!)

N2

(1) Highest e-are in a

strongly bonding

orbital"

1

2"

1*

2

1

4"

3

2

,pz-pz"orbital

(2) As a consequence N2

is very stable,

N2#N2+weakens the

N-N bond. I.E. is

very high

-e-

CO

(1) Highest e-are in a

slightly antibonding

orbital"

1

2"

1*

2

1

4"

3*

2

(higher in energy

than the starting

a.o.s)

(2) As a consequence

CO is not as stable,

CO #CO+actuallyleads to a strong C$O


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What does ionization of N2versus CO have to do with

their activation?

(meaning destroying the molecule or even attachingit 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 "

*3which is

primarily C-based.

:N$N:

Very difficult to

get this lone pair

to donate

(buried in energy!)


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Extrapolation of M.O. Theory Diatomic to Polyatomic

Linear Triatonics like BeH2which can form only"

orbitals

1) In each bonding orbitals, B, The e-density is

continuous over adjacent atoms, in antibondingorbitals, A, there is a node.

2) In each bonding orbital, the electron pair is

spread out (delocalized) over entire molecule.


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BeH2M.O. Diagram

Main

Features

Bonding in H-Be-H

"12"2

2= 4 Bonding electrons Distributed over two

Be-H bonds. So two single bonds.

Be atom

One 2s and three

2p orbitals lie at

much higher

energy than H 1s

and they are close

together due to

lower effectivenuclear char e

Two H atoms

Two equal energy

1s orbitals are

placed in the

diagram.

They are low in

energy due to

higher effectivenuclear charge.


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146

Trigonal Planar Molecules

AB3 BF3, CO32-

, NO3-

Recall, we invoked - bonding in CO32-

and NO3-as

part of resonance structures like these:

Q. But how does M.O. theory account for -

bonding?

More to the point: Can M.O. theory explain the

one bond in the above structure needing to be

in 3 places at the same time?

A. YES!!!


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M.O. Treatment of AB3planar molecules.

For example BF3, NO3-, CO3

2-

Requires two different groups of atomic orbitals to beconsidered.

1) Hybrid Orbitals on central atom, A, and on B that

will be used to make "- bonds (A-B bonds) and

used to house lone pairs (in plane)

2)

Group Orbitals on outer atoms, B, that are madeof pzorbitals (out-of-plane) that can overlap with

the pzorbital 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


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The part of the M.O. Diagram that depicts

the - bonding is:

if BF3then:

each of these is a single bond

based on sp2overlap

and there is 1/3 of a

bond (1 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 thestarting 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.


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Multi-Center Bonding in Electron Deficient Molecules

This happens when you dont have enough electrons tohave a two-electron bond between all adjacent atoms.

Examples (classic ones)

*

There are eight adjacent pairs of atoms in these

molecules but count electrons

*These are not planar

(note that BH3and Al(CH3)3are not really the way the

formulae indicate)

Consider these two resonance forms (canonical forms)

2B 6e-(3 each)

H 6e-

12e

-

You need 16e-to make

8 bonds but you haveonly 12e

-which is only

enough for 6 bonds


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This implies that in each bridge, that one

electron pair is shared between (or

distributed over) two BH bonds.

This would lead to a bond order of for each BH

bridge bond but still result in the other two

bonds being normal 2e-bonds.


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Q. Doesnt this seem a little artificial to you?

A.

Yes. There is a better way to think about this withM.O. theory

B has sp3hybridization for tetrahedral Boron

- BH2has two ordinary bonds made from two of the

four sp3hybrids and the H 1s orbitals.

These BH2fragments are coplanar

- the remaining two sp3hybrids overlap in a

perpendicular orientation with the bridging H atoms


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Formation of three-center two-electron Bonds in B2H6


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Combinations of B sp3hybrids and H1s orbital


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MO Theory Extra Review