MO_theory_stuff.pdf

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DEPARTMENT OF

CHEMISTRYIntroduction to Symmetry and Inor

Spectrosc

3. Using Symmetry to Understand Bonding3.1 Introduction

In addition to its role in predicting vibrational spectra, symmetry can also be applied to understanding bonding

electronic structure and reaction mechanisms. Many of the skills you have acquired for use in spectroscopy artransferable to these areas. This part of the course uses the group theory tools you already know and shows how thecan be developed and applied for the construction of molecular orbital diagrams and for understanding the octet an

eighteen electron rules, hypervalency and ligand-field effects in transition metal chemistry.

The figure below shows a molecular orbital (MO) diagram for H2:

H2has 2 electrons and both go into the bonding orbital.

The bond orderis

bond order = (number of bonding electrons - number of antibonding electrons)

= (2 - 0) = 1

The MO diagram predicts a bond order of 1 (a single bond).

To construct the MO diagram, the following steps are followed:

(1) Divide the molecule into two (or more) fragments- in this case the molecule is made from two H atoms

The atomic orbitals of one fragment are placed on the left hand side and those of the other fragment are placed on thright hand side. Each H only uses a 1s orbital.

(2) Work out the symmetry of the orbitals.

(3) Overlap the orbitals on the left with those on the right to produce the molecular orbitals in the centre. Eac

time a pair of orbitals overlap, a bonding and an anti-bonding orbital is produced.

(4) Fill up the MOs with the valence electrons from the bottom upwards, remembering to put a maximum of

electrons in each level (Pauli principle).

3.2 The octet rule

Example 30: tetrahedral methane

The MO diagram for H2is straightforward as there are only two atoms. For bigger molecules, group theory can be use

to simplify the process:

C

O

NT

E

NT

S

Page 1 of 203. Using Symmetry to Understand Bonding

2/24/2004http://www.hull.ac.uk/php/chsajb/symmetry&spectroscopy/ho_3.html


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Only orbitals with the same symmetry can overlap

(1) Fragments.

All of the H atoms are equivalent and these form one fragment.

The C atom is different to the 4H atoms and this forms the second fragment.

(2) Symmetry:

Working out the symmetry of the carbon orbitals:

A carbon atom has a 2s22p2configuration and uses both its 2s and 2p orbitals in bonding. As the carbon atom is at th

centre of molecule, there is a veryquick way of finding the symmetry of these orbitals.

2px, 2p

yand 2p

zhave the same symmetry as x,yand zrespectively. These are given in the penultimate column of th

character table. (Note that this is the same as for the IR activity!)

px, py, pz=

x, y, z= t

2

The carbon orbitals are said to span t2 they are triply degenerate in a tetrahedral molecule and will all be involved

equally in the bonding.

Finding the symmetry of the 2s orbital is even easier. As a s-orbital is spherical, the C 2s orbital is unchanged by an

of the symmetry operations. It is "totally symmetric" and alwayshas the label given to the top rowin the charactetable:

s= a

1

Working out the symmetry of the H 1s orbitals:

The hydrogen atoms in methane are all equivalent requiring that the electron density on each is the same. This iachieved by finding the symmetry of the allowed combinations of the hydrogen When working out the symmetry o

stretches in Handout 2, stretch

was calculated by noting how many bonds were unchanged by the symmetr

operations.

To work out H1s

, we need to work out how many of the H 1s orbitals are left unchanged by the operations

(Remember that a 1s orbital is just a sphere)

CH4is a tetrahedral ("T

d") molecule:

effect of operation character,

E

all orbitals unchanged:

+4

C3 3 orbitals swapped over.

1 orbital unchanged:




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Compare these numbers with the ones for stretch

in the tetrahedral CCl4 molecule in Example 23.

H1s must b

reduced in the usual way, by eye or using the reduction formula:

H1s= a

1+ t

2

The hydrogen 1s orbitals in methane are said to span a1+ t

2symmetry.

One combination of the hydrogen 1s orbitals has a1symmetry and is singly-degenerate.

Three combinations of the hydrogen 1s orbitals have t2symmetry. They are triply degenerate.

These combinations of the hydrogen orbitals are called

Symmetry Adapted Linear Combinations or SALCs.

Comparison of the symmetries of the carbon and hydrogen orbitals shows a perfect match with an a1and a t

2set from

+1

3C2

all orbitals swapped over:

0

6S4

all orbitals swapped over:

0

6d

two orbitals swapped,

two orbitals unchanged:

+2

E 8C3 3C

2 6S

4 6

d

H1s

4 1 0 0 2




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Using this method of matching the shapes of the orbitals on the central atom, the SALCs for the 4H 1s orbitals in

methane (or indeed any tetrahedral molecule) have been constructed: (see appendix).

(3) Overlap orbitals of the same symmetry.

The carbon 2s orbital and the hydrogen a1SALC can overlap in-phase, to make a bonding molecular orbital, and out-

of-phase, to make an anti-bonding molecular orbital:

Each of the carbon 2p orbitals similarly overlaps with one of the hydrogen t2SALCs to make three degenerate t

bonding and three degenerate t2anti-bonding orbitals:

(4) The full molecular orbital diagram can now be constructed:

(

3)(3)

1t2 2t

2




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There are 8 valence electrons in methane (4 from C and 1 from each H) so the configuration is,

(1a1)2(1t

2)6.

All of the electrons are in bonding orbitals so the bond order is,

bond order = (number of bonding electrons - number of antibonding electrons)

= (8 - 0) = 4

It is important to note that,

each molecular orbital involves the carbon bonding equally to each of the four hydrogen.

each C-H bond is the same,

total bond order = 4 or bond order per C-H = 1

each C-H bond is made up of contributions from four molecular orbitals,

C-H bond is a1+ t

2

the carbon achieves an octet as it is involved in covalent bonding in four doubly occupied orbitals.

Example 31: Square planar methane

Why is CH4is a tetrahedral and not square planar

(D4h

)?




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(1) Fragments.

All of the H atoms are equivalent and these form one fragment.

The C atom is different to the 4H atoms and this forms the second fragment.

(2) Symmetry.

Carbon:

The symmetry of the carbon 2s and 2p orbitals can be obtained straight from the D4h

character table:

Hydrogen:

To work out H1s, we need to work out how many of the H 1s orbitals are left unchanged by the D 4h symmetr

operations:

By eye or using the reduction formula, this breaks downas,

H1s= a

1g+ b

1g+ e

u

One combination of the hydrogen 1s orbitals has a1g

symmetry and is singly-degenerate.

One combination of the hydrogen 1s orbitals has b1g

symmetry and is singly-degenerate.

Two combinations of the hydrogen 1s orbitals have eusymmetry. They are doubly degenerate.

The combinations of the hydrogen 1s orbitals that have these symmetries can again be found by matching with theorbitals of the central atom. The carbon 2s orbital and one of the hydrogen combinations have a

1gsymmetry. The

relative phases of the hydrogen orbitals can then be deduced:

orbital symmetry

2s a1g

the top line of the character table, as always

2px,

2py e

u (x,y) in penultimate column of the character table

2pz a

2u z in penultimate column of the character table

E 2C4 C2 2C2 2C2 i 2S4 h 2v 2d

H1s

4 0 0 2 0 0 0 4 2 0




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The two eucombinations can similarly be found by requiring that they have the same symmetry properties as the

carbon px,and p

y-orbitals:

There are no entries in the penultimate column of the D4h

table for the B1g

row: no p-orbitals on carbon have this

symmetry. However, in the final column, there is the entry x2-y2. The entries in the final column give the symmetrieof the d-orbitals on the central atom (as well as telling us about Raman activity!):

x2

-y2= d

x2

-y2

In Examples 34-36, this is used in the construction of molecular orbital diagrams for transition metal compounds.However, it can also be used here to deduce the b

1gcombination of the hydrogen orbitals: this combination must

match a dx2

-y2orbital on the central atom.

Using this method of matching the shapes of the orbitals on the central atom, the SALCs for the 4H 1s orbitals in asquare planar molecule have been constructed: (see appendix).

(3) and (4) The MO diagram can now be constructed by overlapping the orbitals on thetwo fragments:

the C 2s (a1g

) and the H a1g

combination overlap to make a bonding (1a1g

) and an anti-bonding (2a1g

) orbital

the C 2pxand 2p

yorbitals each overlap with the H e

ucombinations to make a degenerate pair of bonding (1e

u)

and a degenerate pair of antibonding (2eu) orbitals

the C 2pzorbital has a

2usymmetry. There is no matching orbital on the hydrogen atoms. This orbital thus remain




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non-bonding (1a2u

).

carbon has no low lying d-orbitals in its valence shell so the hydrogen b1g

combination also has no orbital to

overlap with and remains non-bonding (1b1g

).

The bond order is now,

bond order = (6 0) = 3

3.3 Hypervalency

Molecules such as PF5, SF

6and XeF

4in which the central main group element appears to expand its octet are known

as hypervalent. Despite the large number of stable compounds which are classified as hypervalent, hypervalency hasremained a controversial subject.

Example 32: XeF4.

The highest occupied molecular orbital is actually the non-bonding 1a2u

orbital.

In the square planar geometry, this carbon p-orbitals is pointing in the wrong direction to

overlap with the hydrogens. In the tetrahedral geometry, all of the carbon p-orbitals canbe used to bond. The bonding (and bond order) is maximized in the tetrahedralgeometry.

XeF4is a square planar (D

4h) molecule in which the

xenon atom makes bonds to four fluorine atoms andso achieves an electron count of:

8 from Xe + 4 (1 for each F) = 12




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Unlike the H atoms treated in Examples 30 and 31, the fluorine atom uses s and p orbitals to bond. To simplify the

molecular orbital diagram, the fluorine atom is taken to be sp3hybridized and to use one sp3hybrid directed at the Xe

to bond:

Working out the SALCs for these hybrids is completely analogous to the procedure shown above for hydrogen 1s

orbitals. Indeed, the SALCs for square planar CH4developed in Example 31 can be used by simply replacing the

hydrogen 1s orbital by a fluorine hybrid. The MO diagram is thus completely analogous:

The 12 valence electrons lead to a configuration,

(1a1g

)2(1eu)4(1a

2u)2(1b

1g)2(2a

1g)2

and a bond order,


or,

bond order per Xe-F = 2/4 =

The bond order is positive indicating that the bond is stable.

Example 33: SF6




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The symmetry of the S 3s and 3p orbitals can be obtained from the Ohcharacter table:

3s a1g

and {3px, 3p

y, 3p

z} t

1u.

As in XeF4, the fluorine atoms are taken to use one sp3hybrid to bond to the sulfur. To find the SALCs,

Fmust be

worked out, in the usual way, by calculating the number of hybrids that are unshifted by each of the symmetry

elements of the Ohpoint group:

These may now be reduced using the reduction formula or "by eye". Both methods give,

{F}= a

1g+ e

g+ t

1u

The a1g

combination must match the symmetry of the sulfur 3s orbital since both have a1g

symmetry:

The three t1u

combinations must each match the symmetry of the sulfur 3px, 3p

yand 3p

zorbitals:

The two egcombinations do not match sulfur 3s or 3p orbitals. However, the last column of the character tables show

that two d-orbitals have this symmetry:

(2z2 x2 y2, x2 y2) are dz2and d

x2

-y2

SF6is an octahedral molecule (O

h) molecule in which

the sulfur atom makes bonds to six fluorine atoms

and so achieves an electron count of:

6 from S + 6 (1 for each F) = 12

E 8C3 6C2 6C4 3C2 i 6S4 8S6 3h 6d

F 6 0 0 2 2 0 0 0 4 2




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The egcombinations can thus be constructed by matching to these two d-orbitals:

The MO diagram can now be constructed:

the fluorine a1g

combination overlaps with sulfur 3s to produce bonding (1a1g

) and anti-bonding (2a1g

molecular orbitals,

the three fluorine t1u

combinations each overlap with the sulfur 3px, 3p

yand 3p

zorbitals to produce bondin

(1t1u

) and anti-bonding (2t1u

) molecular orbitals.

there are no sulfur 3s or 3p orbitals with the correct symmetry to overlap with the fluorine egcombinations an

these remain non-bonding.




13/20

With 12 electrons, the configuration is

(1a1g

)2(1t1u

)6(1eg

)4

The bond order is,


or 4/6 = 2/3 per S-F

Of the 12 electrons, only 8 are actually in molecular orbitals (1a1g

and 1t1u

)which are shared between S and F. 4

electrons are in the 1eglevels to which sulfur makes no contribution. Thus, although sulfur appearsto exceed its octet

and the bonding appearshypervalent,

S actually only achieves an octet,

the bond order is actually the same as for methane.

3.4 Transition metal complexes

The SALCs produced in Examples 31, 32 and 34 can also be used for the ligands in transition metal complexes. 3dtransition metals use 3d, 4s, and 4p orbitals to bond to the ligands.

Example 34: -bonding in octahedral complexes




14/20

The symmetry of the metal orbitals can be obtained straight from the Ohcharacter table:

The MO diagram is very similar to that constructed above for SF6. The key differences are:

the relative energies of the metal orbitals are 4s 3d < 4p

the metal is less electronegative than the ligands: the metal atomic orbitals lie at higher energy than the liganorbitals

the ligand SALCs of egsymmetry can overlap with the e

gd-orbitals on the metal to produce bonding (1e

g) an

anti-bonding (2eg) orbitals

each ligand is a 2e-donor so six ligands contribute 12 electrons, filling the 1a1g

, 1t1u

and 1eglevels.

the metal may contribute additional electrons, e.g. Ti3+d1, Ni2+d8, Cu2+d9.

the metal d-orbitals contribute primarily to the 1t2g

and 2eglevels:

orbital symmetry4s a

1g the top line, as always

4px, 4p

y, 4p

z t

1u (x,y,z) in penultimate column

3dxz

, 3dyz

, 3dxy

t2g

(xz, yz, xy) in final column

3dz

2, 3dx

2

-y

2 eg (2z2 x2- y2, x2 y2) in the final column




15/20

and the additional electrons from the metal ion occupy these (high spin or low spin depending on the ligand

the splitting of the d-orbitals into a set of three (1t2g

) and a set of two (2eg) orbitals is the same as that

obtained by simple crystal field arguments.

Example 35: -bonding in tetrahedral complexes

The symmetry of the metal orbitals can be obtained straight from the Tdcharacter table:

The MO diagram is very similar to that constructed above for CH4:

each ligand is a 2e-donor so four ligands contribute 8 electrons, filling the 1a1

and 1t2levels.

the metal may contribute additional electrons, e.g. Ti3+d1, Ni2+d8, Cu2+d9.

orbital symmetry4s a

1 the top line, as always

4px, 4p

y, 4p

z t

2 (x,y,z) in penultimate column

3dxz

, 3dyz

, 3dxy

t2 (xz, yz, xy) in final column

3dz2, 3d

x2-y

2 e (2z2 x2- y2, x2 y2) in the final column




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the metal d-orbitals contribute primarily to the 1e and 2t2levels:

and the additional electrons from the metal ion occupy these (alwayshigh spin for a tetrahedron).

the splitting of the d-orbitals into a set of three (1t2) and a set of two (2e) orbitals is the same as that obtained

by simple crystal field arguments.

the origin of the reversal of the splitting, as compared to the octahedral case, is clear: in the octahedron thereare no ligand SALCs of t

2gsymmetry and in the tetrahedron there are no ligand SALCs of e symmetry.

Example 36: -bonding in octahedral complexes.

All of the examples above have assumed that the ligands are -donors only. Two types of -bonding are found in

transition metal complexes:

donors, such as halides, OH-and O2-. Ligands of this type have occupied orbitals, usually lone pairs, which

are lower in energy than the metal orbitals:




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acceptors, such as CO, C2H

4and pyridine. Ligands of this type have empty orbitals which are higher in energ

than the metal orbitals:

To construct SALCs of these ligand -orbitals, the same strategy used for the -orbitals is used. This turns out to time-

consuming. It turns out that,

= t1g+ t1g+ t1u+ t2u

It is the t2g

combinations that are of most interest since they have the correct symmetry to overlap with the metal d-

orbitals of t2g

symmetry:

The effect on the MO diagram for -donor ligands is:




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the ligand -donor orbitals are lower in energy than the metal 3d orbitals. The result is a decrease in the

splitting of the d-orbitals.oct

is reduced.

the egand t

2gd-orbitals are both anti-bonding.

The effect on the MO diagram for -acceptor ligands is:

the ligand *-donor orbitals are higher in energy than the metal 3d orbitals. The result is an increase in thesplitting of the d-orbitals:

octis imcreased.

the t2g

d-orbitals are bonding and the eg

d-orbitals are antibonding.

Example 37: the eighteen electron rule

For octahedral complexes with -acceptor ligands, there are nine bonding orbitals:

To fill all of these bonding orbitals requires 9 2 = 18 electrons. For complexes with -acceptor ligands, but notfor

complexes of -donor ligands, the stable electron count is eighteen.

Appendix: Symmetry adapted linear combinations (SALCs) for commongeometries

ML2(linear: Dh):

ML2(bent: C

2v):

-bonding 1a1g, 1t1uand 1eu (see Example 34)

-bonding: 1t2g (see Example 36)

g

u




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ML3(trigonal planar: D

3h):

ML3(pyramidal planar: C

3v):

ML4(tetrahedral T

d):

ML4(square planar D

4h):

ML6(octahedral O

h):

a1 b

1

a1 e

a1 e

a1 t

2(3)

a1g

eu(3) b

1g




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By convention, lower case is used for orbitals

It is a useful exercise to make sure you can do this for yourself

It does not matter than carbon does not use d-orbitals in bonding. It is only the shape that is being used here towork out the H orbital combination.

Calculations suggest that the tetrahedral geometry of methane is due to the bonding rather than electron pair

repulsion although both favour a tetrahedral geometry.

It is a useful exercise to make sure you can do this for yourself

x2+y2+z2 in the A1g

row is actually the formula for a sphere and is not a d-orbital.

a1g

t1u

(3) eg


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