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DEPARTMENT OF CHEMISTRY Introduction to Symmetry and Inorg Spectrosc 3. Using Symmetry to Understand Bonding 3.1 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 are transferable to these areas. This part of the course uses the group theory tools you already know and shows how they can be developed and applied for the construction of molecular orbital diagrams and for understanding the octet and eighteen electron rules, hypervalency and ligand-field effects in transition metal chemistry. The figure below shows a molecular orbital (MO) diagram for H 2 : H 2 has 2 electrons and both go into the bonding orbital. The bond order is bond order = ½ (number of bonding electrons - number of antibonding electrons) = ½ (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 the right 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. Each 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 2 electrons in each level (Pauli principle). 3.2 The octet rule Example 30: tetrahedral methane The MO diagram for H 2 is straightforward as there are only two atoms. For bigger molecules, group theory can be used to simplify the process: C O N T E N T S Page 1 of 20 3. Using Symmetry to Understand Bonding 2/24/2004 http://www.hull.ac.uk/php/chsajb/symmetry&spectroscopy/ho_3.html

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Page 1: DEPARTMENT OF CHEMISTRYcraftychemist.org/.../Notes/Symmetry/UsingSymmetrytoUnderstandB… · Introduction to Symmetry and Inorg Spectrosc 3. Using Symmetry to Understand Bonding 3.1

DEPARTMENT OF CHEMISTRY

Introduction to Symmetry and Inorg

Spectrosc

3. Using Symmetry to Understand Bonding

3.1 Introduction

In addition to its role in predicting vibrational spectra, symmetry can also be applied to understanding bondingelectronic structure and reaction mechanisms. Many of the skills you have acquired for use in spectroscopy aretransferable to these areas. This part of the course uses the group theory tools you already know and shows how theycan be developed and applied for the construction of molecular orbital diagrams and for understanding the octet andeighteen electron rules, hypervalency and ligand-field effects in transition metal chemistry.

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

H2 has 2 electrons and both go into the bonding orbital.

The bond order is

bond order = ½ (number of bonding electrons - number of antibonding electrons)

= ½ (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 atomsThe atomic orbitals of one fragment are placed on the left hand side and those of the other fragment are placed on theright 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. Eachtime 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 2electrons in each level (Pauli principle).

3.2 The octet rule

Example 30: tetrahedral methane

The MO diagram for H2 is straightforward as there are only two atoms. For bigger molecules, group theory can be used

to simplify the process:

C

O

N T E

N T

S

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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 2s22p2 configuration and uses both its 2s and 2p orbitals in bonding. As the carbon atom is at thecentre of molecule, there is a very quick way of finding the symmetry of these orbitals.

2px, 2py and 2pz have the same symmetry as x,y and z respectively. These are given in the penultimate column of the

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

Γpx, py, pz = Γx, y, z = t2

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 anyof the symmetry operations. It is "totally symmetric" and always has the label given to the top row in the charactetable:

Γs = a1

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 isachieved by finding the symmetry of the allowed combinations of the hydrogen When working out the symmetry ostretches in Handout 2, Γstretch was calculated by noting how many bonds were unchanged by the symmetry

operations.

To work out ΓH1s, 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)

CH4 is a tetrahedral ("Td") molecule:

effect of operation character, χ

E

all orbitals unchanged:

+4

C3

3 orbitals swapped over.

1 orbital unchanged:

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Compare these numbers with the ones for Γstretch in the tetrahedral CCl4 molecule in Example 23. ΓH1s must be

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

ΓH1s = a1 + t2

The hydrogen 1s orbitals in methane are said to span a1 + t2 symmetry.

One combination of the hydrogen 1s orbitals has a1 symmetry and is singly-degenerate.

Three combinations of the hydrogen 1s orbitals have t2 symmetry. 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 a1 and a t2 set from

+1

3C2

all orbitals swapped over:

0

6S4

all orbitals swapped over:

0

6σd

two orbitals swapped,

two orbitals unchanged:

+2

E 8C3 3C2 6S4 6σd ΓH1s 4 1 0 0 2

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each fragment.

The carbon 2s orbital and one of the hydrogen combinations have a1 symmetry. The behaviour of both under the

symmetry operations must therefore match. The relative phases of the hydrogen orbitals can then be deduced:

To match the shape of the carbon 2s orbital, the hydrogen orbitals must be combining in phase.

The three t2 combinations can similarly be found by requiring that they have the same symmetry properties as the

carbon px, py and pz-orbitals. For example, to match C 2pz:

To match the two lobes of the p-orbital, the two hydrogen orbitals on the bottom of the cube must be out-of-phase with the two hydrogen orbitals on the top.

This is only one of the three t2 symmetry adapted linear combinations (SALCs). The other combinations have the same

nodal plane as the other two carbon p-orbitals and are shown overleaf. They are clearly equivalent to one another by rotation of the cube and their degeneracy is assured.

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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 a1 SALC 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 t2 SALCs to make three degenerate t2bonding and three degenerate t2 anti-bonding orbitals:

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

(× 3)

(× 3) 1t2 2t2

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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(1t2)6.

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

bond order = ½ (number of bonding electrons - number of antibonding electrons)

= ½ (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 ¼ a1 + ¾ t2

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

Example 31: Square planar methane

Why is CH4 is 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 ΓH1s, we need to work out how many of the H 1s orbitals are left unchanged by the D4h symmetry

operations†:

By eye or using the reduction formula, this breaks down† as,

ΓH1s = a1g + b1g + eu

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 eu symmetry. They are doubly degenerate.

The combinations of the hydrogen 1s orbitals that have these symmetries can again be found by matching with the orbitals of the central atom. The carbon 2s orbital and one of the hydrogen combinations have a1g symmetry. 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 eu (x,y) in penultimate column of the character table

2pz a2u z in penultimate column of the character table

E 2C4 C2 2C2’’ 2C2’’ i 2S4 σh 2σv 2σd

ΓH1s 4 0 0 2 0 0 0 4 2 0

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

carbon px,and py-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 symmetriesof the d-orbitals on the central atom (as well as telling us about Raman activity!):

Γx2-y2 = Γdx2-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 b1g combination of the hydrogen orbitals: this combination must

match a dx2-y2 orbital 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 a square 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 2px and 2py orbitals each overlap with the H eu combinations to make a degenerate pair of bonding (1eu)

and a degenerate pair of antibonding (2eu) orbitals

• the C 2pz orbital has a2u symmetry. There is no matching orbital on the hydrogen atoms. This orbital thus remains

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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 = ½ (6 – 0) = 3

3.3 Hypervalency

Molecules such as PF5, SF6 and XeF4 in 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 has remained 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 can be used to bond. The bonding (and bond order) is maximized in the tetrahedral geometry‡.

XeF4 is a square planar (D4h) molecule in which the

xenon atom makes bonds to four fluorine atoms and so 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 sp3 hybridized and to use one sp3 hybrid directed at the Xeto 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 CH4 developed 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(1a2u)2(1b1g)2(2a1g)2

and a bond order,

bond order = ½ (6 – 2) = 2

or,

bond order per Xe-F = 2/4 = ½

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 Oh character table:

3s – a1g and {3px, 3py, 3pz} – t1u.

As in XeF4, the fluorine atoms are taken to use one sp3 hybrid to bond to the sulfur. To find the SALCs, ΓF must be

worked out, in the usual way†, by calculating the number of hybrids that are unshifted by each of the symmetry elements of the Oh point group:

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

Γ{F}= a1g + eg + t1u

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, 3py and 3pz orbitals:

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

that two d-orbitals have this symmetry‡:

“(2z2 – x2 – y2, x2 – y2)” are dz2 and dx2-y2

SF6 is an octahedral molecule (Oh) 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 3σh 6σd

ΓF 6 0 0 2 2 0 0 0 4 2

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The eg combinations 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 (2a1gmolecular orbitals,

the three fluorine t1u combinations each overlap with the sulfur 3px, 3py and 3pz orbitals to produce bonding

(1t1u) and anti-bonding (2t1u) molecular orbitals.

there are no sulfur 3s or 3p orbitals with the correct symmetry to overlap with the fluorine eg combinations and

these remain non-bonding.

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With 12 electrons, the configuration is

(1a1g)2(1t1u)6(1eg)4

The bond order is,

bond order = ½ (8 – 0) = 4

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 1eg levels to which sulfur makes no contribution. Thus, although sulfur appears to exceed its octet

and the bonding appears hypervalent,

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. 3d transition metals use 3d, 4s, and 4p orbitals to bond to the ligands.

Example 34: σ-bonding in octahedral complexes

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The symmetry of the metal orbitals can be obtained straight from the Oh character 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 ligandorbitals

the ligand SALCs of eg symmetry can overlap with the eg d-orbitals on the metal to produce bonding (1eg) and

anti-bonding (2eg) orbitals

each ligand is a 2e- donor so six ligands contribute 12 electrons, filling the 1a1g, 1t1u and 1eg levels.

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

the metal d-orbitals contribute primarily to the 1t2g and 2eg levels:

orbital symmetry 4s a1g the top line, as always

4px, 4py, 4pz t1u (x,y,z) in penultimate column

3dxz, 3dyz, 3dxy t2g (xz, yz, xy) in final column

3dz2, 3dx

2-y

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

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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: σ-bonding in tetrahedral complexes

The symmetry of the metal orbitals can be obtained straight from the Td character 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 1t2 levels.

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

orbital symmetry 4s a1 the top line, as always

4px, 4py, 4pz t2 (x,y,z) in penultimate column

3dxz, 3dyz, 3dxy t2 (xz, yz, xy) in final column

3dz2, 3dx

2-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 2t2 levels:

and the additional electrons from the metal ion occupy these (always high 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 there are no ligand SALCs of t2g symmetry and in the tetrahedron there are no ligand SALCs of e symmetry.

Example 36: π-bonding in octahedral complexes.

All of the examples above have assumed that the ligands are σ-donors only. Two types of π-bonding are found in transition metal complexes:

π 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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π acceptors, such as CO, C2H4 and pyridine. Ligands of this type have empty orbitals which are higher in energy

than the metal orbitals:

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

Γπ = 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 π-donor ligands is:

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the ligand π-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 eg and t2g d-orbitals are both anti-bonding.

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

the ligand π*-donor orbitals are higher in energy than the metal 3d orbitals. The result is an increase in the splitting of the d-orbitals: ∆oct is imcreased.

the t2g d-orbitals are bonding and the eg d-orbitals are antibonding.

Example 37: the eighteen electron rule

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

To fill all of these bonding orbitals requires 9 × 2 = 18 electrons. For complexes with π-acceptor ligands, but not for complexes of π-donor ligands, the stable electron count is eighteen.

Appendix: Symmetry adapted linear combinations (SALCs) for common geometries

ML2 (linear: D∞h):

ML2 (bent: C2v):

σ-bonding 1a1g, 1t1u and 1eu (see Example 34)

π-bonding: 1t2g (see Example 36)

σg

σu

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ML3 (trigonal planar: D3h):

ML3 (pyramidal planar: C3v):

ML4 (tetrahedral Td):

ML4 (square planar D4h):

ML6 (octahedral Oh):

a1 b1

a1’ e’

a1 e

a1 t2 (× 3)

a1g eu (× 3) b1g

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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 to work 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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